position vector positive definite positively skewed principal axis probability density function probability distribution product rule proposition quadratic formula quadratic function qua[r]
(1)HAESE & HARRIS PUBLICATIONS Specialists in mathematics publishing Mathematics for the international student Mathematics HL (Core) second edition Paul Urban David Martin Robert Haese Sandra Haese Michael Haese Mark Humphries cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 for use with IB Diploma Programme black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_00\001IB_HL-2_00.CDR Friday, 25 January 2008 3:13:17 PM PETERDELL symbol_pp symbol_pp swtimes IB_HL-2ed (2) MATHEMATICS FOR THE INTERNATIONAL STUDENT Mathematics HL (Core) second edition Paul Urban David Martin Robert Haese Sandra Haese Michael Haese Mark Humphries B.Sc.(Hons.),B.Ec B.A.,B.Sc.,M.A.,M.Ed.Admin B.Sc B.Sc B.Sc.(Hons.),Ph.D B.Sc.(Hons.) Haese & Harris Publications Frank Collopy Court, Adelaide Airport, SA 5950, AUSTRALIA Telephone: +61 8355 9444, Fax: + 61 8355 9471 Email: info@haeseandharris.com.au www.haeseandharris.com.au Web: National Library of Australia Card Number & ISBN 978-1-876543-11-2 © Haese & Harris Publications 2008 Published by Raksar Nominees Pty Ltd Frank Collopy Court, Adelaide Airport, SA 5950, AUSTRALIA First Edition Reprinted Second Edition Reprinted 2004 2005 three times (with minor corrections), 2006, 2007 2008 2009 (with minor corrections), 2010 Cartoon artwork by John Martin Artwork by Piotr Poturaj and David Purton Cover design by Piotr Poturaj Computer software by David Purton, Thomas Jansson and Troy Cruickshank Typeset in Australia by Susan Haese (Raksar Nominees) Typeset in Times Roman 10\Qw_ /11\Qw_\\" The textbook and its accompanying CD have been developed independently of the International Baccalaureate Organization (IBO) The textbook and CD are in no way connected with, or endorsed by, the IBO This book is copyright Except as permitted by the Copyright Act (any fair dealing for the purposes of private study, research, criticism or review), no part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the publisher Enquiries to be made to Haese & Harris Publications Copying for educational purposes: Where copies of part or the whole of the book are made under Part VB of the Copyright Act, the law requires that the educational institution or the body that administers it has given a remuneration notice to Copyright Agency Limited (CAL) For information, contact the Copyright Agency Limited Acknowledgements: While every attempt has been made to trace and acknowledge copyright, the authors and publishers apologise for any accidental infringement where copyright has proved untraceable They would be pleased to come to a suitable agreement with the rightful owner cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Disclaimer: All the internet addresses (URL’s) given in this book were valid at the time of printing While the authors and publisher regret any inconvenience that changes of address may cause readers, no responsibility for any such changes can be accepted by either the authors or the publisher black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_00\002IB_HL-2_00.CDR Thursday, 11 March 2010 10:12:20 AM PETER IB_HL-2ed (3) FOREWORD Mathematics for the International Student: Mathematics HL has been written to reflect the syllabus for the two-year IB Diploma Mathematics HL course It is not our intention to define the course Teachers are encouraged to use other resources We have developed the book independently of the International Baccalaureate Organization (IBO) in consultation with many experienced teachers of IB Mathematics The text is not endorsed by the IBO This second edition builds on the strengths of the first edition Many excellent suggestions were received from teachers around the world and these are reflected in the changes In some cases sections have been consolidated to allow for greater efficiency Changes have also been made in response to the introduction of a calculator-free examination paper A large number of questions, including some to challenge even the best students, have been added In particular, the final chapter contains over 200 miscellaneous questions, some of which require the use of a graphics calculator These questions have been included to provide more difficult challenges for students and to give them experience at working with problems that may or may not require the use of a graphics calculator The combination of textbook and interactive Student CD will foster the mathematical development of students in a stimulating way Frequent use of the interactive features on the CD is certain to nurture a much deeper understanding and appreciation of mathematical concepts The book contains many problems from the basic to the advanced, to cater for a wide range of student abilities and interests While some of the exercises are simply designed to build skills, every effort has been made to contextualise problems, so that students can see everyday uses and practical applications of the mathematics they are studying, and appreciate the universality of mathematics Emphasis is placed on the gradual development of concepts with appropriate worked examples, but we have also provided extension material for those who wish to go beyond the scope of the syllabus Some proofs have been included for completeness and interest although they will not be examined For students who may not have a good understanding of the necessary background knowledge for this course, we have provided printable pages of information, examples, exercises and answers on the Student CD To access these pages, simply click on the ‘Background knowledge’ icons when running the CD It is not our intention that each chapter be worked through in full Time constraints will not allow for this Teachers must select exercises carefully, according to the abilities and prior knowledge of their students, to make the most efficient use of time and give as thorough coverage of work as possible Investigations throughout the book will add to the discovery aspect of the course and enhance student understanding and learning Many Investigations could be developed into portfolio assignments Teachers should follow the guidelines for portfolio assignments to ensure they set acceptable portfolio pieces for their students that meet the requirement criteria for the portfolios Review sets appear at the end of each chapter and a suggested order for teaching the two-year course is given at the end of this Foreword The extensive use of graphics calculators and computer packages throughout the book enables students to realise the importance, application and appropriate use of technology No single aspect of technology has been favoured It is as important that students work with a pen and paper as it is that they use their calculator or graphics calculator, or use a spreadsheet or graphing package on computer cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The interactive features of the CD allow immediate access to our own specially designed geometry packages, graphing packages and more Teachers are provided with a quick and easy way to demonstrate concepts, and students can discover for themselves and re-visit when necessary black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_00\003IB_HL-2_00.CDR Friday, 25 January 2008 3:56:01 PM PETERDELL IB_HL-2ed (4) Instructions appropriate to each graphic calculator problem are on the CD and can be printed for students These instructions are written for Texas Instruments and Casio calculators In this changing world of mathematics education, we believe that the contextual approach shown in this book, with the associated use of technology, will enhance the students’ understanding, knowledge and appreciation of mathematics, and its universal application We welcome your feedback Email: info@haeseandharris.com.au Web: www.haeseandharris.com.au PMU DCM RCH SHH PMH MAH ACKNOWLEDGEMENTS The authors and publishers would like to thank all those teachers who have offered advice and encouragement Many of them have read page proofs and made constructive comments and suggestions Particular thanks go to Stephen Hobbs who has given generously of his time in reviewing the first edition and making suggestions for improvement in this second edition Thanks are also due to Dr Andrzej Cichy, Peter Blythe, Brendan Watson, Myrricia Holmann, Jeff Jones, Mark Willis, John Poole and Marjut Mäenpää We acknowledge the contributions of John Owen and Mark Bruce in the preparation of the first edition and we also want to thank others who provided assistance – they include: Cameron Hall, Fran O'Connor, Glenn Smith, Anne Walker, Malcolm Coad, Ian Hilditch, Phil Moore, Julie Wilson, Kerrie Clements, Margie Karbassioun, Brian Johnson, Carolyn Farr, Rupert de Smidt, Terry Swain, Marie-Therese Filippi, Nigel Wheeler, Sarah Locke, Rema George The publishers wish to make it clear that acknowledging these individual does not imply any endorsement of this book by any of them and all responsibility for content rests with the authors and publishers TEACHING THE TWO-YEAR COURSE – A SUGGESTED ORDER Teachers are encouraged to carefully check the BACKGROUND KNOWLEDGE sections supplied on the accompanying CD to ensure that basics have been mastered relatively early in the two-year HL course Some of these topics naturally occur at the beginning of a specific chapter, as indicated in the table of contents Click on the BACKGROUND KNOWLEDGE active icons to access the printable pages on the CD Teachers will have their personal preferences for the order in which the chapters are tackled A suggestion is to work progressively from Chapter through to Chapter 20, but leave Chapters 9, 15 and, possibly, 16 for the second year The remaining chapters can be worked through in order Alternatively, for the first year, students could work progressively from Chapter to Chapter 23 but not necessarily including chapters 7, 15 and 16 Chapter ‘Mathematical Induction’ could also be attempted later, perhaps early in the second year In some parts of the world, the topics of Polynomials, Complex Numbers, 3-D Vector Geometry and Calculus are not usually covered until the final year of school Another approach could be to teach just those topics that are included in the Mathematics SL syllabus in the first year and leave the remaining topics for completion in the second year cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 However, it is acknowledged that there is no single best way for all teachers to work through the syllabus Individual teachers have to consider particular needs of their students and other requirements and preferences that they may have black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_00\004IB_HL-2_00.CDR Friday, 25 January 2008 3:20:02 PM PETERDELL IB_HL-2ed (5) USING THE INTERACTIVE STUDENT CD The CD is ideal for independent study Frequent use will nurture a deeper understanding of Mathematics Students can revisit concepts taught in class and undertake their own revision and practice The CD also has the text of the book, allowing students to leave the textbook at school and keep the CD at home The icon denotes an Interactive Link on the CD Simply ‘click’ the icon to access a range of interactive features: w w w w w w spreadsheets video clips graphing and geometry software graphics calculator instructions computer demonstrations and simulations background knowledge (as printable pages) INTERACTIVE LINK For those who want to make sure they have the prerequisite levels of understanding for this course, printable pages of background information, examples, exercises and answers are provided on the CD Click the ‘Background knowledge’ icon on pages 12 and 248 Graphics calculators: Instructions for using graphics calculators are also given on the CD and can be printed Instructions are given for Texas Instruments and Casio calculators Click on the relevant icon (TI or C) to access printable instructions Examples in the textbook are not always given for both types of calculator Where that occurs, click on the relevant icon to access the instructions for the other type of calculator TI C NOTE ON ACCURACY Students are reminded that in assessment tasks, including examination papers, unless otherwise stated in the question, all numerical answers must be given exactly or to three significant figures HL & SL COMBINED CLASSES Refer to our website www.haeseandharris.com.au for guidance in using this textbook in HL and SL combined classes HL OPTIONS This is a companion to the Mathematics HL (Core) textbook It offers coverage of each of the following options: w w w w Topic –Statistics and probability Topic – Sets, relations and groups Topic 10 – Series and differential equations Topic 11 – Discrete mathematics In addition, coverage of the Geometry option for students undertaking the IB Diploma course Further Mathematics is presented on the CD that accompanies the HL Options book SUPPLEMENTARY BOOKS cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A separate book of WORKED SOLUTIONS gives the fully worked solutions for every question (discussions, investigations and projects excepted) in each chapter of the Mathematics HL (Core) textbook The HL (CORE) EXAMINATION PREPARATION & PRACTICE GUIDE offers additional questions and practice exams to help students prepare for the Mathematics HL examination For more information email info@haeseandharris.com.au black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_00\005IB_HL-2_00.CDR Friday, 25 January 2008 12:53:29 PM PETERDELL IB_HL-2ed (6) TABLE OF CONTENTS TABLE OF CONTENTS SYMBOLS AND NOTATION USED IN THIS BOOK 10 BACKGROUND KNOWLEDGE 12 F G H Graphs of exponential functions Growth and decay The natural exponential ‘e’ Review set 3A Review set 3B Review set 3C LOGARITHMS A B C D E F G H Logarithms Logarithms in base 10 Laws of logarithms Natural logarithms Exponential equations using logarithms The change of base rule Graphs of logarithmic functions Growth and decay Review set 4A Review set 4B Review set 4C Review set 4D GRAPHING AND TRANSFORMING FUNCTIONS 125 A B C D Families of functions Transformations of graphs Simple rational functions Further graphical transformations Review set 5A Review set 5B QUADRATIC EQUATIONS AND FUNCTIONS 143 A B C D E F G H Solving quadratic equations (Review) The discriminant of a quadratic The sum and product of the roots Graphing quadratic functions Finding a quadratic from its graph Where functions meet Problem solving with quadratics Quadratic optimisation Review set 6A Review set 6B Review set 6C Review set 6D Review set 6E 145 149 152 153 161 165 167 170 173 174 175 175 176 to access, ‘click’ active icon on CD Solutions of real quadratics with Δ < Complex numbers Real polynomials Roots, zeros and factors magenta yellow 95 100 75 50 75 25 cyan 95 A B C D 100 78 79 80 84 87 50 Index notation Evaluating powers Index laws Algebraic expansion and factorisation Exponential equations 75 A B C D E COMPLEX NUMBERS AND POLYNOMIALS 25 42 44 46 49 50 51 77 50 EXPONENTIALS 25 54 54 56 59 65 72 74 75 76 53 Number patterns Sequences of numbers Arithmetic sequences Geometric sequences Series Miscellaneous problems Review set 2A Review set 2B Review set 2C 95 SEQUENCES AND SERIES A B C D E F 100 H I J 50 18 21 27 28 32 35 41 75 17 Relations and functions Function notation, domain and range Composite functions, f ± g Sign diagrams Inequalities (inequations) The modulus function The reciprocal function x x Asymptotes of other rational functions Inverse functions Functions which have inverses Review set 1A Review set 1B Review set 1C 25 FUNCTIONS A B C D E F G CD CD CD CD CD CD CD CD CD CD CD CD CD 95 Surds and radicals Scientific notation (Standard form) Number systems and set notation Algebraic simplification Linear equations and inequalities Modulus or absolute value Product expansion Factorisation Formula rearrangement Adding and subtracting algebraic fractions Congruence and similarity Coordinate geometry ANSWERS 100 A B C D E F G H I J K L black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_00\006IB_HL-2_00.CDR Friday, 25 January 2008 12:53:50 PM PETERDELL 88 91 95 98 99 99 101 102 104 106 110 112 114 115 120 122 123 123 124 126 128 133 137 140 141 177 178 180 188 193 IB_HL-2ed (7) Graphing polynomials Theorems for real polynomials Review set 7A Review set 7B Review set 7C 201 208 210 211 212 COUNTING AND THE BINOMIAL EXPANSION 213 A B C D E F G The product principle Counting paths Factorial notation Permutations Combinations Binomial expansions The general binomial expansion Review set 8A Review set 8B MATHEMATICAL INDUCTION A B C The process of induction The principle of mathematical induction Indirect proof (extension) Review set 9A Review set 9B Review set 9C G H I J K L M 214 216 217 219 223 226 229 231 232 A B C D E F G 234 236 244 245 245 246 D A B C D E F G H I J CD 248 250 253 263 266 267 268 11 NON-RIGHT ANGLED TRIANGLE TRIGONOMETRY 269 270 272 277 280 281 12 ADVANCED TRIGONOMETRY 283 cyan magenta A B C D E yellow 50 Vectors Operations with vectors 2-D vectors in component form 3-D coordinate geometry 3-D vectors in component form Algebraic operations with vectors Parallelism Unit vectors The scalar product of two vectors The vector product of two vectors Review set 14A Review set 14B Review set 14C Review set 14D Review set 14E Complex numbers as 2-D vectors Modulus, argument, polar form De Moivre's Theorem Roots of complex numbers Further complex number problems Review set 15A Review set 15B Review set 15C 16 LINES AND PLANES IN SPACE A B 75 25 95 285 288 293 296 297 299 100 50 75 25 95 100 50 75 Observing periodic behaviour The sine function Modelling using sine functions The cosine function The tangent function Trigonometric equations 25 95 100 50 75 25 A B C D E F Matrix structure Matrix operations and definitions The inverse of a × matrix × and larger matrices Solving systems of linear equations Solving systems using row operations Induction with matrices Review set 13A Review set 13B Review set 13C Review set 13D Review set 13E 15 COMPLEX NUMBERS 95 The cosine rule The sine rule Using the sine and cosine rules Review set 11A Review set 11B A B C 305 307 309 310 314 318 318 319 320 321 322 14 VECTORS IN AND DIMENSIONS 10 THE UNIT CIRCLE AND RADIAN MEASURE 247 A B C Using trigonometric models Reciprocal trigonometric functions Trigonometric relationships Compound angle formulae Double angle formulae Trigonometric equations in quadratic form Trigonometric series and products Review set 12A Review set 12B Review set 12C Review set 12D 13 MATRICES 233 BACKGROUND KNOWLEDGE – TRIGONOMETRY WITH RIGHT ANGLED TRIANGLES – Printable pages Radian measure Arc length and sector area The unit circle and the basic trigonometric ratios Areas of triangles Review set 10A Review set 10B Review set 10C Lines in 2-D and 3-D Applications of a line in a plane 100 E F TABLE OF CONTENTS black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_00\007IB_HL-2_00.CDR Friday, 25 January 2008 10:27:06 AM PETERDELL 323 324 326 342 348 350 354 364 366 367 368 369 370 371 372 375 383 388 390 393 398 400 402 407 416 417 418 419 420 421 422 425 438 441 445 445 446 447 449 451 456 IB_HL-2ed (8) 461 466 471 473 477 478 479 481 17 DESCRIPTIVE STATISTICS A B C D E F G L A B C D E F G H 483 Continuous numerical data and histograms Measuring the centre of data Cumulative data Measuring the spread of data Statistics using technology Variance and standard deviation The significance of standard deviation Review set 17A Review set 17B 18 PROBABILITY A B C D E F G H I J K 21 APPLICATIONS OF DIFFERENTIAL CALCULUS 485 489 500 502 510 512 518 520 522 528 532 533 537 541 543 546 549 554 558 668 673 677 679 683 684 685 23 DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES 687 A B 560 562 564 565 566 568 Derivatives of circular functions The derivatives of reciprocal circular functions The derivatives of inverse circular functions Maxima and minima with trigonometry Related rates Review set 23A Review set 23B C D E 20 DIFFERENTIAL CALCULUS 589 cyan magenta A B C D E F G yellow Antidifferentiation The fundamental theorem of calculus Integration Integrating eax+b and (ax + b)n Integrating f (u)u0 (x) by substitution Integrating circular functions Definite integrals Review set 24A Review set 24B Review set 24C 25 APPLICATIONS OF INTEGRATION A B 95 100 50 75 592 595 600 604 607 611 616 618 619 620 25 95 100 50 75 25 95 The derivative function Derivatives at a given x-value Simple rules of differentiation The chain rule Product and quotient rules Tangents and normals Higher derivatives Review set 20A Review set 20B Review set 20C 100 50 75 25 A B C D E F G 24 INTEGRATION Finding areas between curves Motion problems 25 570 574 577 579 586 Limits Finding asymptotes using limits Trigonometric limits Calculation of areas under curves Review set 19 622 623 627 634 642 647 652 661 664 665 666 Exponential e Natural logarithms Derivatives of logarithmic functions Applications Some special exponential functions Review set 22A Review set 22B A B C D E 19 INTRODUCTION TO CALCULUS 569 A B C D 621 22 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS 667 525 Experimental probability Sample space Theoretical probability Compound events Using tree diagrams Sampling with and without replacement Binomial probabilities Sets and Venn diagrams Laws of probability Independent events Probabilities using permutations and combinations Bayes’ theorem Review set 18A Review set 18B Review set 18C Review set 18D Time rate of change General rates of change Motion in a straight line Some curve properties Rational functions Inflections and shape Optimisation Implicit differentiation Review set 21A Review set 21B Review set 21C 95 Relationship between lines Planes and distances Angles in space The intersection of two or more planes Review set 16A Review set 16B Review set 16C Review set 16D 100 C D E F 50 TABLE OF CONTENTS 75 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_00\008IB_HL-2_00.CDR Friday, 25 January 2008 10:28:36 AM PETERDELL 688 693 694 697 699 704 705 707 708 710 715 720 722 724 730 734 735 736 737 738 744 IB_HL-2ed (9) TABLE OF CONTENTS Problem solving by integration Review set 25A Review set 25B Review set 25C 748 752 753 755 26 VOLUMES OF REVOLUTION 757 Solids of revolution Volumes for two defining functions Review set 26 758 762 765 27 FURTHER INTEGRATION AND DIFFERENTIAL EQUATIONS 28 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES 786 788 791 794 801 807 810 812 29 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES A B C D 821 828 830 831 833 magenta yellow 95 100 50 75 25 95 100 933 50 INDEX 75 857 25 814 817 ANSWERS 95 100 50 75 25 30 MISCELLANEOUS QUESTIONS cyan 813 Continuous probability density functions Normal distributions The standard normal distribution (Z-distribution) Applications of the normal distribution Review set 29A Review set 29B 95 E F 785 Discrete random variables Discrete probability distributions Expectation The measures of a discrete random variable The binomial distribution The Poisson distribution Review set 28A Review set 28B 100 A B C D 769 771 773 774 783 784 50 B C D E 768 A 767 1 The integrals of p and x + a2 a ¡ x2 Further integration by substitution Integration by parts Miscellaneous integration Separable differential equations Review set 27A Review set 27B 75 A B 25 C black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_00\009IB_HL-2_00.CDR Friday, 25 January 2008 10:28:19 AM PETERDELL IB_HL-2ed (10) SYMBOLS AND NOTATION USED IN THIS BOOK > or > N the set of positive integers and zero, f0, 1, 2, 3, g Z the set of integers, f0, §1, §2, §3, g Z+ the set of positive integers, f1, 2, 3, g Q R + R is not less than the set of positive rational numbers, fx j x > , x Q g [a, b] the set of real numbers ] a, b [ i the set of complex numbers, fa + bi j a, b R g p ¡1 z a complex number z¤ the complex conjugate of z jzj the modulus of z arg z the argument of z un the common difference of an arithmetic sequence r the common ratio of a geometric sequence Sn the sum of the first n terms of a sequence, u1 + u2 + ::::: + un S1 or S ui ³ ´ the real part of z the imaginary part of z n r the set with elements x1 , x2 , the set of all x such that f is a function under which each element of set A has an image in set B f : x 7! y f is a function under which x is mapped to y the image of x under the function f = is not an element of f ¡1 the inverse function of the function f ? the empty (null) set f ±g the composite function of f and g lim f (x) the limit of f (x) as x tends to a U the universal set [ union \ intersection µ is a subset of A0 p a , na x!a dy dx f (x) a2 , d2 y dx2 the complement of the set A a to the power of (if a > then p a a to the power (if a > then jxj , n p n , p nth root of a f 00 (x) a > 0) dn y dxn square root of a a > 0) f (n) (x) R the n modulus or absolute value of x, that is x for x > x2R ¡x for x < x R ´ Z y dx y dx identity or is equivalent to ¼ is approximately equal to > is greater than magenta a ex yellow the derivative of f (x) with respect to x the second derivative of y with respect to x the second derivative of f (x) with respect to x the nth derivative of y with respect to x the nth deriviative of f (x) with respect to x the indefinite integral of y with respect to x 95 the definite integral of y with respect to x between the limits x = a and x = b exponential function of x logarithm to the base a of x 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 cyan the derivative of y with respect to x b loga x 25 n! r!(n ¡ r)! f (x) n u1 + u2 + ::::: + un is an element of the sum to infinity of a sequence, u1 + u2 + ::::: f : A!B the number of elements in the finite set A fx j the open interval a < x < b the nth term of a sequence or series i=1 Re z n(A) the closed interval a x b d n X Im z fx1 , x2 , g is less than or equal to is not greater than the set of positive real numbers, fx j x > 0, x R g C is less than · or the set of rational numbers + Q < is greater than or equal to black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_00\010IB_HL-2_00.CDR Thursday, 11 March 2010 10:14:51 AM PETER IB_HL-2ed (11) ln x sin, cos, tan ¾ arcsin, arccos, arctan the line segment with end points A and B AB the length of [AB] frequencies with which the observations x1 , x2 , x3 , occur the angle between [CA] and [AB] ¢ABC the triangle whose vertices are A, B and C the vector v v ¡ ! AB the vector represented in magnitude and direction by the directed line segment from A to B ¡! the position vector OA a i, j, k unit vectors in the directions of the Cartesian coordinate axes jaj ¡ ! j AB j the magnitude of vector a ¡ ! the magnitude of AB v²w the scalar product of v and w v£w probability density function of the continuous random variable X E(X) the expected value of the random variable X the vector product of v and w the inverse of the non-singular matrix A detA or jAj the determinant of the square matrix A the variance of the random variable X ¹ population mean ¾ population standard deviation ¾2 population variance x sample mean sn2 sample variance sn standard deviation of the sample sn¡1 unbiased estimate of the population variance B(n, p) binomial distribution with parameters n and p Po(m) Poisson distribution with mean m N(¹, ¾ ) A¡1 probability distribution function P(X = x) of the discrete random variable X f (x) Var (X) the angle at A [ or CA bB CAB normal distribution with mean ¹ and variance ¾ X » B(n, p) the random variable X has a binomial distribution with parameters n and p X » Po(m) the random variable X has a Poisson distribution with mean m X » N(¹, ¾ ) the random variable X has a normal distribution with mean ¹ and variance ¾ magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 100 cyan the identity matrix I 95 f1 , f2 , the line containing points A and B b A 50 probability of the event A given B observations of a variable px [AB] 75 P(A j B) x1 , x2 , the point A in the plane with Cartesian coordinates x and y (AB) 25 probability of event A probability of the event “not A” the reciprocal circular functions A(x, y) P(A) P0 (A) the inverse circular functions csc, sec, cot the natural logarithm of x, loge x the circular functions black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_00\011IB_HL-2_00.CDR Friday, 25 January 2008 11:47:10 AM PETERDELL IB_HL-2ed (12) 12 BACKGROUND KNOWLEDGE AND GEOMETRIC FACTS BACKGROUND KNOWLEDGE Before starting this course you can make sure that you have a good understanding of the necessary background knowledge Click on the icon alongside to obtain a printable set of exercises and answers on this background knowledge BACKGROUND KNOWLEDGE NUMBER SETS Click on the icon to access printable facts about number sets SUMMARY OF CIRCLE PROPERTIES ² A circle is a set of points which are equidistant from a fixed point, which is called its centre circle centre ² The circumference is the distance around the entire circle boundary ² An arc of a circle is any continuous part of the circle chord arc ² A chord of a circle is a line segment joining any two points of a circle ² A semi-circle is a half of a circle diameter ² A diameter of a circle is any chord passing through its centre radius ² A radius of a circle is any line segment joining its centre to any point on the circle tangent cyan magenta yellow 95 100 50 75 25 point of contact 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ² A tangent to a circle is any line which touches the circle in exactly one point black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_00\012IB_HL-2_00.CDR Friday, 25 January 2008 3:38:07 PM PETERDELL IB_HL-2ed (13) BACKGROUND KNOWLEDGE AND GEOMETRIC FACTS 13 Click on the appropriate icon to revisit these well known theorems Name of theorem Statement Diagram B The angle in a semi-circle is a right angle Angle in a semi-circle GEOMETRY PACKAGE A AM = BM O A GEOMETRY PACKAGE M B b = 90o OAT The tangent to a circle is perpendicular to the radius at the point of contact Radius-tangent C O The perpendicular from the centre of a circle to a chord bisects the chord Chords of a circle b = 90o ABC O GEOMETRY PACKAGE A T A AP = BP Tangents from an external point are equal in length Tangents from an external point O P GEOMETRY PACKAGE B C The angle at the centre of a circle is twice the angle on the circle subtended by the same arc Angle at the centre O B D GEOMETRY PACKAGE B magenta C yellow GEOMETRY PACKAGE 95 T 100 50 b = BCA b BAS B 75 25 95 100 50 75 25 The angle between a tangent and a chord at the point of contact is equal to the angle subtended by the chord in the alternate segment 95 100 50 75 25 95 100 50 75 25 cyan b = ACB b ADB C A Angle between a tangent and a chord GEOMETRY PACKAGE A Angles subtended by an arc on the circle are equal in size Angles subtended by the same arc b = 2£ACB b AOB black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_00\013IB_HL-2_00.CDR Friday, 25 January 2008 4:10:57 PM PETERDELL A S IB_HL-2ed (14) 14 BACKGROUND KNOWLEDGE AND GEOMETRIC FACTS SUMMARY OF MEASUREMENT FACTS PERIMETER FORMULAE The length of an arc is a fraction of the circumference of a circle The distance around a closed figure is its perimeter For some shapes we can derive a formula for perimeter The formulae for the most common shapes are given below: b a w l r r d q° c l square rectangle triangle circle P = 4l P = 2(l + w) P = a+ b+ c C =2¼r or C = ¼d arc µ l = ( 360 )2¼r AREA FORMULAE Shape Figure Formula Rectangle Area = length £ width width length Triangle Area = 12 base £ height height base base Parallelogram Area = base £ height height base a Area = Trapezium or Trapezoid h b Circle Area = ¼r2 r Sector Area = magenta µ (360 ) £¼r yellow 95 100 50 75 25 95 r 100 50 75 25 95 100 50 75 25 95 100 50 75 25 q cyan (a +2 b) £h black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_00\014IB_HL-2_00.CDR Friday, 25 January 2008 3:40:12 PM PETERDELL IB_HL-2ed (15) BACKGROUND KNOWLEDGE AND GEOMETRIC FACTS 15 SURFACE AREA FORMULAE RECTANGULAR PRISM c A = 2(ab + bc + ac) b a CYLINDER CONE Object Outer surface area Object Outer surface area Hollow cylinder A = 2¼rh (no ends) Open cone A = ¼rs (no base) hollow r h s r hollow A = 2¼rh +¼r2 (one end) Open can hollow Solid cone r h r A = ¼rs + ¼r2 (solid) s solid A = 2¼rh +2¼r (two ends) Solid cylinder solid h r solid SPHERE cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 Area, A = 4¼r2 95 100 50 75 25 95 100 50 75 25 r black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_00\015IB_HL-2_00.CDR Friday, 25 January 2008 3:41:20 PM PETERDELL IB_HL-2ed (16) 16 BACKGROUND KNOWLEDGE AND GEOMETRIC FACTS VOLUME FORMULAE Object Figure Volume Solids of height uniform cross-section Volume of uniform solid = area of end £ length end height end height height Pyramids and cones Volume of a pyramid or cone = 13 (area of base £ height) h base base r Volume of a sphere = 43 ¼r3 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Spheres black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_00\016IB_HL-2_00.CDR Friday, 25 January 2008 4:01:13 PM PETERDELL IB_HL-2ed (17) Chapter Functions Contents: A B C D E F Relations and functions Function notation, domain and range Composite functions, f ± g Sign diagrams Inequalities (inequations) The modulus function G The reciprocal function x H I J Asymptotes of other rational functions Inverse functions Functions which have inverses x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 1A Review set 1B Review set 1C black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\017IB_HL-2_01.CDR Wednesday, 24 October 2007 11:32:36 AM PETERDELL IB_HL-2ed (18) 18 FUNCTIONS (Chapter 1) A RELATIONS AND FUNCTIONS The charges for parking a car in a short-term car park at an airport are given in the table shown alongside Car park charges Period (h) Charge - hours $5:00 - hours $9:00 - hours $11:00 - hours $13:00 - hours $18:00 - 12 hours $22:00 12 - 24 hours $28:00 There is an obvious relationship between the time spent in the car park and the cost The cost is dependent on the length of time the car is parked Looking at this table we might ask: How much would be charged for exactly one hour? Would it be $5 or $9? To make the situation clear, and to avoid confusion, we could adjust the table and draw a graph We need to indicate that 2-3 hours really means a time over hours up to and including hours, i.e., < t So, we now have Car park charges Period Charge < t hours $5:00 < t hours $9:00 < t hours $11:00 < t 6 hours $13:00 < t hours $18:00 < t 12 hours $22:00 12 < t 24 hours $28:00 In mathematical terms, because we have a relationship between two variables, time and cost, the schedule of charges is an example of a relation A relation may consist of a finite number of ordered pairs, such as f(1, 5), (¡2, 3), (4, 3), (1, 6)g or an infinite number of ordered pairs The parking charges example is clearly the latter as any real value of time (t hours) in the interval < t 24 is represented 30 charge ($) 20 exclusion inclusion 10 time (t) 12 15 18 21 24 The set of possible values of the variable on the horizontal axis is called the domain of the relation ² ² For example: ft: < t 24g is the domain for the car park relation f¡2, 1, 4g is the domain of f(1, 5), (¡2, 3), (4, 3), (1, 6)g The set which describes the possible y-values is called the range of the relation ² ² For example: the range of the car park relation is f5, 9, 11, 13, 18, 22, 28g the range of f(1, 5), (¡2, 3), (4, 3), (1, 6)g is f3, 5, 6g cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 We will now look at relations and functions more formally black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\018IB_HL-2_01.CDR Thursday, 12 July 2007 12:18:36 PM DAVID3 IB_HL-2ed (19) FUNCTIONS (Chapter 1) 19 RELATIONS A relation is any set of points on the Cartesian plane A relation is often expressed in the form of an equation connecting the variables x and y For example y = x + and x = y2 are the equations of two relations These equations generate sets of ordered pairs Their graphs are: y y y=x+3 x -3 x x = y2 However, a relation may not be able to be defined by an equation Below are two examples which show this: y (1) All points in the first quadrant are a relation x > 0, y > (2) y These 13 points form a relation x x FUNCTIONS A function, sometimes called a mapping, is a relation in which no two different ordered pairs have the same x-coordinate (first member) We can see from the above definition that a function is a special type of relation TESTING FOR FUNCTIONS Algebraic Test: If a relation is given as an equation, and the substitution of any value for x results in one and only one value of y, we have a function For example: ² y = 3x ¡ is a function, as for any value of x there is only one value of y ² x = y is not a function since if x = 4, say, then y = §2 Geometric Test or “Vertical Line Test”: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 If we draw all possible vertical lines on the graph of a relation, the relation: ² is a function if each line cuts the graph no more than once ² is not a function if at least one line cuts the graph more than once black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\019IB_HL-2_01.CDR Thursday, 12 July 2007 12:20:16 PM DAVID3 IB_HL-2ed (20) 20 FUNCTIONS (Chapter 1) Example Which of the following relations are functions? a b y c y y x x x a b y c y y x x a function not a function x a function GRAPHICAL NOTE ² If a graph contains a small open circle such as ² If a graph contains a small filled-in circle such as ² If a graph contains an arrow head at an end such as then the graph continues indefinitely in that general direction, or the shape may repeat as it has done previously , this point is not included , this point is included EXERCISE 1A Which of the following sets of ordered pairs are functions? Give reasons b f(1, 3), (3, 2), (1, 7), (¡1, 4)g a f(1, 3), (2, 4), (3, 5), (4, 6)g d f(7, 6), (5, 6), (3, 6), (¡4, 6)g c f(2, ¡1), (2, 0), (2, 3), (2, 11)g f f(0, 0), (0, ¡2), (0, 2), (0, 4)g e f(0, 0), (1, 0), (3, 0), (5, 0)g Use the vertical line test to determine which of the following relations are functions: a b c d y y y x e y x x f g y x h y y y x x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 x black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\020IB_HL-2_01.CDR Friday, 25 January 2008 12:34:06 PM PETERDELL x IB_HL-2ed (21) FUNCTIONS (Chapter 1) 21 Will the graph of a straight line always be a function? Give evidence Give algebraic evidence to show that the relation x2 + y2 = is not a function B FUNCTION NOTATION, DOMAIN AND RANGE Function machines are sometimes used to illustrate how functions behave x For example: So, if is fed into the machine, 2(4) + = 11 comes out I double the input and then add 2x + The above ‘machine’ has been programmed to perform a particular function If f is used to represent that particular function we can write: f is the function that will convert x into 2x + So, f would convert into 2(2) + = and ¡4 into 2(¡4) + = ¡5 This function can be written as: f : x `! 2x + function f such that x is converted into 2x + Two other equivalent forms we use are: f(x) = 2x + or y = 2x + f(x) is the value of y for a given value of x, i.e., y = f (x) So, Notice that for f (x) = 2x + 3, f(2) = 2(2) + = and f(¡4) = 2(¡4) + = ¡5: Consequently, f (2) = indicates that the point (2, 7) lies on the graph of the function f (¡4) = ¡5 indicates that the point (¡4, ¡5) also lies on the graph Likewise y (2,¡7) ƒ(x)¡=¡2x¡+¡3 x Note: ² f (x) is read as “f of x” (-4,-5) ² f is the function which converts x into f (x), i.e., f : x 7! f(x) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ² y = f (x) is sometimes called the image of x black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\021IB_HL-2_01.CDR Thursday, 12 July 2007 12:24:12 PM DAVID3 IB_HL-2ed (22) 22 FUNCTIONS (Chapter 1) Example If f : x 7! 2x2 ¡ 3x, find the value of: a f(5) b f (¡4) f (x) = 2x2 ¡ 3x = 2(5)2 ¡ 3(5) = £ 25 ¡ 15 = 35 freplacing x by (5)g a f(5) b f(¡4) = 2(¡4)2 ¡ 3(¡4) = 2(16) + 12 = 44 freplacing x by (¡4)g Example If f (x) = ¡ x ¡ x2 , find in simplest form: f(¡x) = ¡ (¡x) ¡ (¡x)2 = + x ¡ x2 a b a f(¡x) b f(x + 2) freplacing x by (¡x)g f(x + 2) = ¡ (x + 2) ¡ (x + 2)2 = ¡ x ¡ ¡ [x2 + 4x + 4] = ¡ x ¡ x2 ¡ 4x ¡ = ¡x2 ¡ 5x ¡ freplacing x by (x + 2)g EXERCISE 1B.1 If f : x 7! 3x + 2, find the value of: a f(0) b f (2) c f (¡1) d f(¡5) e f (¡ 13 ) If f : x 7! 3x ¡ x2 + 2, find the value of: a f(0) b f (3) c f (¡3) d f(¡7) e f ( 32 ) If f (x) = ¡ 3x, find in simplest form: a f (a) b f(¡a) c f (a + 3) d f(b ¡ 1) e f (x + 2) f f(x + h) If F (x) = 2x2 + 3x ¡ 1, find in simplest form: a F (x + 4) b F (2 ¡ x) c F (¡x) d F (x2 ) e F (x2 ¡ 1) f F (x + h) If G(x) = 2x + : x¡4 evaluate i G(2) ii G(0) iii G(¡ 12 ) find a value of x where G(x) does not exist find G(x + 2) in simplest form find x if G(x) = ¡3: a b c d cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 f represents a function What is the difference in meaning between f and f(x)? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\022IB_HL-2_01.CDR Friday, 16 February 2007 2:31:44 PM PETERDELL IB_HL-2ed (23) FUNCTIONS (Chapter 1) 23 If the value of a photocopier t years after purchase is given by V (t) = 9650¡860t euros: a find V (4) and state what V (4) means b find t when V (t) = 5780 and explain what this represents c find the original purchase price of the photocopier On the same set of axes draw the graphs of three different functions f (x) such that f(2) = and f (5) = 3: Find a linear function f(x) = ax + b for which f (2) = and f(¡3) = 11 10 Given T (x) = ax2 + bx + c, find a, b and c if T (0) = ¡4, T (1) = ¡2 and T (2) = 6: DOMAIN AND RANGE The domain of a relation is the set of permissible values that x may have The range of a relation is the set of permissible values that y may have For example: All values of x > ¡1 are permissible y (1) So, the domain is fx j x > ¡1g or x [ ¡1, [: x All values of y > ¡3 are permissible (-1,-3) So, the range is fy j y > ¡3g or y [ ¡3, [ y (2) x can take any value (2,¡1) So, the domain is fx j x is in R g or x R x y cannot be > So, the range is fy j y 1g or y ] ¡ 1, ] y (3) x can take all values except x = 2: So, the domain is fx j x 6= 2g y=1 Likewise, the range is fy j y 6= 1g x x=2 The domain and range of a relation are often described using interval notation For example: The domain consists of all real x such that x > and we write this as y fx j x > 3g or x [ 3, [: range (3, 2) the set of all x x magenta yellow 95 100 50 75 25 95 Likewise the range would be fy j y > 2g or y [ 2, [ 100 50 25 95 100 50 75 25 95 100 50 75 25 cyan 75 domain such that x > black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\023IB_HL-2_01.CDR Tuesday, 23 October 2007 4:47:58 PM PETERDELL IB_HL-2ed (24) 24 FUNCTIONS (Chapter 1) For the profit function alongside: ² the domain is fx j x > 0g or x [ 0, [ ² profit ($) 100 range fy j y 100g or y ] ¡ 1, 100 ]: the range is items made (x) 10 domain Intervals have corresponding graphs For example: fx j x > 3g or x [ 3, [ is read “the set of all x such that x is greater than or equal to 3” and has x number line graph fx j x < 2g or x ] ¡1, [ has number line graph fx j ¡2 < x 1g or x ] ¡2, ] has number line graph fx j x or x > 4g has number line graph -2 i.e., x ] ¡1, ] or ] 4, [ Note: a x x for numbers between a and b we write a < x < b or x ] a, b [ : b a x for numbers ‘outside’ a and b we write x < a or x > b i.e., x ] ¡1, a [ È ] b, [ where È means ‘or’ b Example For each of the following graphs state the domain and range: y y a b (4,¡3) x x (8,-2) b magenta 95 fy j y > ¡1g or y [¡1, 1[ 100 50 75 25 95 yellow fx j x is in R g or x R Domain is Range is 100 50 75 25 95 fy j y > ¡2g or y [¡2, 1[ 100 50 25 95 100 50 75 25 Range is cyan fx j x 8g or x ] ¡ 1, 8] Domain is 75 a (2,-1) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\024IB_HL-2_01.CDR Thursday, 12 July 2007 12:29:30 PM DAVID3 IB_HL-2ed (25) FUNCTIONS (Chapter 1) 25 EXERCISE 1B.2 Write down the domain and range for each of the following functions: a f(1, 3), (2, 5), (3, 7)g b f(¡1, 3), (0, 3), (2, 5)g d f(x, y) j x2 + y2 = 4, x Z , y > 0g c f(¡3, 1), (¡2, 1), (¡1, 1), (3, 1)g For each of the following graphs find the domain and range: y y a b c y (5,¡3) (-1,¡1) (0,¡2) x x y = -1 x x=2 d e y y f y (-1,¡2) x (2,¡-2) (1,¡-1) f(x) = x=2 x =-2 (-4,-3) Find the domain and range of each of the following functions: c b f : x 7! a f : x 7! 2x ¡ p p f e y = x2 ¡ d f : x 7! x2 + g x -1 x p 2¡x f(x) = p 2x ¡ h i f(x) = j3x ¡ 1j + y= x¡2 f : x 7! + 5¡x cyan f : x 7! k y = x3 + yellow 50 75 25 95 100 50 75 25 x3 100 magenta 3x ¡ ¡x¡2 x2 f : x 7! x + h y = x3 ¡ 3x2 ¡ 9x + 10 j y = x2 + x¡2 l f : x 7! x4 + 4x3 ¡ 16x + 95 i x+4 x¡2 x f 100 y= 95 g 50 f : x 7! 5x ¡ 3x2 75 e 25 95 100 50 75 25 Use a graphics calculator to help sketch graphs of the following functions Find the domain and range of each p a f(x) = x b f : x 7! x p c f : x 7! ¡ x d y = x ¡ 7x + 10 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\025IB_HL-2_01.CDR Thursday, 12 July 2007 12:30:27 PM DAVID3 IB_HL-2ed (26) 26 FUNCTIONS (Chapter 1) INVESTIGATION FLUID FILLING FUNCTIONS When water is added at a constant rate to a cylindrical container, the depth of water in the container is a linear function of time.¡ This is because the volume of water added is directly proportional to the time taken to add it.¡ If water was not added at a constant rate the direct proportionality would not exist water The depth-time graph for the case of a cylinder would be as shown alongside: depth depth DEMO time The question arises: ‘What changes in appearance of the graph occur for different shaped containers?’ Consider a vase of conical shape depth time What to do: For each of the following containers, draw a ‘depth v time’ graph as water is added: a b c d e Use the water filling demonstration to check your answers to question Write a brief report on the connection between the shape of a vessel and the corresponding shape of its depth-time graph You may wish to discuss this in parts For example, first examine cylindrical containers, then conical, then other shapes Gradients of curves must be included in your report Draw possible containers as in question which have the following ‘depth v time’ graphs: a b c d depth depth depth cyan magenta yellow 95 100 50 75 25 time 95 100 50 75 25 time 95 100 50 75 25 95 100 50 75 25 time depth black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\026IB_HL-2_01.CDR Thursday, 12 July 2007 12:31:30 PM DAVID3 time IB_HL-2ed (27) FUNCTIONS (Chapter 1) 27 COMPOSITE FUNCTIONS, f¡±¡g C Given f : x 7! f (x) and g : x 7! g(x), then the composite function of f and g will convert x into f(g(x)) f ± g is used to represent the composite function of f and g f ± g means “f following g” and (f ± g)(x) = f (g(x)) i.e., f ± g : x 7! f(g(x)) Consider f : x 7! x4 and g : x 7! 2x + f ± g means that g converts x to 2x + and then f converts (2x + 3) to (2x + 3)4 This is illustrated by the two function machines below Notice how f is following g x g-function machine 2x + I double and then add f-function machine I raise the number to the power 2x + Algebraically, if f(x) = x (f ± g)(x) = f (g(x)) = f (2x + 3) = (2x + 3)4 (2!\+\3)V and g(x) = 2x + 3, then fg operates on x firstg ff operates on g(x) nextg and (g ± f )(x) = g(f (x)) = g(x4 ) = 2(x4 ) + = 2x4 + So, in general, f (g(x)) 6= g(f(x)) The ability to break down functions into composite functions is useful in differential calculus Example Given f : x 7! 2x + and g : x 7! ¡ 4x find in simplest form: a (f ± g)(x) b (g ± f )(x) f(x) = 2x + and g(x) = ¡ 4x (f ± g)(x) = f (g(x)) = f (3 ¡ 4x) = 2(3 ¡ 4x) + = ¡ 8x + = ¡ 8x cyan magenta yellow b 95 (g ± f)(x) = g(f(x)) = g(2x + 1) = ¡ 4(2x + 1) = ¡ 8x ¡ = ¡8x ¡ 2(¢) + 1, 2(¤) + 1, 2(3x ¡ 4) + 100 50 75 25 f (¢) = f(¤) = f(3x ¡ 4) = 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Note: If f(x) = 2x + then ) a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\027IB_HL-2_01.CDR Friday, 25 January 2008 12:38:22 PM PETERDELL IB_HL-2ed (28) 28 FUNCTIONS (Chapter 1) If F (x) = (f ± g)(x), the domain of F is the domain of g excluding any values of x such that g(x) = u where f (u) is undefined Note: EXERCISE 1C Given f : x 7! 2x + and g : x 7! ¡ x, find in simplest form: a (f ± g)(x) b (g ± f)(x) c (f ± g)(¡3) Given f : x 7! x2 and g : x 7! ¡ x find (f ± g)(x) and (g ± f )(x) Find also the domain and range of f ± g and g ± f: Given f : x 7! x2 + and g : x 7! ¡ x, find in simplest form: a (f ± g)(x) b (g ± f)(x) c x if (g ± f )(x) = f (x) Functions f and g are defined as follows: f = f(0, 2), (1, 3), (2, 0), (3, 1)g Find a f ± g b g±f g = f(0, 3), (1, 2), (2, 1), (3, 0)g c f ±f f and g are defined as: f = f(0, 3), (1, 0), (2, 1), (3, 2)g g = f(0, 1), (1, 2), (2, 3), (3, 0)g Find f ± g f and g are defined as: f = f(0, 2), (1, 5), (2, 7), (3, 9)g g = f(2, 2), (5, 0), (7, 1), (9, 3)g Find a f ± g b g±f Given f(x) = a x+3 x+2 x+1 , find in simplest form: x¡1 (g ± f )(x) c (g ± g)(x) and g(x) = (f ± g)(x) b In each case, find the domain of the composite function a If ax + b = cx + d for all values of x, show that a = c and b = d Hint: If it is true for all x, it is true for x = and x = b Given f (x) = 2x + and g(x) = ax + b and that (f ± g)(x) = x for all values of x, deduce that a = 12 and b = ¡ 32 c Is the result in b true if (g ± f)(x) = x for all x? D SIGN DIAGRAMS Sometimes we not wish to draw a time-consuming graph of a function but wish to know when the function is positive, negative, zero or undefined A sign diagram enables us to this and is relatively easy to construct A sign diagram consists of: ² ² cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 50 75 25 ² 100 a horizontal line which is really the x-axis positive (+) and negative (¡) signs indicating that the graph is above and below the x-axis respectively critical values, the numbers written below the line, which are the graph’s x-intercepts or where it is undefined black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\028IB_HL-2_01.CDR Thursday, 20 December 2007 9:37:41 AM PETERDELL IB_HL-2ed (29) 29 FUNCTIONS (Chapter 1) Consider the three functions given below y = ¡2(x ¡ 1)2 y = (x + 2)(x ¡ 1) Function y y= y y Graph x x x x -2 + -2 Sign diagram Notice that: ² - + - 1 x - + x x A sign change occurs about a critical value for single factors such as (x + 2) and (x ¡ 1), indicating cutting of the x-axis ² No sign change occurs about the critical value for squared factors such as (x ¡ 1)2 , indicating touching of the x-axis ² shows a function is undefined at x = 0 In general: ² ² when a factor has an odd power there is a change of sign about that critical value when a factor has an even power there is no sign change about that critical value Example a Draw sign diagrams for: b y -2 y x=3 x x Qw_ x=-2 a b - cyan magenta x 95 50 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 -2 + + yellow Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\029IB_HL-2_01.CDR Friday, 13 July 2007 12:41:19 PM DAVID3 100 + 75 - black -2 - + Qw_ x IB_HL-2ed (30) 30 FUNCTIONS (Chapter 1) Example a (x + 3)(x ¡ 1) Draw a sign diagram of: a (x + 3)(x ¡ 1) has critical values of ¡3 and b 2(2x + 5)(3 ¡ x) 2(2x + 5)(3 ¡ x) has critical values of ¡ 52 and b + -3 - -\Tw_ x We try any number > 1, e.g., x = As (5)(1) > we put a + sign here - As the factors are ‘single’ the signs alternate giving: - + -3 x We try any number > 3, e.g., x = As 2(15)(¡2) < we put a ¡ sign here As the factors are ‘single’ the signs alternate giving: + + x -\Tw_ a 12 ¡ 3x2 b ¡4(x ¡ 3)2 x Example Draw a sign diagram of: a 12 ¡ 3x2 = ¡3(x2 ¡ 4) = ¡3(x + 2)(x ¡ 2) which has critical values of ¡2 and b ¡4(x ¡ 3)2 has a critical value of -2 x We try any number > e.g., x = As ¡3(5)(1) is < we put a ¡ sign here As the factors are ‘single’ the signs alternate - cyan magenta 95 yellow Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\030IB_HL-2_01.CDR Friday, 13 July 2007 12:41:40 PM DAVID3 100 50 75 25 50 75 25 95 100 50 75 25 95 100 50 75 25 -2 x 95 + We try any number > e.g., x = As ¡4(1)2 is < we put a ¡ sign here As the factor is ‘squared’ the signs not change 100 - x black x IB_HL-2ed (31) 31 FUNCTIONS (Chapter 1) Example x¡1 : 2x + Draw a sign diagram for x¡1 2x + is zero when x = and undefined when x = ¡ 12 : + So, -\Qw_ x¡1 = 2x + For x = 10, x 21 >0 Since (x ¡ 1) and (2x + 1) are single factors the signs alternate ) sign diagram is + - -\Qw_ + x EXERCISE 1D From the graphs below, draw corresponding sign diagrams: a b c y y -1 x y x x d e y f y -2 y x x -2 x g h i y y x y j -3 x -1 x=-1 y k x=1 l y=2 x=-2 -1 y x x=2 x x x y cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 x=3 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\031IB_HL-2_01.CDR Thursday, 12 July 2007 12:52:26 PM DAVID3 IB_HL-2ed (32) 32 FUNCTIONS (Chapter 1) Draw a d g j m sign diagrams for: (x + 4)(x ¡ 2) ¡(x + 1)(x ¡ 3) x2 ¡ x2 ¡ 3x + ¡ 16x ¡ 6x2 b e h k n x(x ¡ 3) (2x ¡ 1)(3 ¡ x) ¡ x2 ¡ 8x2 ¡2x2 + 9x + c f i l o Draw sign diagrams for: a (x + 2)2 d ¡(x ¡ 4)2 g 4x2 ¡ 4x + b e h (x ¡ 3)2 x2 ¡ 2x + ¡x2 ¡ 6x ¡ c f i Draw sign diagrams for: x+2 a x¡1 x(x + 2) (5 ¡ x)(1 ¡ 2x) 5x ¡ x2 6x2 + x ¡ ¡15x2 ¡ x + ¡(x + 2)2 ¡x2 + 4x ¡ ¡4x2 + 12x ¡ b x x+3 c 2x + 4¡x d 4x ¡ 2¡x e 3x x¡2 f ¡8x 3¡x g (x ¡ 1)2 x h 4x (x + 1)2 i (x + 2)(x ¡ 1) 3¡x j x(x ¡ 1) 2¡x k x2 ¡ ¡x l 3¡x 2x2 ¡ x ¡ m x2 ¡ x+1 n x2 + x o x2 + 2x + x+1 p ¡(x ¡ 3)2 (x2 + 2) x+3 q ¡x2 (x + 2) 5¡x r x2 + (x ¡ 3)2 (x ¡ 1) s x¡5 +3 x+1 t x¡2 ¡4 x+3 u 3x + x ¡ ¡ x¡2 x+3 E INEQUALITIES (INEQUATIONS) 2x ¡ x+3 are examples of inequalities x In this section we aim to find all values of the unknown for which the inequality is true 2x + > 11 ¡ x and GROUP INVESTIGATION SOLVING INEQUALITIES 3x + > was: 1¡x Jon’s method of solving If cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\032IB_HL-2_01.CDR Thursday, 12 July 2007 12:53:59 PM DAVID3 3x + > 4, then 1¡x 3x + > 4(1 ¡ x) 3x + > ¡ 4x ) 7x > ) x > 27 IB_HL-2ed (33) 33 FUNCTIONS (Chapter 1) 3x + = 1¡x However, Sarah pointed out that if x = 5, greater than They concluded that there was something wrong with the method of solution 17 ¡4 = ¡4 14 and ¡4 14 is not x=1 y y=4 A graph also highlighted an error x It seems that the correct answer is 27 < x < Wu_ y=-3 y= Questions: 3x + 1- x At what step was Jon’s method wrong? Suggest an algebraic method which does give the correct answer From the Investigation above you should have concluded that multiplying both sides of an inequality by an unknown can lead to incorrect results We therefore need an alternative method ² ² ² ² To solve inequalities we use these steps: ² ² ² ² Note: if if if if Make the RHS zero by shifting all terms to the LHS Fully factorise the LHS Draw a sign diagram for the LHS Determine the values required from the sign diagram a > b and a > b and a > b and a > b > 0, c R , then a + c > b + c c > 0, then ac > bc c < 0, then ac < bc then a2 > b2 Example 10 a 3x2 + 5x > Solve for x: 3x2 + 5x > ) 3x + 5x ¡ > ) (3x ¡ 1)(x + 2) > a b x2 + < 6x fmaking RHS zerog ffully factorising LHSg + Sign diagram of LHS is -2 x Qe_ x ] ¡ 1, ¡2 ] or x [ 13 , [ So, for LHS > 0, x2 + < 6x ) x ¡ 6x + < ) (x ¡ 3)2 < Sign diagram of LHS is b + fmake RHS zerog ffully factorising LHSg + + x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 So, the inequality is never true black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\033IB_HL-2_01.CDR Thursday, 12 July 2007 1:11:09 PM DAVID3 IB_HL-2ed (34) 34 FUNCTIONS (Chapter 1) Example 11 3x + 61 x¡4 a Solve for x: 3x + 61 x¡4 a b 3x + ¡1 x¡4 µ ¶ µ ¶ 3x + x¡4 ¡1 60 x¡4 x¡4 ) ) 3x + ¡ (x ¡ 4) 60 x¡4 ) + - ) + - x qA_p_ , 1[ Thus, x ] ¡ 1, [ or [ 10 + -3 10 x ) ¡ 10 x ³x´ ¡ 10 60 ) x x ¡ 10x ) 60 x - 2x + 60 x¡4 ) 10 x b x x [ ¡3, [ Example 12 Solve for x: 3x + 3x > x¡1 x+1 ) ) 3x + 3x > x¡1 x+1 3x + 3x ¡ >0 x¡1 x+1 (3x + 1)(x + 1) ¡ 3x(x ¡ 1) >0 (x ¡ 1)(x + 1) 7x + >0 (x ¡ 1)(x + 1) ) Sign diagram of LHS is: -1 + -\Qu_ + x Thus, x ] ¡ 1, ¡ 17 [ or ]1, 1[ EXERCISE 1E cyan magenta 95 yellow Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\034IB_HL-2_01.CDR Friday, 13 July 2007 12:42:36 PM DAVID3 c f i l o r 100 50 75 25 95 50 75 100 (x ¡ 1)2 < x2 > 3x 2x2 > 4x2 ¡ 4x + < 2x2 ¡ 4x + > 12x2 > 5x + b e h k n q 25 95 100 50 75 25 for x: (2 ¡ x)(x + 3) > x2 > x x2 < 2x2 > x + 3x2 > 8(x + 2) + 5x < 6x2 95 100 50 75 25 Solve a d g j m p black (2x + 1)(3 ¡ x) > 3x2 + 2x < x2 + 4x + > 6x2 + 7x < 6x2 + 5x 2x2 + > 9x IB_HL-2ed (35) FUNCTIONS (Chapter 1) Solve for x: x+4 a >0 2x ¡ 2x d >1 x¡3 g > 100 x e h < 2x ¡ x+7 x > x+2 x j m n F f i x2 + 5x 60 x2 ¡ 6x x l o 2x ¡ 2x < x+2 x¡2 q x+3 >0 2x + x+2 < 2x ¡ 1¡x <4 1+x c x2 ¡ 2x >0 x+3 x> x k x3 > x p x+1 <0 4¡x x+2 > ¡3 x¡1 x >5 2x ¡ b 35 x2 61 3x ¡ r THE MODULUS FUNCTION The modulus of a real number x is its distance from on the number line Because the modulus is a distance, it cannot be negative -5 So, the modulus of is 7, which is written as j j = and the modulus of ¡5 is 5, which is written as j ¡5 j = 5: j x j is the distance of x from on the number line Thus, |x| If x > |x| If x < x x ALGEBRAIC DEFINITION ½ The modulus of x, j x j = y = j x j has graph if x > if x < x ¡x y y¡=¡|x| This branch is y = -x, x < This branch is y = x, x > x p p p p 72 = 49 = and (¡5)2 = 25 = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 is an equivalent definition of j x j p x2 95 50 75 25 95 100 50 75 25 Thus j x j = 100 Notice that black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\035IB_HL-2_01.CDR Thursday, 12 July 2007 1:56:20 PM DAVID3 IB_HL-2ed (36) 36 FUNCTIONS (Chapter 1) Example 13 If a = ¡3 and b = find: a j7 + aj b j7 + aj a j ab j j ab j b = j + (¡3) j = j4j =4 ¯ ¯ ¯ a + 2b ¯ c ¯ ¯ ¯ a + 2b ¯ ¯ ¯ = ¯ (¡3)2 + 2(4) ¯ = j9 + 8j = j 17 j = 17 c = j (¡3)(4) j = j¡12 j = 12 Example 14 By replacing j x j with x for x > and ¡x for x < 0, write the following functions without the modulus sign and hence graph each function: a f(x) = x ¡ j x j b f(x) = x j x j a If x < 0, f (x) = x ¡ (¡x) = 2x If x > 0, f (x) = x ¡ x = 0: ½ y = 2x for x < So, we graph y = for x > 0: b If x > 0, f (x) = x(x) = x2 If x < 0, f (x) = x(¡x) = ¡x2 : ½ y = x2 for x > So, we graph y = ¡x2 for x < 0: y y branch y = x branch y = x x branch y = 2x branch y = -x EXERCISE 1F.1 If a = ¡2, b = 3, c = ¡4 find the value of: a jaj b jbj c e ja ¡ bj f jaj ¡jbj g i j a j2 j a2 k If x = ¡3, find the value of: cyan magenta yellow d j ab j ja + bj ¯c¯ ¯ ¯ ¯ ¯ a h jaj + jbj l jcj jaj 95 100 Is j a ¡ b j = j a j ¡ j b j ? 50 b 75 c ¯ ¯ ¯ 2x + ¯ ¯ ¯ ¯ 1¡x ¯ 25 95 100 50 75 j5j ¡ jxj 25 95 100 50 b 75 25 Is j a + b j = j a j + j b j ? a 95 j5 ¡ xj 100 50 75 25 a jajjbj black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\036IB_HL-2_01.CDR Thursday, 12 July 2007 1:57:09 PM DAVID3 d ¯ ¯ ¯ ¡ 2x ¡ x2 ¯ IB_HL-2ed (37) FUNCTIONS (Chapter 1) 37 Copy and complete: j ab j a b 6 ¡2 ¡6 ¡6 ¡2 ¯a¯ ¯ ¯ ¯ ¯ b jajjbj jaj jbj What you suspect? Use the fact that j x j = p x2 j ab j = j a j j b j a to prove that: ¯ a ¯ jaj ¯ ¯ b ¯ ¯= , b 6= b jbj c ja ¡ bj = jb ¡ aj Using j a j = a if a > and ¡a if a < 0, write the following functions without modulus signs and hence graph each function: a y = jx ¡2j b y = jx+ 1j c y = ¡jxj d y = jxj +x e g y = jxj +jx ¡2j ¯ ¯ y = ¯ x2 ¡ ¯ h j jxj x y = jxj ¡ jx ¡1j ¯ ¯ y = ¯ x2 ¡ 2x ¯ y= k f i l y = x ¡ 2jxj ¯ ¯ y = ¯ x2 + ¯ ¯ ¯ y = ¯ x2 + 3x + ¯ MODULUS EQUATIONS From the previous exercise you should have discovered these properties of modulus: ² j x j > for all x ² j x j = x for all x ¯ ¯ ¯ x ¯ jxj ¯ ¯= ¯ y ¯ j y j for all x and y, y 6= ² ² j¡x j = j x j for all x ² j xy j = j x j j y j for all x and y ² j a ¡ b j = j b ¡ a j for all a and b It is clear that j x j = has two solutions, x = and x = ¡2 if j x j = a where a > 0, then x = §a In general, We use this rule to solve modulus equations Example 15 a j 2x + j = cyan magenta yellow 95 50 b 75 25 95 100 50 75 25 j 2x + j = 2x + = §7 ) 2x = ¡ or ¡7 ¡ ) 2x = or ¡10 ) x = or ¡5 95 50 75 25 95 100 50 75 25 ) 100 a b j ¡ 2x j = ¡1 100 Solve for x: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\037IB_HL-2_01.CDR Friday, 16 February 2007 2:33:26 PM PETERDELL j ¡ 2x j = ¡1 has no solution as LHS is never negative IB_HL-2ed (38) 38 FUNCTIONS (Chapter 1) Example 16 ¯ ¯ ¯ 3x + ¯ ¯ ¯ ¯ 1¡x ¯=4 Solve for x: ¯ ¯ ¯ 3x + ¯ ¯ ¯ ¯ 1¡x ¯=4 If 3x + = §4 1¡x ) 3x + = ¡4 1¡x then 3x + = ¡4(1 ¡ x) ) 3x + = ¡4 + 4x ) 6=x 3x + =4 1¡x If then 3x + = 4(1 ¡ x) ) 3x + = ¡ 4x ) 7x = ) x = 27 So, x = or 6: j x j = j b j then x = §b Also notice that if Example 17 j x + j = j 2x ¡ j Solve for x: If j x + j = j 2x ¡ j , x + = §(2x ¡ 3) fusing property aboveg then If x + = 2x ¡ then = x If x + = ¡(2x ¡ 3) then x + = ¡2x + ) 3x = ) x = 23 So, x = or 4: EXERCISE 1F.2 Solve for x: a jxj = d jx ¡ 1j = g j 3x ¡ j = Solve for x: ¯ ¯ ¯ x ¯ ¯=3 a ¯¯ x¡1 ¯ b e h j x j = ¡5 j3 ¡ xj = j ¡ 2x j = c f i jxj = j x + j = ¡1 j ¡ 5x j = 12 b ¯ ¯ ¯ 2x ¡ ¯ ¯ ¯ ¯ x+1 ¯=5 c ¯ ¯ ¯ x+3 ¯ ¯ ¯ ¯ ¡ 3x ¯ = Solve for x: a jx + 1j = j2 ¡ xj b jxj = j5 ¡ xj c j 3x ¡ j = j x + j d j 2x + j = j ¡ x j e j ¡ 4x j = j x ¡ j f j 3x + j = j ¡ x j cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Solve for x using i a graphical method ii an algebraic method: a j x + j = 2x + b j 2x + j = j x j ¡ c j x ¡ j = 25 x + black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\038IB_HL-2_01.CDR Thursday, 12 July 2007 1:58:55 PM DAVID3 IB_HL-2ed (39) FUNCTIONS (Chapter 1) 39 MODULUS INEQUALITIES Notice that if j x j < then x lies between ¡2 and i.e., if j x j < 2, then ¡2 < x < 2: Likewise, if j x j > 2, then x > or x < ¡2: ² ² In general, if j x j < k where k > 0, then ¡k < x < k if j x j > k where k > 0, then x < ¡k or x > k Example 18 a j 3x ¡ j < 10 Solve for x: j 3x ¡ j < 10 ) ¡10 < 3x ¡ < 10 ) ¡3 < 3x < 17 ) ¡ 33 < x < 17 a ) x ] ¡ 1, b b j ¡ 2x j > 17 fadding to each partg [ j ¡ 2x j > ) j 2x ¡ j > fusing j a ¡ b j = j b ¡ a jg ) 2x ¡ > or 2x ¡ ¡4 ) 2x > or 2x ¡1 i.e., x ] ¡ 1, ¡ 12 ] or [ 72 , [ ) x > 72 or x ¡ 12 Notice that a and b above could be solved by a different method In j 3x ¡ j < 10, ) j 3x ¡ j < 100 ) (3x ¡ 7)2 ¡ 102 < [3x ¡ + 10] [3x ¡ ¡ 10] < ) [3x + 3] [3x ¡ 17] < ) + The LHS has sign diagram: Example 19 we see that both sides are non-negative fsquaring both sidesg fjaj2 = a2 for all ag - + -1 Qd_U_ x so x ] ¡ 1, 17 [ Find exactly where j x ¡ j > j ¡ x j j x ¡ j > j ¡ x j are non-negative Both sides of the inequality ) we square both sides to get j x ¡ j2 > j ¡ x j2 ) 4(x ¡ 1)2 ¡ (3 ¡ x)2 > fas j a j2 = a2 for a R g ) [2(x ¡ 1) + (3 ¡ x)] [2(x ¡ 1) ¡ (3 ¡ x)] > ) (x + 1)(3x ¡ 5) > The critical values are x = ¡1, magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 with sign diagram: x ] ¡ 1, ¡1 ] or [ 53 , [ (check with GCD) ) cyan black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\039IB_HL-2_01.CDR Thursday, 12 July 2007 2:02:24 PM DAVID3 + -1 - + Te_ x IB_HL-2ed (40) 40 FUNCTIONS (Chapter 1) Sometimes a graphical solution is easier Example 20 Solve graphically: j ¡ 2x j > x + 1: We draw graphs of y = j ¡ 2x j and y = x + on the same set of axes ( ¡ 2x for ¡ 2x > 0, i.e., x 12 y = j ¡ 2x j = ¡1 + 2x for ¡ 2x < 0, i.e., x > 12 y y = - 2x -1 Now j ¡ 2x j > x + when the graph of y = j ¡ 2x j lies above y = x + 1, ) x < or x > 2, y = x +1 x -1 Qw_ i.e., x ] ¡1, [ or x ] 2, [ EXERCISE 1F.3 Solve for x: a jxj < b jxj > c jx + 3j d jx + 4j > e j 2x ¡ j < f j ¡ 4x j > g j 2x + j < h > jx ¡1j i j ¡ 7x j < j j ¡ 7x j > k j ¡ 3x j l > j3 ¡xj Solve: a jx ¡ 3j e b j 2x ¡ j jxj > j2 ¡ xj Solve graphically: a j 2x ¡ j < x c j 3x + j > ¯ ¯ ¯ x ¯ ¯ ¯>3 g ¯ x¡2¯ j x j j ¡ 2x j f ¯ ¯ c ¯ x2 ¡ x ¯ > b 2x ¡ < j x j Graph the function f (x) = d j ¡ 2x j > ¯ ¯ ¯ 2x + ¯ ¯ ¯>2 h ¯ x¡1 ¯ d jxj ¡ > j4 ¡ xj jxj jxj , and hence find all values of x for which > ¡ 12 x¡2 x¡2 a Draw the graph of y = j x + j + j x + j + j x j + j x ¡ j b P Q O A -5 -2 R B P, Q and R are factories which are 5, and km away from factory O respectively cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A security service wishes to know where it should locate its premises along AB so black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\040IB_HL-2_01.CDR Friday, 16 February 2007 2:38:40 PM PETERDELL IB_HL-2ed (41) 41 FUNCTIONS (Chapter 1) that the total length of cable to the factories is a minimum i Explain why the total length of cable is given by j x + j + j x + j + j x j + j x ¡ j where x is the position of the security service on AB ii Where should the security service set up to minimise the length of cable to all factories? What is the minimum length of cable? iii If a fifth factory at S, located km right of O, also requires the security service, where should the security service locate its premises for minimum cable length? Which of these is true? Give proof a j x + y j j x j + j y j for all x, y G b j x ¡ y j > j x j ¡ j y j for all x, y THE RECIPROCAL FUNCTION x 7! x or f (x) = x x x is defined as the reciprocal function It has graph: Notice that: y y=-x y= x is undefined when x = x exists in the The graph of f (x) = x first and third quadrants only ² y=x f (x) = ² (1,¡1) x (-1,-1) ² is symmetric about y = x x and y = ¡x ² as as as as ² f (x) = f (x) = ! reads “approaches” or “tends to” x ! 1, f (x) ! (from above) x ! ¡1, f (x) ! (from below) x ! (from right), y ! x ! (from left), y ! ¡1 is asymptotic to the x-axis x and to the y-axis The graph gets closer to the axes as it gets further from the origin EXERCISE 1G , g(x) = , h(x) = on the same set of axes x x x Comment on any similarities and differences Sketch the graphs of f(x) = cyan magenta 95 yellow Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\041IB_HL-2_01.CDR Friday, 13 July 2007 12:44:50 PM DAVID3 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Sketch the graphs of f(x) = ¡ , g(x) = ¡ , h(x) = ¡ on the same set of axes x x x Comment on any similarities and differences black IB_HL-2ed (42) 42 FUNCTIONS (Chapter 1) H ASYMPTOTES OF OTHER RATIONAL FUNCTIONS 2x + x¡1 Notice that at x = 1, f(x) is undefined Consider the function f(x) = Its graph is: As the graph approaches the vertical line x = 1, we say that x = is a vertical asymptote y y=2 Notice that: f(1:001) = 3002 and f (0:999) = ¡2998 We write: as x ! (from the left), f (x) ! ¡1 as x ! (from the right), f (x) ! x -\Qw_ -1 or alternatively, as x ! 1¡ , f (x) ! ¡1 as x ! 1+ , f (x) ! x=1 We also notice that f (1000) = 2001 ¡1999 + 2:003 and f (¡1000) = + 1:997 999 ¡1001 This indicates that y = is a horizontal asymptote and we write: as x ! 1, y ! (from above) or as x ! 1, y ! 2+ as x ! ¡1, y ! (from below) as x ! ¡1, y ! 2¡ Notice that as jxj ! 1, f(x) ! The sign diagram of y = 2x + x¡1 + -\Qw_ is - + x and can be used to discuss the function near its vertical asymptote without having to graph the function Now consider the function f(x) = x + + Its graph is: x = is a vertical asymptote At x = 2, f (x) is undefined y As x ! 2¡ , f(x) ! ¡1 as x ! 2+ , f(x) ! So, as x ! 1, f(x) ! x + (from above) as x ! ¡1, f(x) ! x + (from below) x yellow 95 100 50 75 25 95 100 50 75 25 Thus y = x + is an oblique asymptote 95 100 50 75 25 95 100 50 75 25 x=2 magenta ! as jxj ! x¡2 Notice that y=x+1 cyan x¡2 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\042IB_HL-2_01.CDR Thursday, 12 July 2007 2:14:04 PM DAVID3 IB_HL-2ed (43) FUNCTIONS (Chapter 1) DISCUSSION 43 ASYMPTOTES What would be the asymptotes of ² y= 2x2 (x ¡ 1)(x ¡ 4) ² y = x2 + INVESTIGATION ? x FINDING ASYMPTOTES Use the graphing package supplied or a graphics calculator to examine the following functions for asymptotes: GRAPHING PACKAGE b y= 3x + x+2 c y= 3x ¡ (x ¡ 2)(x + 1) 2x x ¡4 e y= 1¡x (x + 2)2 f y= x2 + x¡1 x2 ¡ x2 + h y= x2 ¡ 6x + (x + 1)2 i y = x2 + a y = ¡1 + d y= g y= x¡2 x Further examples of asymptotic behaviour are seen in exponential, logarithmic and some trigonometric functions A function may cross a horizontal or oblique asymptote, but never a vertical asymptote Note: EXERCISE 1H cyan f (x) = ¡ x+1 c f : x 7! x+3 (x + 1)(x ¡ 2) d f (x) = x + x¡3 e y= f y= g f : x 7! x + + h g : x 7! 2x ¡ i y= j y = x2 ¡ magenta yellow 50 75 25 95 100 50 75 25 4x ¡ 4x ¡ 5 100 x2 x¡2 x+1 95 x2 ¡ x2 + 100 b 95 x¡2 50 f : x 7! 75 a 25 95 100 50 75 25 For the following functions: i determine the asymptotes ii discuss the behaviour of the function as it approaches its asymptotes iii sketch the graph of the function iv find the coordinates of all points where the function crosses its asymptotes black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\043IB_HL-2_01.CDR Thursday, 12 July 2007 2:14:43 PM DAVID3 2x2 + x2 ¡ 2x x2 + x IB_HL-2ed (44) 44 FUNCTIONS (Chapter 1) I INVERSE FUNCTIONS The operations of + and ¡, £ and ¥, squaring and finding the square root, are inverse operations as one undoes what the other does p 82 = For example, x + ¡ = x, x £ ¥ = x and A function y = f (x) may or may not have an inverse function If y = f(x) has an inverse function, this new function f ¡1 (x) ² ² ² must indeed be a function, i.e., satisfy the vertical line test must be the reflection of y = f (x) in the line y = x must satisfy the condition that f ¡1 : f (x) 7! x The function y = x, defined as f : x 7! x, is the identity function This means that, for any function f that has an inverse function f ¡1 , f ±f ¡1 and f ¡1 ±f must always equal the identity function So, ( f ± f ¡1 )(x) = (f ¡1 ± f )(x) = x, so the inverse function undoes the effect of the function on x If (x, y) lies on f , then (y, x) lies on f ¡1 So reflecting the function in y = x has the algebraic effect of interchanging x and y For example, f : y = 5x + becomes f ¡1 : x = 5y + Consider: y y=ƒ(x) y = f ¡1 (x) is the inverse of y = f(x) as ² it is also a function ² it is the reflection of y = f(x) in the oblique line y = x y=x y=ƒ -1(x) x y This is the reflection of y = f(x) in y = x, but it is not the inverse function of y = f(x) as it fails the vertical line test.¡ y=ƒ(x) x We say that the function y = f(x) does not have an inverse y=x y = f -1 (x) y=ƒ(x), x¡>¡0 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 y=x (x), y¡>¡0 y=ƒ However, consider the same function y = f (x) but with the domain x > The function does now have an inverse function, as drawn alongside y = f (x) subject to x would also have an inverse function -1 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\044IB_HL-2_01.CDR Thursday, 20 December 2007 9:40:37 AM PETERDELL IB_HL-2ed (45) 45 FUNCTIONS (Chapter 1) Example 21 Consider f : x 7! 2x + a On the same axes, graph f and its inverse function f ¡1 b Find f ¡1 (x) using i coordinate geometry and the slope of f ¡1 (x) from a ii variable interchange ¡1 c Check that (f ± f )(x) = (f ¡1 ± f)(x) = x a b f(x) = 2x + passes through (0, 3) and (2, 7) ) f ¡1 (x) passes through (3, 0) and (7, 2) 2¡0 = 7¡3 y¡0 So, its equation is = x¡3 x¡3 i.e., y = x¡3 ) f ¡1 (x) = i This line has slope y y=ƒ(x) (2,¡7) y=x y=ƒ -1(x) (0,¡3) (7,¡2) (3,¡0) x ii f is y = 2x + 3, so f ¡1 is x = 2y + ) x ¡ = 2y x¡3 x¡3 ) =y i.e., f ¡1 (x) = 2 (f ± f ¡1 )(x) = f (f ¡1 (x)) µ ¶ x¡3 =f µ ¶ x¡3 =2 +3 c (f ¡1 ± f )(x) = f ¡1 (f (x)) = f ¡1 (2x + 3) (2x + 3) ¡ = 2x = =x and =x If ¦ includes point (a, b) then ¦-1 includes point (b, a) EXERCISE 1I For each of the following functions f i on the same axes graph y = x, f and f ¡1 ii find f ¡1 (x) using coordinate geometry and i iii find f ¡1 (x) using variable interchange: a f : x 7! 3x + b f : x 7! x+2 For each of the following functions f ii sketch y = f (x), y = f ¡1 (x) and y = x on the same axes i find f ¡1 (x) iii show that f ¡1 ± f = f ± f ¡1 = x, the identity function: magenta yellow ¡ 2x 95 100 50 75 f : x 7! 25 95 100 50 b 75 25 95 100 50 25 95 100 50 75 25 cyan 75 f : x 7! 2x + a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\045IB_HL-2_01.CDR Thursday, 12 July 2007 3:52:48 PM DAVID3 c f : x 7! x + IB_HL-2ed (46) 46 FUNCTIONS (Chapter 1) Copy the graphs of the following functions and in each case include the graphs of y = x and y = f ¡1 (x) : a b c y y y x x -3 -2 x d e y f y y (2,¡2) x x x a Sketch the graph of f : x 7! x2 ¡ and reflect it in the line y = x b Does f have an inverse function? c Does f with restricted domain x > have an inverse function? Sketch the graph of f : x 7! x3 J and its inverse function f ¡1 (x) FUNCTIONS WHICH HAVE INVERSES It is important to understand the distinction between one-to-one and many-to-one functions A one-to-one function is any function where ² for each x there is only one value of y and ² for each y there is only one value of x Functions that are one-to-one satisfy both the ‘vertical line test’ and the ‘horizontal line test’ ² ² This means that: no vertical line can meet the graph more than once no horizontal line can meet the graph more than once Functions that are not one-to-one are called many-to-one While these functions must satisfy the ‘vertical line test’ they not satisfy the ‘horizontal line test’, i.e., at least one y-value has more than one corresponding x-value cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ² ² If the function y = f (x) is one-to-one, it will have an inverse function y = f ¡1 (x): If a function y = f(x) is many-to-one, it will not have an inverse function Many-to-one functions can have inverse functions for a restricted part of the domain (see Example 22) ² black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\046IB_HL-2_01.CDR Thursday, 12 July 2007 3:54:59 PM DAVID3 IB_HL-2ed (47) FUNCTIONS (Chapter 1) 47 Example 22 Consider f : x 7! x2 a Explain why the function defined above does not have an inverse function b Does f : x 7! x2 where x > have an inverse function? c Find f ¡1 (x) for f : x 7! x2 , x > 0: d Sketch y = f (x), y = x and y = f ¡1 (x) for f in b and f ¡1 in c f : x 7! x2 has domain x R and is many-to-one It does not pass the ‘horizontal line test’ a b y If we restrict the domain to x¡>¡0 or x¡2¡[0,¡1[, or in fact any domain which makes f one-to-one, it satisfies the ‘horizontal line test’ and so has an inverse function y f (x) = x , x > x x f is defined by y = x2 , x > ) f ¡1 is defined by x = y2 , y > p ) y = § x, y > p ) y= x p fas ¡ x is 0g p So, f ¡1 (x) = x c Note: d y @=!X' !>0 @=~`! x y=x , x 6= 0, is called the reciprocal function x It is said to be a self-inverse function as f = f ¡1 This is because the graph of y = is symmetrical about the line y = x x Any function with a graph which is symmetrical about the line y = x must be a self-inverse function The function f(x) = EXERCISE 1J Note: If the domain of a function is the set of all real numbers, then the statement x R will be omitted cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Which of the following functions have inverses? In each of these cases, write down the inverse function b f(¡1, 3), (0, 2), (1, 3)g a f(1, 2), (2, 4), (3, 5)g d f(¡1, ¡1), (0, 0), (1, 1)g c f(2, 1), (¡1, 0), (0, 2), (1, 3)g black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\047IB_HL-2_01.CDR Thursday, 12 July 2007 3:55:22 PM DAVID3 IB_HL-2ed (48) 48 FUNCTIONS (Chapter 1) a Show that f : x 7! x has an inverse function for all x 6= b Find f ¡1 algebraically and show that f is a self-inverse function 3x ¡ , x 6= is a self-inverse function by: x¡3 reference to its graph b using algebra Show that f : x 7! a The ‘horizontal line test’ says that: for a function to have an inverse function, no horizontal line can cut it more than once a Explain why this is a valid test for the existence of an inverse function b Which of the following functions have an inverse function? i ii iii y y y x x -1 -2 x (1,-1) c For the functions in b which not have an inverse, specify domains as wide as possible where each function does have an inverse Consider f : x 7! x2 where x a Find f ¡1 (x) b Sketch y = f(x), y = x and y = f ¡1 (x) on the same set of axes a Explain why f : x 7! x2 ¡ 4x + is a function but does not have an inverse function b Explain why f for x > has an inverse function p c Show that the inverse function of the function in b is f ¡1 (x) = + + x d If the domain of f is restricted to x > 2, state the domain and range of i f ii f ¡1 e Show that f ± f ¡1 = f ¡1 ± f = x, the identity function Given f : x 7! (x + 1)2 + where x > ¡1 : a find the defining equation of f ¡1 b sketch, using technology, the graphs of y = f (x), y = x and y = f ¡1 (x) c state the domain and range of i f ii f ¡1 Consider the functions f : x 7! 2x + and g : x 7! a Find g ¡1 (¡1) Given f : x 7! 5x a find i (f ± g ¡1 )(x) = b Solve for x if p and g : x 7! x : f (2) ii g ¡1 (4) 8¡x b solve the equation (g ¡1 ± f )(x) = 25 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 10 Given f : x 7! 2x and g : x 7! 4x ¡ show that (f ¡1 ± g ¡1 )(x) = (g ± f )¡1 (x) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\048IB_HL-2_01.CDR Thursday, 12 July 2007 3:56:15 PM DAVID3 IB_HL-2ed (49) FUNCTIONS (Chapter 1) 49 11 Which of these functions is a self-inverse function, i.e., f ¡1 (x) = f (x)? a f(x) = 2x b f (x) = x c f(x) = ¡x d f(x) = e f (x) = ¡ x x 12 Show that (f ± f ¡1 )(x) = (f ¡1 ± f )(x) = x for: x+3 a f(x) = 3x + b f(x) = 13 a y A b x c y = f (x) f(x) = p x B is the image of A under a reflection in the line y = x If A is (x, f(x)), what are the coordinates of B under the reflection? Substitute your result from a into y = f ¡1 (x) What result you obtain? Explain how to establish that f (f ¡1 (x)) = x also B y = f -1( x) c REVIEW SET 1A If f(x) = 2x ¡ x2 a f (2) find: b f(¡3) c f(¡ 12 ) For each of the following graphs determine: i the range and domain ii the x and y-intercepts iii whether it is a function iv if it has an inverse function y a b y -1 x -1 x -3 -\Wl_T_ (2,-5) Find a, b and c if f (0) = 5, f (¡2) = 21 and f (3) = ¡4 and f(x) = ax2 +bx+c Draw a sign diagram for: (3x + 2)(4 ¡ x) a b x¡3 x2 + 4x + If f(x) = 2x ¡ and g(x) = x2 + 2, find: Solve for x: x(x + 8) a 65 x+2 b a f (g(x)) b g(f (x)) > x¡1 2x + cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Consider x 7! 2x ¡ a On the same set of axes graph y = x, f and f ¡1 b Find f ¡1 (x) using variable interchange c Show that f ± f ¡1 = f ¡1 ± f = x, the identity function black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\049IB_HL-2_01.CDR Thursday, 12 July 2007 3:58:06 PM DAVID3 IB_HL-2ed (50) 50 FUNCTIONS (Chapter 1) Solve for x: ¯ ¯ ¯ 2x + ¯ ¯ ¯=3 a ¯ x¡2 ¯ j 3x ¡ j > j 2x + j b 4x + x2 + x ¡ a determine the asymptotes b discuss the behaviour of the function as it approaches its asymptotes c sketch the graph x 10 Given f : x 7! 3x + and h : x 7! , show that (f ¡1 ± h¡1 )(x) = (h ± f )¡1 (x) For f : x 7! REVIEW SET 1B If g(x) = x2 ¡ 3x, find in simplest form a b g(x + 1) g(x2 ¡ 2) For each of the following functions f (x) find f ¡1 (x) : + 2x a f(x) = ¡ 4x b f(x) = For each of the following graphs, find the domain and range y y a b y¡=¡(x-1)(x-5) (1,-1) x x x¡=¡2 Copy the following graphs and draw the graph of each inverse function: y y a b 2 x x Draw a sign diagram for: b x+9 +x x+5 b x2 ¡ 3x ¡ >0 x+2 Solve for x: Find an f and a g function given that: p a f(g(x)) = ¡ x2 cyan magenta yellow g(f (x)) = 95 100 50 95 100 75 50 b x¡2 x+1 ¶2 j ¡ 3x j > b 25 95 100 50 75 25 95 100 50 75 25 Solve for x: a j 4x ¡ j = j x + j µ 75 2x2 + x 10 25 a x2 ¡ 6x ¡ 16 x¡3 a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\050IB_HL-2_01.CDR Monday, 26 February 2007 9:47:49 AM PETERDELL IB_HL-2ed (51) FUNCTIONS (Chapter 1) 51 3x ¡ x2 ¡ determine the asymptotes discuss the behaviour of the function as it approaches its asymptotes sketch the graph find the coordinates of all points where the function crosses its asymptotes For f (x) = + a b c d 10 Given h : x 7! (x ¡ 4)2 + 3, x [ 4, [ a find the defining equation of h¡1 b show that h ± h¡1 = h¡1 ± h = x REVIEW SET 1C If h(x) = ¡ 3x: a find in simplest form h(2x ¡ 1) find x if h(2x ¡ 1) = ¡2 b For each of the following graphs find the domain and range: a b y (-3,¡5) y (2,¡5) y=x+2 x x y=-3 (-1,-3) x=1 p If f(x) = ¡ 2x and g(x) = x: a find in simplest form i (f ± g)(x) ii (g ± f )(x) b What is the domain and range of f ± g and g ± f? Solve for x: x2 ¡ <6 x¡2 a Consider f (x) = magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a For what value of x is f (x) meaningless? b Sketch the graph of this function using technology c State the domain and range of the function a Solve graphically: x2 2x + 2x + > x¡1 x+2 j 2x ¡ j > x + x b Graph the function f(x) = and hence find all values of x for jxj + x which > 13 jxj + (x + 2)(x ¡ 3) a Draw a sign diagram for x¡1 x2 + x ¡ b Hence, solve for x: <2 x¡1 cyan b black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\051IB_HL-2_01.CDR Thursday, 12 July 2007 3:59:11 PM DAVID3 IB_HL-2ed (52) 52 FUNCTIONS (Chapter 1) (x ¡ 1)2 a determine the asymptotes b discuss the behaviour of the function as it approaches its asymptotes c sketch the graph For f (x) = x ¡ + Find f ¡1 (x) given that f (x) is: cyan b ¡ 5x magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Sketch the graph of g : x 7! x2 + 6x + Explain why g for x ] ¡ 1, ¡3 ] has an inverse function g ¡1 d Sketch the graph of g ¡1 Find algebraically, the equation of g ¡1 Find the range of g and hence the domain and range of g ¡1 : 95 50 75 25 100 a b c e 10 a 4x + black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_01\052IB_HL-2_01.CDR Monday, February 2007 3:56:55 PM PETERDELL IB_HL-2ed (53) Chapter Sequences and series Contents: A B C D E F Number patterns Sequences of numbers Arithmetic sequences Geometric sequences Series Miscellaneous problems cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 2A Review set 2B Review set 2C black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\053IB_HL-2_02.cdr Wednesday, 24 October 2007 11:33:14 AM PETERDELL IB_HL-2ed (54) 54 SEQUENCES AND SERIES (Chapter 2) A NUMBER PATTERNS ² recognise a pattern in a set of numbers, ² describe the pattern in words, and ² continue the pattern An important skill in mathematics is to: A list of numbers where there is a pattern is called a number sequence The numbers in the sequence are said to be its members or its terms For example, 3, 7, 11, 15, form a number sequence The first term is 3, the second term is 7, the third term is 11, and so on We describe this pattern in words: “The sequence starts at and each term is more than the previous one.” Thus, the fifth term is 19, the sixth term is 23, and so on Example Describe the sequence: 14, 17, 20, 23, and write down the next two terms The sequence starts at 14 and each term is more than the previous term The next two terms are 26 and 29 EXERCISE 2A Write down the first four terms of the sequence if you start with: a and add each time b 45 and subtract each time c and multiply by each time d 96 and divide by each time For a d g each of the following write a description of the sequence and find the next terms: 8, 16, 24, 32, b 2, 5, 8, 11, c 36, 31, 26, 21, 96, 89, 82, 75, e 1, 4, 16, 64, f 2, 6, 18, 54, 480, 240, 120, 60, h 243, 81, 27, 9, i 50 000, 10 000, 2000, 400, Describe the following number patterns and write down the next terms: a 1, 4, 9, 16, b 1, 8, 27, 64, c 2, 6, 12, 20, B SEQUENCES OF NUMBERS st row nd row rd row Consider the illustrated tower of bricks The top row, or first row, has three bricks The second row has four bricks, and the third row has five bricks cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 If un represents the number of bricks in row n (from the top) then u1 = 3, u2 = 4, u3 = 5, u4 = 6, black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\054IB_HL-2_02.CDR Monday, 22 October 2007 3:47:19 PM PETERDELL IB_HL-2ed (55) 55 SEQUENCES AND SERIES (Chapter 2) The number pattern: 3, 4, 5, 6, is called a sequence of numbers This sequence can be specified by: ² Using words ² Using an explicit formula “The top row has three bricks and each successive row under it has one more brick.” un = n + is the general term (or nth term) formula for n = 1, 2, 3, 4, 5, u1 = + = X u3 = + = X Check: Early members of a sequence can be graphed Each term is represented by a dot The dots must not be joined Why? u2 = + = X etc un etc 2 n OPENING PROBLEM A circular stadium consists of sections as illustrated, with aisles in between.¡ The diagram shows the 13 tiers of concrete steps for the final section, Section¡K.¡ Seats are to be placed along every concrete step, with each seat being 0:45¡m wide.¡ AB, the arc at the front of the first row, is 14:4¡m long, while CD, the arc at the back of the back row, is 20:25¡m long 20.25 m For you to consider: D C How wide is each concrete step? Section K What is the length of the arc of the back of Row 1, Row 2, Row 3, etc? 14.4 m How many seats are there in Row 1, A Row Row 2, Row 3, , Row 13? How many sections are there in the stadium? What is the total seating capacity of the stadium? What is the radius of the ‘playing surface’? 13 m B rm to centre of circular stadium To solve problems like the Opening Problem and many others, a detailed study of sequences and their sums (called series) is required NUMBER SEQUENCES A number sequence is a set of numbers defined by a rule that is valid for positive integers cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A number sequence is a function whose domain is the set of positive integers Sequences may be defined in one of the following ways: ² using a formula which represents the general term (or nth term) ² giving a description in words ² listing the first few terms and assuming that the pattern represented continues indefinitely black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_02\055IB_HL-2_02.CDR Thursday, 11 March 2010 10:15:26 AM PETER IB_HL-2ed (56) 56 SEQUENCES AND SERIES (Chapter 2) THE GENERAL TERM un , Tn , tn , An , etc can all be used to represent the general term (or nth term) of a sequence and are defined for n = 1, 2, 3, 4, 5, 6, fun g represents the sequence that can be generated by using un as the nth term fun g is a function, i.e., n 7! un , n Z + For example, f2n + 1g generates the sequence 3, 5, 7, 9, 11, EXERCISE 2B List the first five terms of the sequence: f2ng f2n + 3g a e f2n + 2g f2n + 11g b f c g f2n ¡ 1g f3n + 1g d h f2n ¡ 3g f4n ¡ 3g c f6 £ ( 12 )n g d f(¡2)n g List the first five terms of the sequence: f2n g a f3 £ 2n g b List the first five terms of the sequence f15 ¡ (¡2)n g C ARITHMETIC SEQUENCES An arithmetic sequence is a sequence in which each term differs from the previous one by the same fixed number For example: 2, 5, 8, 11, 14, is arithmetic as ¡ = ¡ = 11 ¡ = 14 ¡ 11, etc 31, 27, 23, 19, is arithmetic as 27 ¡ 31 = 23 ¡ 27 = 19 ¡ 23, etc Likewise, ALGEBRAIC DEFINITION fun g is arithmetic , un +1 ¡ un = d for all positive integers n where d is a constant (the common difference) ² , is read as ‘if and only if’ ² If fun g is arithmetic then un+1 ¡ un is a constant and if un+1 ¡ un is a constant then fun g is arithmetic Note: THE NAME ‘ARITHMETIC’ If a, b and c are any consecutive terms of an arithmetic sequence then b¡a = c¡b 2b = a + c a+c ) b= fequating common differencesg ) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 So, the middle term is the arithmetic mean of the terms on either side of it black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\056IB_HL-2_02.CDR Monday, 22 October 2007 3:52:19 PM PETERDELL IB_HL-2ed (57) SEQUENCES AND SERIES (Chapter 2) 57 THE GENERAL TERM FORMULA Suppose the first term of an arithmetic sequence is u1 and the common difference is d Then u2 = u1 + d, u3 = u1 + 2d, u4 = u1 + 3d, and so on Hence un = u1 + (n ¡ 1)d The coefficient of d is one less than the subscript So, for an arithmetic sequence with first term u1 and common difference d the general term (or nth term) is un = u1 + (n ¡ 1)d Example Consider the sequence 2, 9, 16, 23, 30, a Show that the sequence is arithmetic b Find the formula for the general term un c Find the 100th term of the sequence d Is i 828 ii 2341 a member of the sequence? 9¡2 = 16 ¡ = 23 ¡ 16 = 30 ¡ 23 = a So, assuming that the pattern continues, consecutive terms differ by ) the sequence is arithmetic with u1 = 2, d = b un = u1 + (n ¡ 1)d c If n = 100, u100 = 7(100) ¡ = 695: d i ) ) un = + 7(n ¡ 1) i.e., un = 7n ¡ ii Let un = 2341 ) 7n ¡ = 2341 ) 7n = 2346 ) n = 335 17 Let un = 828 7n ¡ = 828 ) 7n = 833 ) n = 119 ) 828 is a term of the sequence In fact it is the 119th term which is not possible as n must be an integer ) 2341 cannot be a term Example Find k given that 3k + 1, k and ¡3 are consecutive terms of an arithmetic sequence cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Since the terms are consecutive, k ¡ (3k + 1) = ¡3 ¡ k fequating differencesg ) k ¡ 3k ¡ = ¡3 ¡ k ) ¡2k ¡ = ¡3 ¡ k ) ¡1 + = ¡k + 2k ) k=2 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\057IB_HL-2_02.CDR Wednesday, 24 October 2007 12:47:21 PM PETERDELL IB_HL-2ed (58) 58 SEQUENCES AND SERIES (Chapter 2) Example Find the general term un for an arithmetic sequence with u3 = and u8 = ¡17: ) ) u3 = u8 = ¡17 u1 + 2d = u1 + 7d = ¡17 fun = u1 + (n ¡ 1)dg :::: (1) :::: (2) We now solve (1) and (2) simultaneously: ¡u1 ¡ 2d = ¡8 u1 + 7d = ¡17 ) 5d = ¡25 ) d = ¡5 fadding the equationsg u1 + 2(¡5) = ) u1 ¡ 10 = ) u1 = 18 So in (1) Now ) ) ) un un un un Check: u3 = 23 ¡ 5(3) = 23 ¡ 15 =8 X = u1 + (n ¡ 1)d = 18 ¡ 5(n ¡ 1) = 18 ¡ 5n + = 23 ¡ 5n u8 = 23 ¡ 5(8) = 23 ¡ 40 = ¡17 X Example Insert four numbers between and 12 so that all six numbers are in arithmetic sequence If the numbers are 3, + d, + 2d, + 3d, + 4d, 12 then + 5d = 12 ) 5d = ) d = 95 = 1:8 So, we have 3, 4:8, 6:6, 8:4, 10:2, 12 EXERCISE 2C Consider the sequence 6, 17, 28, 39, 50, a Show that the sequence is arithmetic b d c Find its 50th term e Is 761 a member? cyan magenta yellow b d 95 100 50 Find the formula for its general term Is ¡143 a member? 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Consider the sequence 87, 83, 79, 75, a Show that the sequence is arithmetic c Find the 40th term Find the formula for its general term Is 325 a member? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\058IB_HL-2_02.CDR Monday, 22 October 2007 3:56:31 PM PETERDELL IB_HL-2ed (59) SEQUENCES AND SERIES (Chapter 2) 59 A sequence is defined by un = 3n ¡ 2: a Prove that the sequence is arithmetic Hint: Find un+1 ¡ un : c Find the 57th term b Find u1 and d d What is the least term of the sequence which is greater than 450? 71 ¡ 7n : c Find u75 : a Prove that the sequence is arithmetic b Find u1 and d d For what values of n are the terms of the sequence less than ¡200? A sequence is defined by un = Find k given the consecutive arithmetic terms: a 32, k, b k + 1, 2k + 1, 13 c 5, k, k2 ¡ Find the general term un for an arithmetic sequence given that: b u5 = ¡2 and u12 = ¡12 12 a u7 = 41 and u13 = 77 c the seventh term is and the fifteenth term is ¡39 d the eleventh and eighth terms are ¡16 and ¡11 12 respectively a Insert three numbers between and 10 so that all five numbers are in arithmetic sequence b Insert six numbers between ¡1 and 32 so that all eight numbers are in arithmetic sequence Consider the finite arithmetic sequence 36, 35 13 , 34 23 , , ¡30 a Find u1 and d b How many terms does the sequence have? An arithmetic sequence starts 23, 36, 49, 62, What is the first term of the sequence to exceed 100 000? D GEOMETRIC SEQUENCES A sequence is geometric if each term can be obtained from the previous one by multiplying by the same non-zero constant For example: 2, 10, 50, 250, is a geometric sequence as £ = 10 and 10 £ = 50 and 50 £ = 250 Notice that constant 10 = 50 10 = 250 50 = 5, so each term divided by the previous one gives the same Algebraic definition: un +1 = r for all positive integers n un where r is a constant called the common ratio fun g is geometric , cyan magenta yellow 95 100 50 75 is geometric with is geometric with 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 For example: ² 2, 10, 50, 250, ² 2, ¡10, 50, ¡250, black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\059IB_HL-2_02.CDR Monday, 22 October 2007 3:59:16 PM PETERDELL r = r = ¡5 IB_HL-2ed (60) 60 SEQUENCES AND SERIES (Chapter 2) THE NAME ‘GEOMETRIC’ If a, b and c are any consecutive terms of a geometric sequence then b c = a b p p b2 = ac and so b = § ac where ac is the geometric mean of a and c ) THE GENERAL TERM Suppose the first term of a geometric sequence is u1 and the common ratio is r Then u2 = u1 r, u3 = u1 r2 , u4 = u1 r3 , and so on un = u1 rn¡1 Hence The power of r is one less than the subscript for a geometric sequence with first term u1 and common ratio r, the general term (or nth term) is un = u1 rn ¡ So, Example For the sequence 8, 4, 2, 1, 12 , a Show that the sequence is geometric c Hence, find the 12th term as a fraction a b c b Find the general term un = 12 = 12 = 12 = 12 So, assuming the pattern continues, consecutive terms have a common ratio of ) the sequence is geometric with u1 = and r = 12 : ¡ ¢n¡1 ) un = 12 or un = 23 £ (2¡1 )n¡1 un = u1 rn¡1 = 23 £ 2¡n+1 11 u12 = £ ( ) = 23+(¡n+1) = 256 = 24¡n Example k ¡ 1, 2k and 21 ¡ k are consecutive terms of a geometric sequence Find k 2k 21 ¡ k = fequating r’sg k¡1 2k ) 4k2 = (21 ¡ k)(k ¡ 1) ) 4k2 = 21k ¡ 21 ¡ k2 + k 5k2 ¡ 22k + 21 = (5k ¡ 7)(k ¡ 3) = and so k = 75 or Since the terms are geometric, the terms are: magenta yellow X 95 100 50 75 25 50 25 95 100 50 75 25 95 100 50 75 25 cyan 14 98 5, , : the terms are: 2, 6, 18: X If k = 95 If k = 100 Check: 75 ) ) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\060IB_HL-2_02.CDR Wednesday, 24 October 2007 12:40:05 PM PETERDELL fr = 7g fr = 3g IB_HL-2ed (61) SEQUENCES AND SERIES (Chapter 2) 61 Example A geometric sequence has u2 = ¡6 and u5 = 162 Find its general term u2 = u1 r = ¡6 (1) and u5 = u1 r4 = 162 (2) 162 u1 r4 = u1 r ¡6 So, f(2) ¥ (1)g ) r3 = ¡27 p ) r = ¡27 ) r = ¡3 Note: (¡3)n¡1 6= ¡3n¡1 as we not know the value of n If n is odd, then (¡3)n¡1 = 3n¡1 If n is even, then (¡3)n¡1 = ¡3n¡1 ) we cannot simplify the answer u1 (¡3) = ¡6 ) u1 = and so in (1) Thus un = £ (¡3)n¡1 : Example p p Find the first term of the geometric sequence 6, 2, 12, 12 2, exceeds 1400 p Now u1 = and r = p ) un = £ ( 2)n¡1 : which Next we need to find n such that un > 1400 p Using a graphics calculator with Y1 = £ ( 2)^(n ¡ 1), we view a table of values: So, the first term to exceed 1400 is u17 where u17 = 1536 Note: Later we can solve problems like this one using logarithms EXERCISE 2D.1 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 a Show that the sequence 12, ¡6, 3, ¡ 32 , is geometric b Find un and hence find the 13th term (as a fraction) 75 25 a Show that the sequence 5, 10, 20, 40, is geometric b Find un and hence find the 15th term 95 100 50 75 25 For the geometric sequence with first two terms given, find b and c: a 2, 6, b, c, b 10, 5, b, c, c 12, ¡6, b, c, black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\061IB_HL-2_02.CDR Monday, 22 October 2007 4:02:54 PM PETERDELL IB_HL-2ed (62) 62 SEQUENCES AND SERIES (Chapter 2) Show that the sequence 8, ¡6, 4:5, ¡3:375, is geometric and hence find the 10th term as a decimal p p Show that the sequence 8, 2, 4, 2, is geometric Hence find, in simplest form, the general term un Find k given that the following are consecutive terms of a geometric sequence: a 7, k, 28 b k, 3k, 20 ¡ k c k, k + 8, 9k Find the general term un of the geometric sequence which has: a u4 = 24 and u7 = 192 b u3 = and u6 = ¡1 c u7 = 24 and u15 = 384 d u3 = and u7 = a Find the first term of the sequence 2, 6, 18, 54, which exceeds 10 000 p p b Find the first term of the sequence 4, 3, 12, 12 3, which exceeds 4800 c Find the first term of the sequence 12, 6, 3, 1:5, which is less than 0:0001 : COMPOUND INTEREST Consider the following: You invest $1000 in the bank You leave the money in the bank for years You are paid an interest rate of 10% p.a The interest is added to your investment each year An interest rate of 10% p.a is paid, increasing the value of your investment yearly Your percentage increase each year is 10%, so at the end of the year you will have 100% + 10% = 110% of the value at its start This corresponds to a multiplier of 1:1 After one year your investment is worth $1000 £ 1:1 = $1100 After two years it is worth $1100 £ 1:1 = $1000 £ 1:1 £ 1:1 = $1000 £ (1:1)2 = $1210 After three years it is worth $1210 £ 1:1 = $1000 £ (1:1)2 £ 1:1 = $1000 £ (1:1)3 This suggests that if the money is left in your account for n years it would amount to $1000 £ (1:1)n Observe that: u1 u2 u3 u4 = $1000 = u1 £ 1:1 = u1 £ (1:1)2 = u1 £ (1:1)3 = initial investment = amount after year = amount after years = amount after years un+1 = u1 £ (1:1)n = amount after n years un+1 = u1 £ r n In general, we can use the compound interest formula cyan magenta yellow 95 100 50 75 25 95 100 50 n = number of years un+1 = amount after n years 75 25 95 100 50 75 25 u1 = initial investment r = growth multiplier 95 100 50 75 25 where black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\062IB_HL-2_02.CDR Monday, 22 October 2007 4:04:16 PM PETERDELL IB_HL-2ed (63) SEQUENCES AND SERIES (Chapter 2) 63 Example 10 $5000 is invested for years at 7% p.a compound interest, compounded annually What will it amount to at the end of this period? u5 = u1 £ r4 is the amount after years = 5000 £ (1:07)4 ffor a 7% increase 100% becomes 107%g ¼ 6553:98 f5000 So, it amounts to $6553:98 × 1:07 ^ ENTER g Example 11 How much should I invest now if I want the maturing value to be $10¡000 in 4¡years’ time, if I am able to invest at 8:5% p.a.¡ compounded annually? u1 = ?, u5 = 10 000, r = 1:085 fusing un+1 = u1 £ rn g u5 = u1 £ r4 ) 10 000 = u1 £ (1:085)4 10 000 (1:085)4 ) u1 = ) u1 ¼ 7215:74 f10 000 ÷ 1:085 ^ ENTER g So, you should invest $7215:74 now EXERCISE 2D.2 a What will an investment of $3000 at 10% p.a compound interest amount to after years? b What part of this is interest? How much compound interest is earned by investing E20 000 at 12% p.a if the investment is over a year period? a What will an investment of 30 000 Yen at 10% p.a compound interest amount to after years? b What part of this is interest? How much compound interest is earned by investing $80 000 at 9% p.a., if the investment is over a year period? What will an investment of 100 000 Yen amount to after years if it earns 8% p.a compounded twice annually? What will an investment of $45 000 amount to after 21 months if it earns 7:5% p.a compounded quarterly? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 How much money must be invested now if you require $20 000 for a holiday in years’ time and the money can be invested at a fixed rate of 7:5% p.a compounded annually? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\063IB_HL-2_02.CDR Monday, 22 October 2007 4:05:05 PM PETERDELL IB_HL-2ed (64) 64 SEQUENCES AND SERIES (Chapter 2) What initial investment is required to produce a maturing amount of $15 000 in 60 months’ time given that a fixed rate of 5:5% p.a compounded annually is guaranteed? How much should I invest now if I want a maturing amount of E25 000 in years’ time and the money can be invested at a fixed rate of 8% p.a compounded quarterly? 10 What initial investment is required to produce a maturing amount of 40 000 Yen in years’ time if your money can be invested at 9% p.a., compounded monthly? OTHER GEOMETRIC SEQUENCE PROBLEMS Example 12 The initial population of rabbits on a farm was 50 The population increased by 7% each week a How many rabbits were present after: i 15 weeks ii 30 weeks? b How long would it take for the population to reach 500? We notice that u1 = 50 and r = 1:07 u2 = 50 £ 1:07 = the population after week a b un+1 = u1 £ rn ) u16 = 50 £ (1:07)15 ¼ 137:95 i.e., 138 rabbits i ii and u31 = 50 £ (1:07)30 ¼ 380:61 i.e., 381 rabbits un+1 = u1 £ (1:07)n after n weeks So, we need to find when 50 £ (1:07)n = 500 Trial and error on your calculator gives n ¼ 34 weeks or using the Equation Solver gives n ¼ 34:03 or by finding the point of intersection of Y1 = 50 £ 1:07^X and Y2 = 500 on a graphics calculator, the solution is ¼ 34:03 weeks EXERCISE 2D.3 A nest of ants initially consists of 500 ants The population is increasing by 12% each week cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a How many ants will there be after i 10 weeks ii 20 weeks? b Use technology to find how many weeks it will take for the ant population to reach 2000 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\064IB_HL-2_02.CDR Monday, 22 October 2007 4:08:02 PM PETERDELL IB_HL-2ed (65) SEQUENCES AND SERIES (Chapter 2) 65 The animal Eraticus is endangered Since 1992 there has only been one colony remaining and in 1992 the population of the colony was 555 Since then the population has been steadily decreasing at 4:5% per year Find: a the population in the year 2007 b the year in which we would expect the population to have declined to 50 E SERIES A series is the addition of the terms of a sequence, i.e., u1 + u2 + u3 + :::: + un is a series The sum of a series is the result when we perform the addition Given a series which includes the first n terms of a sequence, its sum is Sn = u1 + u2 + u3 + :::: + un Example 13 For the sequence 1, 4, 9, 16, 25, a Write down an expression for Sn b Find Sn for n = 1, 2, 3, and Sn = 12 + 22 + 32 + 42 + :::: + n2 fall terms are perfect squaresg a b S1 S2 S3 S4 S5 =1 = 1+4= = + + = 14 = + + + 16 = 30 = + + + 16 + 25 = 55 SIGMA NOTATION u1 + u2 + u3 + u4 + :::: + un can be written more compactly using sigma notation P , which is called sigma, is the equivalent of capital S in the Greek alphabet n P We write u1 + u2 + u3 + u4 + :::: + un as uk So, k=1 n P uk reads “the sum of all numbers of the form uk where k = 1, 2, 3, , up to n” k=1 Example 14 P a Expand and evaluate: b (k + 1) k=1 P a P k k=1 P k k=1 b (k + 1) k=1 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 =2+3+4+5+6+7+8 = 35 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\065IB_HL-2_02.CDR Monday, 22 October 2007 4:10:24 PM PETERDELL = = 31 32 + + + 16 + 32 IB_HL-2ed (66) 66 SEQUENCES AND SERIES (Chapter 2) n P Note: (ak + bk ) = k=1 n P ak + k=1 n P If c is a constant, n P bk k=1 n P cak = c k=1 ak n P and k=1 c = cn k=1 EXERCISE 2E.1 For the following sequences: ii i write down an expression for Sn a 3, 11, 19, 27, b 42, 37, 32, 27, d 2, 3, 12 , 34 , find S5 c 12, 6, 3, 12 , 1, 12 , 14 , 18 , e f 1, 8, 27, 64, Expand and evaluate: P a (3k ¡ 5) b k=1 P (11 ¡ 2k) P c k=1 k(k + 1) k=1 For un = 3n ¡ 1, write u1 +u2 +u3 +::::+ u20 the sum Show that: n P a c = cn b k=1 a n P cak = c k=1 n P Explain why n P k=1 b n P using sigma notation and evaluate n P k=1 (3k + 4k ¡ 3) = (ak + bk ) = k=1 n P 10 £ 2k¡1 k=1 c ak P d k2 + k=1 n P n P ak + k=1 n P bk k=1 k ¡ 3n k=1 n P n(n + 1) n(n + 1)(2n + 1) k2 = and k=1 k=1 n P (k + 1)(k + 2) find in simplest form Given that k= k=1 Check your answer in the case when n = 10 ARITHMETIC SERIES An arithmetic series is the addition of successive terms of an arithmetic sequence For example: 21, 23, 25, 27, ., 49 is an arithmetic sequence So, 21 + 23 + 25 + 27 + ::::: + 49 is an arithmetic series SUM OF AN ARITHMETIC SERIES If the first term is u1 and the common difference is d, then the terms are: u1 , u1 + d, u1 + 2d, u1 + 3d, etc cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Suppose that un is the final term of an arithmetic series black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\066IB_HL-2_02.CDR Monday, 22 October 2007 4:13:17 PM PETERDELL IB_HL-2ed (67) SEQUENCES AND SERIES (Chapter 2) 67 So, Sn = u1 + (u1 + d) + (u1 + 2d) + :::: + (un ¡ 2d) + (un ¡ d) + un But Sn = un + (un ¡ d) + (un ¡ 2d) + :::: + (u1 + 2d) + (u1 + d) + u1 freversing themg Adding these two equations vertically we get 2Sn = (u1 + un ) + (u1 + un ) + (u1 + un ) + :::: + (u1 + un ) + (u1 + un ) + (u1 + un ) | {z } n of these ) 2Sn = n(u1 + un ) n ) Sn = (u1 + un ) where un = u1 + (n ¡ 1)d Sn = So, n (u1 + un ) Sn = or n (2u1 + (n ¡ 1)d) Example 15 Find the sum of + + 10 + 13 + :::: to 50 terms The series is arithmetic with u1 = 4, d = and n = 50: n So, S50 = 50 fusing Sn = (2u1 + (n ¡ 1)d)g (2 £ + 49 £ 3) = 3875 Example 16 Find the sum of ¡6 + + + 15 + :::: + 141 The series is arithmetic with u1 = ¡6, d = and un = 141: First we need to find n Now un = u1 + (n ¡ 1)d = ) ¡6 + 7(n ¡ 1) = ) 7(n ¡ 1) = ) n¡1 = ) n= 141 141 147 21 22 Using Sn = n (u1 + un ), 22 (¡6 + 141) S22 = = 11 £ 135 = 1485 EXERCISE 2E.2 Find the sum of: b 12 + + 12 + + :::: to 50 terms d 50 + 48 12 + 47 + 45 12 + :::: to 80 terms a + + 11 + 15 + :::: to 20 terms c 100 + 93 + 86 + 79 + :::: to 40 terms Find the sum of: b 50 + 49 12 + 49 + 48 12 + :::: + (¡20) a + + 11 + 14 + :::: + 101 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 c + 10 12 + 13 + 15 12 + :::: + 83 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\067IB_HL-2_02.CDR Monday, 22 October 2007 4:16:18 PM PETERDELL IB_HL-2ed (68) 68 SEQUENCES AND SERIES (Chapter 2) Evaluate these arithmetic series: 10 P a (2k + 5) b 15 P k=1 (k ¡ 50) c k=1 20 P k=1 µ k+3 ¶ An arithmetic series has seven terms The first term is and the last term is 53 Find the sum of the series An arithmetic series has eleven terms The first term is and the last term is ¡27 Find the sum of the series A bricklayer builds a triangular wall with layers of bricks as shown If the bricklayer uses 171 bricks, how many layers did he build? Each section of a soccer stadium has 44 rows with 22 seats in the first row, 23 in the second row, 24 in the third row, and so on How many seats are there in: a row 44 b each section c the stadium which has 25 sections? Find the sum of: b the multiples of between and 1000 a the first 50 multiples of 11 c the integers between and 100 which are not divisible by n(n Prove that the sum of the first n positive integers is + 1) 10 Consider the series of odd numbers + + + + :::: a What is the nth odd number un ? b Prove that “the sum of the first n odd numbers is n2 ” c Check your answer to b by finding S1 , S2 , S3 and S4 11 Find the first two terms of an arithmetic sequence where the sixth term is 21 and the sum of the first seventeen terms is 12 Three consecutive terms of an arithmetic sequence have a sum of 12 and a product of ¡80 Find the terms Hint: Let the terms be x ¡ d, x and x + d 13 Five consecutive terms of an arithmetic sequence have a sum of 40 The product of the middle and the two end terms is 224 Find the terms of the sequence GEOMETRIC SERIES A geometric series is the addition of successive terms of a geometric sequence For example, So, 1, 2, 4, 8, 16, :::: , 1024 is a geometric sequence + + + + 16 + :::: + 1024 is a geometric series SUM OF A GEOMETRIC SERIES cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 If the first term is u1 and the common ratio is r, then the terms are: u1 , u1 r, u1 r2 , u1 r3 , etc black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\068IB_HL-2_02.CDR Monday, 22 October 2007 4:18:05 PM PETERDELL IB_HL-2ed (69) SEQUENCES AND SERIES (Chapter 2) Sn = u1 + u1 r + u1 r2 + u1 r3 + So, u3 u2 and for r 6= 1, Sn = + u1 rn¡2 + u1 rn¡1 :::: u4 u1 (rn ¡ 1) r¡1 un¡1 Sn = or un u1 (1 ¡ rn ) 1¡r If Sn = u1 + u1 r + u1 r2 + u1 r3 + :::: + u1 rn¡2 + u1 rn¡1 Proof: 69 (1) then rSn = (u1 r + u1 r2 + u1 r3 + u1 r4 + :::: + u1 rn¡1 ) + u1 rn ) rSn = (Sn ¡ u1 ) + u1 rn ffrom (1)g ) rSn ¡ Sn = u1 r ¡ u1 n ) Sn (r ¡ 1) = u1 (rn ¡ 1) and so Sn = u1 (rn ¡ 1) u1 (1 ¡ rn ) or for r 6= r¡1 1¡r Discuss the case r = Example 17 Find the sum of + + 18 + 54 + :::: to 12 terms The series is geometric with u1 = 2, r = and n = 12 2(312 ¡ 1) 3¡1 = 531 440 fusing So, S12 = Sn = u1 (rn ¡ 1) g r¡1 Example 18 Find a formula for Sn for ¡ + ¡ + :::: to n terms The series is geometric with u1 = 9, r = ¡ 13 , “n” = n 9(1 ¡ u1 (1 ¡ r ) = 1¡r n So, Sn = ) Sn = 27 (1 This answer cannot be simplified as we not know if n is odd or even (¡ 13 )n ) ¡ (¡ 13 )n ) EXERCISE 2E.3 Find the sum of the following series: a 12 + + + 1:5 + :::: to 10 terms cyan magenta 95 50 75 25 95 50 100 yellow p1 100 d 1¡ + :::: to 15 terms 75 25 95 100 50 75 25 95 100 50 75 25 c ¡ + 12 ¡ p p + + 7 + 49 + :::: to 12 terms b black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\069IB_HL-2_02.CDR Wednesday, 24 October 2007 12:41:24 PM PETERDELL + ¡ p 2 + :::: to 20 terms IB_HL-2ed (70) 70 SEQUENCES AND SERIES (Chapter 2) Find a formula for Sn for: p p a + + 3 + + :::: to n terms b 12 + + + 12 + :::: to n terms c 0:9 + 0:09 + 0:009 + 0:0009 + :::: to n terms d 20 ¡ 10 + ¡ 12 + :::: to n terms Evaluate these geometric series: 10 P £ 2k¡1 b a 12 P k=1 ( 12 )k¡2 c k=1 25 P £ (¡2)k k=1 Each year a salesperson is paid a bonus of $2000 which is banked into the same account It earns a fixed rate of interest of 6% p.a with interest being paid annually The amount at the end of each year in the account is calculated as follows: A0 = 2000 A1 = A0 £ 1:06 + 2000 A2 = A1 £ 1:06 + 2000 etc a Show that A2 = 2000 + 2000 £ 1:06 + 2000 £ (1:06)2 : b Show that A3 = 2000[1 + 1:06 + (1:06)2 + (1:06)3 ]: c Find the total bank balance after 10 years, assuming there are no fees or charges : 2n Find S1 , S2 , S3 , S4 and S5 in fractional form From a guess the formula for Sn u1 (1 ¡ rn ) Find Sn using Sn = 1¡r Comment on Sn as n gets very large What is the relationship between the given diagram and d? Consider Sn = a b c d e + + + 16 + :::: + Qr_ Qw_ qA_y_ Qi_ eA_w_ SUM OF AN INFINITE GEOMETRIC SERIES Sometimes it is necessary to consider Sn = u1 (1 ¡ rn ) 1¡r when n gets very large What happens to Sn in this situation? If j r j > 1, the series is said to be divergent and the sum becomes infinitely large If ¡1 < r < 1, i.e., j r j < 1, then rn approaches for very large n This means that Sn will get closer and closer to u1 1¡r S= We say that the series converges and we write its sum as We call this the limiting sum of the series u1 1¡r for j r j < cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 This result can be used to find the value of recurring decimals black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\070IB_HL-2_02.CDR Wednesday, November 2007 1:48:03 PM PETERDELL IB_HL-2ed (71) SEQUENCES AND SERIES (Chapter 2) 71 Example 19 0:¹7 = Write 0:¹7 as a rational number 10 + 100 + 1000 + 10 000 + :::: which is a geometric series with infinitely many terms ) S= So, 0:¹7 = u1 = 10 = 1¡r ¡ 10 9 EXERCISE 2E.4 Consider 0:¹3 = 10 100 + + 1000 + :::: which is an infinite geometric series a What are i u1 and b Using a, show that 0:¹3 = 13 Write as a rational number: Use S = u1 1¡r ii r? a 0:4 b 0:16 c 0:312 to check your answers to Exercise 2E.3, question 5d Find the sum of each of the following infinite geometric series: a 18 + 12 + + :::::: b 18:9 ¡ 6:3 + 2:1 ¡ :::::: Find each of the following: P a k k=1 P b k=0 ¡ ¢k ¡ 25 Determine whether each of the following series is convergent If so, find the sum of the series If not, find the smallest value of n for which the sum of the first n terms of the series exceeds 100 a 18 ¡ + 4:5 ¡ :::::: b 1:2 + 1:8 + 2:7 + :::::: The sum of the first three terms of a convergent geometric series is 19 The sum of the series is 27 Find the first term and the common ratio The second term of a convergent geometric series is 85 The sum of the series is 10 Show that there are two possible series and find the first term and the common ratio in each case A ball takes second to hit the ground when dropped It then takes 90% of this time to rebound to its new height and this continues until the ball comes to rest a Show that the total time of motion is given by + 2(0:9) + 2(0:9)2 + 2(0:9)3 + ::::: b Find Sn for the series in a c How long does it take for the ball to come to rest? ground cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 This diagram is inaccurate as the motion is really up and down on the same spot It has been separated out to help us visualise what is happening 100 50 75 25 95 100 50 75 25 Note: black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_02\071IB_HL-2_02.CDR Thursday, 11 December 2008 1:58:33 PM TROY IB_HL-2ed (72) 72 SEQUENCES AND SERIES (Chapter 2) F MISCELLANEOUS PROBLEMS EXERCISE 2F Henk starts a new job selling TV sets He hopes to sell 11 sets in the first week, 14 in the next, 17 in the next, and so on in arithmetic sequence In what week does Henk hope to sell his 2000th TV set? A computer is bought for $2795 and depreciates at a rate of 2% per month After how many months will its value reduce to $500? A geometric series has a second term of and the sum of its first three terms is ¡14 Find its fourth term When a ball falls vertically off a table it rebounds 75% of its height after each bounce If it travels a total distance of 490 cm, how high was the table top above the floor? An arithmetic and a geometric sequence both have a first term of and their second terms are equal The 14th term of the arithmetic sequence is three times the third term of the geometric sequence Find the twentieth term of each sequence Evaluate: X a k=1 Find n given that: µ X Find x if k=1 n X a 3x ¶k¡1 12 X b k(k + 1)(k + 2) 100 £ (1:2)k¡3 k=6 b (2k + 3) = 1517 k=1 n X £ 3k¡1 = 177 146 k=1 = n(3n + 11) b Find the twentieth term of the sequence The sum of the first n terms of an arithmetic sequence is a Find its first two terms 10 Mortgage repayments: $8000 is borrowed over a 2-year period at a rate of 12% p.a Quarterly repayments are made and the interest is adjusted each quarter, which means that the amount repaid in the period is deducted and the interest is charged on the new amount owed There are £ = repayments and the interest per quarter is 12% = 3%: At the end of the first quarter the amount owed, A1 , is given by $8000 £ 1:03 ¡ R, where R is the amount of each repayment A2 = A1 £ 1:03 ¡ R At the end of the second quarter the amount owed, A2 , is given by: = ($8000 £ 1:03 ¡ R) £ 1:03 ¡ R = $8000 £ (1:03)2 ¡ 1:03R ¡ R cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a Find a similar expression for the amount owed at the end of the third quarter, A3 b Write down an expression for the amount owed at the end of the 8th quarter, A8 , and hence deduce the value of R Hint: What value we want A8 to have? c If the amount borrowed at adjusted interest conditions is $P, the interest rate is r% per repayment interval, and there r m r ) £ 100 P (1 + 100 are m repayments, show that the : R= r m (1 + 100 ) ¡ amount of each repayment is black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\072IB_HL-2_02.CDR Monday, 22 October 2007 4:34:34 PM PETERDELL IB_HL-2ed (73) SEQUENCES AND SERIES (Chapter 2) 73 VON KOCH’S SNOWFLAKE CURVE INVESTIGATION C2 C1 C3 , C4 , , , In this investigation we consider a limit curve named after the Swedish mathematician Niels Fabian Helge von Koch (1870 - 1924) To draw Von Koch’s Snowflake curve we ² start with an equilateral triangle, C1 ² then divide each side into equal parts ² then on each middle part draw an equilateral triangle ² then delete the side of the smaller triangle which lies on C1 The resulting curve is C2 , and C3 , C4 , C5 , are found by ‘pushing out’ equilateral triangles on each edge of the previous curve as we did with C1 to get C2 We get a sequence of special curves C1 , C2 , C3 , C4 , and Von Koch’s curve is the limiting case when n is infinitely large Your task is to investigate the perimeter and area of Von Koch’s curve What to do: Suppose C1 has a perimeter of units Find the perimeter of C2 , C3 , C4 and C5 becomes Hint: so parts become parts Remembering that Von Koch’s curve is Cn , where n is infinitely large, find the perimeter of Von Koch’s curve Suppose the area of C1 is unit2 Explain why the areas of C2 , C3 , C4 and C5 are A2 = + A4 = + units [1 + + A3 = + 13 [1 + 49 ] units2 ( 49 )2 ] units2 A5 = + 13 [1 + + ( 49 )2 + ( 49 )3 ] units2 Use your calculator to find An where n = 1, 2, 3, 4, 5, 6, 7, etc., giving answers which are as accurate as your calculator permits What you think will be the area within Von Koch’s snowflake curve? Is there anything remarkable about your answers to and 2? Similarly, investigate the sequence of curves obtained by ‘pushing out’ squares on successive curves from the middle third of each side, i.e., the curves C1 , C2 , C3 , C4 , etc C3 C2 C1 , , , cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Region contains holes black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\073IB_HL-2_02.CDR Wednesday, 24 October 2007 12:43:48 PM PETERDELL IB_HL-2ed (74) 74 SEQUENCES AND SERIES (Chapter 2) REVIEW SET 2A List the first four members of the sequences defined by: 3n + a un = 3n¡2 b un = n+3 c un = 2n ¡ (¡3)n A sequence is defined by un = 68 ¡ 5n a Prove that the sequence is arithmetic b Find u1 and d c Find the 37th term d What is the first term of the sequence less than ¡200? a Show that the sequence 3, 12, 48, 192, is geometric b Find un and hence find u9 Find k if 3k, k ¡ and k + are consecutive terms of an arithmetic sequence Find the general term of an arithmetic sequence given that u7 = 31 and u15 = ¡17 Hence, find the value of u34 A sequence is defined by un = 6( 12 )n¡1 a Prove that the sequence is geometric b Find u1 and r c Find the 16th term to significant figures Show that 28, 23, 18, 13, is arithmetic and hence find un and the sum Sn of the first n terms in simplest form Find k given that 4, k and k2 ¡ are consecutive geometric terms Determine the general term of a geometric sequence given that its sixth term is 256 16 and its tenth term is 10 Find the sum of each of the following infinite geometric series: a 1:21 ¡ 1:1 + ¡ :::::: b 14 + + 21 + :::::: 11 x + and x ¡ are the first two terms of a geometric series Find the values of x for which the series converges 12 The sum of the first two terms of a geometric series is 90 The third term is 24 a Show that there are two possible series and find the first term and the common ratio in each case b Show that both series converge and find their respective sums 13 a, b and c are consecutive terms of both an arithmetic and geometric sequence What can be deduced about a, b and c? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 14 x, y and z are consecutive terms of a geometric sequence If x + y + z = 73 and x2 + y + z = 91 , find the values of x, y and z black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\074IB_HL-2_02.CDR Monday, 22 October 2007 4:43:22 PM PETERDELL IB_HL-2ed (75) SEQUENCES AND SERIES (Chapter 2) 75 REVIEW SET 2B a Determine the number of terms in the sequence 24, 23 14 , 22 12 , , ¡36 b Find the value of u35 for the sequence in a c Find the sum of the terms of the sequence in a Insert six numbers between 23 and so that all eight numbers are in arithmetic sequence Find the formula for un , the general term of: a 86, 83, 80, 77, b 34 , 1, 76 , 97 , c 100, 90, 81, 72:9, Note: One of these sequences is neither arithmetic nor geometric Write down the expansion of: Write in the form n P P a k2 k+3 P k=1 k + b k=1 (:::::) : k=1 a b + 11 + 18 + 25 + :::: for n terms Find the sum of: a + + 15 + 21 + :::: to 23 terms Calculate: b k=1 + + 16 + 32 + :::: for n terms b 24 + 12 + + + :::: to 12 terms ¶ µ X 31 ¡ 3k a 15 X 50(0:8)k¡1 k=1 Find the first term of the sequence 5, 10, 20, 40, which exceeds 10 000 What will an investment of E6000 at 7% p.a compound interest amount to after years if the interest is compounded: a annually b quarterly c monthly? 10 Find the sum of each of the following infinite geometric series: p a 18 ¡ 12 + ¡ :::::: b + + + :::::: 11 2x and x ¡ are the first two terms of a convergent geometric series If the sum of the series is 18 , find x, clearly explaining why there is only one possible value 12 Find X ¡ ¢k¡1 25 k=7 13 a, b, c, d and e are consecutive terms of an arithmetic sequence Prove that a + e = b + d = 2c 14 Find the sum of the n consecutive geometric terms which can be inserted between and cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 15 Suppose an arithmetic sequence and a geometric sequence with common ratio r have r2 the same first two terms Show that the third term of the geometric sequence is 2r ¡ times the third term of the arithmetic sequence black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\075IB_HL-2_02.CDR Monday, 22 October 2007 4:46:30 PM PETERDELL IB_HL-2ed (76) 76 SEQUENCES AND SERIES (Chapter 2) REVIEW SET 2C A geometric sequence has u6 = 24 and u11 = 768 Determine the general term of the sequence and hence find: b the sum of the first 15 terms a u17 How many terms of the series 11 + 16 + 21 + 26 + :::: are needed to exceed a sum of 450? Find the first term of the sequence 24, 8, 83 , 89 , which is less than 0:001 a Determine the number of terms in the sequence 128, 64, 32, 16, , b Find the sum of these terms 512 $12 500 is invested in an account which pays 8:25% p.a compounded Find the value of the investment after years if the interest is compounded: a b half-yearly monthly How much should be invested at a fixed rate of 9% p.a compounded interest if you wish it to amount to $20 000 after years with interest paid monthly? In 2004 there were 3000 koalas on Koala Island Since then, the population of koalas on the island has increased by 5% each year a How many koalas were on the island in 2007? b In what year will the population first exceed 5000? A ball bounces from a height of metres and returns to 80% of its previous height on each bounce Find the total distance travelled by the ball until it stops bouncing X a Under what conditions will the series b Find X 50(2x¡1)k¡1 converge? Explain! k=1 50(2x ¡ 1)k¡1 if x = 0:3 : k=1 3n2 + 5n 10 The sum of the first n terms of a sequence is a Find the nth term b Prove that the sequence is arithmetic 11 a, b and c are consecutive terms of an arithmetic sequence Prove that the following are also consecutive terms of an arithmetic sequence: 1 p p and p a b + c, c + a and a + b b p p , p c+ a b+ c a+ b ¡ (111111::::::1) | {z } 2n 1’s n 2’s magenta yellow 95 100 50 75 25 95 and 1111 ¡ 22 = 1089 = 332 :) 100 50 75 25 95 100 50 75 25 95 100 50 75 25 (For example: 11 ¡ = = 32 cyan is a perfect square (22222::::::2) | {z } 12 Show that black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_02\076IB_HL-2_02.CDR Thursday, November 2007 1:54:18 PM PETERDELL IB_HL-2ed (77) Chapter Exponentials Contents: A B C D E F G H Index notation Evaluating powers Index laws Algebraic expansion and factorisation Exponential equations Graphs of exponential functions Growth and decay The natural exponential ‘e’ cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 3A Review set 3B Review set 3C black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_03\077IB_HL-2_03.CDR Wednesday, 24 October 2007 11:33:41 AM PETERDELL IB_HL-2ed (78) 78 EXPONENTIALS (Chapter 3) We often deal with numbers that are repeatedly multiplied together Mathematicians use indices or exponents to represent such expressions For example, £ £ = 53 Indices have many applications in areas such as finance, engineering, physics, electronics, biology and computer science Problems encountered in these areas may involve situations where quantities increase or decrease over time Such problems are often examples of exponential growth or decay OPENING PROBLEM LEGEND OF THE AMBALAPPUZHA PAAL PAYASAM According to Hindu legend, Lord Krishna once appeared as a sage before the king who ruled a region of India, and challenged him to a game of chess The prize if Lord Krishna won was based on the chessboard: that the king would provide him with a single grain of rice for the first square, two grains of rice for the second square, four grains of rice for the third square, and so on doubling the rice on each successive square on the board Lord Krishna of course did win, and the king was most unhappy when he realised he owed more rice than there was in the world Consider the following questions: Is there a function which describes the number of grains of rice on each square? How many grains of rice would there be on the 40th square? Using your knowledge of series, find the total number of grains of rice that the king owed A INDEX NOTATION Rather than writing £ £ £ £ 3, we can write such a product as 35 35 reads “three to the power of five” or “three with index five” Thus 43 = £ £ and 56 = £ £ £ £ £ If n is a positive integer, then an is the product of n factors of a base power, index or exponent an = a £ a £ a £ a £ £ a {z } | n factors EXERCISE 3A List the first six powers of: a b c Copy and complete the values of these common powers: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a 51 = :::: , 52 = :::: , 53 = :::: , 54 = :::: b 61 = :::: , 62 = :::: , 63 = :::: , 64 = :::: c 71 = :::: , 72 = :::: , 73 = :::: , 74 = :::: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_03\078IB_HL-2_03.CDR Tuesday, 23 October 2007 9:31:09 AM PETERDELL IB_HL-2ed (79) EXPONENTIALS (Chapter 3) 79 HISTORICAL NOTE Nicomachus discovered an interesting number pattern involving cubes and sums of odd numbers Nicomachus was born in Roman Syria (now Jerash, Jordan) around 100 AD He wrote in Greek and was a Pythagorean B = 13 + = = 23 + + 11 = 27 = 33 EVALUATING POWERS So far we have only considered positive bases raised to a power We will now briefly look at negative bases Consider the statements below: (¡1)1 (¡1)2 (¡1)3 (¡1)4 (¡2)1 (¡2)2 (¡2)3 (¡2)4 = ¡1 = ¡1 £ ¡1 = = ¡1 £ ¡1 £ ¡1 = ¡1 = ¡1 £ ¡1 £ ¡1 £ ¡1 = = ¡2 = ¡2 £ ¡2 = = ¡2 £ ¡2 £ ¡2 = ¡8 = ¡2 £ ¡2 £ ¡2 £ ¡2 = 16 From the patterns above we can see that: A negative base raised to an odd power is negative A negative base raised to an even power is positive CALCULATOR USE Although different calculators vary in the appearance of keys, they all perform operations of raising to powers in a similar manner Power keys x2 squares the number in the display ^ raises the number in the display to the power ^ raises the number in the display to the power raises the number in the display to the power ¡4 ^ Example a 65 Find, using your calculator: cyan Press: ^ b Press: c Press: ( ¡74 ENTER Answer 7776 ) 625 ^ ENTER ¡2401 ^ ENTER c magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 You will need to check if your calculator uses the same key sequence as in the examples.¡ If not, work out the sequence which gives you the correct answers 75 25 95 100 50 75 25 Note: a b (¡5)4 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_03\079IB_HL-2_03.CDR Tuesday, 23 October 2007 9:34:33 AM PETERDELL IB_HL-2ed (80) 80 EXPONENTIALS (Chapter 3) Example a 5¡2 Find using your calculator, and comment on: a Press: ^ b Press: b Answer 0:04 ENTER ÷ 52 ^ ENTER 0:04 ¡2 The answers indicate that = 2: EXERCISE 3B Simplify, then use a calculator to check your answer: a (¡1)5 b (¡1)6 c (¡1)14 d (¡1)19 e (¡1)8 f ¡18 g ¡(¡1)8 h (¡2)5 i ¡25 j ¡(¡2)6 k (¡5)4 l ¡(¡5)4 Use your calculator to find the value of the following, recording the entire display: a f 47 (¡8)6 b g 74 ¡86 c h ¡55 2:139 d i (¡5)5 ¡2:139 Use your calculator to find the values of the following: a 9¡1 b c 6¡2 91 e 3¡4 f g 170 34 What you notice? Consider 31 , 32 , 33 , 34 , 35 e j 86 (¡2:13)9 d 62 h (0:366)0 Look for a pattern and find the last digit of 3101 What is the last digit of 7217 ? Answer the Opening Problem on page 78 C INDEX LAWS (am )n = amn ² a¡ n and an are reciprocals, i.e., a¡ n = ² ² a0 = for all a 6= magenta yellow 95 100 for all a 6= 75 50 an 25 95 100 50 75 25 95 ² (ab)n = an bn 100 50 25 95 75 ² cyan £a =a n µ ¶n a an ² = n (b 6= 0) b b am = am ¡ n (a 6= 0) an a m 100 50 m +n ² 75 25 The following are laws of indices for m, n Z : black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_03\080IB_HL-2_03.CDR Monday, 15 December 2008 10:25:19 AM TROY IB_HL-2ed (81) EXPONENTIALS (Chapter 3) 81 Example Write as powers of 2: a 16 b a b 16 =2£2£2£2 = 24 16 16 c d £ 2n c = 20 d 24 = 2¡4 = e £ 2n = 22 £ 2n = 22+n 2m 2m e = 2m 23 = 2m¡3 Example a¡3 b2 c¡1 Write without negative indices: a¡3 b2 b2 c = ¡1 c a , a3 fas a¡3 = = c1 g c¡1 Example in non-fractional form 21¡n Write = 2¡(1¡n) 21¡n = 2¡1+n = 2n¡1 EXERCISE 3C.1 Write as powers of 2: a b g h e 32 f j 64 k 128 l 32 128 27 d 27 e f j 243 k 243 c d i 64 c i Write as powers of 3: a b g 81 h 81 Write as a single power of 2: a £ 2a b £ 2b c f 2c g 2m 2¡m h d (2x+1 )2 e 21¡n i 2x+1 2x j 21¡x 27 £ 3d e £ 27t j 9n+1 32n¡1 £ 2t (21¡n )¡1 4x cyan magenta yellow 95 100 50 75 25 i 27t h 95 3y 100 g 50 3y 75 f 25 d £ 9n c 95 27a 100 b 50 £ 3p 75 a 25 95 100 50 75 25 Write as a single power of 3: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_03\081IB_HL-2_03.CDR Tuesday, 23 October 2007 9:40:04 AM PETERDELL 9a 31¡a IB_HL-2ed (82) 82 EXPONENTIALS (Chapter 3) Write without negative indices: a ab¡2 b f a2 b¡1 c¡2 g (ab)¡2 a¡3 c (2ab¡1 )2 d (3a¡2 b)2 e a2 b¡1 c2 h a¡2 b¡3 i 2a¡1 d2 j 12a m¡3 d an b¡m e a¡n a2+n Write in non-fractional form: an a b c b¡n 32¡n RATIONAL INDICES 1 1+1 Notice that a £ a = a 2 = a1 = a p p and a £ a = a also a2 = So, 1 ffor index laws to be obeyedg p a fby direct comparisong Likewise a £ a £ a = a1 = a p p p and a £ a £ a = a suggests a3 = Thus in general, an = p a p n a where p n a reads “the nth root of a” Notice also that a £ a £ a = a2 ³ ´3 ) a3 = a2 ) a3 = m an = In general, fif (am )n = amn is to be usedg p a2 : p n am Example p a p a Write as a single power of 2: p b p b = 23 = c 2£ 22 =2 = 25 95 100 50 75 25 95 50 75 25 95 100 50 75 25 95 100 50 75 25 100 yellow p ¡1 magenta p = (22 ) =2 cyan c black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_03\082IB_HL-2_03.CDR Tuesday, 23 October 2007 9:42:26 AM PETERDELL IB_HL-2ed (83) EXPONENTIALS (Chapter 3) Example 83 Use your calculator to evaluate Calculator: For press: Answer: ÷ ( ^ ) ¼ 2:639 015 ENTER Example ¡2 Without using a calculator, write in simplest rational form: a 83 a ¡2 b 27 = (23 ) = (33 ) 3£ ¡2 3£¡ f(am )n = amn g =2 = 24 = 16 b 27 =3 = 3¡2 = 19 EXERCISE 3C.2 Write as a single power of 2: p a 52 b p p f 2£ 32 g p 2 Write as a single power of 3: p a 33 b p 3 c p 2 d h p ( 2)3 i c p d Write the following in the form ax p p a 37 b 27 c p f p 27 g p p 16 p 3 e j e p p p where a is a prime number and x is rational: p p p 16 d 32 e 49 p 16 h i p 32 j p 49 d d p e 32 j 125 Use your calculator to find: 34 a 28 b c ¡1 Use your calculator to evaluate in three different ways: p p p a b 48 c 27 ¡3 cyan magenta yellow h 95 100 50 ¡4 75 ¡3 16 25 c g 95 95 100 50 75 ¡1 100 83 50 f b 75 42 25 a 25 95 100 50 75 25 Without using a calculator, write in simplest rational form: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_03\083IB_HL-2_03.CDR Tuesday, 23 October 2007 9:46:17 AM PETERDELL d 25 i 27 ¡4 ¡2 IB_HL-2ed (84) 84 EXPONENTIALS (Chapter 3) D ALGEBRAIC EXPANSION AND FACTORISATION (a + b)(c + d) = ac + ad + bc + bd (a + b)(a ¡ b) = a2 ¡ b2 (a + b)2 = a2 + 2ab + b2 (a ¡ b)2 = a2 ¡ 2ab + b2 EXPANSION LAWS Example ¡1 Expand and simplify: x ¡1 x ¡1 =x ¡1 (x + 2x ¡ 3x £x ¡1 + x ¡1 (x + 2x ¡ 3x ) ) ¡1 £ 2x ¡ x ¡1 feach term is £ by x¡ g £ 3x = x1 + 2x0 ¡ 3x¡1 =x+2¡ x fadding indicesg Example 10 a b (ex + e¡x )2 a (2x + 3)(2x + 1) Expand and simplify: (2x + 3)(2x + 1) = 2x £ 2x + 2x + £ 2x + = 22x + £ 2x + = 4x + 22+x + b (ex + e¡x )2 = (ex )2 + 2ex £ e¡x + (e¡x )2 = e2x + 2e0 + e¡2x = e2x + + e¡2x EXERCISE 3D.1 Expand and simplify: cyan magenta ) ¡1 x (x + 2x + 3x ¡1 (5x ¡ 2)(5x ¡ 4) e (3x ¡ 1)2 f (4x + 7)2 h (2x + 3)(2x ¡ 3) i (x + x k (ex ¡ e¡x )2 l (5 ¡ 2¡x )2 yellow black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_03\084IB_HL-2_03.CDR Tuesday, 23 October 2007 12:03:50 PM PETERDELL ) c 95 25 95 100 50 75 25 95 100 50 75 25 j 1 (3x + 2)(3x + 5) 100 (x + 2)(x ¡ 2) (x + )2 x b 50 g x 75 (2x + 3)2 ¡1 i 25 d 5¡x (52x + 5x ) Expand and simplify: a (2x + 1)(2x + 3) f x (x + x h 2¡x (2x + 5) c 3x (2 ¡ 3¡x ) e 95 ex (ex + 2) 100 g 2x (2x + 1) b 50 d x2 (x3 + 2x2 + 1) 75 a (x2 + x + x ) ¡1 ¡1 )(x ¡ x ) IB_HL-2ed (85) EXPONENTIALS (Chapter 3) 85 FACTORISATION AND SIMPLIFICATION Example 11 a 2n+3 + 2n Factorise: 2n+3 + 2n = 2n 23 + 2n = 2n (23 + 1) = 2n £ a c 23n + 22n c 23n + 22n = 22n 2n + 22n = 22n (2n + 1) b 2n+3 + 2n+3 + = 2n 23 + = 8(2n ) + = 8(2n + 1) b Example 12 a 4x ¡ Factorise: b 9x + 4(3x ) + 4x ¡ = (2x )2 ¡ 32 = (2x + 3)(2x ¡ 3) a fdifference of two squaresg 9x + 4(3x ) + = (3x )2 + 4(3x ) + = (3x + 2)2 b fcompare a2 + 4a + 4g fas a2 + 4a + = (a + 2)2 g Example 13 6n 3n a Simplify: 6n 3n a = 4n 6n b or 2n 3n 3n = 6n 3n ¡ ¢n 4n 6n = 2n 2n 2n 3n = 4n 6n ¡ ¢n = = 2n 3n ¡ ¢n b = 2n = 2n or Example 14 yellow 2m+2 ¡ 2m 2m 2m+2 ¡ 2m 2m m 2 ¡ 2m = 2m m (4 ¡ 1) = 2m1 =3 95 50 75 25 b 95 100 50 75 95 50 75 25 95 100 50 75 25 100 magenta 25 3n + 6n 3n n + 2n 3n = 3n n (1 + 2n ) = 3n = + 2n a cyan b 100 3n + 6n 3n a Simplify: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_03\085IB_HL-2_03.CDR Thursday, 25 October 2007 9:47:11 AM PETERDELL c c 2m+3 + 2m 2m+3 + 2m 2m 23 + 2m = 2m (8 + 1) = 91 m =2 IB_HL-2ed (86) 86 EXPONENTIALS (Chapter 3) Example 15 4x + 2x ¡ 20 = Solve for x: 4x + 2x ¡ 20 = ) (2 ) + 2x ¡ 20 = ) (2x ¡ 4)(2x + 5) = ) 2x = or 2x = ¡5 ) 2x = 22 ) x=2 x fcompare a2 + a ¡ 20 = 0g fas a2 + a ¡ 20 = (a ¡ 4)(a + 5)g f2x cannot be negativeg EXERCISE 3D.2 Factorise: a 52x + 5x d 5n+1 ¡ g 3(2n ) + 2n+1 b e h 3n+2 + 3n 6n+2 ¡ 2n+2 + 2n+1 + 2n c f i en + e3n 4n+2 ¡ 16 3n+1 + 2(3n ) + 3n¡1 Factorise: a 9x ¡ d 25 ¡ 4x g 9x + 10(3x ) + 25 b e h 4x ¡ 25 9x ¡ 4x 4x ¡ 14(2x ) + 49 c f i 16 ¡ 9x 4x + 6(2x ) + 25x ¡ 4(5x ) + Factorise: a 4x + 9(2x ) + 18 d 9x + 4(3x ) ¡ b e 4x ¡ 2x ¡ 20 25x + 5x ¡ c f 9x + 9(3x ) + 14 49x ¡ 7x+1 + 12 Simplify: 12n a 6n 35x 7x e b 20a 2a c 6b 2b d 4n 20n f 6a 8a g 5n+1 5n h 5n+1 5 Simplify: a 6m + 2m 2m b 2n + 12n 2n c 8n + 4n 2n d 6n + 12n + 2n e 5n+1 ¡ 5n f 5n+1 ¡ 5n 5n g 2n ¡ 2n¡1 2n h 2n + 2n¡1 2n + 2n+1 i 3n+1 ¡ 3n 3n + 3n¡1 µ Simplify: (n + 1) + (n ¡ 1) n cyan magenta n¡1 95 100 50 75 25 95 yellow µ ¶ 4x ¡ 2x ¡ = 25x ¡ 23(5x ) ¡ 50 = 100 50 b e 25 95 100 50 75 25 95 100 50 75 25 Solve for x: a 4x ¡ 6(2x ) + = d 9x = 3x + n b 75 a n black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_03\086IB_HL-2_03.CDR Tuesday, 23 October 2007 10:14:33 AM PETERDELL ¡3 n c f n+1 ¶ 9x ¡ 12(3x ) + 27 = 49x + = 2(7x ) IB_HL-2ed (87) 87 EXPONENTIALS (Chapter 3) E EXPONENTIAL EQUATIONS An exponential equation is an equation in which the unknown occurs as part of the index or exponent For example: 2x = and 30 £ 3x = are both exponential equations If 2x = 8, then 2x = 23 Thus x = 3, and this is the only solution If ax = ak , then x = k So, if the base numbers are the same, we can equate indices Hence: Example 16 x x+2 a = 16 Solve for x: b 2x = 16 ) 2x = 24 ) x=4 a = Once we have the same base we then equate the indices 27 3x+2 = b 27 ) 3x+2 = 3¡3 ) x + = ¡3 ) x = ¡5 Example 17 4x = a Solve for x: (22 )x = 23 ) a 4x = ¡1 ) (32 )x¡2 = ) ) 32(x¡2) = 3¡1 2x ¡ = ¡1 ) 2x = ) x = 32 ) 22x = 23 ) 2x = ) x = 32 b 9x¡2 = 9x¡2 = b Remember to use the index laws correctly! EXERCISE 3E Solve for x: a 2x = 2 e 2x = i 2x¡2 = b 2x = c 3x = 27 f 3x = g 2x = j 3x+1 = k 2x+1 = 64 27 d 2x = h 2x+1 = l 21¡2x = cyan 9x¡3 = ( 14 )1¡x = yellow 95 n 50 j d 49x = 8x+2 = 32 h 81¡x = k ( 12 )x+1 = l ( 13 )x+2 = o ( 17 )x = 49 p ( 12 )x+1 = 32 100 g 75 25x = 25 50 25 100 magenta 9x = 4x = 8¡x c m f 95 42x¡1 = 8x = 100 i b 75 4x = 95 e 50 4x = 32 75 a 25 95 100 50 75 25 Solve for x: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_03\087IB_HL-2_03.CDR Tuesday, 23 October 2007 10:17:23 AM PETERDELL IB_HL-2ed (88) 88 EXPONENTIALS (Chapter 3) Solve for x: 42x+1 = 81¡x a b 92¡x = ( 13 )2x+1 c 2x £ 81¡x = b £ 2x = 56 c £ 2x+1 = 24 e £ ( 13 )x = 36 f £ ( 12 )x = 20 Solve for x: a £ 2x = 24 12 £ 3¡x = d 4 F GRAPHS OF EXPONENTIAL FUNCTIONS We have learned to deal with bn where n Q , i.e., n is any rational number But what about bn where n R , i.e., where n is not necessarily a rational? To answer this question, we can look at graphs of exponential functions y The most simple general exponential function has the form y = bx where b > 0, b 6= For example, y = 2x is an exponential function Table of values: y = 2x x ¡3 ¡2 ¡1 y 2 ¡10 We notice that for x = ¡10, say, y = ¼ 0:001 -3 -2 -1 x Also when x = ¡50, y = 2¡50 ¼ 8:88 £ 10¡16 So, it appears that as x becomes large and negative, the graph of y = 2x approaches the x-axis from above it We say that y = 2x is ‘asymptotic to the x-axis’, or ‘y = is a horizontal asymptote’ INVESTIGATION EXPONENTIAL GRAPHS We will investigate families of exponential functions GRAPHING PACKAGE TI C What to do: a On the same set of axes, use a graphing package or graphics calculator to graph the following functions: y = 2x , y = 3x , y = 10x , y = (1:3)x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 functions in a are all members of the family y = bx : What effect does changing b values have on the shape of the graph? What is the y-intercept of each graph? What is the horizontal asymptote of each graph? 50 25 95 100 50 75 25 b The i ii iii 75 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_03\088IB_HL-2_03.CDR Wednesday, November 2007 2:35:25 PM PETERDELL IB_HL-2ed (89) EXPONENTIALS (Chapter 3) 89 a On the same set of axes, use a graphing package or graphics calculator to graph the following functions: y = 2x , y = 2x + 1, y = 2x ¡ b The functions in a are all members of the family y = 2x + d where d is a constant i What effect does changing d values have on the position of the graph? ii What effect does changing d values have on the shape of the graph? iii What is the horizontal asymptote of each graph? iv What is the horizontal asymptote of y = 2x + d? c To graph y = 2x + d from y = 2x what transformation is used? a On the same set of axes, use a graphing package or graphics calculator to graph the following functions: y = 2x , y = 2x¡1 , y = 2x+2 , y = 2x¡3 b The i ii iii functions in a are all members of the family y = 2x¡c : What effect does changing c values have on the position of the graph? What effect does changing c values have on the shape of the graph? What is the horizontal asymptote of each graph? c To graph y = 2x¡c from y = 2x , what transformation is used? a On the same set of axes, use a graphing package or graphics calculator to graph the functions y = 2x and y = 2¡x b i What is the y-intercept of each graph? ii What is the horizontal asymptote of each graph? iii What transformation moves y = 2x to y = 2¡x ? a On the same set of axes, use a graphing package or graphics calculator to graph the following functions: i y = 2x , y = £ 2x , y = 12 £ 2x ii y = ¡2x , y = ¡3 £ 2x , y = ¡ 12 £ 2x b The functions in a are all members of the family y = a £ 2x where a is a constant Comment on the effect on the graph when i a > ii a < c What is the horizontal asymptote of each graph? Why? From your investigation you should have discovered that: For the general exponential function y = a £ bx¡c + d I I I I b controls how steeply the graph increases or decreases c controls horizontal translation d controls vertical translation and y = d is the equation of the horizontal asymptote ² If a > 0, b > ² If a > 0, < b < the function is the function is increasing decreasing cyan magenta yellow 95 If a < 0, < b < the function is increasing 100 50 75 25 95 100 50 75 25 95 100 50 ² If a < 0, b > the function is decreasing 75 25 95 100 50 75 25 ² black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_03\089IB_HL-2_03.CDR Thursday, 25 October 2007 9:48:26 AM PETERDELL IB_HL-2ed (90) 90 EXPONENTIALS (Chapter 3) We can sketch reasonably accurate graphs of exponential functions using: ² ² ² All exponential graphs are similar in shape and have a horizontal asymptote the horizontal asymptote the y-intercept two other points, say when x = 2, x = ¡2 Example 18 Sketch the graph of y = 2¡x ¡ 3: For y = 2¡x ¡ the horizontal asymptote is y = ¡3 y When x = 0, y = ¡ =1¡3 = ¡2 x ) the y-intercept is ¡2 When x = 2, y = 2¡2 ¡ = 14 ¡ -2 y¡=¡2 -x-3 H.A y=-3 = ¡2 34 When x = ¡2, y = 22 ¡ = We now have a well-defined meaning for bn where b, n R because simple exponential functions have smooth increasing or decreasing graphs EXERCISE 3F y x Given the graph of y = we can find approximate values of 2x for various x values For example: (see point A) I 21:8 ¼ 3:5 2:3 (see point B) I ¼5 y = 2x B A Use the graph to determine approximate values of: p b 20:8 a 2 or c 21:5 d 2¡1:6 p e p 2¡ f -2 -1 H.A y=0 x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Draw freehand sketches of the following pairs of graphs using your observations from black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_03\090IB_HL-2_03.CDR Tuesday, 23 October 2007 10:51:06 AM PETERDELL IB_HL-2ed (91) EXPONENTIALS (Chapter 3) the previous investigation: a y = 2x and y = 2x ¡ c y = 2x and y = 2x¡2 y = 2x y = 2x b d 91 and y = 2¡x and y = 2(2x ) GRAPHING PACKAGE Check your answers to using technology Draw freehand sketches of the following pairs of graphs: a y = 3x and y = 3¡x b y = 3x and y = 3x + d y = 3x and y = 3x¡1 c y = 3x and y = ¡3x Sketch the graphs of: a y = 2x + y = ¡ 2¡x p Use your GDC to graph the functions in question above and find y when x = 2: y = ¡ 2x b c y = 2¡x + d For the graphs of the functions in question above, discuss the behaviour of y as x ! §1 Hence determine the horizontal asymptotes for each graph G GROWTH AND DECAY In this exercise we will examine situations where quantities are either increasing or decreasing exponentially These situations are known as growth and decay, and occur frequently in the world around us For example, populations of animals, people, and bacteria usually grow in an exponential way Radioactive substances and items that depreciate usually decay exponentially GROWTH Consider a population of 100 mice which under favourable conditions is increasing by 20% each week To increase a quantity by 20%, we multiply it by 120% or 1:2 400 So, if Pn is the population after n weeks, then 200 P0 P1 P2 P3 = 100 fthe original populationg = P0 £ 1:2 = 100 £ 1:2 = P1 £ 1:2 = 100 £ (1:2)2 = P2 £ 1:2 = 100 £ (1:2)3 , etc Pn 300 100 n (weeks) and from this pattern we see that Pn = 100 £ (1:2)n Alternatively: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 This is an example of a geometric sequence and we could find the rule to generate it Clearly r = 1:2 and so as Pn = P0 rn , then Pn = 100 £ (1:2)n for n = 0, 1, 2, 3, black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_03\091IB_HL-2_03.CDR Thursday, 25 October 2007 9:49:35 AM PETERDELL IB_HL-2ed (92) 92 EXPONENTIALS (Chapter 3) Example 19 An entomologist monitoring a grasshopper plague notices that the area affected by the grasshoppers is given by An = 1000 £ 20:2n hectares, where n is the number of weeks after the initial observation a Find the original affected area b Find the affected area after i weeks ii 10 weeks c Find the affected area after 12 weeks d Draw the graph of An against n a A0 = 1000 £ 20 = 1000 £ = 1000 ) original area was 1000 b i c A12 = 1000 £ 20:2£12 = 1000 £ 22:4 fPress: 1000 ¼ 5278 ) after 12 weeks, area affected d is about 5300 A5 = 1000 £ 21 = 2000 i.e., area is 2000 ii A10 = 1000 £ 22 = 4000 i.e., area is 4000 × ^ 2:4 ENTER g A (ha) c 6000 b ii 4000 bi 2000 a n (weeks) 10 12 14 EXERCISE 3G.1 The weight Wt of bacteria in a culture t hours after establishment is given by Wt = 100 £ 20:1t grams Find: a the initial weight b the weight after i hours ii 10 hours iii 24 hours c Sketch the graph of Wt against t using the results of a and b only d Use technology to graph Y1 = 100 £ 20:1X and check your answers to a, b and c A breeding program to ensure the survival of pygmy possums was established with an initial population of 50 (25 pairs) From a previous program, the expected population Pn in n years’ time is given by Pn = P0 £ 20:3n cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 What is the value of P0 ? What is the expected population after: i years ii years iii 10 years? Sketch the graph of Pn against n using a and b only Use technology to graph Y1 = 50 £ 20:3X and check your answers in b 95 100 50 75 25 a b c d black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_03\092IB_HL-2_03.CDR Thursday, 11 March 2010 10:15:54 AM PETER IB_HL-2ed (93) EXPONENTIALS (Chapter 3) 93 The speed Vt of a chemical reaction is given by Vt = V0 £ 20:05t where t is the temperature in o C Find: b the speed at 20o C a the speed at 0o C c the percentage increase in speed at 20o C compared with the speed at 0o C µ ¶ V50 ¡ V20 d Find £ 100% What does this calculation represent? V20 A species of bear is introduced to a large island off Alaska where previously there were no bears pairs of bears were introduced in 1998 It is expected that the population will increase according to Bt = B0 £ 20:18t where t is the time since the introduction b Find the expected bear population in 2018 a Find B0 c Find the expected percentage increase from 2008 to 2018 DECAY Now consider a radioactive substance of original weight 20 grams which decays or reduces by 5% each year The multiplier is now 95% or 0:95 So, if Wn is the weight after n years, then: W0 W1 W2 W3 25 W (grams) n 20 = 20 grams = W0 £ 0:95 = 20 £ 0:95 grams = W1 £ 0:95 = 20 £ (0:95)2 grams = W2 £ 0:95 = 20 £ (0:95)3 grams 15 10 W20 = 20 £ (0:95)20 ¼ 7:2 grams 100 W100 = 20 £ (0:95) 10 ¼ 0:1 grams n (years) 20 and from this pattern we see that Wn = 20 £ (0:95)n Alternatively: Once again we have a geometric sequence with W0 = 20 and r = 0:95, and consequently Wn = 20 £ (0:95)n for n = 0, 1, 2, 3, Example 20 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 When a diesel-electric generator is switched off, the current dies away according to the formula I(t) = 24 £ (0:25)t amps, where t is the time in seconds a Find I(t) when t = 0, 1, and b What current flowed in the generator at the instant when it was switched off? c Plot the graph of I(t) for t > using the information above d Use your graph and/or technology to find how long it takes for the current to reach amps black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_03\093IB_HL-2_03.CDR Thursday, 11 March 2010 10:16:11 AM PETER IB_HL-2ed (94) 94 EXPONENTIALS (Chapter 3) a I(t) = 24 £ (0:25)t amps I(1) = 24 £ (0:25)1 = amps I(0) = 24 £ (0:25)0 = 24 amps b 25 20 15 10 d I(3) = 24 £ (0:25)3 = 0:375 amps ) 24 amps of current flowed When t = 0, I(0) = 24 c I(2) = 24 £ (0:25)2 = 1:5 amps I (amps) t (seconds) From the graph above, the approximate time to reach amps is 1:3 seconds or By finding the point of intersection of Y1 = 24 £ (0:25)^ X and Y2 = on a graphics calculator, the solution is ¼ 1:29 seconds Example 21 The weight of radioactive material remaining after t years is given by Wt = W0 £ 2¡0:001t grams a Find the original weight b Find the percentage remaining after 200 years a When t = 0, W0 = W0 £ 20 = W0 ) W0 is the original weight b When t = 200, W200 = W0 £ 2¡0:001£200 = W0 £ 2¡0:2 ¼ W0 £ 0:8706 ¼ 87:06% of W0 ) 87:1% remains EXERCISE 3G.2 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The weight of a radioactive substance t years after being set aside is given by W (t) = 250 £ (0:998)t grams a How much radioactive substance was put aside? b Determine the weight of the substance after: i 400 years ii 800 years iii 1200 years c Sketch the graph of W (t) for t > 0, using the above information d Use your graph or graphics calculator to find how long it takes for the substance to decay to 125 grams black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_03\094IB_HL-2_03.CDR Tuesday, 23 October 2007 11:01:29 AM PETERDELL IB_HL-2ed (95) EXPONENTIALS (Chapter 3) 95 The temperature T of a liquid which has been placed in a refrigerator is given by T (t) = 100 £ 2¡0:02t o C where t is the time in minutes Find: a the initial temperature b the temperature after: i 15 minutes ii 20 minutes iii 78 minutes c Sketch the graph of T (t) for t > using a and b only The weight Wt grams of radioactive substance remaining after t years is given by Wt = 1000 £ 2¡0:03t grams Find: a the initial weight b the weight after: i 10 years ii 100 years iii 1000 years c Graph Wt against t using a and b only The weight Wt of radioactive uranium remaining after t years is given by the formula Wt = W0 £ 2¡0:0002t grams, t > Find: a the original weight H b the percentage weight loss after 1000 years THE NATURAL EXPONENTIAL ‘e’ We have seen that the simplest exponential functions are of the form f(x)¡=¡bx where ¡1: b¡>¡0, b¡= Below are some graphs of simple exponential functions y=(0.2) x y=5 x y y=2 x y=(0.5) x y=1.2 x x We can see that for all positive values of the base b, the graph is always positive bx > for all b > Hence There are a vast number of possible choices for the base number However, where exponential data is examined in science, engineering, and other areas, the base e ¼ 2:7183 is commonly used cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 e is a special number in mathematics It is irrational like ¼, and just as ¼ is the ratio of a circle’s area to its diameter, e also has a physical meaning We explore this meaning in the following investigation black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_03\095IB_HL-2_03.CDR Thursday, 25 October 2007 9:51:21 AM PETERDELL IB_HL-2ed (96) 96 EXPONENTIALS (Chapter 3) INVESTIGATION CONTINUOUS COMPOUND INTEREST A formula for calculating the amount to which an investment grows is un = u0 (1 + i)n where u0 is the initial amount un is the final amount i is the interest rate per compounding period n is the number of periods or number of times the interest is compounded We will investigate the final value of an investment for various values of n, and allow n to get extremely large What to do: Suppose $1000 is invested for one year at a fixed rate of 6% p.a Use your calculator to find the final amount or maturing value if the interest is paid: a annually (n = 1, i = 6% = 0:06) c monthly d daily b quarterly (n = 4, i = 6% = 0:015) e by the second f by the millisecond Comment on your answers obtained in If r is the percentage rate per year, t is the number of years, and r N is the number of interest payments per year, then i = and n = Nt N ³ r ´Nt This means that the growth formula becomes un = u0 + N ·µ ¶a ¸ rt N If we let a = , show that un = u0 + : r a For continuous compound growth, the number of interest payments per year N gets very large a Explain why a gets very large as N gets very large b Copy and complete the table: Give answers as accurately as technology permits You should have discovered that for very large a values, µ ¶a 1+ ¼ 2:718 281 828 459:::::: a a µ ¶a 1+ a 10 100 1000 10 000 100 000 Now use the e x key of your calculator to find the value of e1 , i.e., press e x = ex or = What you notice? un = u0 ert For continuous growth, cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 where u0 is the initial amount r is the annual percentage rate t is the number of years Use this formula to find the final value if $1000 is invested for years at a fixed rate of 6% p.a., where the interest is calculated continuously black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_03\096IB_HL-2_03.CDR Thursday, 11 March 2010 10:16:41 AM PETER IB_HL-2ed (97) 97 EXPONENTIALS (Chapter 3) From Investigation we observe that: “If interest is paid continuously (instantaneously) then the formula for calculating a compounding amount un = u0 (1 + i)n can be replaced by un = u0 ert , where r is the percentage rate p.a and t is the number of years.” RESEARCH RESEARCHING e y What to do: The ‘bell curve’ which models statistical distributions is shown alongside Research the equation of this curve x p e + = is called Euler’s equation where i is the imaginary number ¡1 Research the significance of this equation i¼ The series f (x) = + x + 12 x2 + 2£3 x x + 2£3£4 x + :::::: has infinitely many terms It has been shown that f(x) = e Check this statement by finding an approximation for f (1) using its first 20 terms EXERCISE 3H e x key on a calculator to find the value of e to as many digits as possible Use the Sketch, on the same set of axes, the graphs of y = 2x , y = ex and y = 3x Comment on any observations GRAPHING PACKAGE Sketch, on the same set of axes, the graphs of y = ex and y = e¡x What is the geometric connection between these two graphs? For the general exponential function y = aekx , what is the y-intercept? Consider y = 2ex a Explain why y can never be < b Find y if: i Find, to significant figures, the value of: b e3 c e0:7 a e2 d Write the following as powers of e: p p e b e e a x = ¡20 ii p e e e¡1 d e2 x = 20 c p e c (e¡0:04 ) d (e¡0:836 ) e4:829 1000e1:2642 d h e¡4:829 0:25e¡3:6742 Simplify: t (e0:36 ) a t (e0:064 ) 16 b Find, to five significant figures, the values of: b e¡2:31 c a e2:31 ¡0:1764 ¡0:6342 e 50e f 80e g t t cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 10 On the same set of axes, sketch and clearly label the graphs of: g : x 7! ex¡2 , h : x 7! ex + f : x 7! ex , State the domain and range of each function black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_03\097IB_HL-2_03.CDR Tuesday, 23 October 2007 11:46:32 AM PETERDELL IB_HL-2ed (98) 98 EXPONENTIALS (Chapter 3) 11 On the same set of axes, sketch and clearly label the graphs of: g : x 7! ¡ex , h : x 7! 10 ¡ ex f : x 7! ex , State the domain and range of each function t 12 The weight of bacteria in a culture is given by W (t) = 2e grams where t is the time in hours after the culture was set to grow a What is the weight of the culture at: i t=0 ii t = 30 t = 12 hours iii t iv t = hours? b Use a to sketch the graph of W (t) = 2e 13 The current flowing in an electrical circuit t seconds after it is switched off is given by I(t) = 75e¡0:15t amps a What current is still flowing in the circuit after: i t = sec ii t = 10 sec? b Use your graphics calculator to sketch ¡0:15t and I = I(t) = 75e c Find how long it would take for the current to fall to amp a Given f : x 7! ex , find the defining equation of f ¡1 b Sketch the graphs of y = ex , y = x and y = f ¡1 (x) on the same set of axes 14 REVIEW SET 3A Simplify: a ¡(¡1)10 b ¡(¡3)3 30 ¡ 3¡1 c Simplify using the index laws: c 5(x2 y)2 (5x2 )2 Write the following as a power of 2: b 16 ¥ 2¡3 a £ 2¡4 c 84 Write without brackets or negative indices: b (ab)¡1 a b¡3 c ab¡1 a a4 b5 £ a2 b2 6xy5 ¥ 9x2 y b 2x¡3 = Find the value of x, without using your calculator: a Evaluate without using a calculator: a 83 b 32 b 9x = 272¡2x ¡2 27 Evaluate, correct to significant figures, using your calculator: a 34 ¡1 b c 27 If f(x) = £ 2x , find the value of: a f (0) b f(3) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 On the same set of axes draw the graphs of a y = 2x b the y-intercept and the equation of the horizontal asymptote black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_03\098IB_HL-2_03.CDR Tuesday, 23 October 2007 11:48:52 AM PETERDELL p 100 c f (¡2) y = 2x ¡ 4, stating IB_HL-2ed (99) EXPONENTIALS (Chapter 3) 99 10 The temperature of a liquid t minutes after it was heated is given by T = 80 £ (0:913)t o C Find: a the initial temperature of the liquid b the temperature after i t = 12 ii t = 24 iii t = 36 minutes c Draw the graph of T against t for t > 0, using the above or technology d Hence, find the time taken for the temperature to reach 25o C REVIEW SET 3B Simplify: ¡(¡2)3 a 5¡1 ¡ 50 b Simplify using the index laws: a (a7 )3 pq £ p3 q b Write as powers of 2: 16 a 2x £ b c Write without brackets or negative indices: b 2(ab)¡2 a x¡2 £ x¡3 Solve for x without using a calculator: Write as powers of 3: a Write as a single power of 3: For y = 3x ¡ : a find y when x = 0, §1, §2 c sketch the graph of y = 3x ¡ 2x+1 = 32 2ab¡2 b d 4x+1 = ¡ ¢x 243 p ( 3)1¡x £ 91¡2x b discuss y as x ! and as x ! ¡1 state the equation of any asymptote b d Without using a calculator, solve for x: 27 c 27 9a a 4x ¥ c a b 81 8ab5 2a4 b4 c 27x = a b 91¡x = 27x+2 10 On the same set of axes, sketch and clearly label the graphs of: f : x 7! ex , g : x 7! ex¡1 , h : x 7! ¡ ex State the domain and range of each function REVIEW SET 3C a Write £ 2n as a power of b Evaluate 7¡1 ¡ 70 c Write ( 23 )¡3 in simplest fractional form µ ¡1 ¶2 2a d Simplify Do not have negative indices or brackets in your answer b2 2x+1 a Write 288 as a product of prime numbers in index form b Simplify 21¡x Write as powers of in simplest form: p a b 5 c p d 25a+3 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_03\099IB_HL-2_03.CDR Tuesday, 23 October 2007 11:57:11 AM PETERDELL IB_HL-2ed (100) 100 EXPONENTIALS (Chapter 3) Simplify: a ¡(¡2)2 b (¡ 12 a¡3 )2 c (¡3b¡1 )¡3 Expand and simplify: a ex (e¡x + ex ) b (2x + 5)2 c (x ¡ 7)(x + 7) Expand and simplify: a (3 ¡ 2a )2 b p p ( x + 2)( x ¡ 2) c 2¡x (22x + 2x ) b £ ( 13 )x = 324 a £ 2x = 192 Solve for x: The weight of a radioactive substance after t years is given by W = 1500 £ (0:993)t grams a Find the original amount of radioactive material b Find the amount of radioactive material remaining after: i 400 years ii 800 years c Sketch the graph of W against t, t > 0, using the above or technology d Hence, find the time taken for the weight to reduce to 100 grams cyan magenta yellow 95 100 50 b discuss y as x ! and as x ! ¡1 d state the equation of any asymptote 75 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 10 For y = 2e¡x + : a find y when x = 0, §1, §2 c sketch the graph of y = 2e¡x + 16a 2b 25 8a = , find b 128 If 4a 2b = and black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_03\100IB_HL-2_03.CDR Tuesday, 23 October 2007 11:57:54 AM PETERDELL IB_HL-2ed (101) Chapter Logarithms Contents: A B C D E F G H Logarithms Logarithms in base 10 Laws of logarithms Natural logarithms Exponential equations using logarithms The change of base rule Graphs of logarithmic functions Growth and decay cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 4A Review set 4B Review set 4C Review set 4D black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\101IB_HL-2_04.CDR Wednesday, 24 October 2007 11:34:00 AM PETERDELL IB_HL-2ed (102) 102 LOGARITHMS (Chapter 4) OPENING PROBLEM Paulo knows that when he invests E12 000 for n years at an interest rate of 8:35% p.a compounded annually, the value of the investment at the end of this period is given by An+1 = 12 000 £ (1:0835)n euros Consider the following questions: What is the value of A1 and what is its interpretation? How would we find the value of the investment after years? If we let n = 2:25, A3:25 = 12 000 £ (1:0835)2:25 : Does the power 2:25 have a meaning? What is the interpretation of the value of A3:25 ? How long would it take for the investment to double in value? What would the graph of An+1 against n look like? After studying the concepts of this chapter, you should be able to answer the questions above A LOGARITHMS HISTORICAL NOTE In the late 16th century, astronomers spent a large part of their working lives doing the complex and tedious calculations of spherical trigonometry needed to understand the movement of celestial bodies A Scotsman, John Napier, discovered a method of simplifying these calculations using logarithms So effective was Napier’s method that it was said he effectively doubled the life of an astronomer by reducing the time required to these calculations Consider the function f : x 7! 10x f ( x) = 10 x or y = 10 x The defining equation of f is f (x) = 10x or y = 10x y y=x Now consider the graph of f and its inverse function f ¡1 f -1 The question arises: ¡1 How can we write f in functional form, or, what is the defining function of f ¡1 ? x x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 As f is defined by y = 10 , f ¡1 is defined by x = 10y finterchanging x and yg black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\102IB_HL-2_04.CDR Wednesday, 24 October 2007 9:26:31 AM PETERDELL IB_HL-2ed (103) 103 LOGARITHMS (Chapter 4) So, y is the exponent to which 10 (the base) is raised in order to get x We write this as y = log10 x and say that So, “y is the logarithm of x in base 10.” ² if f (x) = 10x , then f ¡1 (x) = log10 x ² if f (x) = bx , then f ¡1 (x) = logb x LOGARITHMS IN BASE b If A = bn , b 6= 1, b > 0, we say that n is the logarithm of A in base b, and that A = bn , n = logb A, A > In general: A = bn , n = logb A is a short way of writing: “if A = bn then n = logb A, and if n = logb A then A = bn ” We say that A = bn = 23 means that = log2 and vice versa log5 25 = means that 25 = 52 and vice versa ² ² For example: and n = logb A are equivalent or interchangeable statements If y = bx then x = logb y, and so x = logb bx If x = by then y = logb x, and so x = blogb x provided x > Example a b Write an equivalent exponential statement for log10 1000 = Write an equivalent logarithmic statement for 34 = 81 a From log10 1000 = we deduce that 103 = 1000 b From 34 = 81 we deduce that log3 81 = Example Find: a log10 100 b log2 32 c log5 (0:2) a To find log10 100 we ask “What power must 10 be raised to, to get 100?” As 102 = 100, then log10 100 = b As 25 = 32, then log2 32 = c As 5¡1 = = 0:2, then log5 (0:2) = ¡1 EXERCISE 4A Write an equivalent exponential statement for: b log10 (0:1) = ¡1 a log10 10 000 = cyan yellow 95 100 50 75 25 95 100 50 75 25 95 50 100 magenta log2 ( 14 ) = ¡2 e log2 = 75 25 95 100 50 75 25 d black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\103IB_HL-2_04.CDR Friday, January 2008 9:07:37 AM DAVID3 c f p log10 10 = 12 p log3 27 = 1:5 IB_HL-2ed (104) 104 LOGARITHMS (Chapter 4) Write an equivalent logarithmic statement for: b 2¡3 = 18 a 22 = e 26 = 64 d 72 = 49 Find: a log10 100 000 e log2 64 i log2 (0:125) m b f j q r Use your calculator to find: b log10 25 a log10 152 Solve for x: a log2 x = b log4 x = 10¡2 = 0:01 3¡3 = 27 c g k p log3 log5 25 log4 16 d h l log2 log5 125 log36 o loga an p log8 s log4 t log9 c log10 74 log10 (0:01) log2 128 log9 p log2 p log6 6 n log3 243 ¡ ¢ logt 1t c f c d d logx 81 = log10 0:8 log2 (x ¡ 6) = a Prove that loga an = n b Hence, find: i log4 16 ¶ µ v log2 p ii log2 iii log3 vi p log5 (25 5) vii log3 B ¡1¢ ³ p1 iv ´ viii p log10 100 ¶ µ p log4 2 LOGARITHMS IN BASE 10 10 000 = 104 1000 = 103 100 = 102 10 = 101 = 100 0:1 = 10¡1 0:01 = 10¡2 0:001 = 10¡3 For example, Many positive numbers can be easily written in the form 10x etc p p Numbers like 10, 10 10 and p can also be written in the form 10x as follows: 10 p p p 10 10 10 10 = 101 £ 100:5 = 10 = 10¡ 1:5 = 10 0:5 = 10 = 10¡0:2 In fact, all positive numbers can be written in the form 10x by using logarithms in base 10 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The logarithm in base 10 of a positive number is its power of 10 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\104IB_HL-2_04.CDR Wednesday, 24 October 2007 9:32:05 AM PETERDELL IB_HL-2ed (105) LOGARITHMS (Chapter 4) 105 For example: ² Since 1000 = 103 , we write log10 1000 = or log 1000 = ² Since 0:01 = 10¡2 , we write log10 (0:01) = ¡2 or log(0:01) = ¡2 a = 10 log a In algebraic form, for any a > If no base is indicated we assume it means base 10 Notice that a must be positive since 10 > for all x R x Notice also that log 1000 = log 103 = and log 0:01 = log 10¡2 = ¡2 log 10 x = x give us the useful alternative Example a b Without using a calculator, find: i log 100 Check your answers using technology a i log 100 = log 102 = b i press log 100 ii press log 10 ^ 0:25 ) ) ii p log( 10): p log( 10) = log(10 ) = ii Answer: ENTER Answer: 0:25 ENTER Example Use your calculator to write the following in the form 10x where x is correct to decimal places: a b 800 c 0:08 a b = 10log ¼ 100:9031 800 = 10log 800 ¼ 102:9031 c 0:08 = 10log 0:08 ¼ 10¡1:0969 Example Use your calculator to find: i Explain why log 20 = log + 1: cyan magenta ii log log 20 yellow 95 100 50 75 25 95 b log 20 = log(2 £ 10) ¼ log(100:3010 £ 101 ) fadding indicesg ¼ log 101:3010 ¼ 1:3010 ¼ log + 100 50 75 25 log 20 ¼ 1:3010 fcalculatorg ii log ¼ 0:3010 95 i 100 50 25 95 100 50 75 25 a 75 a b black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\105IB_HL-2_04.CDR Wednesday, 24 October 2007 9:37:11 AM PETERDELL IB_HL-2ed (106) 106 LOGARITHMS (Chapter 4) Example x = 10log x x = 103 x = 1000 a Find x if: a log x = b log x ¼ ¡0:271 ) ) x = 10log x x ¼ 10¡0:271 x ¼ 0:536 b ) ) EXERCISE 4B Without using a calculator, a log 10 000 b p f e log 10 p j i log 100 log 10n m n find: log 0:001 p log( 10) ³ ´ 100 log p 10 c g k log(10a £ 100) o Find using a calculator: a e b log 10 000 p log 100 f p log 10 ³ ´ log p110 c log 0:001 p log 10 10 log 10 ³ ´ log p 10 p log(10 £ 10) µ ¶ 10 log 10m g d h l p log p log 10 10 p log 1000 10 µ a¶ 10 log 10b d log h log p 10 ³ ´ p 10 Use your calculator to write these in the form 10x where x is correct to decimal places: a b 60 c 6000 d 0:6 e 0:006 f 15 g 1500 h 1:5 i 0:15 j 0:000 15 a Use your calculator to find: i log b Explain why log 300 = log + 2: ii log 300 a Use your calculator to find: i log b Explain why log 0:05 = log ¡ 2: ii log 0:05 Find x if: a log x = e log x = 12 b f c g log x = log x = ¡ 12 C log x = log x ¼ 0:8351 d h log x = ¡1 log x ¼ ¡3:1997 LAWS OF LOGARITHMS INVESTIGATION DISCOVERING THE LAWS OF LOGARITHMS What to do: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Use your calculator to find: a log + log b log + log c log + log 20 d log e log 21 f log 80 From your answers, suggest a possible simplification for ln a + ln b black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\106IB_HL-2_04.CDR Wednesday, 24 October 2007 9:42:09 AM PETERDELL IB_HL-2ed (107) 107 LOGARITHMS (Chapter 4) Use your calculator to find: a log ¡ log d log From your answers, suggest a Use your calculator to find: a log d log(23 ) From your answers, suggest a b log 12 ¡ log c log ¡ log e log f log(0:6) possible simplification for ln a ¡ ln b: b log c ¡4 log e log(52 ) f log(3¡4 ) possible simplification for n log a From the investigation, you should have discovered the three important laws of logarithms: ² If A and B are both positive then: ² ² log A + log B = log(AB) µ ¶ A log A ¡ log B = log B n log A = log (An ) More generally, in any base c we have these laws of logarithms: ² If A and B are both positive then: ² ² logc A + logc B = logc (AB) µ ¶ A logc A ¡ logc B = logc B n logc A = logc (An ) These laws are easily established using the first three index laws Example Use the laws of logarithms to write the following as a single logarithm: a log + log b log 24 ¡ log c log ¡ a log 24 ¡ log ¡ ¢ = log 24 = log b log + log = log(5 £ 3) = log 15 c log ¡ = log ¡ log 101 ¡5¢ = log 10 ¡ ¢ = log 12 Example magenta yellow 95 100 50 75 25 log ¡ log = log(72 ) ¡ log(23 ) = log 49 ¡ log ¡ ¢ = log 49 95 50 75 25 95 100 50 75 25 95 100 50 75 25 cyan 100 a Write as a single logarithm in the form log a, a Q a log ¡ log b log ¡ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\107IB_HL-2_04.CDR Wednesday, 24 October 2007 9:45:01 AM PETERDELL b log ¡ = log(32 ) ¡ log 101 = log ¡ log 10 = log(0:9) IB_HL-2ed (108) 108 LOGARITHMS (Chapter 4) Example log log Simplify log log 23 = log log 22 log = log = 32 without using a calculator Example 10 a log Show that: ¡ ¢ a log 19 = ¡2 log ¡1¢ b log 500 ¢ ¡ = log 1000 = log(3¡2 ) = ¡2 log = log 1000 ¡ log = log 103 ¡ log = ¡ log b log 500 = ¡ log EXERCISE 4C.1 Write a d g j m as a single logarithm: log + log log + log + log + log b e h k log ¡ log log + log(0:4) log ¡ log 40 ¡ c f i l log 50 ¡ n ¡ log 50 o Write as a single logarithm or integer: a log + log b log + log d log ¡ log e g ¡ log ¡ log h ¡ log + log 20 log + log 3 Simplify without using a calculator: log log 27 a b log log log log d log 25 log(0:2) e f log ¡ log ¡1¢ log i 2¡ c log log f log log(0:25) c log 40 ¡ log log + log + log log + log ¡ log log ¡ log ¡ log ¡ ¢ log 43 + log + log log ¡ log Check your answers using a calculator Show that: a log = log ¡ ¢ d log 15 = ¡ log b p log = log c log e log = ¡ log f log 5000 = ¡ log 2 ¡1¢ = ¡3 log If p = logb 2, q = logb and r = logb write in terms of p, q and r: cyan magenta yellow 95 c logb 45 f logb (0:2) 100 50 75 25 95 100 50 75 25 95 logb 108 ¡5¢ logb 32 e 100 50 75 25 95 100 50 75 25 d b logb ³ p ´ logb 5 a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\108IB_HL-2_04.CDR Wednesday, 24 October 2007 9:46:51 AM PETERDELL IB_HL-2ed (109) 109 LOGARITHMS (Chapter 4) If log2 P = x, log2 Q = y a log2 (P R) d p log2 (P Q) and log2 R = z b log2 (RQ2 ) µ e write in terms of x, y and z: µ ¶ PR c log2 Q µ 2p ¶ R Q f log2 P3 Q3 p R log2 ¶ If logt M = 1:29 and logt N = 1:72 find: a b logt N µ c logt (MN) logt N2 p M ¶ LOGARITHMIC EQUATIONS Example 11 a y = a2 b Write these as logarithmic equations (in base 10): a ) ) ) c y log y log y log y = a2 b = log(a2 b) = log a2 + log b = log a + log b µ ¶ P = 20 p n b y= 20 n2 20 P =p n y= ) µ c a b3 ³ ´ a log y = log b log y = log a ¡ log b3 log y = log a ¡ log b b ) log P = log a b3 ¶ ) ) and so log P = log 20 ¡ log n Example 12 a log A = log b + log c b log M = log a ¡ Write the following equations without logarithms: a log M = log a ¡ ) log M = log a3 ¡ log 101 µ 3¶ a ) log M = log 10 a3 ) M= 10 b log A = log b + log c ) log A = log b + log c2 ) log A = log(bc2 ) ) A = bc2 EXERCISE 4C.2 magenta 95 c M = ad4 d g y = abx h k S = 200 £ 2t l 100 50 75 50 75 25 95 yellow 25 100 50 25 95 100 50 75 25 cyan j ab c f L= y = 20b3 a Q= n b r a N= b b 100 i a 95 e y = 2x p R=b l 75 Write the following as logarithmic equations (in base 10): black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\109IB_HL-2_04.CDR Wednesday, 24 October 2007 11:00:17 AM PETERDELL p T =5 d 20 F =p n m a y= n b IB_HL-2ed (110) 110 LOGARITHMS (Chapter 4) Write the following equations without logarithms: b log F = log ¡ log t a log D = log e + log d log M = log b + log c c log P = log x e log B = log m ¡ log n f log N = ¡ 13 log p g log P = log x + h log Q = ¡ log x Solve for x: a log3 27 + log3 ( 13 ) = log3 x p c log5 125 ¡ log5 = log5 x e log x + log(x + 1) = log 30 D b log5 x = log5 ¡ log5 (6 ¡ x) d f log20 x = + log20 10 log(x + 2) ¡ log(x ¡ 2) = log NATURAL LOGARITHMS In Chapter we came across the natural exponential e ¼ 2:718 28 If f is the exponential function x 7! ex (i.e., f(x) = ex or y = ex ) then its inverse function, f ¡1 is x = ey or y = loge x y = loge x is the reflection of y = ex in the mirror line y = x ln x is used to represent loge x ln x is called the natural logarithm of x So, ln = ln e0 = Notice that: p ln e = ln e = ln e = ln e1 = ln e2 = µ ¶ and ln = ln e¡1 = ¡1 e ln ex = x and eln x = x In general, ¢x ¡ Also, since ax = eln a = ex ln a , ax = ex ln a Example 13 Use your calculator to write the following in the form ek where k is correct to decimal places: a 50 b 0:005 a b 50 = eln 50 ¼ e3:9120 ln x fusing x = e 0:005 = eln 0:005 ¼ e¡5:2983 g Example 14 magenta b yellow 95 100 50 75 25 95 ln x = 2:17 ) x = e2:17 ) x ¼ 8:76 100 50 25 95 100 50 75 25 95 100 50 75 25 cyan 75 a Find x if: a ln x = 2:17 b ln x = ¡0:384 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\110IB_HL-2_04.CDR Friday, January 2008 9:11:28 AM DAVID3 ln x = ¡0:384 ) x = e¡0:384 ) x ¼ 0:681 IB_HL-2ed (111) 111 LOGARITHMS (Chapter 4) Example 15 Use the laws of logarithms to write the following as a single logarithm: a ln + ln b ln 24 ¡ ln c ln ¡ a ln 24 ¡ ln ¡ ¢ = ln 24 = ln b ln + ln = ln(5 £ 3) = ln 15 ln ¡ = ln ¡ ln e1 ¡ ¢ = ln 5e c Example 16 ln ¡ ln = ln(72 ) ¡ ln(23 ) = ln 49 ¡ ln ¡ ¢ = ln 49 a Use the laws of logarithms to simplify: a ln ¡ ln b ln ¡ Example 17 a Show that: ¡ ¢ a ln 19 = ¡2 ln ln b µ ¡1¢ b ln 500 = ln = ln(3¡2 ) = ¡2 ln b ln 500 ¼ 6:9078 ¡ ln 2 ln ¡ = ln(32 ) ¡ ln e = ln ¡ ln e µ ¶ = ln e 1000 ¶ = ln 1000 ¡ ln ¼ 6:9078 ¡ ln Example 18 Write the following equations without logarithms: a ln A = ln c + b ln M = ln a ¡ a ln M = ln a ¡ ) ln M ¡ ln a = ¡2 ) ln M ¡ ln a3 = ¡2 µ ¶ M ) ln = ¡2 a3 b ln A = ln c + ) ln A ¡ ln c = ) ln A ¡ ln c2 = µ ¶ A ) ln = c M = e¡2 a3 ) M = a3 e¡2 A = e3 c2 ) A = e3 c2 ) ) or M = a3 e2 EXERCISE 4D Without using a calculator find: b ln a ln e3 c p ln e µ d ln e2 ¶ Check your answers to question using a calculator cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Explain why ln(¡2) and ln cannot be found black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\111IB_HL-2_04.CDR Wednesday, 24 October 2007 10:01:25 AM PETERDELL IB_HL-2ed (112) 112 LOGARITHMS (Chapter 4) µ ¶ ea d ln(e ) b ln(e £ e ) c e ln eb x Use your calculator to write these in the form e where x is correct to dec places: a b 60 c 6000 d 0:6 e 0:006 f 15 g 1500 h 1:5 i 0:15 j 0:000 15 Simplify: a ln ea ¢ ¡ ln ea £ eb a Find x if: a ln x = e ln x = ¡5 b f c g ln x = ln x ¼ 0:835 Write as a single logarithm: a ln 15 + ln d ln + ln g + ln j + ln a b ln x = ¡1 ln x ¼ ¡3:2971 d h ln x = ln x ¼ 2:145 ln 15 ¡ ln ln + ln(0:2) ln ¡ ln 20 ¡ c f i l ln 20 ¡ ln ln + ln + ln ln + ln ¡ ln ln 12 ¡ ln ¡ ln Write in the form ln a, a Q : a ln + ln b ln + ln c i ln ¡ ln ¡1¢ ln 27 ¡ ¢ ¡2 ln 14 c ln f b e h k d ln ¡ ln e g ¡ ln h Show that: a ln 27 = ln ¡ ¢ d ln 16 = ¡ ln b f ln + ln ¡ ¢ ¡ ln 12 p ln = 12 ln ³ ´ ln p12 = ¡ 12 ln ¡ 16 ¢ ³e´ = ¡4 ln = ¡5 ln i = ¡ ln ´ ln p = ¡ 15 ln 10 Write the following equations without logarithms: b ln F = ¡ ln p + a ln D = ln x + c ln P = f ln N = ¡ 13 ln g g ln p 5= e h ln ln ¡ 32 ¢ ³ d ln M = ln y + e ln B = ln t ¡ g ln Q ¼ ln x + 2:159 h ln D ¼ 0:4 ln n ¡ 0:6582 E ln ln x EXPONENTIAL EQUATIONS USING LOGARITHMS In Chapter we found solutions to simple exponental equations by creating equal bases and then equating indices However, it is not always easy to make the bases the same In these situations we use logarithms to find the solution Example 19 Solve for x, giving your answer to significant figures: 2x = 30 cyan magenta yellow 95 100 50 75 25 95 ffind the logarithm of each sideg 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ) 2x = 30 log 2x = log 30 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\112IB_HL-2_04.CDR Thursday, 25 October 2007 3:27:22 PM PETERDELL IB_HL-2ed (113) 113 LOGARITHMS (Chapter 4) ) flog an = n log ag x log = log 30 log 30 ) x= log ) x ¼ 4:91 (3 s.f.) Example 20 Solve for t to significant figures: 200 £ 20:04t = 6: 200 £ 20:04t = 6 ) 20:04t = 200 fdividing both sides by 200g 0:04t ) = 0:03 0:04t = log 0:03 ffind the logarithm of each sideg ) log ) 0:04t £ log = log 0:03 flog an = n log ag log 0:03 ) t= ¼ ¡126 (3 s.f.) 0:04 £ log EXERCISE 4E Solve for x, giving your answer correct to significant figures: b 3x = 20 c 4x = 100 a 2x = 10 e 2x = 0:08 f 3x = 0:000 25 d (1:2)x = 1000 ( 12 )x = 0:005 g h ( 34 )x = 10¡4 i (0:99)x = 0:000 01 Find the solution to the following correct to significant figures: a 200 £ 20:25t = 600 b 20 £ 20:06t = 450 c 30 £ 3¡0:25t = d 12 £ 2¡0:05t = 0:12 e 50 £ 5¡0:02t = f 300 £ 20:005t = 1000 To solve exponential equations of the form ex = a we simply use the property: If ex = a then x = ln a This rule is clearly true, because if ex = a then ln ex = ln a ) x = ln a ffinding ln of both sidesg fln ex = xg Example 21 x b e = 21:879 Find x to s.f if: a ex = 30 x e = 21:879 x ) = ln 21:879 cyan magenta yellow 95 100 x ¼ 9:257 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ) 50 b 75 ex = 30 ) x = ln 30 ) x ¼ 3:401 25 a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\113IB_HL-2_04.CDR Wednesday, 24 October 2007 10:06:33 AM PETERDELL c 20e4x = 0:0382 c 20e4x = 0:0382 ) e4x = 0:001 91 ) 4x = ln 0:001 91 ) 4x ¼ ¡6:2607 ) x ¼ ¡1:565 IB_HL-2ed (114) 114 LOGARITHMS (Chapter 4) Solve for x, giving answers correct to significant figures: b ex = 1000 a ex = 10 x e2 = 20 £ e0:06x = 8:312 d g x e = 157:8 50 £ e¡0:03x = 0:816 e h c ex = 0:008 62 f i e 10 = 0:016 82 41:83e0:652x = 1000 x Note: Remember that you may not need to use logs when solving exponential equations It is usually much easier if you can get the same base on both sides F THE CHANGE OF BASE RULE If logb A = x, then bx = A ) logc bx = logc A ) x logc b = logc A logc A ) x= logc b ftaking logarithms in base cg fpower law of logarithmsg logb A = So, logc A logc b Example 22 a letting log2 = x logc A b using the rule logb A = logc b i c = 10 ii c = e Find log2 by: a Let log2 = x ) = 2x ) log 2x = log ) x log = log log ¼ 3:17 ) x= log with: log10 log10 ¼ 3:17 b i log2 = ii log2 = ln ln ¼ 3:17 Example 23 Solve for x: 8x ¡ 5(4x ) = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 8x ¡ 5(4x ) = ) 23x ¡ 5(22x ) = ) 22x (2x ¡ 5) = fas 22x > for all xg ) 2x = ) x = log2 log ¼ 2:32 fCheck this using technology.g ) x= log black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\114IB_HL-2_04.CDR Wednesday, 24 October 2007 10:08:51 AM PETERDELL IB_HL-2ed (115) LOGARITHMS (Chapter 4) 115 EXERCISE 4F log10 A log10 b Use the rule logb A = a b log3 12 2x = 0:051 a c log 12 1250 ln A ln b Use the rule logb A = to find, correct to significant figures: log0:4 (0:006 984) to solve, correct to significant figures: 4x = 213:8 b d log3 (0:067) c 32x+1 = 4:069 Hint: In 2a 2x = 0:051 implies that x = log2 (0:051) Solve for x: a 25x ¡ 3(5x ) = b 8(9x ) ¡ 3x = Solve for x: p a log4 x3 + log2 x = b log16 x5 = log64 125 ¡ log4 p x Find the exact value of x for which 4x £ 54x+3 = 102x+3 G GRAPHS OF LOGARITHMIC FUNCTIONS Consider the general exponential function f : x 7! ax , a > 0, a 6= 1: The defining equation of f is f (x) = ax (or y = ax ) The graph of y = ax is: For < a < 1: For a > 1: y y (-1, ) a (-1, ) a (1, a) 1 (1, a) x x These functions have the horizontal asymptote y = (the x-axis) They have domain R (all real numbers) and range fy : y > 0g or y ] 0, [ Obviously the function y = ax is one-to-one and has an inverse function f ¡1 So, if f is y = ax , then f ¡1 is x = ay , i.e., y = loga x If f (x) = ax cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 The domain of f = the range of f ¡1 The range of f = the domain of f ¡1 ² 95 The domain of f ¡1 is fx : x > 0g or x ] 0, [ The range of f ¡1 is y R 100 50 ² 75 25 95 100 50 75 25 Note: then f ¡ (x) = loga x black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\115IB_HL-2_04.CDR Wednesday, 24 October 2007 10:10:41 AM PETERDELL IB_HL-2ed (116) 116 LOGARITHMS (Chapter 4) LOGARITHMIC GRAPHS The graphs of y = loga x are: For < a < 1: For a > 1: y¡=¡a x y¡=¡a x y y y = log a x (a, 1) y = log a x (1a , - 1) x x y¡=¡x (a, 1) 1 ( 1a , - 1) y=x Note: ² ² ² ² ² both graphs are reflections of y = ax in the line y = x both functions have domain fx: x > 0g or x ] 0, [ we can only find logarithms of positive numbers both graphs have the vertical asymptote x = (the y-axis) for < a < 1, as x ! 1, y ! ¡1 and as x ! (from right), y ! for a > 1, as x ! 1, y ! and as x ! (from right), y ! ¡1 ² to find the domain of loga g(x), we find the solutions of g(x) > Example 24 Consider the function f : x 7! log2 (x ¡ 1) + 1: a Find the domain and range of f b Find any asymptotes and axis intercepts c Sketch the graph of f showing all important features d Find f ¡1 and explain how to verify your answer a x ¡ > when x > b As x ! from the right, y ! ¡1 ) x = is the vertical asymptote As x ! 1, y ! 1: When x = 0, y is undefined ) there is no y-intercept ) x ¡ = 2¡1 ) x = 12 When y = 0, log2 (x ¡ 1) = ¡1 So, the domain is x ] 1, [ and the range is y2R So, the x-intercept is 12 c f is defined by y = log2 (x ¡ 1) + ) f is defined by x = log2 (y ¡ 1) + ) x ¡ = log2 (y ¡ 1) yellow -2 10 x x=1 is V.A 95 50 -2 75 25 95 100 50 75 25 95 50 75 25 95 100 50 75 25 100 magenta (5,¡3) ¡1 cyan y = log (x - 1) + 100 d y To graph using your calculator we will need to change the base log(x ¡ 1) So, we graph y = +1 log black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\116IB_HL-2_04.CDR Wednesday, 24 October 2007 10:18:33 AM PETERDELL IB_HL-2ed (117) 117 LOGARITHMS (Chapter 4) ) y ¡ = 2x¡1 ) y = 2x¡1 + ) f ¡1 (x) = 2x¡1 + which has a H.A of y = X Its domain is x R , range is y ] 1, [ Graphics calculator tip: When graphing f, f ¡1 and y = x on the same axes, it is best to set the scale so that y = x makes a 45o angle with both axes Why? To ensure the graphs are not distorted, use a square window Recall that: ² ² inverse functions are formed by interchanging x and y y = f ¡1 (x) is the reflection of y = f (x) in the line y = x Example 25 Given f : x 7! ex¡3 a find the defining equation of f ¡1 b sketch the graphs of f and f ¡1 on the same set of axes c state the domain and range of f and f ¡1 d find any asymptotes and intercepts f (x) = ex¡3 ) f is x = ey¡3 ) y ¡ = ln x ) y = + ln x a b y ¡1 c domain range d y = + ln x (1, 3) f ¡1 x>0 y2R f x2R y>0 y = e x -3 (3, 1) x y=x For f: HA is y = 0, for f ¡1 : VA is x = ¡3 For f: y-int is (0, e ), for f ¡1 : x-int is (e¡3 , 0) EXERCISE 4G.1 For the following functions f : i Find the domain and range ii Find any asymptotes and axes intercepts iii Sketch the graph of y = f (x) showing all important features iv Solve f(x) = ¡1 algebraically and check the solution on your graph v Find f ¡1 and explain how to verify your answer magenta yellow 95 f : x 7! ¡ log3 (x + 1) f : x 7! ¡ log5 (x ¡ 2) f :x! log2 (x2 ¡ 3x ¡ 4) 100 50 75 25 b d f 95 100 50 75 25 95 100 50 25 95 100 50 75 25 cyan 75 f : x 7! log3 (x + 1) f : x 7! log5 (x ¡ 2) ¡ f :x! ¡ log2 x2 a c e black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\117IB_HL-2_04.CDR Wednesday, 24 October 2007 10:24:41 AM PETERDELL IB_HL-2ed (118) 118 LOGARITHMS (Chapter 4) For the following functions: i find the defining equation of f ¡1 ii sketch the graphs of f and f ¡1 on the same set of axes iii state the domain and range of f and f ¡1 iv Find any asymptotes a c f : x 7! ex + f : x 7! ln x ¡ where x > Given f : x 7! e2x a (f ¡1 f : x 7! ex+1 ¡ f : x 7! ln(x ¡ 1) + where x > b d and g : x 7! 2x ¡ 1, find the defining equations of: ± g)(x) (g ± f )¡1 (x) b Consider the graphs A and B One of them is the graph of y = ln x and the other is the graph of y = ln(x ¡ 2): a Identify which is which Give evidence for your answer b Redraw the graphs on a new set of axes and add to them the graph of y = ln(x + 2): y A B x c Find the equation of the vertical asymptote for each graph Kelly said that in order to graph y = ln(x2 ), you could first graph y = ln x and then double the distances away from the x-axis Connecting these points will give the graph of y = ln x2 y Is she correct? Give evidence y = ln&x2* y = ln x x For the function f : x 7! ex+3 + a Find the defining equation for f ¡1 b Find the values of x for which: i f (x) < 2:1 ii f (x) < 2:01 iii f(x) < 2:001 iv f(x) < 2:0001 and hence conjecture the horizontal asymptote for the graph of f c Determine the horizontal asymptote of f (x) by discussing the behaviour of f (x) as x ! §1: d Hence, determine the vertical asymptote and the domain of f ¡1 FURTHER INEQUALITIES We have seen on many occasions what it means to solve an equation Usually the equations have been presented to us in one of the forms f (x) = or f (x) = g(x): cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Simply, it means we must find all possible values of the pronumeral, x in this case, that make the equation true black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\118IB_HL-2_04.CDR Wednesday, 24 October 2007 10:26:20 AM PETERDELL IB_HL-2ed (119) 119 LOGARITHMS (Chapter 4) For equations of the form f (x) = 0, we can graph f (x) and then find where the graph meets the x-axis For equations of the form f(x) = g(x), we can either graph f (x) and g(x) separately and find the x-coordinate(s) of their point(s) of intersection, or, we can graph y = f (x) ¡ g(x) and find where this graph meets the x-axis We can use these same principles in order to solve inequalities Note that when solving inequalities, only real number solutions are possible Example 26 a ex = 2x2 + x + Solve for x: b ex > 2x2 + x + We graph f(x) = ex and g(x) = 2x2 + x + 1: a y¡=¡2xX¡+¡x¡+¡1 Using technology we find the points of intersection of f (x) and g(x): ) x = and x ¼ 3:21 are solutions » 3.21 We need to be sure that the graphs will not meet again We could graph f (x) ¡ g(x) and find where the graph meets the x-axis Note: y¡=¡ex y y=ƒ(x)-g(x) y x¡=¡0 x x¡»¡3.21 x Using the same graphs as above, we seek values of x for which f(x) ¡ g(x) > This is where the graph of f (x) either meets or is higher than the graph of g(x) The solution is x = or x > 3:21 x We could even graph the solution set: b »3.21 and could describe it as x = or x [ 3:21, [: EXERCISE 4G.2 Solve for x: a x2 > ex b x3 < e¡x c ¡ x > ln x State the domain of f : x 7! x2 ln x Hence find where f (x) 0: a Use technology to sketch the graph of f : x 7! 2 ¡ e2x ¡x+1 : x b State the domain and range of this function cyan magenta yellow ¡x+1 95 > x 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 c Hence find all x R for which e2x black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_04\119IB_HL-2_04.CDR Friday, 12 December 2008 11:51:11 AM TROY IB_HL-2ed (120) 120 LOGARITHMS (Chapter 4) H GROWTH AND DECAY In Chapter we showed how exponential functions can be used to model a variety of growth and decay situations These included the growth of populations and the decay of radioactive substances In this section we consider more growth and decay problems, focussing particularly on how logarithms can be used in their solution POPULATION GROWTH Example 27 A farmer monitoring an insect plague notices that the area affected by the insects is given by An = 1000 £ 20:7n hectares, where n is the number of weeks after the initial observation a Draw an accurate graph of An against n and use your graph to estimate the time taken for the affected area to reach 5000 b Check your answer to a using logarithms and using suitable technology a b An When An = 5000, 1000 £ 20:7n = 5000 6000 ) 20:7n = 5000 ) log 20:7n = log 4000 ) 0:7n log = log log ) n= 2000 0:7 £ log ¼3.3 ) n ¼ 3:32 ) it takes about weeks and n (weeks) more days Using technology we find the intersection of y = 1000 £ 20:7x and y = 5000 This confirms n ¼ 3:32 EXERCISE 4H.1 The weight Wt of bacteria in a culture t hours after establishment is given by Wt = 20 £ 20:15t grams Find the time for the weight of the culture to reach: a 30 grams b 100 grams The mass Mt of bacteria in a culture t hours after establishment is given by Mt = 25 £ e0:1t grams Find the time for the mass of the culture to reach: a 50 grams b 100 grams cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A biologist monitoring a fire ant infestation notices that the area affected by the ants is given by An = 2000 £ e0:57n hectares, where n is the number of weeks after the initial observation a Draw an accurate graph of An against n and use your graph to estimate the time taken for the infested area to reach 10 000 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\120IB_HL-2_04.CDR Wednesday, November 2007 2:41:30 PM PETERDELL IB_HL-2ed (121) 121 LOGARITHMS (Chapter 4) b Find the answer to a using logarithms c Check your answer to b using suitable technology FINANCIAL GROWTH Suppose an amount u1 is invested at a rate of r% each compounding period In this case the value of the investment after n periods is given by un+1 = u1 £ rn In order to find n, the period of the investment, we need to use logarithms Example 28 Iryna has E5000 to invest in an account that pays 5:2% p.a interest compounded annually How long will it take for her investment to reach E20 000? Now un+1 = u1 £ rn ) 20 000 = 5000 £ (1:052)n ) (1:052)n = ) log(1:052)n = log ) n £ log 1:052 = log log ) n= ¼ 27:3 years log 1:052 un+1 = 20 000 after n years u1 = 5000 r = 105:2% = 1:052 ) it will take at least 28 years EXERCISE 4H.2 A house is expected to increase in value at an average rate of 7:5% p.a If the house is worth $160 000 now, how long is it expected to take for the value to reach $250 000? Thabo has $10 000 to invest in an account that pays 4:8% p.a compounded annually How long will it take for his investment to grow to $15 000? Dien invests $15 000 at 8:4% p.a compounded monthly He will withdraw his money when it reaches $25 000, at which time he plans to travel The formula un+1 = u1 £ rn can be used to calculate the time needed where n is the time in months a Explain why r = 1:007 b After how many months can he withdraw the money? Revisit the Opening Problem on page 102 and answer the questions posed DECAY EXERCISE 4H.3 The mass Mt of radioactive substance remaining after t years is given by Mt = 1000 £ e¡0:04t grams Find the time taken for the mass to: a halve b reach 25 grams c reach 1% of its original value cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A man jumps from an aeroplane and his speed of descent is given by V = 50(1¡e¡0:2t ) m s¡1 where t is the time in seconds Find the time taken for his speed to reach 40 m s¡1 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\121IB_HL-2_04.CDR Wednesday, 24 October 2007 10:50:43 AM PETERDELL IB_HL-2ed (122) 122 LOGARITHMS (Chapter 4) The temperature T of a liquid which has been placed in a refrigerator is given by T = + 96 £ e¡0:03t o C, where t is the time in minutes Find the time required for the temperature to reach: a 25o C b 5o C The weight Wt of radioactive substance remaining after t years is given by Wt = 1000 £ 2¡0:04t grams Find the time taken for the weight to: a halve b reach 20 grams c reach 1% of its original value The weight W (t) of radioactive uranium remaining after t years is given by the formula W (t) = W0 £ 2¡0:0002t grams, t > Find the time taken for the original weight to fall to: a 25% of its original value b 0:1% of its original value The current I flowing in a transistor radio t seconds after it is switched off is given by I = I0 £ 2¡0:02t amps Find the time taken for the current to drop to 10% of its original value A parachutist jumps from the basket of a stationary hot air balloon His speed of descent is given by V = 50(1 ¡ 2¡0:2t ) m s¡1 where t is the time in seconds Find the time taken for his speed to reach 40 m s¡1 REVIEW SET 4A Find the following without using a calculator Show all working a log4 64 b log2 256 c log2 (0:25) d log25 e f log6 g log81 h log9 (0:1) i log27 j p Without using a calculator, find: a log 10 Find x if: log2 x = ¡3 a Write as logarithmic equations: b log p 10 b log5 x ¼ 2:743 a P = £ bx c b log2 k ¼ 1:699 + x log A ¼ log B ¡ 2:602 b log(10a £ 10b+1 ) log3 x ¼ ¡3:145 m= Write the following equations without logarithms: a c c n3 p2 loga Q = loga P + loga R Solve for x, giving your answer correct to significant figures: b 20 £ 22x+1 = 500 a 5x = 7 The a b c d log8 p logk k ¡ t weight of radioactive substance after t years is Wt = 2500 £ 3000 grams Find the initial weight Find the time taken for the weight to reduce to 30% of its original value Find the percentage weight loss after 1500 years Sketch the graph of Wt against t Solve for x: 16x ¡ £ 8x = Solve the equation log3 (10x2 ¡ x ¡ 2) = + log3 x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 10 Find the exact value of a which satisfies the equation 53a £ 42a+1 = 103a+2 ln x Give your answer in the form where x, y Z : ln y black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_04\122IB_HL-2_04.CDR Thursday, 11 March 2010 10:17:02 AM PETER IB_HL-2ed (123) 123 LOGARITHMS (Chapter 4) REVIEW SET 4B Without using a calculator, find the base 10 logarithms of: p 1000 a 10 p 10 b 10a 10¡b c Solve for x: a log x = b log3 (x + 2) = 1:732 c log2 ( Write as a single logarithm: a log 16 + log b log2 16 ¡ log2 c + log4 x ) = ¡0:671 10 Write the following equations without logarithms: a log T = log x ¡ log y b log2 K = log2 n + 3x = 300 a Solve for x: b 30 £ 51¡x = 0:15 c log2 t 3x+2 = 21¡x If A = log2 and B = log2 3, write the following in terms of A and B: p b log2 54 c log2 (8 3) d log2 (20:25) e log2 (0:¹8) a log2 36 For a b c d e the function g : x 7! log3 (x + 2) ¡ : Find the domain and range Find any asymptotes and axes intercepts for the graph of the function Sketch the graph of y = g(x): Find g ¡1 Explain how to verify your answer for g¡1 Sketch the graphs of g, g ¡1 and y = x on the same axes Solve exactly for a in the equation log4 a5 + log2 a = log8 625: A straight line has equation y = mx + c Its gradient is ¡2 and it passes through ¢ ¡ the point 1, log5 25 a Find the equation of the line b If y = log5 M , find an expression for M in terms of x c Hence, find the value of x when M = 25 y 4x £ 2y = 16 and 8x = 2 10 Solve simultaneously for x and y: p 11 Solve log8 x2 + = 13 REVIEW SET 4C a On the same set of axes sketch and clearly label graphs of: g : x 7! e¡x and h : x 7! ¡e¡x f : x 7! ex , b What is the geometric connection between: i f and g Sketch on the same set of axes the graphs of y = ex ii g and h? and y = 3ex A particle moves in a straight line such that its displacement from the origin O is ¡t given by s(t) = 120t ¡ 40e metres, where t is the time in seconds, t > a Find the position of the particle at i t = ii ¡t cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b Hence sketch the graph of s(t) = 120t ¡ 40e black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\123IB_HL-2_04.CDR Wednesday, November 2007 2:50:38 PM PETERDELL t = iii t = 20: for t > IB_HL-2ed (124) 124 LOGARITHMS (Chapter 4) Without using a calculator, find: Simplify: ln(e2x ) a ln(e5 ) a b ln(e2 ex ) b Write as a single logarithm: a ln + ln b ln 60 ¡ ln 20 c c ln p ln( e) ³e´ ln µ ¶ e ex ln 200 ¡ ln + ln d ln + ln c Write in the form a ln k where a and k are positive whole numbers and k is prime: a ln 32 b ln 125 c ln 729 Solve for x, giving answers correct to significant figures: ex = 400 a Solve 12(2x ) = + x e2x+1 = 11 b 25e = 750 c e2x = 7ex ¡ 12 d 10 giving your answer in the form m + log2 n, m, n Z 2x REVIEW SET 4D On the same set of axes, sketch and clearly label the graphs of: g : x 7! e¡x , h : x 7! e¡x ¡ f : x 7! ex , State the domain and range of each function Sketch on the same set of axes, the graphs of y = ex and y = e3x Without using a calculator, find: µ ¶ p a ln(e e) b ln e Write in the form ex : Simplify: a ln + ln a c b 20 b µ ln e p e5 ¶ 3000 c c ln ¡ ln ¡ ln 0:075 d ln 81 Write the following equations without logarithms: a ln P = 1:5 ln Q + ln T b ln M = 1:2 ¡ 0:5 ln N Consider g : x 7! 2ex ¡ a Find the defining equation of g ¡1 b Sketch the graphs of g and g¡1 on the same set of axes c State the domain and range of g and g ¡1 The weight Wt grams of radioactive substance remaining after t weeks is given by ¡ t 20 Wt = 8000 £ e a halve grams Find the time for the weight to: b reach 1000 g c reach 0:1% of its original value cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The function f is defined for x > by f (x) = ln(x2 ¡ 16) ¡ ln x ¡ ln(x ¡ 4) µ ¶ x+a a Express f(x) in the form ln , stating the value of a Z : x b Find an expression for f ¡1 (x) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_04\124IB_HL-2_04.CDR Wednesday, 24 October 2007 10:58:22 AM PETERDELL IB_HL-2ed (125) Chapter Graphing and transforming functions Contents: A B C D Families of functions Transformations of graphs Simple rational functions Further graphical transformations cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 5A Review set 5B black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_05\125IB_HL-2_05.CDR Wednesday, 24 October 2007 11:40:00 AM PETERDELL IB_HL-2ed (126) 126 GRAPHING AND TRANSFORMING FUNCTIONS (Chapter 5) A FAMILIES OF FUNCTIONS There are several families of functions that you are already familiar with These include: Name General form Function notation Linear f (x) = ax + b, a 6= f : x 7! ax + b, a 6= Quadratic f (x) = ax2 + bx + c, a 6= f : x 7! ax2 + bx + c, a 6= Cubic f (x) = ax3 + bx2 + cx + d, a 6= f : x 7! ax3 + bx2 + cx + d, a 6= Absolute value f (x) = jxj f : x 7! jxj Exponential f (x) = a , a > 0, a 6= f : x 7! ax , a > 0, a 6= Logarithmic f (x) = loge x or f(x) = ln x f : x 7! ln x Reciprocal f (x) = x k , x 6= x f : x 7! k , x 6= x These families of functions have different and distinctive graphs We can compare them by considering important graphical features such as: ² the axes intercepts (where the graph cuts the x and y-axes) ² slopes ² turning points (maxima and minima) ² values of x where the function does not exist ² the presence of asymptotes (lines or curves that the graph approaches) INVESTIGATION FUNCTION FAMILIES In this investigation you are encouraged to use the graphing package supplied.¡ Click on the icon to access this package GRAPHING PACKAGE What to do: From the menu, graph on the same set of axes: y = 2x + 1, y = 2x + 3, y = 2x ¡ Comment on all lines of the form y = 2x + b From the menu, graph on the same set of axes: y = x + 2, y = 2x + 2, y = 4x + 2, y = ¡x + 2, y = ¡ 12 x + Comment on all lines of the form y = ax + On the same set of axes graph: y = x2 , y = 2x2 , y = 12 x2 , y = ¡x2 , y = ¡3x2 , y = ¡ 15 x2 Comment on all functions of the form y = ax2 , a 6= cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 On the same set of axes graph: y = x2 , y = (x ¡ 1)2 + 2, y = (x + 1)2 ¡ 3, y = (x ¡ 2)2 ¡ and other functions of the form y = (x ¡ h)2 + k of your choice Comment on the functions of this form black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_05\126IB_HL-2_05.CDR Wednesday, 24 October 2007 11:46:55 AM PETERDELL IB_HL-2ed (127) 127 GRAPHING AND TRANSFORMING FUNCTIONS (Chapter 5) On a b c the same set of axes, graph these absolute value functions: y = j x j, y = j x j, y = j 2x j y = j x j, y = j x j + 2, y = j x j ¡ y = j x j, y = j x ¡ j, y = j x + j, y = j x ¡ j + Write a brief report on your discoveries On the same set of axes, graph these functions: 10 ¡1 ¡2 ¡5 b y= , y= , y= a y= , y= , y= x x x x x x 1 1 1 c y= , y= , y= d y = , y = + 2, y = ¡ x x¡2 x+3 x x x e y= 2 , y= + 2, y = ¡1 x x¡1 x+2 Write a brief report on your discoveries Example If f (x) = x2 , find in simplest form: ³x´ a f(2x) b f c f (x) + ³x´ b f c f (x) + a f(2x) = 2x2 + ³ x ´2 = (2x) = = 4x2 = d f(x + 3) ¡ d f(x + 3) ¡ = (x + 3)2 ¡ = x2 + 6x + ¡ = x2 + 6x + x2 EXERCISE 5A If f(x) = x, find in simplest form: a b f(2x) f (x) + c f (x) d f(x) + c f (x + 1) d f(x + 1) ¡ If f(x) = x3 , find in simplest form: a b f(4x) f(2x) Note: (x + 1)3 = x3 + 3x2 + 3x + See the binomial theorem, Chapter If f(x) = 2x , find in simplest form: a f (¡x) + c f (x ¡ 2) + d f(x) + , find in simplest form: x f(¡x) b f ( 12 x) c f(x) + d f(x ¡ 1) + b f(2x) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 25 95 100 50 75 25 a 75 If f(x) = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_05\127IB_HL-2_05.CDR Wednesday, 24 October 2007 11:48:13 AM PETERDELL IB_HL-2ed (128) 128 GRAPHING AND TRANSFORMING FUNCTIONS (Chapter 5) For the following questions, use the graphing package or your graphics calculator to graph and find the key features of the functions GRAPHING PACKAGE Consider f : x 7! 2x + or y = 2x + a Graph the function b Find algebraically, the: i x-axis intercept ii y-axis intercept c Use technology to check the axes intercepts found in b iii slope Consider f : x 7! (x ¡ 2)2 ¡ a Graph the function b Find algebraically the x and y axes intercepts c Use technology to check that: i the x-axis intercepts are ¡1 and ii iii the vertex is (2, ¡9) the y-intercept is ¡5 Consider f : x 7! 2x3 ¡ 9x2 + 12x ¡ a Graph the function ii the y-intercept is ¡5 b Check that: i the x-intercepts are and 12 iii the minimum turning point is at (2, ¡1) iv the maximum turning point is at (1, 0) Sketch the graph of y = j x j Note: j x j = x if x > and j x j = ¡x if x < Consider f : x 7! 2x Graph the function and check these key features: a as x ! 1, 2x ! b as x ! ¡1, 2x ! (from above) c the y-intercept is d 2x is > for all x ! reads ‘approaches’ or ‘tends to’ 10 Consider f : x 7! loge x Graph the function and then check that: a as x ! 1, ln x ! b as x ! (from the right), ln x ! ¡1 c ln x only exists if x > d the x-intercept is e the y-axis is an asymptote B TRANSFORMATIONS OF GRAPHS cyan magenta yellow 95 100 50 75 25 95 100 50 y = f(¡x) 75 ² 25 y = ¡f (x) ² y = f(kx), k is a positive constant 95 ² 100 y = p f (x), p is a positive constant 50 ² 75 y = f(x ¡ a), a is a constant 25 ² y = f (x) + b, b is a constant 95 ² 100 50 75 25 In the next exercise you should discover the graphical connection between y = f(x) and functions of the form: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_05\128IB_HL-2_05.CDR Wednesday, 24 October 2007 11:53:44 AM PETERDELL IB_HL-2ed (129) 129 GRAPHING AND TRANSFORMING FUNCTIONS (Chapter 5) TYPES y = f(x) + b AND y = f(x ¡ a) EXERCISE 5B.1 a Sketch the graph of f(x) = x2 b On the same set of axes sketch the graphs of: ii y = f (x) ¡ 3, i.e., y = x2 ¡ i y = f(x) + 2, i.e., y = x2 + c What is the connection between the graphs of y = f(x) and y = f (x) + b if: i b>0 ii b < 0? For each of the following functions f, sketch on the same set of axes y = f (x), y = f (x) + and y = f(x) ¡ a f(x) = j x j b f (x) = 2x c f (x) = x3 d f (x) = x Summarise your observations by describing the graphical transformation of y = f(x) as it becomes y = f (x) + b a On the same set of axes, graph: f(x) = x2 , y = f(x ¡ 3) and y = f(x + 2) b What is the connection between the graphs of y = f(x) and y = f (x ¡ a) if: i a>0 ii a < 0? For each of the following functions f, sketch on the same set of axes the graphs of y = f (x), y = f(x ¡ 1) and y = f (x + 2) a f(x) = j x j b f (x) = x3 c f (x) = ln x d f (x) = x Summarise your observations by describing the geometrical transformation of y = f(x) as it becomes y = f (x ¡ a) For each of the following functions sketch: y = f (x), y = f(x ¡ 2) + and y = f (x + 1) ¡ on the same set of axes b f(x) = ex c f(x) = a f(x) = x2 x Copy these functions and then draw the graph of y = f(x ¡ 2) ¡ 3: a b y y x x Given f(x) = x2 is transformed to g(x) = (x ¡ 3)2 + : a find the images of the following points on f(x) : i (0, 0) ii where x = ¡3 iii where x = b find the points on f (x) which correspond to the following points on g(x) : cyan yellow iii 95 100 50 75 (¡2, 27) 25 95 100 50 75 25 95 magenta ii (1, 6) 100 50 75 25 95 100 50 75 25 i black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_05\129IB_HL-2_05.CDR Wednesday, 24 October 2007 11:55:56 AM PETERDELL (1 12 , 14 ) IB_HL-2ed (130) 130 GRAPHING AND TRANSFORMING FUNCTIONS (Chapter 5) TYPES y = p f(x) , p > AND y = f(kx) , k > EXERCISE 5B.2 Sketch on the same set of axes, the graphs of y = f (x), y = f (x) and y = f (x) for each of: a f(x) = x2 b f(x) = j x j c d f(x) = ex e f(x) = ln x f f(x) = x3 f(x) = x Sketch on the same set of axes, the graphs of y = f (x), y = 12 f(x) and y = 14 f (x) for each of: a f(x) = x2 f(x) = x3 b c f(x) = ex Using and 2, summarise your observations by describing the graphical transformation of y = f (x) to y = p f(x) for p > Sketch on the same set of axes, the graphs of y = f (x) and y = f(2x) for each of: b y = (x ¡ 1)2 c y = (x + 3)2 a y = x2 Sketch on the same set of axes, the graphs of y = f (x) and y = f(3x) for each of: c y = ex a y=x b y = x2 ¡ ¢ Sketch on the same set of axes, the graphs of y = f (x) and y = f x2 for each of: a y = x2 b c y = 2x y = (x + 2)2 Using 4, and 6, summarise your observations by describing the graphical transformation of y = f (x) to y = f (kx) for k > Consider the function f : x 7! x2 On the same set of axes sketch the graphs of: a y = f(x), y = f(x ¡ 2) + and y = f(x + 1) ¡ ¡ ¢ ¡ ¢ ¡ ¢ b y = f(x), y = f(x ¡ 3), y = f x2 ¡ , y = f x2 ¡ and y = f x2 ¡ + c y = f(x) and y = f(2x + 5) + a Given that the following points lie on y = f (x), find the corresponding points on the image function y = 3f (2x) : i (3, ¡5) ii (1, 2) iii (¡2, 1) b Find the points on y = f (x) which are moved to the following points under the transformation y = 3f (2x) : i (2, 1) ii (¡3, 2) iii (¡7, 3) 10 The function y = f(x) is transformed to the function g(x) = + 2f ( 12 x + 1): cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a Fully describe the transformation that maps f(x) onto g(x) b Using a, find the image points of the following points on f(x): i (1, ¡3) ii (2, 1) iii (¡1, ¡2) c Find the points on f (x) which correspond to the following points on g(x): i (¡2, ¡5) ii (1, ¡1) iii (5, 0) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_05\130IB_HL-2_05.CDR Wednesday, 24 October 2007 11:59:23 AM PETERDELL IB_HL-2ed (131) 131 GRAPHING AND TRANSFORMING FUNCTIONS (Chapter 5) TYPES y = ¡f(x) AND y = f(¡x) EXERCISE 5B.3 On the same set of axes, sketch the graphs a y = 3x and y = ¡3x c y = x2 and y = ¡x2 e y = x3 ¡ and y = ¡x3 + of: b y = ex and y = ¡ex d y = ln x and y = ¡ ln x f y = 2(x + 1)2 and y = ¡2(x + 1)2 Based on question 1, what transformation moves y = f (x) to y = ¡f (x)? a Find f (¡x) for: i ii f (x) = 2x + f(x) = x2 + 2x + iii f (x) = j x ¡ j iii f (x) = j x ¡ j b Graph y = f (x) and y = f (¡x) for: i ii f (x) = 2x + f(x) = x2 + 2x + Based on question 3, what transformation moves y = f (x) to y = f (¡x)? The function y = f (x) is transformed to g(x) = ¡f (x): a Find the points on g(x) corresponding to the following points on f (x): i (3, 0) ii (2, ¡1) iii (¡3, 2) have been transformed to the following points on g(x): b Find the points on f (x) that i (7, ¡1) ii (¡5, 0) iii (¡3, ¡2) The function y = f (x) is transformed to h(x) = f(¡x): a Find the image points on i (2, ¡1) ii b Find the points on f (x) i (5, ¡4) ii h(x) for the following points on f(x): (0, 3) iii (¡1, 2) corresponding to the following points on h(x): (0, 3) iii (2, 3) A function y = f(x) is transformed to the function y = ¡f (¡x) = g(x): a Describe the nature of the transformation b If (3, ¡7) lies on y = f(x), find the transformed point on g(x) c Find the point on f (x) that transforms to the point (¡5, ¡1) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 For y = f (x ¡ a) + b, the graph is translated horizontally a units and £ ¤ vertically b units We say it is translated by the vector ab : continued next page 95 I 100 For y = f (x ¡ a), the effect of a is to translate the graph horizontally through a units ² If a > it moves to the right ² If a < it moves to the left 50 I 75 For y = f (x) + b, the effect of b is to translate the graph vertically through b units ² If b > it moves upwards ² If b < it moves downwards 25 I 95 100 50 75 25 Summary of graphical transformations on y = f (x) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_05\131IB_HL-2_05.CDR Wednesday, 24 October 2007 12:04:48 PM PETERDELL IB_HL-2ed (132) 132 GRAPHING AND TRANSFORMING FUNCTIONS (Chapter 5) continued from previous page I For y = p f (x), p > 0, the effect of p is to vertically stretch the graph by a factor of p ² If p > it moves points of y = f (x) further away from the x-axis ² If < p < it moves points of y = f (x) closer to the x-axis I For y = f (kx), k > 0, the effect of k is to horizontally compress the graph by a factor of k ² If k > it moves points of y = f(x) closer to the y-axis ² If < k < it moves points of y = f (x) further away from the y-axis I I For y = ¡f (x), the effect is to reflect y = f (x) in the x-axis For y = f(¡x), the effect is to reflect y = f (x) in the y-axis Note: Stretching by a factor of p1 is equivalent to compressing by a factor of p For example, y = 13 f (x) indicates that y = f(x) is vertically stretched by a factor of 13 , or compressed by a factor of Likewise, y = f ( 14 x) indicates that y = f (x) is horizontally compressed by a factor of 14 , or stretched by a factor of EXERCISE 5B.4 Copy the following graphs for y = f(x) and sketch the graphs of y = ¡f(x) on the same axes a b c y y y x x x Given the following graphs of y = f (x), sketch graphs of y = f (¡x) : a b c y y y y=1 x x x x=2 The scales on the graphs below are the same Match each equation to its graph y = x4 y = 2x4 B b y cyan magenta 95 100 50 75 x 25 95 50 75 25 100 yellow D d y x 95 100 50 75 25 95 100 50 75 25 x y = 12 x4 C c y A a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_05\132IB_HL-2_05.CDR Wednesday, 24 October 2007 12:06:44 PM PETERDELL y = 6x4 y x IB_HL-2ed (133) GRAPHING AND TRANSFORMING FUNCTIONS (Chapter 5) For the graph of y = f(x) given, draw sketches of: a y = 2f (x) b y = 12 f (x) c y = f (x + 2) d y = f (2x) e y = f ( x) y y c x For the graph of y = g(x) given, draw sketches of: a y = g(x) + b y = ¡g(x) g(x) -2 133 d y = g(¡x) y = g(x + 1) x y For the graph of y = h(x) given, draw sketches of: a y = h(x) + b y = 12 h(x) ¡ ¢ c y = h(¡x) d y = h x2 y=h(x) -1 -2 C x (2,-2) SIMPLE RATIONAL FUNCTIONS ax + b d , x 6= ¡ cx + d c is called a simple rational function Any function x 7! where a, b, c and d are constants These functions are characterised by the presence of both a horizontal asymptote (HA) and a vertical asymptote (VA) Any graph of a simple rational function can be obtained from the reciprocal function x 7! x by a combination of transformations including: ² magenta x yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 x stretches and compressions (vertical and/or horizontal) 95 100 50 75 25 y= a translation (vertical and/or horizontal) ² cyan y black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_05\133IB_HL-2_05.CDR Wednesday, 24 October 2007 12:07:45 PM PETERDELL IB_HL-2ed (134) 134 GRAPHING AND TRANSFORMING FUNCTIONS (Chapter 5) Example a Find the function y = g(x) that results when transforming the reciprocal function, x 7! by: a vertical stretch with factor then a horizontal x h i : compression with factor 3, then a translation of ¡2 b c Find the asymptotes of each function found in a Is the function found in a a self inverse function? Explain a Under a vertical stretch with factor 2, Under a horizontal compression with factor 3, h i , Under a translation of ¡2 ) f (x) = x µ ¶ becomes x x becomes 3x (3x) ¡2 3(x ¡ 3) becomes f2f(x)g ff (3x)g ff (x ¡ 3) ¡ 2g 2(3x ¡ 9) ¡6x + 20 ¡2= ¡ = 3x ¡ 3x ¡ 3x ¡ 3x ¡ are: VA x = 0, HA y = x h i g ) for the new function VA is x = 3, HA is y = ¡2 fas translated ¡2 b The asymptotes of y = c y From a graphics calculator the graph is found as shown y=x It is not symmetrical about y = x Hence, it is not a self inverse function x y=-2 x=3 Note: f(x) = k x is a vertical stretch of x 7! x with factor k EXERCISE 5C cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 50 75 25 100 ax + b , the function that results when x 7! a Find, in the form y = cx + d x is transformed by: ii a horizontal stretch of factor i a vertical stretch of factor 12 iii a horizontal translation of ¡3 iv a vertical translation of v all of i, ii, iii and iv ax + b b Find the domain and range of y = as found in a v cx + d black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_05\134IB_HL-2_05.CDR Wednesday, 24 October 2007 12:09:26 PM PETERDELL IB_HL-2ed (135) 135 GRAPHING AND TRANSFORMING FUNCTIONS (Chapter 5) Example 2x ¡ , find: x+1 b how to transform the function to give x 7! x For the function f : x 7! a the asymptotes a 2x ¡ 2(x + 1) ¡ 8 ¡8 = =2¡ = +2 x+1 x+1 x+1 x+1 £ ¤ ¡8 from f (x) = This represents a translation of ¡1 which has x VA x = and HA y = f (x) = So, f (x) = Note: 2x ¡ x+1 has VA x = ¡1 and HA y = ² 2x ¡ is undefined when x = ¡1 x+1 ² as jxj ! 1, f (x) ! ² the domain of f (x) = the range of f(x) = b To get f(x) = 2x ¡ x+1 2x ¡ x+1 2x ¡ x+1 is fy: y 6= 2g from f (x) = x we µ ¶ becomes = g f x x x 8 f becomes ¡ g x x ¡8 ¡8 f becomes + 2g x x+1 vertically stretch by a factor of then reflect in the x-axis £ ¡1 ¤ then translate by is fx: x 6= ¡1g So, to the opposite we h i , then reflect in the x-axis, then vertically stretch by factor 18 translate by ¡2 2x ¡ 2(x ¡ 1) ¡ 2x ¡ 8 becomes y = ¡2= ¡2 =¡ x+1 (x ¡ 1) + x x 8 8 then ¡ becomes and becomes = x x x (8x) x Check: y = cyan magenta yellow 3x ¡ x+1 95 50 75 25 95 100 50 b f : x 7! 75 25 2x + x¡1 95 100 50 75 25 95 100 50 75 25 a f : x 7! x 2x + f : x 7! 2¡x how to transform f(x) to give the function x 7! 100 For these functions find: i the asymptotes ii black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_05\135IB_HL-2_05.CDR Wednesday, 24 October 2007 12:10:34 PM PETERDELL c IB_HL-2ed (136) 136 GRAPHING AND TRANSFORMING FUNCTIONS (Chapter 5) Example a Find the asymptotes of y = f (x): 4x + b Find the axes intercepts x¡2 i VA ii HA Discuss the behaviour of f near its Sketch the graph of the function 4x + to x 7! Describe the transformations which move x 7! x x¡2 Consider f(x) = c d e a y= 4(x ¡ 2) + 11 11 4x + = =4+ x¡2 x¡2 x¡2 fwhere y is undefinedg fas jxj ! 1, y ! 4g So, the function has VA x = and has HA y = b When x = 0, y = ¡2 ) y-intercept is ¡1 12 when y = 0, 4x + = x = ¡ 34 c ) x-intercept is ¡ 34 i as x ! (from the left), y ! ¡1 ii as x ! ¡1, y ! (from below) as x ! (from the right), y ! as x ! 1, y ! (from above) y d e y=4 x -\Er_ 11 becomes under a vertical x x stretch with factor 11, and then 11 4x + becomes under a x x¡2 £ ¤ translation of 24 x=2 -1\Qw_ For the following functions: i find the asymptotes ii find the axes intercepts iii discuss the graph’s behaviour near its VA and its HA iv sketch the graph to the given function v Describe the transformations which move x 7! x 2x + 3 2x ¡ 5x ¡ a y= b y= c y= d y= x+1 x¡2 3¡x 2x + cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 In order to remove noxious weeds from her property Helga sprays with a weedicide The chemical is slow acting and the number of weeds per hectare remaining after t days is 100 modelled by N = 20 + weeds/ha t+2 a How many weeds per were alive before the spraying? b How many weeds will be alive after days? c How long will it take for the number of weeds still alive to be 40/ha? d Sketch the graph of N against t e According to the model, is the spraying going to eradicate all weeds? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_05\136IB_HL-2_05.CDR Wednesday, 24 October 2007 12:12:55 PM PETERDELL IB_HL-2ed (137) 137 GRAPHING AND TRANSFORMING FUNCTIONS (Chapter 5) Graphics calculator tips: ² To find the zeros of a function y = f (x), simply graph the function and find its x-intercepts This is equivalent to finding the roots or solutions of the equation f(x) = 0: ² To check that you have found the correct asymptotes (VA and HA): I Try to find y for the x-asymptote value It should be undefined I Try to find y for large x values, e.g., §109 It should give a value close to the y-asymptote value D FURTHER GRAPHICAL TRANSFORMATIONS In this exercise you should discover the graphical connection between y = f(x) and functions of the form y = , y = j f(x) j and y = f (j x j): f (x) THE RECIPROCAL FUNCTION y = f (x) Example Graph on the same set of axes: a y = x ¡ and y = x¡2 a y b y = x2 b y = x-2 and y = x2 y y = x2 y= x x-2 y= -\Qw_ -2 x2 x x=2 EXERCISE 5D.1 Graph on the same set of axes: ¡1 a y = ¡x2 and y = x y = (x ¡ 1)(x ¡ 3) and y = b (x ¡ 1)(x ¡ 3) Invariant points are points which not move under a transformation Show that if y = f(x) is transformed to y = , invariant points occur at y = §1 f(x) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Check your results in question for invariant points black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_05\137IB_HL-2_05.CDR Wednesday, 24 October 2007 12:20:12 PM PETERDELL IB_HL-2ed (138) 138 GRAPHING AND TRANSFORMING FUNCTIONS (Chapter 5) DISCUSSION THE GRAPHICAL CONNECTION BETWEEN y = f (x) AND y = f (x) True or false? Discuss: ² ² ² ² The zeros of f (x) become VA values of and the VA values of f(x) f(x) become the zeros of f (x) and the minimum Maximum values of f(x) become minimum values of f(x) values of f (x) become maximum values of f(x) 1 > also and when f (x) < 0, < also When f (x) > 0, f (x) f(x) 1 ! §1 and when ! 0, f(x) ! §1: When f (x) ! 0, f (x) f(x) THE MODULUS FUNCTIONS y = jf (x)j AND y = f (jxj) Example Draw the graph of f(x) = 3x(x ¡ 2) and on the same axes draw the graphs of: a y = jf(x)j b y = f (jxj) ( a y = jf (x)j = ( f(x) if f (x) > ¡f (x) if f (x) < b y = f (jxj) = This means the graph is unchanged for f(x)¡>¡0, reflected in the x-axis for f(x)¡<¡0: f(x) if x > f(¡x) if x < This means the graph is unchanged if x¡>¡0, reflected in the y-axis if x¡<¡0: y y -2 x -1 x y=ƒ(x) -3 DISCUSSION THE GRAPHICAL CONNECTION BETWEEN y = f (x), y = j f (x) j AND y = f (j x j) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 The graph of y = j f (x) j can be obtained from the graph of y = f (x) by reflecting in the x-axis (Try a y = x2 b y = ex ¡ c y = ln(x + 2) ) 25 ² y = j f(x) j ) y = f (x) for f(x) > and y = ¡f (x) for f(x) < 95 ² 100 50 75 25 True or false? Discuss: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_05\138IB_HL-2_05.CDR Wednesday, 24 October 2007 12:23:35 PM PETERDELL IB_HL-2ed (139) 139 GRAPHING AND TRANSFORMING FUNCTIONS (Chapter 5) ² Points on the y-axis are invariant for f(x) ! j f(x) j ² The graph of y = f (j x j) can be obtained from the graph of y = f (x) by reflecting in the y-axis ² y = f (j x j) ) ² Points on the x-axis are invariant for f (x) ! f (j x j) y = f (x) for x > and y = f (¡x) for x < 0: EXERCISE 5D.2 Draw y = x(x + 2) and on the same set of axes graph: a y = j f (x) j b y = f (j x j) : Copy the following graphs for y = f(x) and on the same axes graph y = f (x) a b c y y y y¡=¡¦(x) y¡=¡¦(x) y¡=¡¦(x) y=1 x x x x=4 (1\Qw_\'-2) Copy the following graphs for y = f(x) and on the same axes graph y = j f (x) j : a b c y y y y¡=¡¦(x) y¡=¡¦(x) y¡=¡¦(x) -2 x y=2 x x -2 x=4 Repeat question 3, but this time graph y = f (j x j) instead of y = j f (x) j Suppose the function f (x) is transformed to j f(x) j : For the following points on f (x) find the image points on j f (x) j : a (3, 0) b (5, ¡2) c (0, 7) d (2, 2) Suppose the function f (x) is transformed to f (j x j) : a For the following points on f (x) find the image points: i (0, 3) ii (1, 3) iii (7, ¡4) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b For the points on f (j x j) find the points on f(x) that have been transformed: i (0, 3) ii (¡1, 3) iii (10, ¡8) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_05\139IB_HL-2_05.CDR Friday, January 2008 9:12:38 AM DAVID3 IB_HL-2ed (140) 140 GRAPHING AND TRANSFORMING FUNCTIONS (Chapter 5) REVIEW SET 5A If f (x) = x2 ¡ 2x, find in simplest form: a f (3) b f(¡2) c f (2x) d f (¡x) If f (x) = ¡ x ¡ x2 , find in simplest form: 3f(x) ¡ e ³x´ e 2f(x) ¡ f(¡x) Consider the function f : x 7! x2 On the same set of axes graph: a y = f(x) b y = f (x + 2) c y = 2f(x + 2) d y = 2f(x + 2) ¡ a b f (4) c f(¡1) f (x ¡ 1) d f The graph of f(x) = 3x3 ¡ 2x2 + x + is translated to its image g(x) by the £1¤ Write the equation of g(x) in the form g(x) = ax3 + bx2 + cx + d vector ¡2 Consider f(x) = (x + 1)2 ¡ a b c d Use your calculator to help graph the function Find algebraically i the x-intercepts ii the y-intercept What are the coordinates of the vertex of the function? Use your calculator to check your answers to b and c Consider f : x 7! 2¡x a Use your calculator to help graph the function b True or false: i as x ! 1, 2¡x ! ii iv iii the y-intercept is 12 as x ! ¡1, 2¡x ! 2¡x > for all x? Sketch the graph of f (x) = ¡x2 , and on the same set of axes sketch the graph of: a y = f(¡x) b y = ¡f(x) c y = f (2x) d y = f (x ¡ 2) The graph of y1 = f (x) is shown alongside The x-axis is a tangent to f (x) at x = a and f(x) cuts the x-axis at x = b y On the same diagram sketch the graph of y2 = f (x ¡ c) where < c < b ¡ a Indicate the coordinates of the points of intersection of y2 with the x-axis a cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The graph of f (x) is drawn alongside a Copy the graph of f (x) and draw ¦(x) the graph of j f (x) j on the same set of axes 1 b Find the y-intercept of f (x) c Show on the diagram the points that -1 (2,-1) are invariant for the function : f (x) on the same set of axes d Draw the graph of y = f (x) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_05\140IB_HL-2_05.CDR Thursday, 25 October 2007 3:17:48 PM PETERDELL x b (4,¡2) x (3,-1) IB_HL-2ed (141) 141 GRAPHING AND TRANSFORMING FUNCTIONS (Chapter 5) 10 a If f (x) = x + 2, find the equation of the function F obtained by stretching the function f vertically by a factor of and compressing the function horizontally by a factor of 2, followed by a translation of 12 horizontally and ¡3 vertically i Check that the point (1, 3) remains invariant under the transformation described in part a b ii What happens to the points (0, 2) and (¡1, 1) under the transformation given in part a? iii Show that the points in ii lie on the graph of y = F (x): y 11 The graph of f(x) = x2 is transformed to the graph of g(x) by a reflection and a translation as illustrated in the diagram alongside Find the equation of y = g(x) in the form ax2 + bx + c y=¦(x) V(-3,¡2) x y=g(x) REVIEW SET 5B , find in simplest form: x ³x´ f (¡4) b f(2x) c f If f (x) = a 4f (x + 2) ¡ d Consider f(x) : x 7! 3x ¡ a Sketch the function f: b Find algebraically the i x-intercept ii y-intercept iii slope c i Find y when x = 0:3 ii Find x when y = 0:7 For what values of x, where a is a positive real number, is j x ¡ a j = j x j ¡ a ? c , x 6= ¡c, c > x+c a On a set of axes like those shown, sketch the graph of f(x) Label clearly any points of intersection with the axes and any asymptotes b On the same set of axes, sketch the graph of f (x) Let f(x) = Label clearly any points of intersection with the axes For the graph of y = f(x), sketch graphs of: a y = f(¡x) b y = ¡f(x) c y = f(x + 2) d y = f(x) + (-3, 3) y -c c y x (5, 8) y¡=¡¦(x) x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 (2,-1) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_05\141IB_HL-2_05.CDR Thursday, 25 October 2007 3:26:33 PM PETERDELL IB_HL-2ed (142) 142 GRAPHING AND TRANSFORMING FUNCTIONS (Chapter 5) The graph of y = f(x) is given On the same set of axes graph each pair of functions: a y = f (x) and y = f (x ¡ 2) + 1 b y = f (x) and y = f (x) y y¡=¡¦(x) (2,-2) c y = f (x) and y = j f (x) j x x¡=¡4 a Find the equation of the function y = f(x) that results from transforming the function x 7! by: a reflection in the y-axis, then a vertical stretch of x £ ¤ factor 3, then a horizontal compression of factor 2, then a translation of 11 : b Sketch y = f (x) and state its domain and range c Does y = f(x) have an inverse function? Explain d Is the function f a self-inverse function? Give graphical and algebraic evidence 2x ¡ 3x + Your discussion should include: the asymptote equations, axes intercepts, and what happens near the asymptotes Discuss and sketch the graph of y = y The graph alongside is that of a rational relation ax + b : Find its equation in the form y = cx + d x=3 x y = -2 10 For each of the following functions: h i i find y = f (x), the result of a translation ¡2 : ii Sketch the original function and its translated function on the same set of axes iii Clearly state any asymptotes of each function iv State the domain and range of each function b y = log4 x y = 2x a magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 cyan a Sketch the graph of f (x) = ¡2x + 3, clearly showing the axis intercepts b Find the invariant points for the graph of y = : f (x) c State the equation of the vertical asymptote and find the y-intercept of the graph of y = f(x) on the same axes as in part a, showing clearly d Sketch the graph of y = f (x) the information found in parts b and c e On a new pair of axes, sketch the graphs of y = j f (x) j and y = f (j x j) showing clearly all important features 11 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_05\142IB_HL-2_05.CDR Wednesday, 24 October 2007 12:33:25 PM PETERDELL IB_HL-2ed (143) Chapter Quadratic equations and functions A B C D E F G H Contents: Solving quadratic equations (Review) The discriminant of a quadratic The sum and product of the roots Graphing quadratic functions Finding a quadratic from its graph Where functions meet Problem solving with quadratics Quadratic optimisation cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 6A Review set 6C Review set 6E black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\143IB_HL-2_06.CDR Monday, 29 October 2007 11:30:32 AM PETERDELL Review set 6B Review set 6D IB_HL-2ed (144) 144 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) QUADRATICS A quadratic equation is an equation of the form ax2 + bx + c = where a, b and c are constants, a 6= A quadratic function is a function of the form f (x) = ax2 + bx + c, a 6= Alternatively, it can be written as f : x 7! ax2 + bx + c, a 6= Quadratic functions are members of the family of polynomials f : x 7! ax3 + bx2 + cx + d, a 6= is a cubic polynomial f : x 7! ax4 + bx3 + cx2 + dx + e, a 6= is a quartic polynomial HISTORICAL NOTE Galileo Galilei (1564 - 1642) was born in Pisa, Tuscany He was a philosopher who played a significant role in the scientific revolution of that time Within his research he conducted a series of experiments on the paths of projectiles, attempting to find a mathematical description of falling bodies Two of Galileo’s experiments consisted of rolling a ball down a grooved ramp that was placed at a fixed height above the floor and inclined at a fixed angle to the horizontal In one experiment the ball left the end of the ramp and descended to the floor In the second, a horizontal shelf was placed at the end of the ramp, and the ball travelled along this shelf before descending to the floor In each experiment Galileo altered the release height (h) of the ball and measured the distance (d) the ball travelled before landing The units of measurement were called ‘punti’ (points) In both experiments Galileo found that once the ball left the ramp or shelf, its path was parabolic and could therefore be modelled by a quadratic function Galileo OPENING PROBLEM Farmer Brown wishes to construct a rectangular pen for her chickens using her barn for one of the sides The other three sides will be constructed out of 80 m of chicken wire as illustrated She can make the pen long and thin, or short and fat, or any rectangular shape in between, as long as she uses exactly 80 m of chicken wire However, what she really wants to know is which rectangular shape will give her chickens the maximum area pen barn Can you: ² decide on a suitable variable to use and construct an area function in terms of this variable cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ² use the area function to find the maximum area that the pen may have and the shape of the pen when this occurs? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\144IB_HL-2_06.CDR Wednesday, 24 October 2007 1:55:01 PM PETERDELL IB_HL-2ed (145) 145 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) A SOLVING QUADRATIC EQUATIONS (REVIEW) Acme Leather Jacket Co makes and sells x leather jackets each week and their profit function is given by P = ¡12:5x2 + 550x ¡ 2125 dollars How many jackets must be made and sold each week in order to obtain a weekly profit of $3000? Clearly we need to solve the equation: ¡12:5x2 + 550x ¡ 2125 = 3000 i.e., 12:5x2 ¡ 550x + 5125 = This equation is of the form ax2 + bx + c = and is thus a quadratic equation To solve quadratic equations we can: ² factorise the quadratic and use the Null Factor law: “if ab = then a = or b = 0” ² complete the square ² use the quadratic formula ² use technology Definition: The roots or solutions of ax2 + bx + c = are the values of x which satisfy the equation, i.e., make it true For example, x = is a root of x2 ¡ 3x + = since, when x = 2, x2 ¡ 3x + = (2)2 ¡ 3(2) + = ¡ + = X SOLVING BY FACTORISATION Step Step Step Step 1: 2: 3: 4: Make one side of the equation by transferring all terms to one side Fully factorise the other side Use the ‘Null Factor law’: “if ab = then a = or b = 0” Solve the resulting elementary linear equations Example a 3x2 + 5x = cyan magenta yellow 95 50 b 75 25 95 100 50 75 25 95 100 50 25 95 100 50 75 25 ) 3x2 + 5x = x(3x + 5) = ) x = or 3x + = ) x = or x = ¡ 53 75 a b x2 = 5x + x2 = 5x + ) x2 ¡ 5x ¡ = ) (x ¡ 6)(x + 1) = ) x ¡ = or x + = ) x = or ¡1 100 Solve for x: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\145IB_HL-2_06.CDR Friday, 14 December 2007 10:35:07 AM PETERDELL IB_HL-2ed (146) 146 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) Example a 4x2 + = 4x Solve for x: a b 6x2 = 11x + 10 4x2 + = 4x ) 4x2 ¡ 4x + = ) (2x ¡ 1)2 = ) x= Example b 6x2 = 11x + 10 ) 6x2 ¡ 11x ¡ 10 = ) (2x ¡ 5)(3x + 2) = ) x = 52 or ¡ 23 = ¡7 x ) x(3x + ) = ¡7x x 3x + Solve for x: = ¡7 x 3x + fmultiply both sides by xg fexpand the bracketsg ) 3x2 + = ¡7x fmake the RHS 0g ) 3x2 + 7x + = ) (x + 2)(3x + 1) = ffactorisingg ) x = ¡2 or ¡ EXERCISE 6A.1 Solve the following by factorisation: b 6x2 + 2x = a 4x2 + 7x = e 3x2 = 8x d 2x ¡ 11x = g x2 ¡ 5x + = h x2 = 2x + k x2 + x = 12 j + x = 6x c f i l 3x2 ¡ 7x = 9x = 6x2 x2 + 21 = 10x x2 + 8x = 33 Solve the following by factorisation: b 2x2 ¡ 13x ¡ = a 9x2 ¡ 12x + = e 2x2 + = 5x d 3x + 5x = h 4x2 + 4x = g 3x2 = 10x + k 7x2 + 6x = j 12x = 11x + 15 c f i l 3x2 = 16x + 12 3x2 = 4x + 4x2 = 11x + 15x2 + 2x = 56 x+ e 2x ¡ f x+3 =¡ 1¡x x cyan magenta yellow 95 =3 x 100 95 100 50 75 25 95 100 50 75 = ¡1 x 50 d 75 ¡ 4x2 = 3(2x + 1) + 25 c (x + 2)(1 ¡ x) = ¡4 25 b (x + 1)2 = 2x2 ¡ 5x + 11 95 a 100 50 75 25 Solve for x: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\146IB_HL-2_06.CDR Wednesday, 24 October 2007 2:03:34 PM PETERDELL IB_HL-2ed (147) 147 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) SOLVING BY ‘COMPLETING THE SQUARE’ As you would be aware by now, not all quadratics factorise easily For example, x2 + 4x + cannot be factorised by using a simple factorisation approach This means that we need a different approach in order to solve x2 + 4x + = One way is to use the ‘completing the square’ technique Equations of the form ax2 + bx + c = can be converted to the form (x + p)2 = q from which the solutions are easy to obtain Notice that if X = a, pthen X = § a Example a (x + 2)2 = Solve exactly for x: a (x + 2)2 = p ) x+2 = § p ) x = ¡2 § b (x ¡ 1)2 = ¡5 (x ¡ 1)2 = ¡5 has no real solutions since the perfect square (x ¡ 1)2 cannot be negative b Example x2 + 4x + = Solve for exact values of x: x2 + 4x + = ) x2 + 4x = ¡1 x2 + 4x + 22 = ¡1 + 22 ) (x + 2)2 = p ) x+2 = § p ) x = ¡2 § ) fput the constant on the RHSg fcompleting the squareg ffactorisingg The squared number we add to both sides is ¶2 µ coefficient of x EXERCISE 6A.2 Solve for exact values of x: (x + 5)2 = (x ¡ 8)2 = (x + 1)2 + = 11 a d g b e h (x + 6)2 = ¡11 2(x + 3)2 = 10 (2x + 1)2 = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Solve for exact values of x by completing the square: b x2 + 6x + = a x2 ¡ 4x + = e x2 + 6x + = d x2 = 4x + h x2 + 10 = 8x g x + 6x = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\147IB_HL-2_06.CDR Wednesday, 24 October 2007 2:07:25 PM PETERDELL c f i (x ¡ 4)2 = 3(x ¡ 2)2 = 18 (1 ¡ 3x)2 ¡ = c f i x2 ¡ 14x + 46 = x2 = 2x + x2 + 6x = ¡11 IB_HL-2ed (148) 148 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) If the coefficient of x2 is not 1, we first divide throughout to make it For example, 2x2 + 10x + = ¡3x2 + 12x + = becomes x2 + 5x + becomes x2 ¡ 4x ¡ 3 Solve for exact values of x by completing the square: b 2x2 ¡ 10x + = a 2x2 + 4x + = e 5x2 ¡ 15x + = d 3x2 = 6x + =0 =0 c f 3x2 + 12x + = 4x2 + 4x = THE QUADRATIC FORMULA Many quadratic equations cannot be solved by factorising, and completing the square can be rather tedious Consequently, the quadratic formula has been developed This formula is: If ax + bx + c = 0, then x = Proof: If ax2 + bx + c = 0, b c then x2 + x + = a a b c ) x2 + x =¡ a a µ ¶2 µ ¶2 b b b c =¡ + ) x2 + x + a 2a a 2a µ ¶2 b2 ¡ 4ac b = ) x+ 2a 4a2 r b2 ¡ 4ac b ) x+ =§ 2a 4a2 p ¡b § b2 ¡ 4ac ) x= 2a ¡b § p b2 ¡ 4ac 2a fdividing each term by a, as a 6= 0g fcompleting the square on LHSg For example, consider the Acme Leather Jacket Co equation from page 145 We need to solve: 12:5x2 ¡ 550x + 5125 = Trying to factorise so in this case a = 12:5, b = ¡550, c = 5125 this equation or p using ‘completing 550 § (¡550)2 ¡ 4(12:5)(5125) the square’ would ) x = not be easy 2(12:5) p 550 § 46 250 = 25 ¼ 30:60 or 13:40 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 However, x needs to be a whole number, so x = 13 or 31 would produce a profit of around $3000 each week black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\148IB_HL-2_06.CDR Wednesday, 24 October 2007 2:09:03 PM PETERDELL IB_HL-2ed (149) 149 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) Example a x2 ¡ 2x ¡ = Solve for x: a b 2x2 + 3x ¡ = x2 ¡ 2x ¡ = has a = 1, b = ¡2, c = ¡6 p ¡(¡2) § (¡2)2 ¡ 4(1)(¡6) ) x= 2(1) p § + 24 ) x= p § 28 ) x= p 2§2 ) x= p ) x = 1§ p p Solutions are: + and ¡ 7: b 2x2 + 3x ¡ = has a = 2, b = 3, c = ¡6 p ¡3 § 32 ¡ 4(2)(¡6) ) x= 2(2) p ¡3 § + 48 ) x= p ¡3 § 57 ) x= Solutions are: p ¡3 + 57 and p ¡3 ¡ 57 : EXERCISE 6A.3 Use the quadratic formula to solve exactly for x: b x2 + 6x + = a x2 ¡ 4x ¡ = e x2 ¡ 4x + = d x2 + 4x = p h (3x + 1)2 = ¡2x g x2 ¡ 2x + = c f x2 + = 4x 2x2 ¡ 2x ¡ = i (x + 3)(2x + 1) = c (x ¡ 2)2 = + x f 2x ¡ Use the quadratic formula to solve exactly for x: a (x+2)(x¡1) = 2¡3x b (2x + 1)2 = ¡ x d x¡1 = 2x + 2¡x e x¡ B =1 x =3 x THE DISCRIMINANT OF A QUADRATIC In the quadratic formula, the quantity b2 ¡ 4ac under the square root sign is called the discriminant cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The symbol delta ¢ is used to represent the discriminant, so ¢ = b2 ¡ 4ac p ¡b § ¢ The quadratic formula becomes x = where ¢ replaces b2 ¡ 4ac 2a black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_06\149IB_HL-2_06.CDR Wednesday, January 2009 9:10:44 AM TROY IB_HL-2ed (150) 150 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) Note: ¡b x= is the only solution (a repeated or double root) 2a p ¢ is a positive real number, so there are two distinct real p p ¡b + ¢ ¡b ¡ ¢ roots: and 2a 2a p ¢ is not a real number and so there are no real roots ² If ¢ = 0, ² If ¢ > 0, ² If ¢ < 0, ² If a, b and c are rational and ¢ is a perfect square then the equation has two rational roots which can be found by factorisation Example Use the discriminant to determine the nature of the roots of: b 3x2 ¡ 4x ¡ = a 2x2 ¡ 2x + = a ¢ = b2 ¡ 4ac = (¡2)2 ¡ 4(2)(3) = ¡20 which is < ) there are no real roots ¢ = b2 ¡ 4ac = (¡4)2 ¡ 4(3)(¡2) = 40 which is > 40 is not a perfect square so there are distinct irrational roots b Example For x2 ¡ 2x + m = 0, find ¢ and hence find the values of m for which the equation has: a a repeated root b distinct real roots c no real roots x2 ¡ 2x + m = has a = 1, b = ¡2 and c = m ) ¢ = b2 ¡ 4ac = (¡2)2 ¡ 4(1)(m) = ¡ 4m a b For a repeated root ¢=0 ) ¡ 4m = ) = 4m ) m=1 repeated distinct real roots Notice: c For distinct real roots ¢>0 ) ¡ 4m > ) ¡4m > ¡4 ) m<1 For no real roots ¢<0 ) ¡ 4m < ) ¡4m < ¡4 ) m>1 imaginary roots cyan magenta yellow 95 100 50 Discriminant value ¢>0 ¢=0 ¢>0 ¢<0 75 25 95 100 50 75 25 95 100 50 75 25 Roots of quadratic two real distinct roots two identical real roots (repeated) two real roots no real roots 95 100 50 75 25 Summary: m values black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\150IB_HL-2_06.CDR Wednesday, 24 October 2007 2:16:14 PM PETERDELL IB_HL-2ed (151) 151 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) Example For the equation kx2 + [k + 3]x = diagram for it Hence, find the value of a two distinct real roots c a repeated root For kx2 + [k + 3]x ¡ = 0, find the discriminant ¢ and draw a sign k for which the equation has: b two real roots d no real roots a = k, b = k + 3, c = ¡1 So, ¢ = b ¡ 4ac = (k + 3)2 ¡ 4(k)(¡1) = k + 6k + + 4k = k + 10k + = (k + 9)(k + 1) a b c d For For For For and has sign diagram: + -9 two distinct real roots, two real roots, a repeated root, no real roots, ) ) ) ) ¢>0 ¢>0 ¢=0 ¢<0 - + -1 k k < ¡9 or k > ¡1: k ¡9 or k > ¡1: k = ¡9 or k = ¡1: ¡9 < k < ¡1: EXERCISE 6B By using the discriminant only, state the nature of the solutions of: p b x2 + 3x + = c 3x2 + 2x ¡ = a x2 + 7x ¡ = e x2 + x + = f 16x2 ¡ 8x + = d 5x2 + 4x ¡ = By using the discriminant only, determine which of the following quadratic equations have rational roots which can be found by factorisation b 2x2 ¡ 7x ¡ = c 3x2 + 4x + = a 6x2 ¡ 5x ¡ = 2 e 4x ¡ 3x + = f 8x2 + 2x ¡ = d 6x ¡ 47x ¡ = For the following quadratic equations, determine ¢ in simplest form and draw a sign diagram for it Hence find the value of m for which the equation has: i a repeated root ii two distinct real roots iii no real roots 2 b mx + 3x + = c mx2 ¡ 3x + = a x + 4x + m = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 For the following quadratic equations, find the discriminant ¢ and hence draw a sign diagram for it Find all k values for which the equation has: i two distinct real roots ii two real roots iii a repeated root iv no real roots b kx2 ¡ 2x + k = a 2x2 + kx ¡ k = d 2x2 + [k ¡ 2]x + = c x2 + [k + 2]x + = f [k + 1]x2 + kx + k = e x2 + [3k ¡ 1]x + [2k + 10] = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\151IB_HL-2_06.CDR Monday, 29 October 2007 11:32:45 AM PETERDELL IB_HL-2ed (152) 152 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) C THE SUM AND PRODUCT OF THE ROOTS If ax2 + bx + c = has roots ® and ¯ then ® + ¯ = ¡ b a and ®¯ = c a For example: if ® and ¯ are the roots of 2x2 ¡ 2x ¡ = then ® + ¯ = and ®¯ = ¡ 12 Proof: Method (Quadratic formula) p p ¡b + ¢ ¡b ¡ ¢ Let ® = , ¯= 2a 2a p p ¡b + ¢ ¡ b ¡ ¢ ) ®+¯ = 2a Method As ax2 + bx + c = a(x ¡ ®)(x ¡ ¯), ax2 + bx + c = a(x2 ¡ [® + ¯]x + ®¯) b c ) x2 + x + = x2 ¡ [® + ¯]x + ®¯ a a ¡2b b =¡ 2a a à p ! p !à ¡b ¡ ¢ ¡b + ¢ and ®¯ = 2a 2a Equating coefficients, b c ®+¯ =¡ and ®¯ = a a = = b2 ¡ ¢ b2 ¡ (b2 ¡ 4ac) 4ac c = = = 4a 4a2 4a a Example 10 Find the sum and product of the roots of 25x2 ¡ 20x + = Check your answer by solving the quadratic b 20 = = a 25 c and ®¯ = = a 25 If ® and ¯ are the roots then ® + ¯ = ¡ Check: 25x2 ¡ 20x + = has roots p p p p 20 § 400 ¡ 4(25)(1) 20 § 300 20 § 10 2§ = = = 50 50 50 p p 2+ 2¡ These have sum = + = X 5 à p ! p !à 2¡ 4¡3 2+ = = X and product = 5 25 25 EXERCISE 6C Find the sum and product of the roots of: cyan magenta yellow 95 c 100 50 75 25 95 100 50 75 25 x2 + 11x ¡ 13 = b 95 100 50 75 25 3x2 ¡ 2x + = 95 100 50 75 25 a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\152IB_HL-2_06.CDR Wednesday, 24 October 2007 2:21:10 PM PETERDELL 5x2 ¡ 6x ¡ 14 = IB_HL-2ed (153) 153 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) The equation kx2 ¡ (1 + k)x + (3k + 2) = is such that the sum of its roots is twice their product Find k and the two roots ax2 ¡ 6x + a ¡ = 0, a 6= has one root which is double the other a Let the roots be ® and 2® and find two equations involving ® b Find a and the two roots of the quadratic equation kx2 + (k ¡ 8)x + (1 ¡ k) = has one root which is two more than the other Find k and the two roots The roots of the equation x2 ¡ 6x + = are ® and ¯ 1 Find the simplest quadratic equation with roots ® + and ¯ + ¯ ® The roots of 2x2 ¡ 3x ¡ = are p and q Find all quadratic equations with roots p2 + q and q2 + p kx2 + [k + 2]x ¡ = has roots which are real and positive Find the possible values that k may have D GRAPHING QUADRATIC FUNCTIONS REVIEW OF TERMINOLOGY The equation of a quadratic function is given by y = ax2 + bx + c, where a 6= axis of symmetry y The graph of a quadratic function is called a parabola The point where the graph ‘turns’ is called the vertex If the graph opens upward, the y-coordinate of the vertex is the minimum and the graph is concave upwards parabola x zero If the graph opens downward, the y-coordinate of the vertex is the maximum and the graph is concave downwards zero y-intercept minimum The vertical line that passes through the vertex is called the axis of symmetry All parabolas are symmetrical about the axis of symmetry vertex The point where the graph crosses the y-axis is the y-intercept cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The points (if they exist) where the graph crosses the x-axis are called the x-intercepts, and correspond to the zeros of the function black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\153IB_HL-2_06.CDR Wednesday, November 2007 2:59:58 PM PETERDELL IB_HL-2ed (154) 154 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) GRAPHING y = a(x¡®)(x¡¯) INVESTIGATION This investigation is best done using a graphing package or graphics calculator GRAPHING PACKAGE TI C What to do: a Use technology to help you to sketch: y = (x ¡ 1)(x ¡ 3), y = 2(x ¡ 1)(x ¡ 3), y = ¡(x ¡ 1)(x ¡ 3), y = ¡3(x ¡ 1)(x ¡ 3) and y = ¡ 12 (x ¡ 1)(x ¡ 3) b Find the x-intercepts for each function in a c What is the geometrical significance of a in y = a(x ¡ 1)(x ¡ 3)? a Use technology to help you to sketch: y = 2(x ¡ 1)(x ¡ 4), y = 2(x ¡ 3)(x ¡ 5), y = 2x(x + 5) and y = 2(x + 2)(x + 4) y = 2(x + 1)(x ¡ 2), b Find the x-intercepts for each function in a c What is the geometrical significance of ® and ¯ in y = 2(x ¡ ®)(x ¡ ¯)? a Use technology to help you to sketch: y = 2(x ¡ 3)2 , y = 2(x ¡ 1)2 , y = 2(x + 2)2 , y = 2x2 b Find the x-intercepts for each function in a c What is the geometrical significance of ® in y = 2(x ¡ ®)2 ? Copy and complete: ² If a quadratic has factorisation y = a(x ¡ ®)(x ¡ ¯) it the x-axis at ² If a quadratic has factorisation y = a(x ¡ ®)2 it the x-axis at GRAPHING y = a(x¡h) 2+k INVESTIGATION This investigation is also best done using technology GRAPHING PACKAGE What to do: a Use technology to help you to sketch: y = (x ¡ 3)2 + 2, y = 2(x ¡ 3)2 + 2, y = ¡2(x ¡ 3)2 + 2, and y = ¡ (x ¡ 3)2 + y = ¡(x ¡ 3) + b Find the coordinates of the vertex for each function in a c What is the geometrical significance of a in y = a(x ¡ 3)2 + 2? a Use technology to help you to sketch: y = 2(x ¡ 2)2 + 4, y = 2(x ¡ 1)2 + 3, y = 2(x + 2)2 ¡ y = 2(x + 1)2 + 4, TI C y = 2(x ¡ 3)2 + 1, and y = 2(x + 3)2 ¡ b Find the coordinates of the vertex for each function in a c What is the geometrical significance of h and k in y = 2(x ¡ h)2 + k? cyan magenta yellow 95 100 50 then its vertex has coordinates 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Copy and complete: If a quadratic is in the form y = a(x ¡ h)2 + k black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\154IB_HL-2_06.CDR Wednesday, 24 October 2007 2:29:16 PM PETERDELL IB_HL-2ed (155) 155 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) From Investigations and you should have discovered that: The coefficient of x2 (which is a) controls the degree of width of the graph and whether it opens upwards or downwards ² I a > produces or concave up a < produces or concave down I If ¡1 < a < 1, a 6= the graph is wider than y = x2 : If a < ¡1 or a > the graph is narrower than y = x2 : Quadratic form, a 6= Graph ² y = a(x ¡ ®)(x ¡ ¯) ®, ¯ are real Facts x-intercepts are ® and ¯ a x b x= ² y = a(x ¡ ®)2 ® is real a+b x=a touches x-axis at ® axis of symmetry is x = ® vertex is (®, 0) x V (a, 0) ² y = a(x ¡ h)2 + k axis of symmetry is x = ®+¯ ³ ´ ®+¯ ®+¯ vertex is , f( ) axis of symmetry is x = h vertex is (h, k) x=h V (h, k) ² y = ax2 + bx + c (general quadratic form) p -b - ¢ 2a p x -b + ¢ 2a x= -b 2a axis of symmetry is ¡b x= 2a x-intercepts for ¢ > are p ¡b § ¢ 2a where ¢ = b2 ¡ 4ac cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 p p b ¡b ¡ ¢ ¡b + ¢ Note: ¡ is the average of and irrespective of the sign of ¢ 2a 2a 2a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\155IB_HL-2_06.CDR Monday, 29 October 2007 11:36:22 AM PETERDELL IB_HL-2ed (156) 156 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) SKETCHING GRAPHS USING KEY FACTS Example 11 Using axes intercepts only, sketch the graphs of: a y = 2(x + 3)(x ¡ 1) b y = ¡2(x ¡ 1)(x ¡ 2) a y = 2(x + 3)(x ¡ 1) has x-intercepts ¡3, y = ¡2(x ¡ 1)(x ¡ 2) has x-intercepts 1, b When x = 0, y = 2(3)(¡1) = ¡6 When x = 0, y = ¡2(¡1)(¡2) = ¡4 ) y-intercept is ¡6 ) y-intercept is ¡4 y y = 12 (x + 2)2 c y = 12 (x + 2)2 touches x-axis at ¡2 When x = 0, y = 12 (2)2 =2 ) y-intercept is y y -3 c x x -4 -2 -6 x Example 12 Use the vertex, axis of symmetry and y-intercept to graph y = ¡2(x + 1)2 + 4: The vertex is (¡1, 4) V(-1, 4) y The axis of symmetry is x = ¡1: When x = 0, y = ¡2(1)2 + =2 ) a<0 x shape !\=\-1 Example 13 cyan magenta yellow 95 100 50 the coordinates of the vertex Hence, sketch the graph 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 For the quadratic y = 2x2 + 6x ¡ 3, find: a the equation of the axis of symmetry b c the axes intercepts d black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\156IB_HL-2_06.CDR Wednesday, 24 October 2007 2:38:06 PM PETERDELL IB_HL-2ed (157) 157 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) For y = 2x2 + 6x ¡ 3, a = 2, b = 6, c = ¡3 ¡b ¡6 = = ¡ 32 2a a b a>0 ) shape When x = ¡ 32 , y = 2(¡ 32 )2 + 6(¡ 32 ) ¡ ) axis of symmetry is x = ¡ 32 = ¡7 12 fsimplifyingg ) vertex is (¡ 32 , ¡7 12 ) When x = 0, y = ¡3 ) y-intercept is ¡3 c d When y = 0, 2x2 + 6x ¡ = p ¡6 § 36 ¡ 4(2)(¡3) ) x= ) x ¼ ¡3:44 or 0:44 x =- 23 y x » -3.44 -3 » 0.44 V(-1 21 , - 21 ) Example 14 Determine the coordinates of the vertex of y = 2x2 ¡ 8x + 1: y = 2x2 ¡ 8x + has a = 2, b = ¡8, c = ¡b ¡(¡8) and so = =2 2a 2£2 The vertex is called the maximum turning point or the minimum turning point depending on whether the graph is concave down or concave up ) equation of axis of symmetry is x = and when x = 2, y = 2(2)2 ¡ 8(2) + = ¡7 ) the vertex has coordinates (2, ¡7) EXERCISE 6D.1 Using axes intercepts only, sketch the graphs of: a y = (x ¡ 4)(x + 2) b y = ¡(x ¡ 4)(x + 2) d y = ¡3x(x + 4) e y = 2(x + 3)2 c f y = 2(x + 3)(x + 5) y = ¡ 14 (x + 2)2 c f y = ¡2(x ¡ 1)2 ¡ y = ¡ 10 (x + 2)2 ¡ What is the axis of symmetry of each graph in question 1? Use the vertex, axis of symmetry and y-intercept to graph: b y = 2(x + 2)2 + a y = (x ¡ 1)2 + e y = ¡ 13 (x ¡ 1)2 + d y = (x ¡ 3) + Find the turning point or vertex for the following quadratic functions: magenta yellow 95 100 50 75 25 y = x2 + 2x ¡ y = 2x2 + 8x ¡ y = 2x2 ¡ 10x + 95 b e h 100 50 75 25 95 100 50 25 95 100 50 75 25 cyan 75 y = x2 ¡ 4x + y = ¡3x2 + y = 2x2 + 6x ¡ a d g black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\157IB_HL-2_06.CDR Monday, 29 October 2007 11:37:07 AM PETERDELL c f i y = 2x2 + y = ¡x2 ¡ 4x ¡ y = ¡ 12 x2 + x ¡ IB_HL-2ed (158) 158 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) Find a d g j the x-intercepts for: y = x2 ¡ y = x2 + x ¡ 12 y = ¡2x2 ¡ 4x ¡ y = x2 + 4x ¡ b e h k y y y y = 2x2 ¡ = 4x ¡ x2 = 4x2 ¡ 24x + 36 = x2 ¡ 6x ¡ For the following quadratics, find: i the equation of the axis of symmetry iii the axes intercepts, if they exist a d g y = x2 ¡ 2x + y = ¡x2 + 3x ¡ y = 6x ¡ x2 ii iv y y y y = x2 + 7x + 10 = ¡x2 ¡ 6x ¡ = x2 ¡ 4x + = x2 + 8x + 11 the coordinates of the vertex Hence, sketch the graph y = x2 + 4x ¡ y = ¡3x2 + 4x ¡ y = ¡x2 ¡ 6x ¡ b e h c f i l c f i y = 2x2 ¡ 5x + y = ¡2x2 + x + y = ¡ 14 x2 + 2x + SKETCHING GRAPHS BY COMPLETING THE SQUARE If we wish to find the vertex of a quadratic given in general form y = ax2 + bx + c then one approach is to convert it to the form y = a(x ¡ h)2 + k where we can read off the vertex (h, k) To this we may choose to ‘complete the square’ Example 15 Write y = x2 + 4x + in the form y = (x ¡ h)2 + k by completing the square Hence sketch y = x2 + 4x + 3, stating the coordinates of the vertex y = x2 + 4x + y = x2 + 4x + 22 + ¡ 22 y = (x + 2)2 ¡ ) ) shift units left -1 shift unit down -1 @\=\!X\+\4!\+3 Vertex is (¡2, ¡1) and y-intercept is -2 @\=\!X -2 vertex (-2'\-1) Example 16 Convert y = 3x2 ¡ 4x + into the form y = a(x ¡ h)2 + k by ‘completing the square’ Hence, write down the coordinates of its vertex and sketch the graph of the function y = 3x2 ¡ 4x + = 3[x2 ¡ 43 x + 13 ] ftake out a factor of 3g = 3[x2 ¡ 2( 23 )x + ( 23 )2 ¡ ( 23 )2 + 13 ] cyan magenta fcomplete the squareg yellow 95 100 50 75 25 95 100 50 75 fwrite as a perfect squareg 25 + 3] 95 ¡ 100 50 75 25 95 100 50 75 25 = 3[(x ¡ 2 3) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\158IB_HL-2_06.CDR Monday, 29 October 2007 11:37:31 AM PETERDELL IB_HL-2ed (159) 159 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) = 3[(x ¡ 23 )2 ¡ 19 ] = 3(x ¡ 2 3) ¡ @ ! = We_ fexpand to desired formg So the vertex is ( 23 , ¡ 13 ) The y-intercept is -0.5 -1 ! 1.5 V(We_ ,-Qe_) We can use technology to confirm this For example: EXERCISE 6D.2 Write the following quadratics in the form y = (x ¡ h)2 + k by ‘completing the square’ Hence sketch each function, stating the vertex: b y = x2 + 4x ¡ c y = x2 ¡ 4x a y = x2 ¡ 2x + e y = x2 + 5x ¡ f y = x2 ¡ 3x + d y = x2 + 3x h y = x2 + 8x ¡ i y = x2 ¡ 5x + g y = x2 ¡ 6x + For each i ii iii iv v of the following quadratics: convert into the form y = a(x ¡ h)2 + k state the coordinates of the vertex find the y-intercept Hence, sketch the graph of the quadratic Use technology to check your answer y = 2x2 + 4x + y = 2x2 ¡ 6x + y = ¡x2 + 4x + a c e b d f by ‘completing the square’ a is always the factor to be ‘taken out’ y = 2x2 ¡ 8x + y = 3x2 ¡ 6x + y = ¡2x2 ¡ 5x + 3 Use your graphing package or graphics calculator to graph each of the following functions Hence write each function in the form y¡=¡a(x¡¡¡h)2¡+¡k y = x2 ¡ 4x + y = 2x2 + 6x ¡ a d b e y = x2 + 6x + y = ¡2x2 ¡ 10x + GRAPHING PACKAGE TI C c f y = ¡x2 + 4x + y = 3x2 ¡ 9x ¡ THE DISCRIMINANT AND THE QUADRATIC GRAPH cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Consider the graphs of: y = x2 ¡ 2x + 3, y = x2 ¡ 2x + 1, y = x2 ¡ 2x ¡ All of these curves have the same axis of symmetry: x = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\159IB_HL-2_06.CDR Monday, 29 October 2007 11:37:43 AM PETERDELL IB_HL-2ed (160) 160 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) y = x2 ¡ 2x + y = x2 ¡ 2x + y y = x2 ¡ 2x ¡ y -1 ¢ = b2 ¡ 4ac = (¡2)2 ¡ 4(1)(3) = ¡8 ¢ = b2 ¡ 4ac = (¡2)2 ¡ 4(1)(1) =0 ¢<0 ¢=0 does not cut the x-axis touches the x-axis ¢ = b2 ¡ 4ac = (¡2)2 ¡ 4(1)(¡3) = 16 ¢>0 ² ² ² The discriminant ¢ determines if the graph: x -3 x x cuts the x-axis twice does not cut the x-axis touches the x-axis cuts the x-axis twice (¢ < 0) (¢ = 0) (¢ > 0) Example 17 Use the discriminant to determine the relationship between the graph and b y = ¡2x2 + 5x + the x-axis for: a y = x2 + 3x + a b a = 1, b = 3, c = ) ¢ = b2 ¡ 4ac = ¡ 4(1)(4) = ¡7 which is < The graph does not cut the x-axis a > ) concave up x It lies entirely above the x-axis a = ¡2, b = 5, c = ) ¢ = b2 ¡ 4ac = 25 ¡ 4(¡2)(1) = 33 which is > a < ) concave down x ) the graph cuts the x-axis twice POSITIVE DEFINITE AND NEGATIVE DEFINITE QUADRATICS Positive definite quadratics are quadratics which are positive for all values of x, i.e., ax2 + bx + c > for all x R Negative definite quadratics are quadratics which are negative for all values of x, i.e., ax2 + bx + c < for all x R magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 x ² A quadratic is positive definite if a > and ¢ < ² A quadratic is negative definite if a < and ¢ < Tests: cyan x black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\160IB_HL-2_06.CDR Wednesday, November 2007 3:13:18 PM PETERDELL IB_HL-2ed (161) 161 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) EXERCISE 6D.3 Use the discriminant to determine the relationship between the graph and x-axis for: p b y = x2 + 2x + c y = ¡2x2 + 3x + a y = x2 + 7x ¡ e y = ¡x2 + x + f y = 9x2 + 6x + d y = 6x2 + 5x ¡ Show that: a x2 ¡ 3x + > for all x c 2x2 ¡ 4x + is positive definite b d 4x ¡ x2 ¡ < for all x ¡2x2 + 3x ¡ is negative definite Explain why 3x2 + kx ¡ is never positive definite for any value of k Under what conditions is 2x2 + kx + positive definite? E FINDING A QUADRATIC FROM ITS GRAPH If we are given sufficient information on or about a graph we can determine the quadratic function in whatever form is required Example 18 Find the equation of the quadratic with graph: a b y y -1 x a Since the x-intercepts are ¡1 and 3, y = a(x + 1)(x ¡ 3), a < But when x = 0, y = ) = a(1)(¡3) ) a = ¡1 So, y = ¡(x + 1)(x ¡ 3): b x Since it touches at 2, y = a(x ¡ 2)2 , a > But when x = 0, y = ) = a(¡2)2 ) a=2 So, y = 2(x ¡ 2)2 : Example 19 The axis of symmetry is x = 1, so the other x-intercept is Find the equation of the quadratic with graph: ) y 16 But when x = 0, y = 16 ) 16 = a(2)(¡4) ) a = ¡2 x magenta yellow 95 100 50 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 cyan 75 ) the quadratic is y = ¡2(x + 2)(x ¡ 4) x=1 25 -2 y = a(x + 2)(x ¡ 4) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\161IB_HL-2_06.CDR Wednesday, 24 October 2007 3:11:45 PM PETERDELL IB_HL-2ed (162) 162 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) Example 20 Find, in the form y = ax2 + bx + c, the equation of the quadratic whose graph cuts the x-axis at and ¡3 and passes through the point (2, ¡20) Since the x-intercepts are and ¡3, the equation is y = a(x ¡ 4)(x + 3) where a 6= But when x = 2, y = ¡20 ) ¡20 = a(2 ¡ 4)(2 + 3) ) ¡20 = a(¡2)(5) ) a=2 ) the equation is y = 2(x ¡ 4)(x + 3), or y = 2x2 ¡ 2x ¡ 24 Example 21 a b y Find the equation of the quadratic given its graph is: y V(-4, 2) 16 x -2 x V(3,-2) For a vertex (3, ¡2) the quadratic has the form y = a(x ¡ 3)2 ¡ But when x = 0, y = 16 ) 16 = a(¡3)2 ¡ ) 16 = 9a ¡ ) 9a = 18 ) a=2 So, y = 2(x ¡ 3)2 ¡ a b For a vertex (¡4, 2) the quadratic has the form y = a(x + 4)2 + But when x = ¡2, y = ) = a(2)2 + ) 4a = ¡2 ) a = ¡ 12 So, y = ¡ 12 (x + 4)2 + EXERCISE 6E Find the equation of the quadratic with graph: a b y y c d x x e y f magenta -2 yellow 95 100 50 75 25 95 100 50 75 25 y x x 95 100 50 75 25 95 100 50 75 25 x x -3 -1 12 y cyan y black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\162IB_HL-2_06.CDR Wednesday, 24 October 2007 3:16:50 PM PETERDELL IB_HL-2ed (163) 163 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) Find the quadratic with graph: a b y c y y 12 x -4 -12 x x x =-3 x =-1 x=3 Find, in the form y = ax2 + bx + c, the equation of the quadratic whose graph: cuts the x-axis at and 1, and passes through (2, ¡9) cuts the x-axis at and ¡ 12 , and passes through (3, ¡14) touches the x-axis at and passes through (¡2, ¡25) touches the x-axis at ¡2 and passes through (¡1, 4) cuts the x-axis at 3, passes through (5, 12) and has axis of symmetry x = cuts the x-axis at 5, passes through (2, 5) and has axis of symmetry x = a b c d e f If V is the vertex, find the equation of the quadratic given its graph is: y y y a b c V(3,¡8) V(2,¡4) x x x V(2,-1) y d y e &Ew_ ' Qw_* (3,¡1) x y f V(2,¡3) x x V(4,-6) V&Qw_ '-\wwE_* INVESTIGATION FINDING QUADRATIC FUNCTIONS y¡=¡2x2¡+¡3x¡+¡7 is a quadratic function from which the following table of values is obtained: x y 12 21 34 51 72 Consider adding two further rows to this table: a row called ¢1 which gives differences between successive y-values, and a row called ¢2 which gives differences between successive ¢1 -values 21 yellow 25 95 100 50 75 25 95 50 75 25 95 100 50 75 25 100 magenta 13 9-5 cyan 34 51 17 72 21 34-21 95 ¢1 ¢2 12 72-51 100 50 x y 75 So, we have: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\163IB_HL-2_06.CDR Wednesday, 24 October 2007 3:28:58 PM PETERDELL IB_HL-2ed (164) 164 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) This table is known as a difference table What to do: Construct difference tables (for x = 0, 1, 2, 3, 4, 5) for each of the following quadratic functions: a y = x2 + 4x + b y = 3x2 ¡ 4x c y = 5x ¡ x2 d y = 4x2 ¡ 5x + 2 What you notice about the ¢2 row for the quadratic functions in 1? Consider the general quadratic y = ax2 + bx + c, a 6= a Copy and complete the following difference table: x y ° c a+b+c 4a + 2b + c :::::: :::::: :::::: ° :::::: :::::: :::::: :::::: ¢1 ° :::::: :::::: :::::: ¢2 b Comment on the ¢2 row c What can the encircled numbers be used for? Use what you have noticed in to determine, if possible, the quadratic functions with the following tables of values: a x y c x y 1 2 ¡1 15 26 ¡8 ¡18 b x y 10 d x y 18 32 52 ¡1 ¡7 ¡15 Cutting up Pizzas Given a pizza, we wish to determine the maximum number of pieces into which it can be cut using n cuts across it For example, for n = we have i.e., pieces for n = we have a Copy and complete: Number of cuts, n Maximum number of pieces, Pn i.e., pieces 3 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b Complete ¢1 and ¢2 rows and hence determine (if possible) a quadratic formula for Pn c For a huge pizza with 12 cuts across it, find the maximum number of pieces resulting black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\164IB_HL-2_06.CDR Wednesday, 24 October 2007 3:34:43 PM PETERDELL IB_HL-2ed (165) 165 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) F WHERE FUNCTIONS MEET Consider the graphs of a quadratic function and a linear function on the same set of axes Notice that we could have: cutting (2 points of intersection) touching (1 point of intersection) missing (no points of intersection) If the graphs meet, the coordinates of the points of intersection of the graphs of the two functions can be found by solving the two equations simultaneously Example 22 Find the coordinates of the points of intersection of the graphs with equations y = x2 ¡ x ¡ 18 and y = x ¡ 3: y = x2 ¡ x ¡ 18 meets y = x ¡ where x2 ¡ x ¡ 18 = x ¡ fRHS = 0g ) x2 ¡ 2x ¡ 15 = ) (x ¡ 5)(x + 3) = ffactorisingg ) x = or ¡3 Substituting into y = x ¡ 3, when x = 5, y = and when x = ¡3, y = ¡6 ) the graphs meet at (5, 2) and (¡3, ¡6) Example 23 y = 2x + c is a tangent to y = 2x2 ¡ 3x + Find c y = 2x + c meets y = 2x2 ¡ 3x + where 2x2 ¡ 3x + = 2x + c ) 2x ¡ 5x + (4 ¡ c) = Now this quadratic has ¢ = since the graphs touch ) (¡5)2 ¡ 4(2)(4 ¡ c) = ) 8(4 ¡ c) = 25 ) ¡ c = 18 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ) c= black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\165IB_HL-2_06.CDR Wednesday, 24 October 2007 3:38:36 PM PETERDELL IB_HL-2ed (166) 166 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) EXERCISE 6F Find the coordinates of the point(s) of intersection of the graphs with equations: b y = ¡x2 + 3x + and y = 2x ¡ a y = x2 ¡ 2x + and y = x + c y = x2 ¡ 4x + and y = 2x ¡ d y = ¡x2 + 4x ¡ and y = 5x ¡ Use a graphing package or a GDC to find the coordinates of the points of intersection (to decimal places) of the graphs with equations: a c GRAPHING PACKAGE TI C y = x2 ¡ 3x + and y = x + y = ¡x2 ¡ 2x + and y = x + y = x2 ¡ 5x + and y = x ¡ y = ¡x2 + 4x ¡ and y = 5x ¡ b d Find, by algebraic means, the points of intersection of the graphs with equations: b y = x2 + 2x ¡ and y = x ¡ a y = x2 and y = x + c y = 2x2 ¡ x + and y = + x + x2 d xy = and y = x + Use technology to check your solutions to the questions in Find possible values of c for which the lines y = 3x + c are tangents to the parabola with equation y = x2 ¡ 5x + 7: Find the values of m for which the lines y = mx ¡ are tangents to the curve with equation y = x2 ¡ 4x + 2: Find the slopes of the lines with y-intercept (0, 1) that are tangents to the curve y = 3x2 + 5x + a For what values of c the lines y = x + c never meet the parabola with equation y = 2x2 ¡ 3x ¡ 7? b Choose one of the values of c found in part a above and sketch the graphs using technology to illustrate that these curves never meet THE PARABOLA y2 = 4ax INVESTIGATION y A parabola is defined as the locus of all points which are equidistant from a fixed point called the focus and a fixed line called the directrix P(x,¡y) N x Suppose the focus is F(a, 0) and the directrix is the vertical line x = ¡a F(a,¡0) x¡=¡-a What to do: Suggest a reason or reasons why we would let the focus be at (a, 0) and the directrix be the line x¡=¡¡a PRINTABLE GRAPH PAPER Use the circular-linear graph paper provided to graph the parabola which has focus F(2, 0) and directrix x¡=¡¡2 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Use the definition given above to show that the equation of the parabola is y2 = 4ax black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\166IB_HL-2_06.CDR Wednesday, 23 January 2008 3:32:16 PM PETERDELL IB_HL-2ed (167) QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) Let y = mx + c be a tangent to y = 4ax y at the point P Use quadratic theory to show that: a a = mc µ ¶ y¡=¡mx¡+¡c a 2a , b P is at m m Suppose a ray of light comes in parallel to the axis of symmetry (the x-axis) It strikes a parabolic mirror at P and is reflected to cut the x-axis at R(k, 0) ®1 = ®2 by the Reflection Principle 167 P x y az P ax Q R x a Deduce that triangle PQR is isosceles b Hence, deduce that k = a Clearly state what special result follows from b List real life examples of where the result in has been utilised G PROBLEM SOLVING WITH QUADRATICS When solving some problems algebraically, a quadratic equation results We are generally only interested in any real solutions which result If the resulting quadratic equation has no real roots then the problem has no real solution Any answer we obtain must be checked to see if it is reasonable For example: ² if we are finding a length then it must be positive and we reject any negative solutions ² if we are finding ‘how many people are present’ then clearly a fractional answer would be unacceptable General problem solving method: Step 1: Step 2: Step 3: Step 4: If the information is given in words, translate it into algebra using a pronumeral such as x for the unknown Write down the resulting equation Solve the equation by a suitable method Examine the solutions carefully to see if they are acceptable Give your answer in a sentence Example 24 A rectangle has length cm longer than its width Its area is 42 cm2 Find its width cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 If the width is x cm, then the length is (x + 3) cm x cm Therefore x(x + 3) = 42 fequating areasg ) x + 3x ¡ 42 = (x + 3) cm ) x ¼ ¡8:15 or 5:15 fusing technologyg We reject the negative solution as lengths are positive ) width ¼ 5:15 cm black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\167IB_HL-2_06.CDR Monday, 29 October 2007 11:54:27 AM PETERDELL IB_HL-2ed (168) 168 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) Example 25 Is it possible to bend a 12 cm length of wire to form the legs of a right angled triangle with area 20 cm2 ? becomes 12 cm x(12 Area, A = x) = 20 ) ) x(12 ¡ x) = 40 ) 12x ¡ x2 ¡ 40 = x(12 ¡ area 20 cm ¡ x) x 12 - x p ¡16 ) x ¡ 12x + 40 = which gives x = There are no real solutions, indicating the impossibility 12 § Example 26 A wall is 12 m long It is timber panelled using vertical sheets of panelling of equal width If the sheets had been 0:2 m wider, less sheets would have been required What is the width of the timber panelling used? Let x m be the width of each panel 12 ) is the number of sheets needed x ¢ ¡ Now if the sheets are x + 15 m in width µ ¶ 12 ¡ sheets are needed x xm 12 m ¶ 12 ¡ = 12 x 12 ¡ = 12 ) 12 ¡ 2x + 5x 12 ¡ =0 ) ¡2x + 5x ) ¡10x2 + 12 ¡ 2x = ) 5x2 + x ¡ = ) (5x + 6)(x ¡ 1) = ) x = ¡ 65 or ) each sheet is m wide ¢ ¡ So, x + 15 µ flength of wallg fexpanding LHSg f£ each term by 5xg f¥ each term by ¡2g where x > EXERCISE 6G cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Two integers differ by 12 and the sum of their squares is 74 Find the integers black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\168IB_HL-2_06.CDR Monday, 29 October 2007 11:55:43 AM PETERDELL IB_HL-2ed (169) QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) 169 The sum of a number and its reciprocal is 15 Find the number The sum of a natural number and its square is 210 Find the number The product of two consecutive even numbers is 360 Find the numbers The product of two consecutive odd numbers is 255 Find the numbers n The number of diagonals of an n-sided polygon is given by the formula D = (n¡3) A polygon has 90 diagonals How many sides does it have? The length of a rectangle is cm longer than its width Find its width given that its area is 26 cm2 A rectangular box has a square base, and its height is cm longer than the length of each side of its base a If each side of its base has length x cm, show that its total surface area is given by A = 6x2 + 4x cm2 b If the total surface area is 240 cm2 , find the dimensions of the box x cm An open box contains 80 cm3 It is made from a square piece of tinplate with cm squares cut from each of its corners Find the dimensions of the original piece of tinplate cm 10 Is it possible to bend a 20 cm length of wire into the shape of a rectangle which has an area of 30 cm2 ? 11 The golden rectangle is the rectangle defined by the following statement: The golden rectangle can be divided into a square and a smaller rectangle by a line which is parallel to its shorter sides, the smaller rectangle being similar to the original rectangle A Y B D X C Thus, if ABCD is the golden rectangle, ADXY is a square and BCXY is similar to ABCD AB The ratio of for the golden rectangle is called the golden ratio AD p 1+ Hint: Let AB = x units and AD = unit Show that the golden ratio is 12 A triangular paddock has a road AB forming its longest side AB is km long The fences AC and CB are at right angles If BC is 400 m longer than AC, find the area of the paddock in hectares A km B C cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 13 Find the width of a uniform concrete path placed around a 30 m by 40 m rectangular lawn, given that the concrete has area one quarter of the lawn black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\169IB_HL-2_06.CDR Friday, 14 December 2007 11:11:23 AM PETERDELL IB_HL-2ed (170) 170 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) 14 Chuong and Hassan both drive 40 km from home to work each day One day Chuong said to Hassan, “If you drive home at your usual speed, I will average 40 km h¡1 faster than you and arrive home in 20 minutes less time.” Find Hassan’s speed 15 If the average speed of a small aeroplane had been 120 km h¡1 less, it would have taken a half an hour longer to fly 1000 km Find the speed of the plane 16 Two trains travel a 160 km track each day The express travels 10 km h¡1 faster and takes 30 minutes less than the normal train Find the speed of the express 17 A group of elderly citizens chartered a bus for $160 However, at the last minute of them fell ill and had to miss the trip As a consequence, the other citizens had to pay an extra $1 each How many elderly citizens went on the trip? 18 A tunnel is parabolic in shape with dimensions shown: A truck carrying a wide load is 4:8 m high and 3:9 m wide, and needs to pass through the tunnel Determine whether the truck will fit 8m 6m parabolic arch vertical supports B 19 AB is the longest vertical support of a bridge which contains a parabolic arch The vertical supports are 10 m apart The arch meets the vertical end supports m above the road 70 m 6¡m 10 m A roadway a If axes are drawn on the diagram of the bridge above, with x-axis the road and y-axis on AB, find the equation of the parabolic arch in the form y = ax2 + c b Hence, determine the lengths of all other vertical supports H QUADRATIC OPTIMISATION For a quadratic function y = ax2 + bx + c, we have already seen that: magenta 95 100 25 50 75 25 95 yellow if a < 0, the maximum value of y occurs b at x = ¡ 2a ymin 95 100 50 75 25 95 100 50 75 25 value of y occurs b at x = ¡ 2a cyan ² b 2a 50 x =¡ 75 if a > 0, the minimum 100 ² black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\170IB_HL-2_06.CDR Friday, 14 December 2007 11:11:33 AM PETERDELL ymax x =¡ b 2a IB_HL-2ed (171) 171 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) The process of finding the maximum or minimum value of a function is called optimisation ² ² Optimisation is a very useful tool when looking at such issues as: maximising profits minimising costs Example 27 Find the maximum or minimum value of the following quadratics, and the b y = + 3x ¡ 2x2 corresponding value of x: a y = x2 + x ¡ a For y = x2 + x ¡ a = 1, b = 1, c = ¡3: For y = ¡2x2 + 3x + a = ¡2, b = 3, c = 3: b As a > 0, shape is As a < 0, shape is ) the minimum value occurs ¡b when x = = ¡ 12 2a ) the maximum value occurs ¡b ¡3 when x = = = 34 2a ¡4 and y = (¡ 12 )2 + (¡ 12 ) ¡ and y = ¡2( 34 )2 + 3( 34 ) + = ¡3 14 = 18 ) the minimum value of y is ¡3 14 ) the maximum value of y is 18 occurring when x = ¡ 12 : occurring when x = 34 : Example 28 A vegetable gardener has 40 m of fencing to enclose a rectangular garden plot where one side is an existing brick wall If the two equal sides are x m long: a show that the area enclosed is given by A = x(40 ¡ 2x) m2 b find the dimensions of the vegetable garden of maximum area xm X Y xm a Side XY = 40 ¡ 2x m Now area = length £ width ) A = x(40 ¡ 2x) m2 : b A = 40x ¡ 2x2 = ¡2x2 + 40x is a quadratic in x, with a = ¡2, b = 40, c = 0: As a < 0, shape is Z xm brick wall ¡b ¡40 = = 10 2a ¡4 ) the area is maximised when YZ = 10 m and XY = 20 m cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The max area occurs when x = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\171IB_HL-2_06.CDR Wednesday, 24 October 2007 4:37:49 PM PETERDELL IB_HL-2ed (172) 172 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) EXERCISE 6H Find the maximum or minimum values of the following quadratics, and the corresponding values of x: b y = ¡ 2x ¡ x2 c y = + 2x ¡ 3x2 a y = x2 ¡ 2x d y = 2x2 + x ¡ e y = 4x2 ¡ x + f y = 7x ¡ 2x2 The profit in manufacturing x refrigerators per day, is given by the profit relation P = ¡3x2 + 240x ¡ 800 dollars How many refrigerators should be made each day to maximise the total profit? What is the maximum profit? A rectangular plot is enclosed by 200 m of fencing and has an area of A square metres Show that: a A = 100x ¡ x2 where x m is the length of one of its sides b the area is maximised when the rectangle is a square xm A rectangular paddock to enclose horses is to be made with one side being a straight water drain If 1000 m of fencing is available for the other sides, what dimensions should be used for the paddock so that it encloses the maximum possible area? ym 1800 m of fencing is available to fence six identical pig pens as shown in the diagram a Explain why 9x + 8y = 1800: b Show that the total area of each pen is given by A = ¡ 98 x2 + 225x m2 xm c If the area enclosed is to be maximised, what is the shape of each pen? a If 500 m of fencing is available to make rectangular pens of identical shape, find the dimensions that maximise the area of each pen if the plan is: y y = xX - 3x The graphs of y = x2 ¡ 3x and y = 2x ¡ x2 are illustrated a Prove that the graphs meet where x = and x = 12 b Find the maximum vertical separation between the curves for x 12 x b y = 2x - xX cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Infinitely many rectangles may be inscribed B A within the right angled triangle shown alongside cm One of them is illustrated a Let AB = x cm and BC = y cm D C Use similar triangles to find y in terms of x cm b Find the dimensions of rectangle ABCD of maximum area black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\172IB_HL-2_06.CDR Wednesday, 24 October 2007 4:43:41 PM PETERDELL IB_HL-2ed (173) QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) 173 A manufacturer of pot-belly stoves has the following situation to consider If x are made per week, each one will cost (50 + 400 x ) dollars and the total receipts per week for selling them would be (550x ¡ 2x ) dollars How many pot-belly stoves should be made per week in order to maximise profits? ¡1 ¢ 10 The total cost of producing x toasters per day is given by C = 10 x + 20x + 25 euros, and the selling price of each toaster is (44 ¡ 15 x) euros How many toasters should be produced each day in order to maximise the total profit? 11 A manufacturer of barbeques knows that if x of them are made each week then each one will cost (60 + 800 x ) pounds and the total receipts per week will be (1000x ¡ 3x2 ) pounds How many barbeques should be made per week to maximise profits? 12 The points P1 (a1 , b1 ), P2 (a2 , b2 ), P3 (a3 , b3 ), , Pn (an , bn ) are obtained by experiment These points are believed to be close to linear through the origin O(0, 0) To find the equation of the ‘line of best fit’ through the origin we decide to minimise y Mn Pc Mv Pz Mx Mc Pn Pv Px Mz (P1 M1 )2 +(P2 M2 )2 +(P3 M3 )2 +::::::+(Pn Mn )2 where [ Pi Mi ] is the vertical line segment connecting each point Pi with the line x Find the slope of the ‘line of best fit’ in terms of and bi (i = 1, 2, 3, 4, , n) 13 Write f (x) = (x ¡ a ¡ b)(x ¡ a + b)(x + a ¡ b)(x + a + b) in expanded form and hence determine the least value of f (x) Assume that a and b are real constants 14 By considering f (x) = (a1 x ¡ b1 )2 + (a2 x ¡ b2 )2 , use quadratic theory to prove the p p Cauchy-Schwarz inequality: j a1 b1 + a2 b2 j a + a b + b 2 15 b1 , c1 , b2 and c2 are real numbers such that b1 b2 = 2(c1 + c2 ) Show that at least one of the equations x2 + b1 x + c1 = 0, x2 + b2 x + c2 = has two real roots REVIEW SET 6A cyan b state the equation of the axis of symmetry d find the y-intercept f use technology to check your answers magenta yellow 95 100 50 75 25 95 100 50 75 25 95 y = 12 (x ¡ 2)2 ¡ 4: state the equation of the axis of symmetry find the coordinates of the vertex c find the y-intercept sketch the graph of the function e use technology to check your answers 100 50 25 95 100 50 75 25 For a b d y = ¡2(x + 2)(x ¡ 1): state the x-intercepts find the coordinates of the vertex sketch the graph of the function 75 For a c e black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\173IB_HL-2_06.CDR Thursday, 25 October 2007 9:04:45 AM PETERDELL IB_HL-2ed (174) 174 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) For a b d y = 2x2 + 6x ¡ 3: convert into the form y = a(x ¡ h)2 + k by ‘completing the square’ state the coordinates of the vertex c find the y-intercept sketch the graph e use technology to check your answers Solve the following equations, giving exact answers: a x2 ¡ 11x = 60 b 3x2 ¡ x ¡ 10 = c 3x2 ¡ 12x = c 2x2 ¡ 7x + = Solve the following equations: x2 + 10 = 7x a b x+ 12 =7 x Solve the following equation by completing the square: x2 + 7x ¡ = Solve the following using the quadratic formula: a x2 ¡ 7x + = b 2x2 ¡ 5x + = a For what values of c the lines with equations y = 3x + c intersect the parabola y = x2 + x ¡ in two distinct points? b Choose one such value of c from part a and find the points of intersection The roots of 2x2 ¡ 3x = are ® and ¯ Find the simplest quadratic equation 1 which has roots and ¯ ® REVIEW SET 6B Draw the graph of y = ¡x2 + 2x Find the equation of the axis of symmetry and the vertex of y = ¡3x2 + 8x + 7: Use the discriminant only to determine the number of solutions to: a 3x2 ¡ 5x + = b ¡2x2 ¡ 4x + = Find the maximum or minimum value of the relation y = ¡2x2 + 4x + and the value of x for which the maximum or minimum occurs Find the points of intersection of y = x2 ¡ 3x and y = 3x2 ¡ 5x ¡ 24 For what values of k does the graph of y = ¡2x2 + 5x + k not cut the x-axis? C D existing wall 60 m of chicken wire is available to construct a chicken enclosure against an existing wall The enclosure is to be rectangular a If BC = x m, show that the area of rectangle ABCD is given by A = (30x ¡ 12 x2 ) m2 b Find the dimensions of the enclosure which will maximise the area enclosed xm B A For what values of m are the lines y = mx ¡ 10 tangents to the parabola y = 3x2 + 7x + 2? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 One of the roots of kx2 + (1 ¡ 3k)x + (k ¡ 6) = is the negative reciprocal of the other root Find k and the two roots black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_06\174IB_HL-2_06.CDR Thursday, 11 March 2010 10:22:35 AM PETER IB_HL-2ed (175) QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) 175 REVIEW SET 6C x2 + 5x + = Solve using the quadratic formula: a x2 ¡ 5x ¡ = 3x2 + 11x ¡ = b 2x2 ¡ 7x ¡ = Use technology to solve: a (x ¡ 2)(x + 1) = 3x ¡ b 2x ¡ = x Using the discriminant only, determine the nature of the solutions of: b 3x2 ¡ 24x + 48 = a 2x2 ¡ 5x ¡ = Solve the following equations: a b Find the values of m for which 2x2 ¡ 3x + m = has: a a repeated root b two distinct real roots If AB is the same length as CD, BC is cm shorter than AB, and BE is cm in length, find the length of AB c no real roots D E A C B Find the length of the hypotenuse of a right angled triangle with one leg cm longer than the other and the hypotenuse cm longer than the longer leg Find the y-intercept of the line with slope ¡3 that is tangential to the parabola y = 2x2 ¡ 5x + ax2 + [3 ¡ a]x ¡ = has roots which are real and positive What values can a have? REVIEW SET 6D Use the vertex, axis of symmetry and y-intercept to graph: b y = ¡ 12 (x + 4)2 + a y = (x ¡ 2)2 ¡ For the quadratic y = 2x2 + 4x ¡ 1, find: a the equation of the axis of symmetry b c the axes intercepts d the coordinates of the vertex Hence sketch the graph Use the discriminant only to find the relationship between the graph and the x-axis for: b y = ¡3x2 ¡ 7x + a y = 2x2 + 3x ¡ Determine if the quadratic functions are positive definite, negative definite or neither: b y = 3x2 + x + 11 a y = ¡2x2 + 3x + Find the equation of the quadratic relation with graph: a y -3 b y (2, 25) x -27 x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The sum of a number and its reciprocal is 30 Find the number black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\175IB_HL-2_06.CDR Wednesday, November 2007 3:14:20 PM PETERDELL IB_HL-2ed (176) 176 QUADRATIC EQUATIONS AND FUNCTIONS (Chapter 6) An open square container is made by cutting cm square pieces out of a piece of tinplate If the capacity is 120 cm3 , find the size of the original piece of tinplate Find the points where y = ¡x2 ¡ 5x + and y = x2 + 3x + 11 meet Show that no line with a y-intercept of (0, 10) will ever be tangential to the curve with equation y = 3x2 + 7x ¡ 1 10 Find all quadratic equations which have roots m ¡ and n ¡ given that m n m and n are the roots of 3x ¡ 2x ¡ = REVIEW SET 6E Find the equation of the quadratic relation with graph: a b y y 18 x -3 c (2,-20) x d y y x -2 -3 x¡=¡4 x Find an expression for a quadratic which cuts the x-axis at and ¡2 and has y-intercept 24 Give your answer in the form y = ax2 + bx + c Find, in the form y = ax2 + bx + c, the equation of the quadratic whose graph: a touches the x-axis at and passes through (2, 12) b has vertex (¡4, 1) and passes through (1, 11) Find the maximum or minimum value of the following quadratics, and the correspondb y = ¡2x2 ¡ 5x + ing value of x: a y = 3x2 + 4x + For what values of k would the graph of y = x2 ¡ 2x + k cut the x-axis twice? Check your answer(s) using technology ym 600 m of fencing are used to construct xm rectangular animal pens as shown 600 ¡ 8x a Show that y = b Find the area A of each pen in terms of x c Find the dimensions of each pen if each pen is to have maximum area d What is the maximum area of each pen? Show that the lines with equations y = ¡5x + k y = x2 ¡ 3x + c if and only if c ¡ k = are tangents to the parabola cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 4x2 ¡ 3x ¡ = has roots p, q Find all quadratic equations with roots p3 and q black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_06\176IB_HL-2_06.CDR Thursday, 25 October 2007 9:14:09 AM PETERDELL IB_HL-2ed (177) Chapter Complex numbers and polynomials A Solutions of real quadratics with ¢ < B Complex numbers C Real polynomials D Roots, zeros and factors E Graphing polynomials F Theorems for real polynomials Contents: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 7A Review set 7B Review set 7C black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\177IB_HL-2_07.CDR Thursday, 25 October 2007 3:09:51 PM PETERDELL IB_HL-2ed (178) 178 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) A SOLUTIONS OF REAL QUADRATICS WITH ¢ < In Chapter 6, we determined that: If ax2 + bx + c = 0, a 6= and a, b, c R , then the solutions or roots are found p ¡b § ¢ using the formula x = where ¢ = b2 ¡ 4ac is known as the discriminant 2a ² ¢>0 ² ¢=0 ² ¢<0 We also observed that if: we have two real distinct solutions we have two real identical solutions we have no real solutions However, it is in fact possible to write down two solutions for the case where ¢ < To this we need imaginary numbers p In 1572, Rafael Bombelli defined the imaginary number i = ¡1 It is called ‘imaginary’ because we cannot place it on a number line With i defined, we can write down solutions for quadratic equations with ¢ < They are called complex solutions because they include a real and an imaginary part Any number of the form a + bi where a and b are real and i = called a complex number p ¡1 is Example a x2 = ¡4 Solve the quadratic equations: a x2 = ¡4 p ) x = § ¡4 p p ) x = § ¡1 ) b z2 + z + = ¢ = ¡ = ¡7 p ¡1 § ¡7 Now z = fquadratic formulag p p ¡1 § 7i ) z= = ¡ 12 § 27 i b x = §2i In Example above, notice that ¢ < in both cases In each case we have found two complex solutions of the form a + bi, where a and b are real HISTORICAL NOTE cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 18th century mathematicians enjoyed playing with these new ‘imaginary’ numbers, but they were regarded as little more than interesting curiosities until the work of Gauss (1777 1855), the German mathematician, astronomer and physicist For centuries mathematicians attempted to find a method of trisecting an angle using a compass and straight edge Gauss put an end to this when he used complex numbers to prove the impossibility of such a construction From his systematic use of complex numbers and the special results, he was able to convince mathematicians of their usefulness black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\178IB_HL-2_07.CDR Thursday, 25 October 2007 10:11:32 AM PETERDELL IB_HL-2ed (179) COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) 179 Early last century Steinmetz (an American engineer) used complex numbers to solve electrical problems, illustrating that complex numbers did have a practical application Complex numbers are now used extensively in electronics, engineering, and various scientific fields, especially physics Example a Write as a product of linear factors: a x2 + b x2 + 11 x2 + = x2 ¡ 4i2 b = (x + 2i)(x ¡ 2i) x2 + 11 = x2 ¡ 11i2 p p = (x + i 11)(x ¡ i 11) Example a Solve for x: a x2 + = b x3 + 2x = x2 + = b x3 + 2x = ) x2 ¡ 9i2 = ) x(x2 + 2) = (x + 3i)(x ¡ 3i) = ) x(x2 ¡ 2i2 ) = p p ) x = §3i x(x + i 2)(x ¡ i 2) = p ) x = or §i Example x ¡ 4x + 13 = Solve for x: x2 ¡ 4x + 13 = ) ) ) ) p 16 ¡ 4(1)(13) x= p § ¡36 x= § 6i x= x = + 3i or ¡ 3i 4§ Example x4 + x2 = ) x4 + x2 ¡ = ) (x2 + 3)(x2 ¡ 2) = p p p p (x + i 3)(x ¡ i 3)(x + 2)(x ¡ 2) = p p ) x = §i or § Solve for x: x4 + x2 = ) EXERCISE 7A cyan magenta yellow 95 d 100 50 25 75 q ¡ 14 c 95 100 50 p ¡64 75 25 95 100 50 75 25 95 100 50 75 25 Write in terms of i: p a ¡9 b black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\179IB_HL-2_07.CDR Thursday, 25 October 2007 10:14:33 AM PETERDELL p ¡5 e p ¡8 IB_HL-2ed (180) 180 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) Write as a product of linear factors: b x2 + a x2 ¡ f 4x2 + e 4x2 ¡ j x3 + x i x3 ¡ x c g k x2 ¡ 2x2 ¡ x4 ¡ d h l x2 + 2x2 + x4 ¡ 16 Solve for x: a x2 ¡ 25 = e 4x2 ¡ = i x3 ¡ 3x = c g k x2 ¡ = x3 ¡ 4x = x4 ¡ = d h l x2 + = x3 + 4x = x4 = 81 b f j x2 + 25 = 4x2 + = x3 + 3x = Solve for x: a x2 ¡ 10x + 29 = 2x2 + = 6x d e x2 + 6x + 25 = p x2 ¡ 3x + = f b e x4 = x2 + x4 + = 2x2 c f b Solve for x: a x4 + 2x2 = d x4 + 9x2 + 14 = B c x2 + 14x + 50 = 2x + = x x4 + 5x2 = 36 x4 + 2x2 + = COMPLEX NUMBERS FORMAL DEFINITION OF A COMPLEX NUMBER (IN CARTESIAN FORM) Any number of the form a + bi where a and b are real and i = complex number p ¡1 is called a Notice that real numbers are complex numbers in the special case where b = A complex number of the form bi where b 6= is called purely imaginary THE ‘SUM OF TWO SQUARES’ a2 + b2 = a2 ¡ b2 i2 fas i2 = ¡1g = (a + bi)(a ¡ bi) Notice that: a2 ¡ b2 = (a + b)(a ¡ b) a2 + b2 = (a + bi)(a ¡ bi) Compare: and fthe difference of two squares factorisationg fthe sum of two squares factorisationg REAL AND IMAGINARY PARTS OF COMPLEX NUMBERS If we write z = a + bi where a and b are real then: ² a is the real part of z and we write ² b is the imaginary part of z and we write cyan magenta yellow Im(z) = 95 100 50 p Im(z) = ¡ 2: 75 25 95 and 100 Re(z) = 50 and 75 Re(z) = 25 95 100 50 75 if z = + 3i, p if z = ¡ 2i, 25 95 100 50 75 25 So, a = Re(z), b = Im(z): black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\180IB_HL-2_07.CDR Thursday, 25 October 2007 10:20:04 AM PETERDELL IB_HL-2ed (181) 181 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) OPERATIONS WITH COMPLEX NUMBERS Notice that: for radicals, (2 + p p p p 3) + (4 + 3) = (2 + 4) + (1 + 2) = + 3 and for complex numbers, (2 + i) + (4 + 2i) = (2 + 4) + (1 + 2)i = + 3i p p p p p p (2 + 3)(4 + 3) = + + + 2( 3)2 = + + Also, notice that and (2 + i)(4 + 2i) = + 4i + 4i + 2i2 = + 8i ¡2 In fact the operations with complex numbers are identical to those with radicals, but with p p i2 = ¡1 rather than ( 2)2 = or ( 3)2 = So, we can add, subtract, multiply and divide complex numbers in the same way we perform these operations with radicals: (a + bi) + (c + di) = (a + c) + (b + d)i (a + bi) ¡ (c + di) = (a ¡ c) + (b ¡ d)i (a + bi)(c + di) = ac + adi + bci + bdi2 µ ¶µ ¶ a + bi a + bi c ¡ di ac ¡ adi + bci ¡ bdi2 = = c + di c + di c ¡ di c2 + d2 addition subtraction multiplication division Notice how division can be performed using a multiplication technique to obtain a real number in the denominator Example If z = + 2i and w = ¡ i find: a z+w b z¡w a z¡w = (3 + 2i) ¡ (4 ¡ i) = + 2i ¡ + i = ¡1 + 3i b z+w = (3 + 2i) + (4 ¡ i) =7+i c zw c zw = (3 + 2i)(4 ¡ i) = 12 ¡ 3i + 8i ¡ 2i2 = 12 + 5i + = 14 + 5i Example z + 2i = w 4¡i µ ¶µ ¶ + 2i 4+i = 4¡i 4+i If z = + 2i and w = ¡ i z find in the form a + bi, w where a and b are real 12 + 3i + 8i + 2i2 16 ¡ i2 10 + 11i = 17 = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\181IB_HL-2_07.CDR Thursday, 25 October 2007 10:25:11 AM PETERDELL 10 17 + 11 17 i IB_HL-2ed (182) 182 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) EXERCISE 7B.1 Copy and complete: z + 2i 5¡i Re(z) Im(z) z ¡3 + 4i ¡7 ¡ 2i ¡11i p i Re(z) Im(z) If z = ¡ 2i and w = + i, find in simplest form: a z+w b 2z c iw e 2z ¡ 3w f zw g w2 d h z¡w z2 For z = + i and w = ¡2 + 3i, find in simplest form: a z + 2w b z2 c z3 e w2 f zw g z2 w d h iz izw Simplify in for n = 0, 1, 2, 3, 4, 5, 6, 7, 8, and also for n = ¡1, ¡2, ¡3, ¡4, and ¡5 Hence, simplify i4n+3 where n is any integer Write (1 + i)4 in simplest form and hence find (1 + i)101 in simplest form Suppose (a + bi)2 = ¡16 ¡ 30i where a and b are real Find the possible values of a and b, given that a > For z = ¡ i and w = + 3i, find in the form a + bi where a and b are real: z i w a b c d z ¡2 w z iz Simplify: i ¡ 2i a i(2 ¡ i) ¡ 2i b If z = + i and w = ¡1 + 2i, find: a Im(4z ¡ 3w) b Re(zw) c Im(iz ) c ¡ 2¡i 2+i d Re ³z´ w 10 Check your answers to questions to and to using technology EQUALITY OF COMPLEX NUMBERS Two complex numbers are equal when their real parts are equal and their imaginary parts are equal a + bi = c + di , a = c and b = d cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Suppose b 6= d Now if a + bi = c + di where a, b, c and d are real, then bi ¡ di = c ¡ a ) i(b ¡ d) = c ¡ a c¡a fas b 6= dg ) i= b¡d and this is false as the RHS is real and the LHS is imaginary Thus, the supposition is false and hence b = d and furthermore a = c 95 100 50 75 25 Proof: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\182IB_HL-2_07.CDR Monday, 12 November 2007 9:17:21 AM PETERDELL IB_HL-2ed (183) COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) 183 Example If (x + yi)(2 ¡ i) = ¡i and x, y are real, determine the values of x and y ¡i 2¡i µ ¶µ ¶ ¡i 2+i i.e., x + yi = 2¡i 2+i If (x + yi)(2 ¡ i) = ¡i, then x + yi = ¡2i ¡ i2 ¡ 2i = 4+1 x + yi = ¡ i ) x + yi = ) and so x = 15 , y = ¡ 25 Example Find real numbers x and y for which (x + 2i)(1 ¡ i) = + yi (x + 2i)(1 ¡ i) = + yi x ¡ xi + 2i + = + yi ) [x + 2] + [2 ¡ x]i = + yi x + = and ¡ x = y fequating real and imaginary partsg ) x = and y = ¡1 ) ) EXERCISE 7B.2 Find real numbers x and y such that: a 2x + 3yi = ¡x ¡ 6i c (x + yi)(2 ¡ i) = + i b d Find x and y if x, y R and: a 2(x + yi) = x ¡ yi c (x + i)(3 ¡ iy) = + 13i b d x2 + xi = ¡ 2i (3 + 2i)(x + yi) = ¡i (x + 2i)(y ¡ i) = ¡4 ¡ 7i (x + yi)(2 + i) = 2x ¡ (y + 1)i p Write z in the form a + bi where a, b R and i = ¡1, if the complex number z satisfies the equation 3z + 17i = iz + 11 p z= The complex number z is a solution of the equation + ¡ 2i: 1+i Express z in the form a + bi where a and b Z Find the real values of m and n for which 3(m + ni) = n ¡ 2mi ¡ (1 ¡ 2i) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 3i Express z = p + in the form a + bi where a, b R , giving the exact 2¡i values of the real and imaginary parts of z black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\183IB_HL-2_07.CDR Thursday, 25 October 2007 10:37:18 AM PETERDELL IB_HL-2ed (184) 184 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) COMPLEX CONJUGATES Complex numbers a + bi and a ¡ bi are called complex conjugates If z = a + bi we write its conjugate as z ¤ = a ¡ bi Recall from page 181 that the conjugate is important for division: zw¤ z z w¤ = which makes the denominator real = w w w¤ ww¤ Complex conjugates appear as the solutions of real quadratic equations of the form ax2 + bx + c = where the discriminant ¢ = b2 ¡ 4ac is negative For example: ² ² x2 ¡ 2x + = has and has and x2 + = ¢ = (¡2)2 ¡ 4(1)(5) = ¡16 the solutions are x = + 2i and ¡ 2i ¢ = 02 ¡ 4(1)(4) = ¡16 the solutions are x = 2i and ¡2i ² Quadratics with real coefficients are called real quadratics This does not necessarily mean that its zeros are real ² If a quadratic equation has rational p coefficients and an irrational root of p the form c + d n, then c ¡ d n is also a root These roots are radical conjugates ² If a real quadratic equation has ¢ < and c + di is a complex root then c ¡ di is also a root These roots are complex conjugates Note: If c + di and c ¡ di are roots of a quadratic equation, then the quadratic equation is a(x2 ¡ 2cx + (c2 + d2 )) = for some constant a 6= Theorem: Proof: The sum of the roots = 2c and the product = (c + di)(c ¡ di) = c2 + d2 ) x2 ¡ (sum) x + (product) = ) x2 ¡ 2cx + (c2 + d2 ) = Alternatively: If c + di and c ¡ di are roots then (x ¡ [c + di])(x ¡ [c ¡ di]) = ) (x ¡ c ¡ di)(x ¡ c + di) = ) (x ¡ c)2 ¡ d2 i2 = ) x2 ¡ 2cx + c2 + d2 = In general, a(x2 ¡ 2cx + c2 + d2 ) = for some constant a 6= Example 10 Find all quadratic equations with real coefficients having ¡ 2i as a root As ¡ 2i is a root, + 2i is also a root Sum of roots = ¡ 2i + + 2i Product of roots = (1 ¡ 2i)(1 + 2i) =2 =1+4 =5 So, as x ¡ (sum) x + (product) = 0, cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a(x2 ¡ 2x + 5) = 0, a 6= gives all possible equations black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\184IB_HL-2_07.CDR Monday, 29 October 2007 12:16:03 PM PETERDELL IB_HL-2ed (185) 185 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) Note: The sum of complex conjugates c + di and c ¡ di is 2c which is real The product is (c + di)(c ¡ di) = c2 + d2 which is also real Example 11 Find exact values of a and b if p + i is a root of x2 + ax + b = 0, a, b R Since a and b are real, the quadratic has real coefficients p ¡ i is also a root ) p p p ) sum of roots = + i + ¡ i = 2 p p product of roots = ( + i)( ¡ i) = + = p Thus a = ¡2 and b = EXERCISE 7B.3 Find all quadratic equations with real coefficients and roots of: a 3§i b § 3i c ¡2 § 5i p p f and ¡ g §i e 2§ d p 2§i h ¡6 § i Find exact values of a and b if: a + i is a root of x2 + ax + b = 0, where a and b are real p b ¡ is a root of x2 + ax + b = 0, where a and b are rational c a + is a root of x2 + 4x + b = where a and b are real [Careful!] INVESTIGATION PROPERTIES OF CONJUGATES The purpose of this investigation is to discover any properties that complex conjugates might have What to do: Given z1 = ¡ i and z2 = + i find: a z1¤ b z2¤ c (z1¤ )¤ d (z2¤ )¤ g (z1 ¡ z2 )¤ â đô m z12 h z1¤ ¡ z2¤ i (z1 z2 )¤ â đô o z23 n (z1¤ )2 j z1¤ z2¤ p (z2¤ )3 e (z1 +z2 )¤ µ ¶¤ z1 k z2 f z1¤ + z2¤ z¤ l 1¤ z2 Repeat with z1 and z2 of your choice From and formulate possible rules of conjugates PROPERTIES OF CONJUGATES From Investigation you should have found the following rules for complex conjugates: cyan magenta yellow 95 100 50 75 25 95 100 95 100 50 75 25 95 100 50 75 25 ² 50 (z1 + z2 ) ¤ = z1¤ + z2¤ and (z1 ¡ z2 ) ¤ = z1¤ ¡ z2¤ µ ¶¤ z1 z¤ ¤ ¤ ¤ (z1 z2 ) = z1 £ z2 and = 1¤ , z2 6= z2 z2 75 ² 25 (z ¤ ) ¤ = z ² black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\185IB_HL-2_07.CDR Thursday, 25 October 2007 10:40:14 AM PETERDELL IB_HL-2ed (186) 186 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) ² (z n ) ¤ = (z ¤ )n ² z +z¤ for integers n = 1, and and zz ¤ are real Example 12 Show that (z1 + z2 ) ¤ = z1¤ + z2¤ for all complex numbers z1 and z2 z1¤ = a ¡ bi and z2¤ = c ¡ di ) Let z1 = a + bi and z2 = c + di (z1 + z2 ) ¤ = (a + c) ¡ (b + d)i = a + c ¡ bi ¡ di = a ¡ bi + c ¡ di = z1¤ + z2¤ ) Now z1 + z2 = (a + c) + (b + d)i Example 13 Show that (z1 z2 ) ¤ = z1¤ £ z2¤ for all complex numbers z1 and z2 Let z1 = a + bi and z2 = c + di ) z1 z2 = (a + bi)(c + di) = ac + adi + bci + bdi2 = [ac ¡ bd] + i[ad + bc] Thus (z1 z2 ) ¤ = [ac ¡ bd] ¡ i[ad + bc] (1) Now z1¤ £ z2¤ = (a ¡ bi)(c ¡ di) = ac ¡ adi ¡ bci + bdi2 = [ac ¡ bd] ¡ i[ad + bc] (2) From (1) and (2), (z1 z2 ) ¤ = z1¤ £ z2¤ EXERCISE 7B.4 Show that (z1 ¡ z2 ) ¤ = z1¤ ¡ z2¤ for all complex numbers z1 and z2 Simplify the expression (w ¤ ¡ z) ¤ ¡ (w ¡ 2z ¤ ) using the properties of conjugates It is known that a complex number z satisfies the equation z ¤ = ¡z Show that z is either purely imaginary or zero If z1 = a + bi and z2 = c + di: µ ¶¤ z1 z¤ z1 (in form X + Y i) b show that = 1¤ for all z1 and z2 6= a find z2 z2 z2 µ ¶¤ µ ¶¤ z1 z1 z1¤ = ¤ is to start with £ z2¤ An easier way of proving z2 z2 z2 Show how this can be done, remembering we have already proved that “the conjugate of a product is the product of the conjugates” in Example 13 cyan magenta yellow 95 is purely imaginary or zero 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Prove that for all complex numbers z and w: b zw ¤ ¡ z ¤ w a zw ¤ + z ¤ w is always real black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\186IB_HL-2_07.CDR Thursday, 25 October 2007 10:41:33 AM PETERDELL IB_HL-2ed (187) COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) 187 a If z = a + bi find z in the form X + Y i â đô b Hence, show that z = (z ¤ )2 for all complex numbers z c Repeat a and b but for z instead of z w= a z¡1 where z = a + bi Find the conditions under which: z¤ +1 w is real b w is purely imaginary CONJUGATE GENERALISATIONS Notice that (z1 + z2 + z3 ) ¤ = (z1 + z2 ) ¤ + z3¤ ftreating z1 + z2 as one complex numberg (1) = z1¤ + z2¤ + z3¤ Likewise (z1 + z2 + z3 + z4 ) ¤ = (z1 + z2 + z3 ) ¤ + z4¤ = z1¤ + z2¤ + z3¤ + z4¤ ffrom (1)g Since there is no reason why this process cannot continue for the conjugate of 5, 6, 7, complex numbers we generalise to: (z1 + z2 + z3 + :::::: + zn ) ¤ = z1¤ + z2¤ + z3¤ + :::::: + zn¤ : The process of obtaining the general case from observing the simpler cases when n = 1, 2, 3, 4, is called mathematical induction Proof by the Principle of Mathematical Induction is an exercise that could be undertaken after the completion of Chapter This is a more formal treatment and constitutes a proper proof EXERCISE 7B.5 a b c d Assuming Show that What is the What is the (z1 z2 ) ¤ = z1¤ z2¤ , explain why (z1 z2 z3 ) ¤ = z1¤ z2¤ z3¤ (z1 z2 z3 z4 ) ¤ = z1¤ z2¤ z3¤ z4¤ from a inductive generalisation of your results in a and b? result of letting all zi values be equal to z in c? cyan magenta yellow 95 100 50 75 25 95 for all positive integers n 100 50 75 (z1 + z2 + z3 + ::::: + zn ) ¤ = z1¤ + z2¤ + z3¤ + ::::: + zn¤ and (z1 z2 z3 ::::zn ) ¤ = z1¤ z2¤ z3¤ :::: zn¤ 25 ² (z n ) ¤ = (z ¤ )n ² 95 If z is any complex number then z + z ¤ is real and zz ¤ is real (z ¤ ) ¤ = z If z1 and z2 are any complex numbers then I (z1 ¡ z2 ) ¤ = z1¤ ¡ z2¤ I (z1 + z2 ) ¤ = z1¤ + z2¤ µ ¶¤ z1 z¤ ¤ ¤ ¤ I (z1 z2 ) = z1 z2 I = 1¤ z2 z2 100 50 ² ² ² 75 25 95 100 50 75 25 SUMMARY OF CONJUGATE DISCOVERIES black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\187IB_HL-2_07.CDR Thursday, 25 October 2007 10:44:24 AM PETERDELL IB_HL-2ed (188) 188 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) C REAL POLYNOMIALS Up to this point we have studied linear and quadratic polynomial functions at some depth, with perhaps occasional reference to cubic and quartic polynomials We now turn our attention to general polynomials with real coefficients The degree of a polynomial is its highest power of the variable Polynomials ax + b, a 6= ax2 + bx + c, a 6= ax3 + bx2 + cx + d, a 6= ax4 + bx3 + cx2 + dx + e, a 6= Degree Name linear quadratic cubic quartic a is the leading coefficient and the term not containing the variable x is the constant term A real polynomial has all its coefficients as real numbers (They not p contain i where i = ¡1:) OPERATIONS WITH POLYNOMIALS ADDITION AND SUBTRACTION To add (or subtract) two polynomials we add (or subtract) ‘like terms’ Example 14 If P (x) = x3 ¡ 2x2 + 3x ¡ and Q(x) = 2x3 + x2 ¡ 11 find: a P (x) + Q(x) b P (x) ¡ Q(x) a It is a good idea to place brackets around expressions which are subtracted b P (x) ¡ Q(x) P (x) + Q(x) = x3 ¡ 2x2 + 3x ¡ ¡ [2x3 + x2 ¡ 11] = x ¡ 2x + 3x ¡ ¡ 11 + 2x3 + x2 = x3 ¡ 2x2 + 3x ¡ ¡ 2x3 ¡ x2 + 11 = 3x3 ¡ x2 + 3x ¡ 16 = ¡x3 ¡ 3x2 + 3x + SCALAR MULTIPLICATION To multiply a polynomial by a scalar (constant) we multiply each term by the scalar b cyan magenta yellow b ¡2P (x) 95 100 50 ¡2P (x) = ¡2(x4 ¡ 2x3 + 4x + 7) = ¡2x4 + 4x3 ¡ 8x ¡ 14 75 95 100 50 75 25 95 100 50 75 25 3P (x) = 3(x4 ¡ 2x3 + 4x + 7) = 3x4 ¡ 6x3 + 12x + 21 25 a a 3P (x) 95 If P (x) = x4 ¡ 2x3 + 4x + find: 100 50 75 25 Example 15 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\188IB_HL-2_07.CDR Thursday, 25 October 2007 10:50:38 AM PETERDELL IB_HL-2ed (189) 189 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) POLYNOMIAL MULTIPLICATION To multiply two polynomials, we multiply every term of the first polynomial by every term of the second polynomial and then collect like terms Example 16 If P (x) = x3 ¡ 2x + and Q(x) = 2x2 + 3x ¡ 5, find P (x) Q(x) P (x)Q(x) = (x3 ¡ 2x + 4)(2x2 + 3x ¡ 5) = x3 (2x2 + 3x ¡ 5) ¡ 2x(2x2 + 3x ¡ 5) + 4(2x2 + 3x ¡ 5) = 2x5 + 3x4 ¡ 5x3 ¡ 4x3 ¡ 6x2 + 10x + 8x2 + 12x ¡ 20 = 2x5 + 3x4 ¡ 9x3 + 2x2 + 22x ¡ 20 EXERCISE 7C.1 If P (x) = x2 + 2x + and Q(x) = 4x2 + 5x + 6, find in simplest form: a 3P (x) b P (x) + Q(x) c P (x) ¡ 2Q(x) d P (x)Q(x) If f(x) = x2 ¡ x + and g(x) = x3 ¡ 3x + 5, find in simplest form: a f(x) + g(x) b g(x) ¡ f(x) c 2f (x) + 3g(x) d g(x) + xf(x) e f(x)g(x) f [f (x)]2 Expand and simplify: a (x2 ¡ 2x + 3)(2x + 1) c (x + 2)3 e (2x ¡ 1)4 b d f (x ¡ 1)2 (x2 + 3x ¡ 2) (2x2 ¡ x + 3)2 (3x ¡ 2)2 (2x + 1)(x ¡ 4) NOTE ON SYNTHETIC MULTIPLICATION (OPTIONAL) Polynomial multiplication can be performed using the coefficients only For example, for (x3 + 2x ¡ 5)(2x + 3) we detach coefficients and multiply It is different from the ordinary multiplication of large numbers because we sometimes have negative coefficients and because we not carry tens into the next column 3 x4 x3 x2 2 ¡10 ¡4 x coefficients of x3 + 2x ¡ coefficients of 2x + ¡5 ¡15 So (x3 + 2x ¡ 5)(2x + 3) = 2x4 + 3x3 + 4x2 ¡ 4x ¡ 15 ¡15 constants cyan magenta yellow 50 75 25 b d 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find the following products: a (2x2 ¡ 3x + 5)(3x ¡ 1) c (2x2 + 3x + 2)(5 ¡ x) 95 2 £ 4 (4x2 ¡ x + 2)(2x + 5) (x ¡ 2)2 (2x + 1) 100 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\189IB_HL-2_07.CDR Thursday, 25 October 2007 10:52:49 AM PETERDELL IB_HL-2ed (190) 190 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) e g (x2 ¡ 3x + 2)(2x2 + 4x ¡ 1) (x2 ¡ x + 3)2 h (2x2 + x ¡ 4)2 f i (3x2 ¡ x + 2)(5x2 + 2x ¡ 3) (2x + 5)3 j (x3 + x2 ¡ 2)2 DIVISION OF POLYNOMIALS The division of polynomials is a more difficult process We can divide a polynomial by another polynomial using an algorithm similar to that used for division of whole numbers The division process is only sensible if we divide a polynomial of degree n by another of degree n or less DIVISION BY LINEARS Consider (2x2 + 3x + 4)(x + 2) + If we expand this expression we get (2x2 + 3x + 4)(x + 2) + = 2x3 + 7x2 + 10x + 15 Now consider 2x3 + 7x2 + 10x + 15 divided by x + (2x2 + 3x + 4)(x + 2) + 2x3 + 7x2 + 10x + 15 = x+2 x+2 i.e., ) 2x3 + 7x2 + 10x + 15 (2x2 + 3x + 4)(x + 2) = + x+2 x+2 x+2 ) 2x3 + 7x2 + 10x + 15 + + 3x + = 2x {z } | x+2 x+2 remainder divisor quotient DIVISION ALGORITHM Division may be performed directly using the following algorithm: 2x2 + x+2 2x3 + 7x2 + ¡ (2x3 + 4x2 ) 3x2 + ¡ (3x2 + Step 1: What we multiply x by to get 2x3 ? The answer is 2x2 , and 2x2 (x + 2) = 2x3 + 4x2 3x + 10x + 15 10x 6x) 4x + 15 ¡ (4x + 8) cyan Step 3: Bring down the 10x to obtain 3x2 + 10x Return to Step with the question: “What must we multiply x by to get 3x2 ? ” The answer is 3x, and 3x(x + 2) = 3x2 + 6x etc constants 15 The result can also be achieved by leaving out the variable, as shown alongside magenta yellow 95 100 50 75 25 95 100 50 75 25 15 8) 95 50 10 6) ¡ (4 xs 10 75 25 4) ¡ (3 95 ¡ (2 100 50 75 25 x2 s 100 x3 s Step 2: Subtract 2x3 + 4x2 from 2x3 + 7x2 The answer is 3x2 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\190IB_HL-2_07.CDR Monday, 29 October 2007 12:16:25 PM PETERDELL IB_HL-2ed (191) 191 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) Either way, 2x3 + 7x2 + 10x + 15 = 2x2 + 3x + + , x+2 x+2 where x + 2x2 + 3x + and is called the divisor, is called the quotient, is called the remainder if P (x) is divided by ax + b until a constant remainder R is obtained, In general, P (x) R = Q(x) + ax + b ax + b where ax + b Q(x) R is the divisor, is the quotient, and is the remainder Notice that P (x) = Q(x) £ (ax + b) + R: Example 17 x3 ¡ x2 ¡ 3x ¡ x¡3 Find the quotient and remainder for x2 + 2x + x¡3 x3 ¡ x2 ¡ 3x ¡ ¡ (x3 ¡ 3x2 ) ) quotient is x2 + 2x + and remainder is 2x2 ¡ 3x ¡ (2x2 ¡ 6x) 3x ¡ ¡ (3x ¡ 9) x3 ¡ x2 ¡ 3x ¡ = x2 + 2x + + x¡3 x¡3 Thus So, x3 ¡ x2 ¡ 3x ¡ = (x2 + 2x + 3)(x ¡ 3) + 4: (Check by expanding and simplifying the RHS.) Example 18 x4 + 2x2 ¡ : x+3 x3 ¡ 3x2 + x+3 x4 + 0x3 + 2x2 + ¡ (x4 + 3x3 ) ¡ 3x3 + 2x2 ¡(¡ 3x3 ¡ 9x2 ) 11x2 + ¡ (11x2 + Notice the insertion of 0x3 and 0x Why? 11x ¡ 33 0x ¡ = x3 ¡ 3x2 + 11x ¡ 33 + magenta yellow 98 x+3 So, x4 + 2x ¡ = (x3 ¡ 3x2 + 11x ¡ 33)(x + 3) + 98 95 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ¡ 33x ¡ ¡ (¡ 33x ¡ 99) 98 cyan x4 + 2x ¡ x+3 ) 0x 33x) 100 Perform the division black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\191IB_HL-2_07.CDR Thursday, 25 October 2007 11:43:39 AM PETERDELL IB_HL-2ed (192) 192 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) EXERCISE 7C.2 Find the quotient and remainder for: x2 + 2x ¡ x+2 a Perform the divisions: x2 ¡ 3x + a x¡4 2x3 + 3x2 ¡ 3x ¡ 2x + d Perform the divisions: x2 + a x¡2 x3 + 2x2 ¡ 5x + x¡1 d b x2 ¡ 5x + x¡1 c 2x3 + 6x2 ¡ 4x + x¡2 b x2 + 4x ¡ 11 x+3 c 2x2 ¡ 7x + x¡2 e 3x3 + 11x2 + 8x + 3x ¡ f 2x4 ¡ x3 ¡ x2 + 7x + 2x + b 2x2 + 3x x+1 c 3x2 + 2x ¡ x+2 e 2x3 ¡ x x+4 f x3 + x2 ¡ x¡2 DIVISION BY QUADRATICS As with division by linears we can use the division algorithm to divide polynomials by quadratics The division process stops when the remainder has degree less than that of the divisor, P (x) ex + f i.e., = Q(x) + 2 a + bx + c ax + bx + c The remainder will be linear if e 6= or constant if e = Example 19 Find the quotient and remainder for x4 + 4x3 ¡ x + x2 ¡ x + x2 +5x +4 x2 ¡ x + x4 +4x3 +0x2 ¡x +1 ¡(x4 ¡x3 + x2 ) 5x3 ¡ x2 ¡x ¡(5x3 ¡5x2 +5x) 4x2 ¡6x +1 ¡(4x2 ¡4x +4) ) quotient is x2 + 5x + and remainder is ¡2x ¡ So, x4 + 4x3 ¡ x + = (x2 ¡ x + 1)(x2 + 5x + 4) ¡ 2x ¡ ¡2x ¡3 EXERCISE 7C.3 Find the quotient and remainder for: magenta yellow 95 50 75 25 100 3x3 + x ¡ x2 + c 95 100 50 75 25 3x2 ¡ x x2 ¡ b 95 100 50 75 25 95 100 50 75 25 cyan x3 + 2x2 + x ¡ x2 + x + a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\192IB_HL-2_07.CDR Thursday, 25 October 2007 11:45:14 AM PETERDELL d x¡4 x2 + 2x ¡ IB_HL-2ed (193) 193 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) Carry out the following divisions and also write each in the form P (x) = D(x)Q(x) + R(x): a x2 ¡ x + x2 + x + b d 2x3 ¡ x + (x ¡ 1)2 e x3 +2 c x4 + 3x2 + x ¡ x2 ¡ x + x4 (x + 1)2 f x4 ¡ 2x3 + x + (x ¡ 1)(x + 2) x2 P (x) = (x ¡ 2)(x2 + 2x + 3) + What is the quotient and remainder when P (x) is divided by x ¡ 2? Given that f (x) = (x ¡ 1)(x + 2)(x2 ¡ 3x + 5) + 15 ¡ 10x, find the quotient and remainder when f (x) is divided by x2 + x ¡ PRINTABLE SECTION SYNTHETIC DIVISION (OPTIONAL) Click on the icon for an exercise involving a synthetic division process for the division of a polynomial by a linear D ROOTS, ZEROS AND FACTORS ROOTS AND ZEROS A zero of a polynomial is a value of the variable which makes the polynomial equal to zero The roots of a polynomial equation are values of the variable which satisfy or are solutions to the equation The roots of P (x) = are the zeros of P (x) and the x-intercepts of the graph of y = P (x) If x = 2, x3 + 2x2 ¡ 3x ¡ 10 = + ¡ ¡ 10 =0 ) is a zero of x3 + 2x2 ¡ 3x ¡ 10 and is a root of x3 + 2x2 ¡ 3x ¡ 10 = 0: ® is a zero of polynomial P (x) , ® is a root (or solution) of P (x) = Note: P (®) = , P (®) = Example 20 a x2 ¡ 4x + 53 Find the zeros of: a b z + 3z b We wish to find x such that x2 ¡ 4x + 53 = p § 16 ¡ 4(1)(53) ) x= § 14i ) x= = § 7i We wish to find z such that z + 3z = ) z(z + 3) = p p ) z(z + i 3)(z ¡ i 3) = p ) z = or §i cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 If P (x) = (x + 1)(2x ¡ 1)(x + 2), then (x + 1), (2x ¡ 1) and (x + 2) are its linear factors black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\193IB_HL-2_07.CDR Thursday, 25 October 2007 11:58:19 AM PETERDELL IB_HL-2ed (194) 194 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) Likewise P (x) = (x + 3)2 (2x + 3) has been factorised into linear factors, one of which is repeated In general, (x¡®) is a factor of polynomial P (x) , there exists a polynomial Q(x) such that P (x) = (x ¡ ®)Q(x): Example 21 a z + 4z + Factorise: a b 2z + 5z ¡ 3z ? z + 4z + is zero when b p p ¡4 § 16 ¡ 4(1)(9) = ¡2 § i ) z= p p ) z + 4z + = (z ¡ [¡2 + i 5])(z ¡ [¡2 ¡ i 5]) p p = (z + ¡ i 5)(z + + i 5) Example 22 Find all cubic polynomials with zeros The zeros ¡3 § 2i have 2, 2z + 5z ¡ 3z = z(2z + 5z ¡ 3) = z(2z ¡ 1)(z + 3) ¡3 § 2i: sum = ¡3 + 2i ¡ ¡ 2i = ¡6 and product = (¡3 + 2i)(¡3 ¡ 2i) = 13 and ) come from the quadratic factor z + 6z + 13 comes from the linear factor 2z ¡ ) P (z) = a(2z ¡ 1)(z + 6z + 13), a 6= Example 23 Find all quartic polynomials with zeros of 2, ¡ 13 , ¡1 § The zeros ¡1 § p p p ¡ ¡ = ¡2 and p p product = (¡1 + 5)(¡1 ¡ 5) = ¡4 p have sum = ¡1 + and ) come from the quadratic factor z + 2z ¡ The zeros and ¡ 13 come from the linear factors (z ¡ 2) and (3z + 1) ) P (z) = a(z ¡ 2)(3z + 1)(z + 2z ¡ 4), a 6= cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 ¡2z(z ¡ 2z + 2) = z = 3z + 10 95 c f 100 (2x + 1)(x2 + 3) = z + 5z = 50 b e 75 Find the roots of: a 5x2 = 3x + d x3 = 5x 25 z ¡ 6z + z + 4z ¡ c f x2 + 6x + 10 z + 2z 95 b e 100 50 Find the zeros of: a 2x2 ¡ 5x ¡ 12 d x3 ¡ 4x 75 25 EXERCISE 7D.1 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\194IB_HL-2_07.CDR Thursday, 25 October 2007 12:00:36 PM PETERDELL IB_HL-2ed (195) 195 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) Find the linear factors of: a 2x2 ¡ 7x ¡ 15 d 6z ¡ z ¡ 2z b e z ¡ 6z + 16 z ¡ 6z + c f x3 + 2x2 ¡ 4x z4 ¡ z2 ¡ If P (x) = a(x ¡ ®)(x ¡ ¯)(x ¡ °) then ®, ¯ and ° are its zeros Check that the above statement is correct by finding P (®), P (¯) and P (°) Find all cubic polynomials with zeros of: a §2, b ¡2, §i c Find all quartic polynomials with zeros of: p p b 2, ¡1, §i c a §1, § 3, ¡1 § i d ¡1, ¡2 § p § 3, § i d 2§ p p 5, ¡2 § 3i POLYNOMIAL EQUALITY Polynomials are equal if and only if they generate the same y-value for each x-value This means that graphs of equal polynomials should be identical Two polynomials are equal if and only if they have the same degree (order) and corresponding terms have equal coefficients For example, if 2x3 + 3x2 ¡ 4x + = ax3 + bx2 + cx + d, then a = 2, b = 3, c = ¡4 and d = EQUATING COEFFICIENTS If we know that two polynomials are equal then we can ‘equate coefficients’ in order to find unknown coefficients Example 24 Find constants a, b and c given that: 6x3 + 7x2 ¡ 19x + = (2x ¡ 1)(ax2 + bx + c) for all x If 6x3 + 7x2 ¡ 19x + = (2x ¡ 1)(ax2 + bx + c) then 6x3 + 7x2 ¡ 19x + = 2ax3 + 2bx2 + 2cx ¡ ax2 ¡ bx ¡ c ) 6x3 + 7x2 ¡ 19x + = 2ax3 + [2b ¡ a]x2 + [2c ¡ b]x ¡ c Since this is true for all x, we equate coefficients ) |2a{z = 6}, 2b a = 7}, |2c ¡ b{z= ¡19} and |7 ={z¡c} | ¡{z x2 s xs constants x3 s cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a = and c = ¡7 and consequently |2b ¡ = and {z¡ 14 ¡ b = ¡19} b=5 in both equations So, a = 3, b = and c = ¡7 ) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\195IB_HL-2_07.cdr Thursday, 25 October 2007 12:04:16 PM PETERDELL IB_HL-2ed (196) 196 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) Example 25 Find a and b if z + = (z + az + 3)(z + bz + 3) for all z z + = (z + az + 3)(z + bz + 3) for all z z + = z + bz + 3z +az + abz + 3az +3z + 3bz + ) ) z + = z + [a + b]z + [ab + 6]z + [3a + 3b]z + a + b = (1) < ab + = (2) Equating coefficients gives : 3a + 3b = (3) for all z When solving more equations than unknowns simultaneously, we must check that any solutions fit all equations.¡ If they not, there are no solutions fz sg fz sg fz sg From (1) and (3) we see that b = ¡a So, in (2) a(¡a) + = ) a2 = p p ) a = § and so b = ¨ p p p p ) a = 6, b = ¡ or a = ¡ 6, b = Example 26 x + is a factor of P (x) = x3 + ax2 ¡ 7x + Find a and the other factors Since x + is a factor, x3 + ax2 ¡ 7x + = (x + 3)(x2 + bx + 2) x3 for some constant b = x3 + bx2 + 2x + 3x2 + 3bx + = x3 + [b + 3]x2 + [3b + 2]x + 3b + = ¡7 and a = b + ) b = ¡3 and ) a = Equating coefficients gives P (x) = (x + 3)(x2 ¡ 3x + 2) = (x + 3)(x ¡ 1)(x ¡ 2) ) Example 27 2x + and x ¡ are factors of 2x4 + ax3 ¡ 3x2 + bx + Find a and b and all zeros of the polynomial Since 2x + and x ¡ are factors, 2x4 + ax3 ¡ 3x2 + bx + = (2x + 3)(x ¡ 1)(a quadratic) = (2x2 + x ¡ 3)(x2 + cx ¡ 1) for some c cyan magenta yellow 95 100 50 75 25 2x4 ¡3 = ¡2 + c ¡ ) c=2 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Equating coefficients of x2 gives: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\196IB_HL-2_07.CDR Thursday, 25 October 2007 12:17:06 PM PETERDELL IB_HL-2ed (197) COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) Equating coefficients of x3 : Equating coefficients of x: 197 a = 2c + = + = b = ¡1 ¡ 3c ) b = ¡1 ¡ = ¡7 ) P (x) = (2x + 3)(x ¡ 1)(x2 + 2x ¡ 1) which has zeros of: p p p ¡2 § ¡ 4(1)(¡1) ¡2 § 2 ¡ , and = = ¡1 § 2 p ) the zeros are ¡ , and ¡1 § EXERCISE 7D.2 Find constants a, b and c given that: a 2x2 + 4x + = ax2 + [2b ¡ 6]x + c for all x b 2x3 ¡ x2 + = (x ¡ 1)2 (2x + a) + bx + c for all x Find a and b if: a z + = (z + az + 2)(z + bz + 2) for all z b 2z + 5z + 4z + 7z + = (z + az + 2)(2z + bz + 3) for all z Show that z + 64 can be factorised into two real quadratic factors of the form z + az + and z + bz + 8, but cannot be factorised into two real quadratic factors of the form z + az + 16 and z + bz + 4 Find real numbers a and b such that x4 ¡ 4x2 + 8x ¡ = (x2 + ax + 2)(x2 + bx ¡ 2), and hence solve the equation x4 + 8x = 4x2 + a 2z ¡ is a factor of 2z ¡ z + az ¡ Find a and all zeros of the cubic b 3z + is a factor of 3z ¡ z + [a + 1]z + a Find a and all the zeros of the cubic a Both 2x + and x ¡ are factors of P (x) = 2x4 + ax3 + bx2 ¡ 12x ¡ Find a and b and all zeros of P (x) b x + and 2x ¡ are factors of 2x4 + ax3 + bx2 + ax + Find a and b and hence determine all zeros of the quartic a x3 + 3x2 ¡ 9x + c has two identical linear factors Prove that c is either or ¡27 and factorise the cubic into linear factors in each case b 3x3 + 4x2 ¡ x + m has two identical linear factors Find m and find the zeros of the polynomial in all possible cases THE REMAINDER THEOREM Consider the cubic polynomial P (x) = x3 + 5x2 ¡ 11x + If we divide P (x) by x ¡ 2, we find that: x3 + 5x2 ¡ 11x + = x2 + 7x + + x¡2 x¡2 remainder cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 so on division by x ¡ 2, the remainder is 9: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\197IB_HL-2_07.CDR Thursday, 25 October 2007 12:33:09 PM PETERDELL IB_HL-2ed (198) 198 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) P (2) = + 20 ¡ 22 + = 9, which is the remainder Notice also that By considering other examples like the one above we formulate the Remainder theorem THE REMAINDER THEOREM When a polynomial P (x) is divided by x ¡ k until a constant remainder R is obtained then R = P (k) P (x) = Q(x)(x ¡ k) + R P (k) = Q(k) £ + R P (k) = R By the division algorithm, Now, letting x = k, ) Proof: Example 28 Use the Remainder theorem to find the remainder when x4 ¡ 3x3 + x ¡ is divided by x + If P (x) = x4 ¡ 3x3 + x ¡ 4, then P (¡2) = (¡2)4 ¡ 3(¡2)3 + (¡2) ¡ = 16 + 24 ¡ ¡ = 34 ) when P (x) is divided by x + 2, the remainder is 34 fRemainder theoremg Example 29 When P (x) is divided by x2 ¡ 3x + the quotient is x2 + x ¡ and the remainder R(x) is unknown However, when P (x) is divided by x ¡ the remainder is 29 and when divided by x + the remainder is ¡16 Find R(x) in the form ax + b As the divisor is x2 ¡ 3x + 7, +b P (x) = (x2 + x ¡ 1) (x2 ¡ 3x + 7) + ax{z } }| } | | {z {z Q(x) D(x) R(x) But P (2) = 29 and P (¡1) = ¡16 fRemainder theoremg ) (22 + ¡ 1)(22 ¡ + 7) + 2a + b = 29 and ((¡1) + (¡1) ¡ 1)((¡1)2 ¡ 3(¡1) + 7) + (¡a + b) = ¡16 ) (5)(5) + 2a + b = 29 (¡1)(11) ¡ a + b = ¡16 ) 2a + b = ¡a + b = ¡5 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Solving these gives a = and b = ¡2, so R(x) = 3x ¡ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\198IB_HL-2_07.CDR Thursday, 25 October 2007 1:26:41 PM PETERDELL IB_HL-2ed (199) 199 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) When using the Remainder theorem, it is important to realise that ² P (x) = (x + 2)Q(x) + ² P (¡2) = ² ‘P (x) divided by x + leaves a remainder of 3’ are all equivalent statements EXERCISE 7D.3 Write two equivalent statements for: b If P (x) = (x + 3)Q(x) ¡ 8, then a If P (2) = 7, then c If P (x) divided by x ¡ has a remainder of 11 then Without performing division, find the remainder when: a x3 + 2x2 ¡ 7x + is divided by x ¡ b x4 ¡ 2x2 + 3x ¡ is divided by x + Find a given that: a when x3 ¡ 2x + a is divided by x ¡ 2, the remainder is b when 2x3 + x2 + ax ¡ is divided by x + 1, the remainder is ¡8 Find a and b given that when x3 + 2x2 + ax + b is divided by x ¡ the remainder is 4, and when divided by x + the remainder is 16 2xn +ax2 ¡6 leaves a remainder of ¡7 when divided by x¡1, and 129 when divided by x + Find a and n given that n Z + When P (z) is divided by z ¡ 3z + the remainder is 4z ¡ Find the remainder when P (z) is divided by: a z¡1 b z ¡ When P (z) is divided by z + the remainder is ¡8 and when divided by z ¡ the remainder is Find the remainder when P (z) is divided by (z ¡ 3)(z + 1) ¶ µ P (b) ¡ P (a) If P (x) is divided by (x ¡ a)(x ¡ b), £ (x ¡ a) + P (a) prove that the remainder is: b¡a THE FACTOR THEOREM , k is a zero of P (x) (x ¡ k) is a factor of P (x) Proof: k is a zero of P (x) , P (k) = ,R=0 , P (x) = Q(x)(x ¡ k) , (x ¡ k) is a factor of P (x) fdefinition of a zerog fRemainder theoremg fdivision algorithmg fdefinition of factorg cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The Factor theorem says that if is a zero of P (x) then (x ¡ 2) is a factor of P (x) and vice versa black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\199IB_HL-2_07.CDR Thursday, 25 October 2007 1:35:12 PM PETERDELL IB_HL-2ed (200) 200 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) Example 30 Find k given that x ¡ is a factor of x3 + kx2 ¡ 3x + Hence, fully factorise x3 + kx2 ¡ 3x + Let P (x) = x3 + kx2 ¡ 3x + By the Factor theorem, as x ¡ is a factor then P (2) = ) (2)3 + k(2)2 ¡ 3(2) + = ) + 4k = and so k = ¡2 Now x3 ¡ 2x2 ¡ 3x + = (x ¡ 2)(x2 + ax ¡ 3) for some constant a Equating coefficients of x2 gives: Equating coefficients of x gives: ¡2 = ¡2 + a ¡3 = ¡2a ¡ x3 ¡ 2x2 ¡ 3x + = (x ¡ 2)(x2 ¡ 3) p p = (x ¡ 2)(x + 3)(x ¡ 3) ) or Using synthetic division ) i.e., a = i.e., a = k ¡3 2k + 4k + k+2 2k + 4k + P (2) = 4k + and since P (2) = 0, k = ¡2 Now P (x) = (x ¡ 2)(x2 + [k + 2]x + [2k + 1]) = (x ¡ 2)(x2 ¡ 3) p p = (x ¡ 2)(x + 3)(x ¡ 3) EXERCISE 7D.4 Find k and hence factorise the polynomial if: a 2x3 + x2 + kx ¡ has a factor of x + b x4 ¡ 3x3 ¡ kx2 + 6x has a factor of x ¡ Find a and b given that 2x3 + ax2 + bx + has factors of x ¡ and x + a is a zero of P (z) = z ¡ z + [k ¡ 5]z + [k2 ¡ 7] Find k and hence find all zeros of P (z) b Show that z ¡ is a factor of P (z) = z + mz2 + (3m ¡ 2)z ¡ 10m ¡ for all values of m For what values of m is (z ¡ 2)2 a factor of P (z)? a Consider P (x) = x3 ¡ a3 where a is real i Find P (a) What is the significance of this result? ii Factorise x3 ¡ a3 as the product of a real linear and a quadratic factor b Now consider P (x) = x3 + a3 , where a is real i Find P (¡a) What is the significance of this result? ii Factorise x3 + a3 as the product of a real linear and a quadratic factor cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 50 75 25 100 a Prove that “x + is a factor of xn + , n is odd.” b Find the real number a such that x ¡ ¡ a is a factor of P (x) = x3 ¡ 3ax ¡ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\200IB_HL-2_07.CDR Thursday, 25 October 2007 1:39:17 PM PETERDELL IB_HL-2ed (201) 201 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) E GRAPHING POLYNOMIALS In this section we are obviously only concerned with graphing real polynomials Remember that these are polynomials for which all coefficients are real Use of a graphics calculator or the graphing package provided will help in this section INVESTIGATION CUBIC GRAPHS Every cubic polynomial can be categorised into one of four types: Type Type Type Type 1: 2: 3: 4: Three real, distinct zeros: P (x) = a(x ¡ ®)(x ¡ ¯)(x ¡ °), a 6= Two real zeros, one repeated: P (x) = a(x ¡ ®)2 (x ¡ ¯), a = One real zero repeated three times: P (x) = a(x ¡ ®) , a 6= One real and two imaginary zeros: P (x) = (x ¡ ®)(ax2 + bx + c), ¢ = b2 ¡ 4ac < 0, a 6= What to do: Experiment with the graphs of Type cubics Clearly state the effect of changing both the size and sign of a What is the geometrical significance of ®, ¯, and °? Experiment with the graphs of Type cubics What is the geometrical significance of the squared factor? Experiment with the graphs of Type cubics Do not forget to GRAPHING PACKAGE consider a¡>¡0 and a¡<¡0 What is the geometrical significance of ®? Experiment with the graphs of Type cubics What is the geometrical significance of ® and the quadratic factor which has imaginary zeros? From Investigation you should have discovered that: ² If a > 0, the graph has shape or If a < it is or ² All cubics are continuous smooth curves ² Every cubic polynomial must cut the x-axis at least once, and so has at least one real zero ² For a cubic of the form P (x) = a(x ¡ ®)(x ¡ ¯)(x ¡ °) the graph has three distinct x-intercepts corresponding to the three real, distinct zeros ®, ¯ and ° The graph crosses over or cuts the x-axis at these points, as shown opposite ² For a cubic of the form P (x) = a(x ¡ ®)2 (x ¡ ¯) the graph touches the x-axis at the repeated zero ® and cuts it at the other x-intercept ¯, as shown opposite cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ² For a cubic of the form P (x) = a(x ¡ ®)3 , the graph has only one x-intercept, ® The graph is horizontal at this point, and the x-axis is a tangent to the curve, even though the curve crosses over it black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\201IB_HL-2_07.CDR Monday, 29 October 2007 12:29:26 PM PETERDELL x x x IB_HL-2ed (202) 202 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) ² For a cubic of the form P (x) = (x ¡ ®)(ax2 + bx + c) where ¢ < 0, there is only one x-intercept, ® The graph cuts the x-axis at this point The other two zeros are imaginary and so not show up on the graph x Example 31 y y a Find the equation of the cubic with graph: b x -1 -3 x We_ -8 The x-intercepts are ¡1, 2, ) y = a(x + 1)(x ¡ 2)(x ¡ 4) But when x = 0, y = ¡8 ) a(1)(¡2)(¡4) = ¡8 ) a = ¡1 a Touching at 23 indicates a squared factor (3x ¡ 2)2 Other x-intercept is ¡3, so y = a(3x ¡ 2)2 (x + 3) b But when x = 0, y = So, a(¡2)2 (3) = and ) a = So, y = ¡(x + 1)(x ¡ 2)(x ¡ 4) ² Note: So, y = 12 (3x ¡ 2)2 (x + 3) If an x-intercept is not given, use P (x) = (x ¡ k)2 (ax + b) : | {z } most general form of a linear x Using P (x) = a(x ¡ k) (x + b) is more complicated ² ? If there is clearly only one x-intercept and that is given, use P (x) = (x ¡ k) (ax2 + bx + c) : | {z } k x k most general form of a quadratic What can you say about this quadratic? Either graph is possible Example 32 Find the equation of the cubic which cuts the x-axis at 2, ¡3 and ¡4 and passes through the point (1, ¡40) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The zeros are 2, ¡3 and ¡4, so y = a(x ¡ 2)(x + 3)(x + 4), a 6= But when x = 1, y = ¡40 ) a(¡1)(4)(5) = ¡40 ) ¡20a = ¡40 ) a=2 So, the equation is y = 2(x ¡ 2)(x + 3)(x + 4): black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\202IB_HL-2_07.CDR Monday, 29 October 2007 12:32:27 PM PETERDELL IB_HL-2ed (203) 203 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) EXERCISE 7E.1 What is the geometrical significance of: a a single factor in P (x), such as (x ¡ ®) b a squared factor in P (x), such as (x ¡ ®)2 c a cubed factor in P (x), such as (x ¡ ®)3 ? Find the equation of the cubic with graph: a b y c y y -4 12 \Qw_ -\Qw_ x x -3 x -1 -12 d e y f y y -3 -5 -2 x -2 -\Qw_ x -12 -5 x -4 Find the equation of the cubic whose graph: a cuts the x-axis at 3, 1, ¡2 and passes through (2, ¡4) b cuts the x-axis at ¡2, and 12 and passes through (¡3, ¡21) c touches the x-axis at 1, cuts the x-axis at ¡2 and passes through (4, 54) d touches the x-axis at ¡ 23 , cuts the x-axis at and passes through (¡1, ¡5) Match the given graphs to the corresponding cubic function: b y = ¡(x + 1)(x ¡ 2)(x ¡ 4) a y = 2(x ¡ 1)(x + 2)(x + 4) c y = (x ¡ 1)(x ¡ 2)(x + 4) d y = ¡2(x ¡ 1)(x + 2)(x + 4) e y = ¡(x ¡ 1)(x + 2)(x + 4) f y = 2(x ¡ 1)(x ¡ 2)(x + 4) A B C y y y x -4 -1 x 16 x -8 -4 D E y F y y 16 x -2 -4 x -2 -4 -2 x -4 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 -16 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\203IB_HL-2_07.CDR Thursday, 25 October 2007 2:04:32 PM PETERDELL IB_HL-2ed (204) 204 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) Find the equation of a real cubic polynomial which cuts: a the x-axis at 12 and ¡3, cuts the y-axis at 30 and passes through (1, ¡20) b the x-axis at 1, touches the x-axis at ¡2 and cuts the y-axis at (0, 8) c the x-axis at 2, the y-axis at ¡4 and passes through (1, ¡1) and (¡1, ¡21) INVESTIGATION QUARTIC GRAPHS There are considerably more possible factor types to consider for quartic functions We will consider quartics containing certain types of factors GRAPHING PACKAGE What to do: Experiment with quartics which have: a four different real linear factors b a squared real linear factor and two different real linear factors c two squared real linear factors d a cubed factor and one real linear factor e a real linear factor raised to the fourth power f one real quadratic factor with ¢ < and two real linear factors g two real quadratic factors each with ¢ < From Investigation you should have discovered that: For a quartic polynomial in which a is the coefficient of x4 : ² I If a > the graph opens upwards I If a < the graph opens downwards ² If a quartic with a > is fully factorised into real linear factors, for: I a single factor (x ¡ ®), the graph cuts the x-axis at ® e.g I a squared factor (x ¡ ®)2 , the graph touches the x-axis at ® e.g y a y a x x I a cubed factor (x ¡ ®)3 , the graph cuts the x-axis at ®, but is ‘flat’ at ® e.g y I a quadruple factor (x ¡ ®)4 , the graph touches the x-axis but is ‘flat’ at that point a e.g cyan magenta yellow 95 100 50 75 25 95 y a x 100 50 75 25 95 100 50 75 25 95 100 50 75 25 x black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\204IB_HL-2_07.CDR Monday, 12 November 2007 9:18:09 AM PETERDELL IB_HL-2ed (205) 205 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) ² ² If a quartic with a > has one real quadratic factor with ¢ < we could have: If a quartic with a > has two real quadratic factors both with ¢ < we have: y The graph does not meet the x-axis at all y x x Example 33 The graph touches the x-axis at ¡1 and cuts it at ¡3 and 3, so y = a(x + 1)2 (x + 3)(x ¡ 3) But when x = 0, y = ¡3 Find the equation of the quartic with graph: y -1 -3 ) ¡3 = a(1)2 (3)(¡3) ) ¡3 = ¡9a ) a = 13 x -3 ) y = 13 (x + 1)2 (x + 3)(x ¡ 3) Example 34 Find the quartic which touches the x-axis at 2, cuts it at ¡3, and also passes through (1, ¡12) and (3, 6) (x ¡ 2)2 is a factor as the graph touches the x-axis at (x + 3) is a factor as the graph cuts the x-axis at ¡3 So P (x) = (x ¡ 2)2 (x + 3)(ax + b) where a and b are constants ) (¡1)2 (4)(a + b) = ¡12 ) 12 (6)(3a + b) = Now P (1) = ¡12, and P (3) = 6, i.e., a + b = ¡3 (1) i.e., 3a + b = (2) Solving (1) and (2) simultaneously gives a = 2, b = ¡5 ) P (x) = (x ¡ 2)2 (x + 3)(2x ¡ 5) EXERCISE 7E.2 Find the equation of the quartic with graph: a b y -3 c y -1 x We_ cyan magenta -2 -1 x -6 x yellow 95 100 50 75 25 95 100 50 -16 75 25 95 100 50 75 25 95 100 50 75 25 -1 y black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\205IB_HL-2_07.CDR Thursday, 25 October 2007 2:13:44 PM PETERDELL IB_HL-2ed (206) 206 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) d e y f y -1 -3 Ew_ -1 y (-3' 54) x x -9 -2 -16 x Match the given graphs to the corresponding quartic functions: b y = ¡2(x ¡ 1)2 (x + 1)(x + 3) a y = (x ¡ 1)2 (x + 1)(x + 3) d y = (x ¡ 1)(x + 1)2 (x ¡ 3) c y = (x ¡ 1)(x + 1)2 (x + 3) f y = ¡(x ¡ 1)(x + 1)(x ¡ 3)2 e y = ¡ 13 (x ¡ 1)(x + 1)(x + 3)2 A B C y y -1 -3 D -3 x -1 E y -1 x -3 -1 F y -1 y x y x -3 x -1 x Find the equation of the quartic whose graph: a cuts the x-axis at ¡4 and 12 , touches it at 2, and passes through the point (1, 5) b touches the x-axis at 23 and ¡3, and passes through the point (¡4, 49) c cuts the x-axis at § 12 and §2, and passes through the point (1, ¡18) d touches the x-axis at 1, cuts the y-axis at ¡1, and passes through the points (¡1, ¡4) and (2, 15) DISCUSSION GENERAL POLYNOMIALS What happens to P (x) = an xn + an¡1 xn¡1 + ::::: + a1 x + a0 , an = a polynomial of degree n, n N as jxj ! 1, i.e., as x ! ¡1 and as x ! +1? Notice that as jxj ! 1, the term an xn dominates the value of P (x) and the values of the other terms become insignificant P (x) ! +1, as x ! +1 and P (x) ! +1, as x ! ¡1 So, if an > and n is even, magenta yellow 95 100 50 75 25 95 100 50 75 an > and n is odd an < and n is even an < and n is odd 25 95 100 50 75 25 95 100 50 75 25 cyan ² ² ² Discuss the cases where: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\206IB_HL-2_07.CDR Thursday, 25 October 2007 2:18:46 PM PETERDELL IB_HL-2ed (207) COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) 207 We have already seen that every real cubic polynomial must cut the x-axis at least once, and so has at least one real zero If the exact value of the zero is difficult to find, we can use technology to help us We can then factorise the cubic as a linear factor times a quadratic, and if necessary use the quadratic formula to find the other zeros This method is particularly useful if we have one rational zero and two irrational zeros that are radical conjugates Example 35 Find all zeros of P (x) = 3x3 ¡ 14x2 + 5x + Using the calculator we search for any rational zero In this case x ¼ 0:666 667 or 0:6 indicates x = 23 is a zero and ) (3x ¡ 2) is a factor So, 3x3 ¡ 14x2 + 5x + = (3x ¡ 2)(x2 + ax ¡ 1) = 3x3 + [3a ¡ 2]x2 + [¡3 ¡ 2a]x + 3a ¡ = ¡14 and ¡3 ¡ 2a = ) 3a = ¡12 and ¡2a = ) a = ¡4 ) P (x) = (3x ¡ 2)(x2 ¡ 4x ¡ 1) p which has zeros 23 and § fquadratic formulag Equating coefficients: TI C GRAPHING PACKAGE Example 36 Find all roots of 6x3 + 13x2 + 20x + = x ¼ ¡0:166 666 67 = 16 is a zero, so (6x + 1) is a factor ) (6x + 1)(x2 + ax + 3) = for some constant a Equating coefficients of x2 : + 6a = 13 ) 6a = 12 ) a = Equating coefficients of x: a + 18 = 20 X p ) (6x + 1)(x + 2x + 3) = and x = ¡ 16 or ¡1 § i fquadratic formulag For a quartic polynomial P (x) we first need to establish if there are any x-intercepts at all If there are not then the polynomial must have four complex zeros If there are x-intercepts then we can try to identify linear or quadratic factors EXERCISE 7E.3 cyan magenta yellow 95 x3 ¡ 3x2 + 4x ¡ 2x3 ¡ x2 + 20x ¡ 10 x4 ¡ 6x3 + 22x2 ¡ 48x + 40 100 50 75 25 b d f 95 100 50 75 25 95 100 50 all zeros of: x3 ¡ 3x2 ¡ 3x + 2x3 ¡ 3x2 ¡ 4x ¡ 35 4x4 ¡ 4x3 ¡ 25x2 + x + 75 25 95 100 50 75 25 Find a c e black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\207IB_HL-2_07.CDR Tuesday, 29 January 2008 9:23:43 AM PETERDELL IB_HL-2ed (208) 208 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) Find a c e the roots of: x3 + 2x2 + 3x + = x3 ¡ 6x2 + 12x ¡ = x4 ¡ x3 ¡ 9x2 + 11x + = Factorise into linear factors: a x3 ¡ 3x2 + 4x ¡ c 2x3 ¡ 9x2 + 6x ¡ e 4x3 ¡ 8x2 + x + g 2x4 ¡ 3x3 + 5x2 + 6x ¡ b d f 2x3 + 3x2 ¡ 3x ¡ = 2x3 + 18 = 5x2 + 9x 2x4 ¡ 13x3 + 27x2 = 13x + 15 b d f h x3 + 3x2 + 4x + 12 x3 ¡ 4x2 + 9x ¡ 10 3x4 + 4x3 + 5x2 + 12x ¡ 12 2x3 + 5x2 + 8x + 20 The following cubics will not factorise Find their zeros using technology b x3 + x2 ¡ 7x ¡ a x3 + 2x2 ¡ 6x ¡ F THEOREMS FOR REAL POLYNOMIALS The following theorems are formal statements of discoveries we have made: ² Unique Factorisation theorem Every real polynomial of degree n can be factorised into n complex linear factors, some of which may be repeated Factor theorem k is a zero of P (x) , (x ¡ k) is a factor of P (x) Every real polynomial can be expressed as a product of real linear and real irreducible quadratic factors (where ¢ < 0) If p + qi (q 6= 0) is a zero of a real polynomial then its complex conjugate p ¡ qi is also a zero Every real polynomial of odd degree has at least one real zero All real polynomials of degree n have n zeros, some of which may be repeated These zeros are real and/or complex zeros that occur in conjugate pairs ² ² ² ² ² Example 37 If ¡3 + i is a zero of P (x) = ax3 + 9x2 + ax ¡ 30 where a is real, find a and hence find all zeros of the cubic As P (x) is real, both ¡3 + i and ¡3 ¡ i are zeros These have sum of ¡6 and product of (¡3 + i)(¡3 ¡ i) = 10, so the zeros ¡3 § i come from the quadratic x2 + 6x + 10 Consequently, ax3 + 9x2 + ax ¡ 30 = (x2 + 6x + 10)(ax ¡ 3) -30 axC cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 To find a we equate coefficients of x2 and x ) = 6a ¡ and a = 10a ¡ 18 and a = in both cases ) a = and the other two zeros are ¡3 ¡ i and 32 the linear factor is (ax ¡ 3) i.e., (2x ¡ 3) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\208IB_HL-2_07.CDR Monday, 12 November 2007 9:26:06 AM PETERDELL IB_HL-2ed (209) COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) 209 Example 38 One zero of ax3 + [a + 1]x2 + 10x + 15, a R , is purely imaginary Find a and the zeros of the polynomial Let the purely imaginary zero be bi, b 6= P (x) is real since its coefficients are all real, and so ¡bi is also a zero For bi and ¡bi, their sum = and their product = ¡b2 i2 = b2 ) these two zeros come from x2 + b2 15 So, ax3 + [a + 1]x2 + 10x + 15 = (x2 + b2 )(ax + ) b axC · = ax3 + 15 ¸ 15 x + b2 ax + 15 b2 15 (1) and b2 a = 10 (2) b2 ) b2 a + b2 = 15 fusing (1)g ) 10 + b2 = 15 fusing (2)g p ) b2 = and so b = § Consequently a + = In (2), as b2 = 5, 5a = 10 ) a = 15 or 2x + b2 p a = and the zeros are §i 5, ¡ 32 The linear factor is ax + ) EXERCISE 7F Find all third degree real polynomials with zeros of ¡ 12 and ¡ 3i p(x) is a real cubic polynomial in which p(1) = p(2 + i) = and p(0) = ¡20 Find p(x) in expanded form ¡ 3i is a zero of P (z) = z + pz + q where p and q are real Using conjugate pairs, find p and q and the other two zeros Check your answer by solving for p and q using P (2 ¡ 3i) = + i is a root of z ¡ 2z + az + bz + 10 = 0, where a and b are real Find a and b and the other roots of the equation One zero of P (z) = z + az + 3z + is purely imaginary If a is real, find a and hence factorise P (z) into linear factors At least one zero of P (x) = 3x3 + kx2 + 15x + 10 is purely imaginary Given that k is real, find k and hence resolve P (x) into a product of linear factors cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A scientist working for Crash Test Barriers, Inc is trying to design a crash test barrier whose ideal characteristics are shown graphically below The independent variable t is the time after impact, measured in milliseconds, such that t 700 The black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\209IB_HL-2_07.CDR Monday, 29 October 2007 12:33:06 PM PETERDELL IB_HL-2ed (210) 210 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) dependent variable is the distance that the barrier has been depressed because of the impact, measured in millimetres ¦¡(t) 120 100 80 60 40 20 t (ms) 200 400 600 800 a The equation for this graph is of the form f (t) = kt(t ¡ a)2 , t 700 From the graph, what is the value of a? What does it represent? b If the ideal crash barrier is depressed by 85 mm after 100 milliseconds, find the value of k, and hence find the equation of the graph given c What is the maximum amount of depression, and when does it occur? In the last year (starting 1st January), the volume of water (in megalitres) in a particular dam after t months could be described by the model V (t) = ¡t3 + 30t2 ¡ 131t + 250 The dam authority rules that if the volume falls below 100 ML, irrigation is prohibited During which months, if any, was irrigation prohibited in the last twelve months? Include in your answer a neat sketch of xm any graphs you may have used 10 m A ladder of length 10 metres is leaning up against a wall so that it is just touching a cube shaped box of edge length one metre, that is resting on the ground against the wall What height up the wall does the ladder reach? 1m REVIEW SET 7A Find real numbers a and b such that: a a + ib = b (1 ¡ 2i)(a + bi) = ¡5 ¡ 10i c (a + 2i)(1 + bi) = 17 ¡ 19i If z = + i and w = ¡2 ¡ i, find in simplest form: z¤ a 2z ¡ 3w b w c z3 Find the exact values of the real and imaginary parts of z if z = p p + i+ Find a complex number z such that 2z ¡ = iz ¡ i Write your answer in the form z = a + bi where a, b R Prove that zw¤ ¡ z ¤ w is purely imaginary or zero for all complex numbers z and w z+1 Given w = ¤ where z = a+bi, a, b R , write w in the form x+yi where z +1 x, y R and hence determine the conditions under which w is purely imaginary (3x3 + 2x ¡ 5)(4x ¡ 3) cyan magenta yellow b 100 95 (2x2 ¡ x + 3)2 x3 (x + 2)(x + 3) b 75 50 x3 x+2 25 95 a 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Carry out the following divisions: a Expand and simplify: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_07\210IB_HL-2_07.CDR Thursday, 25 October 2007 2:38:26 PM PETERDELL IB_HL-2ed (211) COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) 211 State and prove the Remainder theorem 10 ¡2 + bi is a solution to z + az + [3 + a] = Find a and b given that they are real 11 Find all zeros of 2z ¡ 5z + 13z ¡ 4z ¡ 12 Factorise z + 2z ¡ 2z + into linear factors p p 13 Find a quartic polynomial with rational coefficients having ¡ i and + as two of its zeros 14 If f (x) = x3 ¡ 3x2 ¡ 9x + b has (x ¡ k)2 as a factor, show that there are two possible values of k For each of these two values of k, find the corresponding value for b and hence solve f (x) = 15 Find k if the line with equation y = 2x + k x2 + y + 8x ¡ 4y + = does not meet the circle with equation 16 When P (x) = xn + 3x2 + kx + is divided by x + the remainder is 12 When P (x) is divided by x ¡ the remainder is Find k and n given that 34 < n < 38 17 If ® and ¯ are two of the roots of x3 ¡ x + = 0, show that ®¯ is a root of x3 + x2 ¡ = Hint: Let x3 ¡ x + = (x ¡ ®)(x ¡ ¯)(x ¡ °) REVIEW SET 7B p Find x, y Z such that z = x + yi satisfies the equation z = ¡ ¡ 2i 2¡i p Without using a calculator, find ¡ 12i Check your answer using a calculator Prove that if z is a complex number then both z + z ¤ and zz ¤ are real If z = + i and w = ¡ 2i find 2w¤ ¡ iz: ¡ 3i Find rationals a and b such that = + 2i: 2a + bi a + is a root of x2 ¡ 6x + b = where a, b R Explain why b has two possible values Find a in each case Find the remainder when x47 ¡ 3x26 + 5x3 + 11 is divided by x + A quartic polynomial P¡(x) has graph y = P¡(x) which touches the x-axis at (¡2, 0), cuts it at (1, 0), cuts the y-axis at (0, 12), and passes through (2, 80) Find an expression for P¡(x) in factored form and hence sketch the graph of y = P¡(x) If P (x) has remainder when divided by x ¡ and remainder ¡13 when divided by x + 2, find the remainder when P (x) is divided by x2 ¡ x ¡ p 10 Find all polynomials of least degree with rational coefficients that have ¡ i and p ¡ as two zeros cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 11 Factorise 2z + z + 10z + as a product of linear factors black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_07\211IB_HL-2_07.CDR Friday, 12 December 2008 11:55:54 AM TROY IB_HL-2ed (212) 212 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 7) 12 Find the general form of all polynomials of least degree which are real and have zeros of + i and ¡1 + 3i 13 ¡ 2i is a zero of z + kz + 32z + 3k ¡ 1, where k is real Find k and all zeros of the quartic 14 Find all the zeros of the polynomial z + 2z + 6z + 8z + given that one of the zeros is purely imaginary 15 When a polynomial P (x) is divided by x2 ¡ 3x + the remainder is 2x + Find the remainder when P (x) is divided by x ¡ REVIEW SET 7C Find real numbers x and y such that (3x + 2yi)(1 ¡ i) = (3y + 1)i ¡ x Solve the equation: z + iz + 10 = 6z Prove that zw¤ + z ¤ w is real for all complex numbers z and w Find real x and y such that: a x + iy = b (3 ¡ 2i)(x + i) = 17 + yi c (x + iy)2 = x ¡ iy z and w are non-real complex numbers with the property that both z + w and zw are real Prove that z ¤ = w p z= Find z if + + 5i ¡ 2i Find the remainder when 2x17 + 5x10 ¡ 7x3 + is divided by x ¡ ¡ i is a zero of 2z + az + 62z + [a ¡ 5], where a is real Find a and the other two zeros Find, in general form, all polynomials of least degree which are real, and have zeros of: p b ¡ i and ¡3 ¡ i: a i and 12 10 P (x) = 2x3 +7x2 +kx¡k is the product of linear factors, of which are identical Show that k can take distinct values and resolve P (x) into linear factors when k takes the largest of these values 11 Find all roots of 2z ¡ 3z + 2z = 6z + 12 Suppose k is real For what values of k does z + az + kz + ka = have: a one real root b real roots? 13 3x + and x ¡ are factors of 6x3 + ax2 ¡ 4ax + b Find a and b 14 Find the exact values of k for which the line y = x ¡ k is a tangent to the circle with equation (x ¡ 2)2 + (y + 3)2 = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 15 Find the quotient and remainder when x4 +3x3 ¡7x2 +11x¡1 is divided by x2 +2 Hence, find a and b for which x4 + 3x3 ¡ 7x2 + (2 + a)x + b is exactly divisible by x2 + black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_07\212IB_HL-2_07.CDR Friday, 12 December 2008 12:00:36 PM TROY IB_HL-2ed (213) Chapter Counting and the binomial expansion Contents: The product principle Counting paths Factorial notation Permutations Combinations Binomial expansions The general binomial expansion A B C D E F G cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 8A Review set 8B black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_08\213IB_HL-2_08.CDR Thursday, 25 October 2007 4:33:05 PM PETERDELL IB_HL-2ed (214) 214 COUNTING AND THE BINOMIAL EXPANSION (Chapter 8) OPENING PROBLEM At an IB Mathematics Teachers’ Conference there are 273 delegates present The organising committee consists of 10 people ² If each committee member shakes hands with every other committee member, how many handshakes take place? Can a 10-sided convex polygon be used to solve this problem? ² If all 273 delegates shake hands with all other delegates, how many handshakes take place now? The Opening Problem is an example of a counting problem The following exercises will help us to solve counting problems without having to list and count the possibilities one by one To this we will examine: ² the product principle ² counting permutations ² counting combinations A THE PRODUCT PRINCIPLE Suppose that there are three towns A, B and C and that different roads could be taken from A to B and two different roads from B to C A We can show this in a diagram: C B How many different pathways are there from A to C going through B? If we take road 1, there are two alternative roads to complete our trip If we take road 2, there are two alternative roads road road to complete our trip, and so on So, there are + + + = £ different pathways A C B Notice that the corresponds to the number of roads from A to B and the corresponds to the number of roads from B to C road road C A Similarly, for: D B there would be £ £ = 24 different pathways from A to D passing through B and C THE PRODUCT PRINCIPLE cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 If there are m different ways of performing an operation and for each of these there are n different ways of performing a second independent operation, then there are mn different ways of performing the two operations in succession black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_08\214IB_HL-2_08.CDR Tuesday, 29 January 2008 9:04:05 AM PETERDELL IB_HL-2ed (215) 215 COUNTING AND THE BINOMIAL EXPANSION (Chapter 8) The product principle can be extended to three or more successive operations Example P S R Q It is possible to take five different paths from Pauline’s to Quinton’s, different paths from Quinton’s to Reiko’s and different paths from Reiko’s to Sam’s How many different pathways could be taken from Pauline’s to Sam’s via Quinton’s and Reiko’s? The total number of different pathways = £ £ = 60 fproduct principleg EXERCISE 8A The illustration shows the possible map routes for a bus service which goes from P to S through both Q and R P How many different routes are possible? Q R S It is decided to label the vertices of a rectangle with the letters A, B, C and D In how many ways is this possible if: a they are to be in clockwise alphabetical order b they are to be in alphabetical order c they are to be in random order? The figure alongside is box-shaped and made of wire An ant crawls along the wire from A to B How many different paths of shortest length lead from A to B? B A In how many different ways can the top two positions be filled in a table tennis competition of teams? A football competition is organised between teams In how many ways is it possible to fill the top places in order of premiership points obtained? How many 3-digit numbers can be formed using the digits 2, 3, 4, and 6: a as often as desired b once only? How many different alpha-numeric plates for motor car registration can be made if the first places are English alphabet letters and those remaining are digits from to 9? cyan magenta yellow 95 b letters be mailed into mail boxes 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 In how many ways can: a letters be mailed into mail boxes c letters be mailed into mail boxes? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_08\215IB_HL-2_08.CDR Friday, 26 October 2007 9:32:31 AM PETERDELL IB_HL-2ed (216) 216 COUNTING AND THE BINOMIAL EXPANSION (Chapter 8) B COUNTING PATHS Consider the following road system leading from P to Q: B From A to Q there are paths P From B to Q there are £ = paths From C to Q there are paths Thus, from P to Q there are + + = 11 paths I Notice that: G E Q F C when going from B to G, we go from B to E and then from E to G, and we multiply the possibilities, when going from P to Q, we must first go from P to A, or P to B or P to C, and we add the possibilities I Consequently: D A ² ² the word and suggests multiplying the possibilities the word or suggests adding the possibilities Example C B A How many different paths lead from P to Q? E D Going from P to A to or from P to D to or from P to D to So, we have + + = 12 Q F P I H G B to C to Q there are £ = paths E to F to Q there are paths G to H to I to Q there are £ = paths different paths EXERCISE 8B How many different paths lead from P to Q? a b Q P P c d Q magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 cyan Q P P Q black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_08\216IB_HL-2_08.CDR Tuesday, 29 January 2008 11:00:00 AM PETERDELL IB_HL-2ed (217) 217 COUNTING AND THE BINOMIAL EXPANSION (Chapter 8) C FACTORIAL NOTATION In problems involving counting, products of consecutive positive integers are common For example, 8£7£6 or £ £ £ £ £ FACTORIAL NOTATION For convenience, we introduce factorial numbers to represent the products of consecutive positive integers For example, the product £ £ £ £ £ can be written as 6! for n > 1, n! is the product of the first n positive integers In general, n! = n(n ¡ 1)(n ¡ 2)(n ¡ 3):::: £ £ £ n! is read “n factorial” Notice that £ £ can be written using factorial numbers only as 8£7£6£5£4£3£2£1 8! 8£7£6= = 5£4£3£2£1 5! PROPERTIES OF FACTORIAL NUMBERS n! = n £ (n ¡ 1)! for n > The factorial rule is which can be extended to n! = n(n ¡ 1)(n ¡ 2)! and so on Using the factorial rule with n = 1, we have 1! = £ 0! 0! = We hence define Example a 4! What integer is equal to: b 5! 3! b 5£4£3£2£1 5! = = £ = 20 3! 3£2£1 a 4! = £ £ £ = 24 c 7! 7£6£5£4£3£2£1 = = 35 4! £ 3! 4£3£2£1£3£2£1 c 7! ? 4! £ 3! Example cyan b 10 £ £ £ 4£3£2£1 10 £ £ £ £ £ £ £ £ £ 10! = 6£5£4£3£2£1 6! magenta yellow 95 100 50 75 25 95 100 50 75 25 95 10 £ £ £ 10 £ £ £ £ £ £ £ £ £ 10! = = 4£3£2£1 4£3£2£1£6£5£4£3£2£1 4! £ 6! 100 b 50 10 £ £ £ = 75 a 25 95 100 50 75 25 Express in factorial form: a 10 £ £ £ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_08\217IB_HL-2_08.CDR Tuesday, 29 January 2008 9:07:52 AM PETERDELL IB_HL-2ed (218) 218 COUNTING AND THE BINOMIAL EXPANSION (Chapter 8) Example Write the following sums and differences as a product by factorising: a 8! + 6! b 10! ¡ 9! + 8! a 10! ¡ 9! + 8! = 10 £ £ 8! ¡ £ 8! + 8! = 8!(90 ¡ + 1) = 8! £ 82 b 8! + 6! = £ £ 6! + 6! = 6!(8 £ + 1) = 6! £ 57 Example 7! ¡ 6! Simplify 7! ¡ 6! £ 6! ¡ 6! = 6!(7 ¡ 1) = using factorisation = 6! EXERCISE 8C Find n! for n = 0, 1, 2, 3, ::::, 10 Simplify without using a calculator: 6! 6! 6! a b c 5! 4! 7! Simplify: n! (n ¡ 1)! a 13 £ 12 £ 11 3£2£1 d 100! 99! e (n + 2)! n! b Express in factorial form: a 7£6£5 4! 6! d f 7! 5! £ 2! (n + 1)! (n ¡ 1)! c b 10 £ c 11 £ 10 £ £ £ e 6£5£4 f 4£3£2£1 20 £ 19 £ 18 £ 17 Write as a product using factorisation: a 5! + 4! b 11! ¡ 10! e 9! + 8! + 7! f 7! ¡ 6! + 8! c g d h 6! + 8! 12! ¡ £ 11! 12! ¡ 10! £ 9! + £ 8! cyan magenta yellow 95 100 50 75 25 (n + 2)! + (n + 1)! n+3 h 95 n! ¡ (n ¡ 1)! n¡1 100 g 50 n! + (n ¡ 1)! (n ¡ 1)! 75 f 25 6! + 5! ¡ 4! 4! e 10! ¡ 9! 9! 95 d 100 10! ¡ 8! 89 50 c 75 10! + 9! 11 25 b 12! ¡ 11! 11 95 a 100 50 75 25 Simplify using factorisation: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_08\218IB_HL-2_08.CDR Friday, 26 October 2007 10:03:51 AM PETERDELL IB_HL-2ed (219) 219 COUNTING AND THE BINOMIAL EXPANSION (Chapter 8) D PERMUTATIONS A permutation of a group of symbols is any arrangement of those symbols in a definite order For example, BAC is a permutation on the symbols A, B and C in which all three of them are used We say the symbols are “taken at a time” Notice that ABC, ACB, BAC, BCA, CAB, CBA are all the different permutations on the symbols A, B and C taken at a time In this exercise we are concerned with the listing of all permutations, and then learning to count how many permutations there are without having to list them all Example List all the permutations on the symbols P, Q and R when they are taken: a at a time b at a time c at a time a b P, Q, R PQ PR QP QR c RP RQ QPR QRP PQR PRQ RPQ RQP Example List all permutations on the symbols W, X, Y and Z taken at a time WXYZ XWYZ YWXZ ZWXY WXZY XWZY YWZX ZWYX WYXZ XYWZ YXWZ ZXWY WYZX XYZW YXZW ZXYW WZXY XZYW YZWX ZYWX WZYX XZWY YZXW ZYXW i.e., 24 of them For large numbers of symbols listing the complete set of permutations is absurd However, we can still count them in the following way In Example there were positions to fill: 1st 2nd 3rd 4th In the 1st position, any of the symbols could be used This leaves any of symbols to go in the 2nd position, which leaves any of symbols to go in the 3rd position 1st 2nd 3rd 4th 1st 2nd 3rd 4th The remaining symbol must go in the 4th position 1st 2nd 3rd 4th cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 So, the total number of permutations = £ £ £ = 24 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_08\219IB_HL-2_08.CDR Tuesday, 29 January 2008 9:15:22 AM PETERDELL fproduct principleg IB_HL-2ed (220) 220 COUNTING AND THE BINOMIAL EXPANSION (Chapter 8) Example If a chess association has 16 teams, in how many different ways could the top positions be filled on the competition ladder? Any of the 16 teams could fill the ‘top’ position Any of the remaining 15 teams could fill the 2nd position Any of the remaining 14 teams could fill the 3rd position Any of the remaining teams could fill the 8th position i.e., 16 1st 15 2nd 14 3rd 13 4th 12 5th 11 6th 10 7th 8th ) total number = 16 £ 15 £ 14 £ 13 £ 12 £ 11 £ 10 £ = 518 918 400 Example 10 Suppose you have the alphabet blocks A, B, C, D and E and they are placed in a row For example you could have: D A E C B a How many different permutations could you have? b How many permutations end in C? c How many permutations have the form A B ? d How many begin and end with a vowel, i.e., A or E? a There are letters taken at a time ) total number = £ £ £ £ = 120 b 1 C must be in the last position (1 way) and the other letters could go into the remaining places in 4! ways any others here C here c ) total number = £ 4! = 24 ways 1 A A goes into place, B goes into place and the remaining letters go into the remaining places in 3! ways B ) total number = £ £ 3! = ways d 1 A or E could go into the 1st position, and after that one is placed, the other one goes into the last position A or E remainder of A or E The remaining letters could be arranged in 3! ways in the remaining positions cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ) total number = £ £ 3! = 12: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_08\220IB_HL-2_08.CDR Wednesday, November 2007 3:57:21 PM PETERDELL IB_HL-2ed (221) 221 COUNTING AND THE BINOMIAL EXPANSION (Chapter 8) Example 11 There are different books arranged in a row on a shelf In how many ways can two of the books, A and B be together? Method A B B A £ A £ B £ £ £ £ £ £ £ £ £ £ £ £ 1: We could £ £ £ £ £ £ B £ £ A £ £ A B £ B A £ £ A B £ B A £ £ A £ £ B have any of the following locations for A and B £ > > > £ > > If we consider any one > > £ > > of these, the remaining > £ > > = books could be placed £ 10 of these £ in 4! different orderings > > > £ > ) total number of ways > > £ > > = 10 £ 4! = 240: > > B > > ; A Method 2: A and B can be put together in 2! ways (i.e., AB or BA) Now consider this pairing as one book (effectively tying a string around them) which together with the other books can be ordered in 5! different ways ) total number = 2! £ 5! = 240: EXERCISE 8D List the set of all permutations on the symbols W, X, Y and Z taken: c three at a time a at a time b two at a time Note: Example has them taken at a time List the set of all permutations on the symbols A, B, C, D and E taken: a at a time b at a time In how many ways can: a different books be arranged on a shelf b different paintings, from a collection of 8, be chosen and in a row c a signal consisting of coloured flags be made if there are 10 different flags to choose from? Suppose you have different coloured flags How many different signals could you make using: a flags only b flags only c or flags? How many different permutations of the letters A, B, C, D, E and F are there if each letter can be used once only? How many of these: a end in ED b begin with F and end with A c begin and end with a vowel (i.e., A or E)? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 How many 3-digit numbers can be constructed from the digits 1, 2, 3, 4, 5, and if each digit may be used: a as often as desired b only once c once only and the number is odd? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_08\221IB_HL-2_08.CDR Friday, 26 October 2007 10:22:13 AM PETERDELL IB_HL-2ed (222) 222 COUNTING AND THE BINOMIAL EXPANSION (Chapter 8) In how many ways can boys and girls be arranged in a row of seats? In how many of these ways the boys and girls alternate? 3-digit numbers are constructed from the digits 0, 1, 2, 3, 4, 5, 6, 7, and using each digit at most once How many such numbers: a can be constructed b end in c end in d are divisible by 5? In how many ways can different books be arranged on a shelf if: a there are no restrictions b books X and Y must be together c books X and Y are never together? 10 A group of 10 students sit randomly in a row of 10 chairs In how many ways can this be done if: a there are no restrictions b students A, B and C are always seated together? 11 How many three-digit numbers, in which no two digits are the same, can be made using the digits 0, 1, 3, 5, if: a there are no restrictions b the numbers must be less than 500 c the numbers must be even and greater than 300? 12 How many different arrangements of four letters chosen from the letters of the word MONDAY are possible if: a there are no restrictions b at least one vowel (A or O) must be used c no two vowels are adjacent? 13 Nine boxes are each labelled with a different whole number from to Five people are allowed to take one box each In how many different ways can this be done if: a there are no restrictions b the first three people decide that they will take even numbered boxes? 14 Alice has booked ten adjacent front-row seats for a basketball game for herself and nine friends Altogether, there are five boys and five girls a Assuming they all arrive, how many different arrangements are there if: i there are no restrictions ii boys and girls are to sit alternately? b Due to a severe snowstorm, only five of Alice’s friends are able to join her for the game How many different ways are there of seating in the 10 seats if: i there are no restrictions ii any two of Alice’s friends are to sit next to her? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 15 At a restaurant, a rectangular table seats eight people, four on each of the longer sides Eight diners sit at the table How many different seating arrangements are there if: a there are no restrictions b two particular people wish to sit directly opposite each other c two particular people wish to sit on the same side of the table, next to each other? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_08\222IB_HL-2_08.CDR Friday, 26 October 2007 10:36:34 AM PETERDELL IB_HL-2ed (223) 223 COUNTING AND THE BINOMIAL EXPANSION (Chapter 8) INVESTIGATION PERMUTATIONS IN A CIRCLE There are permutations on the symbols A, B and C in a line These are: ABC ACB BAC BCA CAB CBA However in a circle there are only different permutations on these symbols These are: B C and A C Notice that as they are the only possibilities with different right-hand and lefthand neighbours A B A B C C A are the same cyclic permutations B B C A What to do: Draw diagrams showing different cyclic permutations for: a one symbol: A b two symbols: A and B c three symbols: A, B and C d four symbols: A, B, C and D Copy and complete: Number of symbols Permutations in a line Permutations in a circle = 3! = 2! If there are n symbols to be permuted in a circle, how many different orderings are possible? E COMBINATIONS A combination is a selection of objects without regard to order or arrangement For example, the possible teams of people selected from A, B, C, D and E are ABC BCD CDE ABD BCE ABE BDE ACD ACE ADE i.e., 10 different combinations Crn is the number of combinations on n distinct or different symbols taken r at a time the number up for selection the number of positions needed to be filled ¡ ¢ may also be written as n Cr or as the binomial coefficient nr Crn Crn From the teams example above we know that C35 = 10: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 However, we know the number of permutations of three people from the possibilities is £ £ = 60, so why is this answer or 3! times larger than C35 ? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_08\223IB_HL-2_08.CDR Friday, 26 October 2007 10:43:29 AM PETERDELL IB_HL-2ed (224) 224 COUNTING AND THE BINOMIAL EXPANSION (Chapter 8) This can be seen if we consider one of these teams, ABC say There are 3! ways in which the members of team ABC can be placed in a definite order, i.e., ABC, ACB, BAC, BCA, CAB, CBA and if this is done for all 10 possible teams we get all possible permutations of the people taken at a time So, £ £ = C35 £ 3! ) C35 = 5£4£3 3£2£1 or 5! 3! £ 2! n (n ¡ 1)(n ¡ 2) (n ¡ r + 3) (n ¡ r + 2) (n ¡ r + 1) n! = In general, Crn = r (r ¡ 1) (r ¡ 2) r!(n ¡ r)! | {z } | {z } Factor form Factorial form Values of Crn can be calculated from your calculator For example, to find C310 : TI-83: Press 10 MATH to select 3:n Cr from the PRB menu, then ENTER Casio: Press 10 OPTN F3 (PROB) F3 (n Cr) F6 EXE Example 12 How many different teams of can be selected from a squad of if: a there are no restrictions b the teams must include the captain? a There are players up for selection and we want any of them This can be done in C47 = 35 ways b If the captain must be included and we need any of the other 6, this can be done in C11 £ C36 = 20 ways Example 13 A committee of is chosen from men and women How many different committees can be chosen if: a there are no restrictions b there must be of each sex c at least one of each sex is needed? a For no restrictions there are + = 13 people up for selection and we want any of them ) total number = C413 = 715: b The men can be chosen in C27 ways and the women can be chosen in C26 ways ) total number = C27 £ C26 = 315: Total number = number with (3 M and W) or (2 M and W) or (1 M and W) = C37 £ C16 + C27 £ C26 + C17 £ C36 = 665 magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 cyan total number = C413 ¡ C47 £ C06 ¡ C07 £ C46 : Alternatively, c black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_08\224IB_HL-2_08.CDR Tuesday, 29 January 2008 9:15:40 AM PETERDELL IB_HL-2ed (225) COUNTING AND THE BINOMIAL EXPANSION (Chapter 8) 225 EXERCISE 8E Evaluate using factor form: a C18 b Check each answer using your calculator C28 c C38 d C68 e C88 : In question you probably noticed that C28 = C68 n Prove using factorial form that this statement is true In general, Crn = Cn¡r ³ ´ ¡9¢ = Find k if k k¡1 : List the different teams of that can be chosen from a squad of (named A, B, C, D, E and F) Check that the formula for Crn gives the total number of teams How many different teams of 11 can be chosen from a squad of 17? Candidates for an examination are required to questions out of In how many ways can this be done? If question was compulsory, how many selections would be possible? How many different committees of can be selected from 13? How many of these committees consist of the president and others? How many different teams of can be selected from a squad of 12? How many of these teams contain: a the captain and vice-captain b exactly one of the captain or the vice-captain? A team of is selected from a squad of 15 of the players are certainties who must be included, and another must be excluded because of injury In how many ways can this be done? 10 In how many ways can people be selected from 10 if: a one person is always in the selection b are excluded from every selection c is always included and are always excluded? 11 A committee of is chosen from 10 men and women Determine the number of ways of selecting the committee if: a there are no restrictions b it must contain men and women d it must contain at least men c it must contain all men e it must contain at least one of each sex 12 A committee of is chosen from doctors, dentists and others Determine the number of ways of selecting the committee if it is to contain: a doctors and dentist b doctors c at least one of the two professions 13 How many diagonals has a 20-sided convex polygon? 14 There are 12 distinct points A, B, C, D, , L on a circle Lines are drawn between each pair of points a How many lines i are there in total ii pass through B? b How many triangles i are determined by the lines ii have one vertex B? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 15 How many 4-digit numbers can be constructed where the digits are in ascending order from left to right? Note: You cannot start with black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_08\225IB_HL-2_08.CDR Friday, 26 October 2007 10:59:06 AM PETERDELL IB_HL-2ed (226) 226 16 COUNTING AND THE BINOMIAL EXPANSION (Chapter 8) a Give an example which demonstrates that: C05 £ C46 + C15 £ C36 + C25 £ C26 + C35 £ C16 + C45 £ C06 = C411 : b Copy and complete: n n m + C2m £Cr¡2 + :::: + Cr¡1 £C1n + Crm £C0n = ::::: C0m £Crn + C1m £Cr¡1 17 In how many ways can 12 people be divided into: a two equal groups b three equal groups? A 18 Line A contains 10 points and line B contains points If all points on line A are joined to all points on line B, determine the maximum number of points of intersection between A and B B Q 19 10 points are located on [PQ], on [QR] and on [RP] All possible lines connecting these 27 points are drawn Determine the maximum number of points of intersection of these lines which lie within triangle PQR F R P BINOMIAL EXPANSIONS The sum a + b is called a binomial as it contains two terms Any expression of the form (a + b)n is called a power of a binomial Consider the following algebraic expansions of the powers (a + b)n (a + b)1 = a + b (a + b)2 = a2 + 2ab + b2 (a + b)3 = (a + b)(a + b)2 = (a + b)(a2 + 2ab + b2 ) = a3 + 2a2 b + ab2 + a2 b + 2ab2 + b3 = a3 + 3a2 b + 3ab2 + b3 We say that: a2 + 2ab + b2 a3 + 3a2 b + 3ab2 + b3 is the binomial expansion of is the binomial expansion of (a + b)2 (a + b)3 THE BINOMIAL EXPANSION OF (a¡+¡b)n, n¡>¡4 INVESTIGATION What to do: Expand (a + b)4 in the same way as for (a + b)3 Similarly, expand algebraically (a + b) expansion of (a + b)4 from Expand (a + b)6 above using your answer for the using your expansion for (a + b)5 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The (a + b)3 = a3 + 3a2 b + 3ab2 + b3 expansion contains terms: a3 , 3a2 b, 3ab2 and b3 The coefficients of these terms are: 3 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_08\226IB_HL-2_08.CDR Friday, 26 October 2007 11:03:42 AM PETERDELL IB_HL-2ed (227) 227 COUNTING AND THE BINOMIAL EXPANSION (Chapter 8) a What can be said about the powers of a and b in each term of the expansion of (a + b)n for n = 0, 1, 2, 3, 4, and 6? b Write down the triangle of coefficients to row 6: n=0 n=1 n=2 n=3 1 1 3 etc row This triangle of coefficients is called Pascal’s triangle Investigate: a the predictability of each row from the previous one b a formula for finding the sum of the numbers in the nth row of Pascal’s triangle Use your results from to predict the elements of the 7th row of Pascal’s triangle and hence write down the binomial expansion of (a + b)7 Check your result algebraically by using (a+b)7 = (a+b)(a+b)6 and your results from above (a + b)4 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 = a4 + 4a3 b1 + 6a2 b2 + 4a1 b3 + b4 From Investigation we obtained Notice that: ² As we look from left to right across the expansion, the powers of a decrease by and the powers of b increase by The sum of the powers of a and b in each term of the expansion is The number of terms in the expansion is + = ² ² In general, for the expansion of (a + b)n ² where n = 1, 2, 3, 4, 5, : As we look from left to right across the expansion, the powers of a decrease by whilst the powers of b increase by The sum of the powers of a and b in each term of the expansion is n The number of terms in the expansion is n + ² ² The expansion (a + b)3 = a3 + 3a2 b + 3ab2 + b3 can be used to expand other cubes Example 14 Using (a + b)3 = a3 + 3a2 b + 3ab2 + b3 , find the binomial expansion of: b (x ¡ 5)3 a (2x + 3)3 a In the expansion of (a + b)3 we substitute a = (2x) and b = (3) (2x + 3) = (2x) + 3(2x)2 (3) + 3(2x)1 (3)2 + (3)3 ) = 8x3 + 36x2 + 54x + 27 on simplifying b This time we substitute a = (x) and b = (¡5) (x ¡ 5)3 = (x)3 + 3(x)2 (¡5) + 3(x)(¡5)2 + (¡5)3 ) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 = x3 ¡ 15x2 + 75x ¡ 125 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_08\227IB_HL-2_08.CDR Friday, 26 October 2007 11:39:07 AM PETERDELL IB_HL-2ed (228) 228 COUNTING AND THE BINOMIAL EXPANSION (Chapter 8) Example 15 ¢5 ¡ b binomial expansion of x ¡ x2 Find the: a 5th row of Pascal’s triangle a the 0th row, for (a + b)0 the 1st row, for (a + b)1 1 1 1 3 1 1 10 10 b the 5th row So, (a + b)5 = a5 + 5a4 b + 10a3 b2 + 10a2 b3 + 5ab4 + b5 ¡ ¢ and we let a = (x) and b = ¡2 x ¡ ¢ ¡ ¢ ¡ ¢2 ¡ ¢3 5 ¡2 ) x ¡ x = (x) + 5(x) x + 10(x)3 ¡2 + 10(x)2 ¡2 x x ¡ ¡2 ¢4 ¡ ¡2 ¢5 + 5(x) x + x = x5 ¡ 10x3 + 40x ¡ 80 80 32 + 3¡ x x x EXERCISE 8F Use the binomial expansion of (a + b)3 a (x + 1)3 b to expand and simplify: (3x ¡ 1)3 (2x + 5)3 c Use (a + b)4 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 a (x ¡ 2)4 b (2x + 3)4 c d to expand and simplify: ¢4 ¢4 ¡ ¡ x + x1 d 2x ¡ x1 a Write down the 6th row of Pascal’s triangle b Find the binomial expansion of: ii (2x ¡ 1)6 i (x + 2)6 Expand and simplify: p a (1 + 2)3 b (1 + ¢3 ¡ 2x + x1 iii p 5) c ¢6 ¡ x + x1 (2 ¡ p 2) a Expand (2 + x)6 b Use the expansion of a to find the value of (2:01)6 Expand and simplify (2x + 3)(x + 1)4 Find the coefficient of: cyan magenta yellow 95 100 50 75 25 b a3 b3 in the expansion of (2a + 3b)6 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a a3 b2 in the expansion of (3a + b)5 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_08\228IB_HL-2_08.CDR Wednesday, November 2007 3:57:31 PM PETERDELL IB_HL-2ed (229) COUNTING AND THE BINOMIAL EXPANSION (Chapter 8) G 229 THE GENERAL BINOMIAL EXPANSION ¡ ¢ ¡ ¢ ¡ ¢ ¡ n ¢ n¡1 ¡n¢ n ab (a + b)n = n0 an + n1 an¡1 b + n2 an¡2 b2 + ::::: + n¡1 + n b ¡n¢ where r = Crn is the binomial coefficient of an¡r br and r = 0, 1, 2, 3, , n The general term, or (r + 1)th term is Tr +1 = ¡n ¢ n ¡ r r a b r Example 16 Write down the first and last terms of the expansion of ¢12 ¡ 2x + x1 ¢12 ¡ ¢ ¡12¢ ¡ ¢ ¡ ¡ ¢ 11 10 2x + x1 = (2x)12 + 12 + :::::: (2x) x + (2x) x ¡ ¢11 ¡12¢ ¡ ¢12 ¡ ¢ + 12 x :::::: + 12 11 (2x) x Example 17 Find the 7th term of ¡ ¢14 3x ¡ x42 a = (3x), b = ¡ ¡4 ¢ x2 and n = 14 ¡n¢ n¡r r b , we let r = r a ¡ ¢6 ¡14¢ T7 = (3x)8 ¡4 x2 So, as Tr+1 = Do not simplify ) Example 18 In the expansion of ¡ ¢12 x + x , find: a the coefficient of x6 a = (x2 ), b = a ¡4¢ b the constant term ¡12¢ 12¡r ¡ ¢r r (x ) x ¡12¢ 24¡2r 4r = r x xr ¡12¢ r 24¡3r = r x ) and n = 12 x Tr+1 = b If 24 ¡ 3r = then 3r = 24 ) r=8 ¡ ¢ ) T9 = 12 x If 24 ¡ 3r = then 3r = 18 ) r=6 ¡ ¢ 6 ) T7 = 12 x cyan magenta yellow 95 100 50 75 25 ) the constant term is ¡12¢ or 32 440 320 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ) the coefficient of x6 is ¡12¢ or 784 704 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_08\229IB_HL-2_08.CDR Friday, 26 October 2007 11:47:04 AM PETERDELL IB_HL-2ed (230) 230 COUNTING AND THE BINOMIAL EXPANSION (Chapter 8) Example 19 (x + 3)(2x ¡ 1)6 ¡¢ ¡¢ = (x + 3)[(2x)6 + 61 (2x)5 (¡1) + 62 (2x)4 (¡1)2 + ::::] ¡¢ ¡¢ = (x + 3)(26 x6 ¡ 61 25 x5 + 62 24 x4 ¡ ::::) Find the coefficient of x5 in the expansion of (x + 3)(2x ¡ 1)6 (2) (1) ¡6¢ from (1) 2 x ¡6¢ 5 and ¡3 x from (2) ¡¢ ¡¢ ) the coefficient of x5 is 62 24 ¡ 61 25 = ¡336 So, the terms containing x5 are EXERCISE 8G Write down the first three and last two terms of the binomial expansion of: ¢15 ¢20 ¡ ¡ a (1 + 2x)11 b 3x + x2 c 2x ¡ x3 Without simplifying, find: a the 6th term of (2x + 5)15 ¢17 ¡ c the 10th term of x ¡ x2 the 4th term of d Find the coefficient of: a x10 in the expansion of (3 + 2x2 )10 ¢12 ¡ c x12 in the expansion of 2x2 ¡ x1 Find the constant term in: ¡ a the expansion of x + ¡ ¢9 x +x ¡ ¢21 the 9th term of 2x ¡ x b ¢ 15 x2 ¢6 ¡ x3 in the expansion of 2x2 ¡ x3 b ¡ the expansion of x ¡ b ¢ x2 a Write down the first rows of Pascal’s triangle b What is the sum of the numbers in: i row ii row iii row iv row v row 5? c Copy and complete: The sum of the numbers in row n of Pascal’s triangle is ¡ ¢ ¡ ¢ ¡ ¢ ¡ n ¢ n¡1 ¡n¢ n x + n x d Show that (1 + x)n = n0 + n1 x + n2 x2 + :::: + n¡1 ¡ n ¢ ¡n¢ ¡n¢ ¡n¢ ¡n¢ Hence deduce that + + + :::: + n¡1 + n = 2n a Find the coefficient of x5 in the expansion of (x + 2)(x2 + 1)8 b Find the coefficient of x6 in the expansion of (2 ¡ x)(3x + 1)9 ¢5 ¡ Expand + 2x ¡ x2 in ascending powers of x as far as the term containing x4 a Show that ¡n¢ = n and ¡n¢ = n(n ¡ 1) are true statements b The third term of (1 + x)n is 36x2 Find the fourth term cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 c If (1 + kx)n = ¡ 12x + 60x2 ¡ ::::: , find the values of k and n black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_08\230IB_HL-2_08.CDR Friday, 26 October 2007 11:50:16 AM PETERDELL IB_HL-2ed (231) COUNTING AND THE BINOMIAL EXPANSION (Chapter 8) Find a if the coefficient of x11 in the expansion of (x2 + 10 From the binomial expansion of (1 + x)n , deduce that: ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ a n0 ¡ n1 + n2 ¡ n3 + + (¡1)n nn = ¡ ¢ ¡2n+1¢ ¡2n+1¢ ¡ ¢ b 2n+1 + + + + 2n+1 = 4n n 11 By considering the binomial expansion of (1 + x)n , find 231 10 ) is 15 ax n X 2r ¡n¢ r : r=0 2 12 (x ¡ 3x + 1) = x ¡ 6x + 11x ¡ 6x + and the sum of its coefficients is ¡ + 11 ¡ + which is What is the sum of the coefficients of (x3 + 2x2 + 3x ¡ 7)100 ? 13 By considering (1 + x)n (1 + x)n = (1 + x)2n , show that ¡n¢2 ¡n¢2 ¡n¢2 ¡n¢2 ¡ ¢2 ¡ ¢ + + + + + nn = 2n n 14 ¡ ¢ = n n¡1 r¡1 ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ b Hence show that n1 + n2 + n3 + n4 + + n nn = n2n¡1 a Prove that r ¡n¢ r c Suppose numbers Pr are defined by ¡ ¢ Pr = nr pr (1 ¡ p)n¡r for r = 0, 1, 2, 3, , n n X i Prove that ii Pr = r=0 n X rPr = np r=1 REVIEW SET 8A Alpha-numeric number plates have two letters followed by four digits How many plates are possible if: a there are no restrictions b the first letter must be a vowel c no letter or digit may be repeated? Ten points are located on a 2-dimensional plane If no three points are collinear: a how many line segments joining two points can be drawn b how many different triangles can be drawn by connecting all 10 points with line segments in any possible way? Simplify: a n! (n ¡ 2)! b n! + (n + 1)! n! a How many committees of five can be selected from eight men and seven women? b How many of the committees contain two men and three women? c How many contain at least one man? Eight people enter a room and each person shakes hands with every other person How many hand shakes are made? cyan magenta yellow 95 (x ¡ 2y)3 100 50 a 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Use the binomial expansion to find: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_08\231IB_HL-2_08.CDR Friday, 26 October 2007 12:01:16 PM PETERDELL b (3x + 2)4 IB_HL-2ed (232) 232 COUNTING AND THE BINOMIAL EXPANSION (Chapter 8) A team of five is chosen from six men and four women a How many different teams are possible with no restrictions? b How many contain at least one of each sex? The letters P, Q, R, S and T are to be arranged in a row How many of the possible arrangements: a end with T b begin with P and end with T? Find the coefficient of x3 in the expansion of (2x + 5)6 ¡ ¢6 10 Find the constant term in the expansion of 2x2 ¡ x1 : 11 The first three terms in the expansion of (1 + kx)n , in ascending powers of x, are ¡ 4x + 15 x Find k and n 12 Eight people enter a room and sit at random in a row of eight chairs In how many ways can the sisters Cathy, Robyn and Jane sit together in the row? a How many three digit numbers can be formed using the digits to 9? b How many of these numbers are divisible by 5? 13 REVIEW SET 8B A team of eight is chosen from 11 men and women How many different teams are possible if there are: a no restrictions b four of each sex on the team c at least two women on the team? A four digit number is constructed using the digits 0, 1, 2, 3, , once only a How many numbers are possible? b How many are divisible by 5? Use Pascal’s triangle to expand (a + b)6 ¢6 ¡ Hence, find the binomial expansion of: a (x ¡ 3)6 b + x1 p p Expand and simplify ( 3+2)5 giving your answer in the form a+b 3, a, b Z Use the expansion of (4 + x)3 to find the exact value of (4:02)3 ¢8 ¡ Find the constant term in the expansion of 3x2 + x1 ¢12 ¡ Find the coefficient of x¡6 in the expansion of 2x ¡ x32 Find the coefficient of x5 in the expansion of (2x + 3)(x ¡ 2)6 ¢9 ¡ Find the possible values of a if the coefficient of x3 in 2x + ax1 ¡ 3x ¡ 10 Find the term independent of x in the expansion of is 288 ¢ x2 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 11 Find the possible values of q if the constant terms in the expansions of ¡ ¢8 ¢4 ¡ x + xq3 and x3 + xq3 are equal black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_08\232IB_HL-2_08.CDR Thursday, 11 March 2010 10:23:15 AM PETER IB_HL-2ed (233) Chapter Mathematical induction Contents: A B C The process of induction The principle of mathematical induction Indirect proof (extension) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 9A Review set 9B Review set 9C black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_09\233IB_HL-2_09.CDR Friday, 26 October 2007 12:34:11 PM PETERDELL IB_HL-2ed (234) 234 MATHEMATICAL INDUCTION (Chapter 9) A THE PROCESS OF INDUCTION The process of formulating a general result from a close examination of the simplest cases is called mathematical induction For example, the the the the 2=2£1 4=2£2 6=2£3 8=2£4 first positive even number is second positive even number is third positive even number is fourth positive even number is and from these results we induce that the nth positive even number is £ n or 2n The statement that “the nth positive even number is 2n” is a summary of the observations of the simple cases n = 1, 2, 3, and is a statement which we believe is true We call such a statement a conjecture or proposition We need to prove a conjecture before we can regard it as a fact Consider the following argument for finding the sum of the first n odd numbers: = = 12 + = = 22 + + = = 32 + + + = 16 = 42 1| + +{z + + 9} = 25 = 52 of these It seems that “the sum of the first n odd numbers is n2 ” This pattern may continue or it may not We require proof of the fact for all positive integers n A formal statement of our conjecture may be: “ 1| + + +{z7 + + :::::} = n2 n of these for all n Z + ” The nth odd number is (2n ¡ 1), so we could also write the proposition as: “1 + + + + + ::::: + (2n ¡ 1) = n2 for all n Z + ” One direct proof of the proposition is to note that the series is arithmetic with u1 = 1, d = and “n” = n n (2(1) + (n ¡ 1)2) n = £ 2n fusing Hence Sn = n (2u1 + (n ¡ 1)d)g cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 = n2 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_09\234IB_HL-2_09.CDR Friday, 26 October 2007 1:59:46 PM PETERDELL IB_HL-2ed (235) 235 MATHEMATICAL INDUCTION (Chapter 9) Example By examining the cases n = 1, 2, and 4, make a conjecture about the sum 1 1 + 2£3 + 3£4 + 4£5 + ::::: + n(n+1) : of Sn = 1£2 S1 = 1£2 = S2 = 1£2 + 2£3 S3 = 1£2 S4 = 1£2 ² Note: = = + 2£3 + 3£4 = + 2£3 + 3£4 + 4£5 + + 12 = = 4 + 20 = If the result in Example is true, then: 1 1 1£2 + 2£3 + 3£4 + 4£5 + ::::: + 1000£1001 = From these results we conjecture that: n Sn = : n+1 1000 1001 fcase n = 1000g The great Swiss mathematician Euler proposed that P (n) = n2 + n + 41 was a formula for generating prime numbers People who read his statement probably checked it for n = 1, 2, 3, 4, 5, ., 10 and agreed with him However, it was found to be incorrect as, for example P (41) = 412 + 41 + 41 = 41(41 + + 1) = 41 £ 43, a composite So, not all propositions are true ² EXERCISE 9A By examining the following using substitutions like n = 1, 2, 3, 4, ., complete the proposition or conjecture a The nth term of the sequence 3, 7, 11, 15, 19, is for n = 1, 2, 3, 4, 3n > + 2n for for 11n ¡ is divisible by + + + + 10 + ::::: + 2n = for for 1! + £ 2! + £ 3! + £ 4! + ::::: + n £ n! = n f + + + + ::::: + = for 2! 3! 4! 5! (n + 1)! for g 7n + is divisible by ³ ´ ¢ ¡ ¢ ¡ ¢ ¡ h ¡ 12 ¡ 13 ¡ 14 ::::: ¡ n+1 = for b c d e 1 + + + ::::: to n terms = £ 5 £ 8 £ 11 i for cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 n points are placed inside a triangle Non-intersecting line segments are drawn connecting the vertices of the triangle and the points within it, to partition the given triangle into smaller triangles Make a proposition concerning the number of triangles obtained in the general case black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_09\235IB_HL-2_09.CDR Friday, 26 October 2007 2:06:38 PM PETERDELL n=1 n=2 IB_HL-2ed (236) 236 MATHEMATICAL INDUCTION (Chapter 9) B THE PRINCIPLE OF MATHEMATICAL INDUCTION PROPOSITION NOTATION We use Pn to represent a proposition which is defined for every integer n where n > a, a Z For example, in the case of Example 1, our proposition Pn is 1 1 n “ + + ::::: + = ” for all n Z + 1£2 2£3 3£4 n(n + 1) n+1 Notice that P1 is “ 1 1 = ” and P2 is “ + = ” 1£2 1£2 2£3 and Pk is “ 1 1 k + + + ::::: + = ” 1£2 2£3 3£4 k(k + 1) k+1 THE PRINCIPLE OF MATHEMATICAL INDUCTION Suppose Pn is a proposition which is defined for every integer n > a, a Z ² Pa is true, and ² Pk+1 is true whenever Pk is true, then Pn is true for all n > a Now if This means that for a = 1, say, if the two above conditions hold, then the truth of P1 implies that P2 is true, which implies that P3 is true, which implies that P4 is true, and so on The principle of mathematical induction constitutes a formal proof that a particular proposition is true One can liken the principle of mathematical induction to the domino effect We imagine an infinite set of dominoes all lined up Provided that: ² the first one topples to the right, and ² the (k + 1)th domino will topple if the kth domino topples, then eventually all will topple DEMO SUMS OF SERIES Example Prove that n X i2 = i=1 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find 12 + 22 + 32 + 42 + ::::: + 1002 : 95 100 50 75 25 b n(n + 1)(2n + 1) for all n Z + a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_09\236IB_HL-2_09.CDR Friday, 26 October 2007 2:14:18 PM PETERDELL IB_HL-2ed (237) MATHEMATICAL INDUCTION (Chapter 9) a 237 n(n + 1)(2n + 1) ” for all n Z + Proof: (By the principle of mathematical induction) Pn is: “12 + 22 + 32 + 42 + ::::: + n2 = (1) LHS = 12 = and RHS = If n = 1, ) (2) P1 is true 1£2£3 =1 If Pk is true, then k X i2 = 12 + 22 + 32 + 42 + ::::: + k = i=1 k(k + 1)(2k + 1) 12 + 22 + 32 + 42 + ::::: + k2 + (k + 1)2 Thus k(k + 1)(2k + 1) + (k + 1)2 fusing Pk g 6 k(k + 1)(2k + 1) = + (k + 1)2 £ Always look 6 for common (k + 1)[k(2k + 1) + 6(k + 1)] factors = (k + 1)(2k2 + k + 6k + 6) = = (k + 1)(2k2 + 7k + 6) (k + 1)(k + 2)(2k + 3) = (k + 1)([k + 1] + 1)(2[k + 1] + 1) = = Thus Pk+1 is true whenever Pk is true Since P1 is true, fPrinciple of mathematical inductiong Pn is true for all n Z + : b 12 + 22 + 32 + 42 + ::::: + 1002 = 100 £ 101 £ 201 = 338 350 fas n = 100g EXERCISE 9B Use the principle of mathematical induction to prove that the following propositions (conjectures) are true for all positive integers n: n n X X n(n + 1) n(n + 1)(n + 2) i= i(i + 1) = b a i=1 i=1 n X c i=1 n(n + 1)(2n + 13) 3i(i + 4) = d n X i=1 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Note: You should remember the result from a: n(n + 1) + + + :::::: + n = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_09\237IB_HL-2_09.CDR Tuesday, 29 January 2008 9:43:09 AM PETERDELL i3 = n2 (n + 1)2 for all n in Z + IB_HL-2ed (238) 238 MATHEMATICAL INDUCTION (Chapter 9) Prove that the following proposition is true for all positive integers n: n X i £ 2i¡1 = (n ¡ 1) £ 2n + i=1 Example Prove that: n X n = (3i ¡ 1)(3i + 2) 6n + i=1 Pn is: “ for all n Z + : 1 n + + :::::: + = ” for all n Z + 2£5 5£8 (3n ¡ 1)(3n + 2) 6n + Proof: (By the principle of mathematical induction) 1 and RHS = = 10 = (1) If n = 1, LHS = 2£5 6£1+4 ) P1 is true 10 (2) If Pk is true, then k X i=1 1 1 k = + + :::::: + = (3i ¡ 1)(3i + 2) 2£5 5£8 (3k ¡ 1)(3k + 2) 6k + 1 1 + + :::::: + + 2£5 5£8 (3k ¡ 1)(3k + 2) (3k + 2)(3k + 5) Now = k + 6k + (3k + 2)(3k + 5) fusing Pk g k + 2(3k + 2) (3k + 2)(3k + 5) µ ¶ µ ¶ 3k + k = £ + £ 2(3k + 2) 3k + (3k + 2)(3k + 5) = = 3k2 + 5k + 2(3k + 2)(3k + 5) = (3k + 2)(k + 1) 2(3k + 2)(3k + 5) = k+1 6k + 10 = [k + 1] 6[k + 1] + cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Thus P1 is true, and Pk+1 is true whenever Pk is true fPrinciple of mathematical inductiong ) Pn is true for all n Z + : black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_09\238IB_HL-2_09.CDR Friday, November 2007 10:24:10 AM PETERDELL IB_HL-2ed (239) 239 MATHEMATICAL INDUCTION (Chapter 9) Prove that the following propositions are true for n Z + : 1 1 n a + + + ::::: + = 1£2 2£3 3£4 n(n + 1) n+1 1 1 and hence find + + + ::::: + 10 £ 11 11 £ 12 12 £ 13 20 £ 21 1 n(n + 3) b + + ::::: + = 1£2£3 2£3£4 n(n + 1)(n + 2) 4(n + 1)(n + 2) Prove that the following propositions are true for n Z + : a £ 1! + £ 2! + £ 3! + £ 4! + ::::: + n £ n! = (n + 1)! ¡ where n! is the product of the first n positive integers n (n + 1)! ¡ 1 + + + + ::::: + = , and hence find the sum 2! 3! 4! 5! (n + 1)! (n + 1)! + + + + ::::: + in rational form 2! 3! 4! 5! 10! b Prove that the following conjecture is true: n + 2(n ¡ 1) + 3(n ¡ 2) + ::::: + (n ¡ 2)3 + (n ¡ 1)2 + n = for all integers n > 1: n(n + 1)(n + 2) 1£6+2£5+3£4+4£3+5£2+6£1 = £ + £ + £ + £ + £ + (1 + + + + + 6) Hint: DIVISIBILITY Consider the expression 4n + for n = 0, 1, 2, 3, 4, 5, 40 + = = £ 41 + = = £ 42 + = 18 = £ 43 + = 66 = £ 22 44 + = 258 = £ 86 We observe that each of the answers is divisible by and so we make the conjecture “4n + is divisible by for all n Z + ” This proposition may or may not be true If it is true, then we should be able to prove it using the principle of mathematical induction 4n + can be proven to be divisible by by using the binomial expansion We observe that Note: 4n + = (1 + 3)n + ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ n ¢ n¡1 ¡n¢ n = 1n + n1 + n2 32 + n3 33 + n4 34 + :::::: + n¡1 + n +2 ¡n¢ ¡n¢ ¡n¢ ¡n¢ ¡ n ¢ n¡1 ¡n¢ n = + + + 3 + + :::::: + n¡1 + n ³ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ n ¢ n¡2 ¡n¢ n¡1 ´ = + n1 + n2 + n3 32 + n4 33 + :::::: + n¡1 + n cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 where the contents of the brackets is an integer black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_09\239IB_HL-2_09.CDR Friday, 26 October 2007 2:43:17 PM PETERDELL IB_HL-2ed (240) 240 MATHEMATICAL INDUCTION (Chapter 9) Example Prove that 4n + is divisible by for n Z , n > Pn is: “4n + is divisible by 3” for n Z , n > Proof: (By the principle of mathematical induction) ) P0 is true (1) If n = 0, 40 + = = £ If Pk is true, then 4k + = 3A where A is an integer Now 4k+1 + = 41 4k + = 4(3A ¡ 2) + fusing Pk g = 12A ¡ + = 12A ¡ = 3(4A ¡ 2) where 4A ¡ is an integer as A Z (2) Thus 4k+1 + is divisible by if 4k + is divisible by Hence P0 is true and Pk+1 is true whenever Pk is true ) Pn is true for all n Z , n > fPrinciple of mathematical inductiong Use the principle of mathematical induction to prove that: a b c d n3 + 2n is divisible by for all positive integers n n(n2 + 5) is divisible by for all integers n Z + 6n ¡ is divisible by for all integers n > 7n ¡ 4n ¡ 3n is divisible by 12 for all n Z + SEQUENCES Example A sequence is defined by: u1 = and un+1 = 2un + for all n Z + : Prove that un = 2n ¡ for all n Z + : Pn is: “if u1 = and un+1 = 2un + for all n Z + , then un = 2n ¡ 1.” Proof: (By the principle of mathematical induction) (1) If n = 1, u1 = 21 ¡ = ¡ = which is true and so P1 is true (2) If Pk is true, then uk = 2k ¡ and uk+1 = 2uk + = 2(2k ¡ 1) + fusing Pk g = 2k+1 ¡ + = 2k+1 ¡ ) Pk+1 is also true cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Thus, P1 is true and Pk+1 is true whenever Pk is true, so Pn is true for all n Z + fPrinciple of mathematical inductiong black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_09\240IB_HL-2_09.CDR Friday, 26 October 2007 2:48:55 PM PETERDELL IB_HL-2ed (241) MATHEMATICAL INDUCTION (Chapter 9) 241 Use the principle of mathematical induction to prove these propositions: a If sequence fun g is defined by: u1 = and un+1 = un + 8n + for all n Z + , then un = 4n2 + n b If the first term of a sequence is and subsequent terms are defined by the recursion formula un+1 = + 3un , then un = 2(3n¡1 ) ¡ un c If a sequence is defined by: u1 = and un+1 = for all n Z + , 2(n + 1) 22¡n then un = n! d If a sequence is defined by: u1 = and un+1 = un + (¡1)n (n + 1)2 for all (¡1)n¡1 n(n + 1) n Z + , then un = A sequence is defined by u1 = and un+1 = un + (2n + 1) for all n Z + By finding un for n = 2, and 4, conjecture a formula for un in terms of n only Prove that your conjecture is true using the principle of mathematical induction for all n Z + (2n + 1)(2n + 3) By finding un for n = 2, and 4, conjecture a formula for un in terms of n only Prove that your conjecture is true using the principle of mathematical induction p p 10 (2 + 3)n = An + Bn for all n Z + , where An and Bn are integers A sequence is defined by u1 = a b c d and un+1 = un + Find An and Bn for n = 1, 2, and Without using induction, show that An+1 = 2An + 3Bn and Bn+1 = An + 2Bn Calculate (An )2 ¡ 3(Bn )2 for n = 1, 2, and and hence conjecture a result Prove that your conjecture is true 11 Prove that un = 2n ¡ (¡1)n is an odd number for all n Z + 12 Another form of the principle of mathematical induction is: If Pn is a proposition defined for all n Z + , and if (1) P1 and P2 are true, and (2) Pk+2 is true whenever Pk and Pk+1 are true, then Pn is true for all n Z + Use this form to prove that: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a If a sequence un is defined by u1 = 11, u2 = 37 and un+2 = 5un+1 ¡ 6un for all n Z + , then un = 5(3n ) ¡ 2n+1 p p b If un = (3 + 5)n + (3 ¡ 5)n where n Z + , then un is a multiple of 2n : Hint: First find a and b such that un+2 = aun+1 + bun black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_09\241IB_HL-2_09.CDR Friday, 26 October 2007 2:59:54 PM PETERDELL IB_HL-2ed (242) 242 MATHEMATICAL INDUCTION (Chapter 9) OTHER APPLICATIONS Proof by the principle of mathematical induction is used in several other areas of mathematics For example, in establishing truths dealing with: ² ² ² ² ² ² inequalities differential calculus matrices geometrical generalisations products complex numbers You will find some proofs with these topics in the appropriate chapters later in the book Example Prove that a convex n-sided polygon has n(n ¡ 3) diagonals for all n > Pn is: “A convex n-sided polygon has n(n ¡ 3) diagonals for all n > 3” Proof: (By the principle of mathematical induction) (1) If n = we have a triangle and 12 £ £ (¡3) = There are diagonals ) P3 is true (2) If Pk is true, a convex k-sided polygon has If we label the vertices 1, 2, 3, 4, 5, ., k ¡ 1, k and k + as an additional vertex then k(k ¡ 3) diagonals k-4 k-3 k-2 k-1 k k+1 ¡ 2} +1 Pk+1 = Pk + k | {z the line from to k was once a side and is now a diagonal the number of diagonals from k + to the vertices 2, 3, 4, 5, ., k ¡ Try it for k¡=3 ¡ ,¡4, ¡ Pk+1 = 12 k(k ¡ 3) + k ¡ ) = 12 k(k ¡ 3) + 22 (k ¡ 1) = 12 [k2 ¡ 3k + 2k ¡ 2] = 12 [k2 ¡ k ¡ 2] = 12 (k + 1)(k ¡ 2) = 12 (k + 1)([k + 1] ¡ 3) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Thus P3 is true and Pk+1 is true whenever Pk is true ) Pn is true for all n > fPrinciple of mathematical inductiong black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_09\242IB_HL-2_09.CDR Friday, November 2007 10:26:58 AM PETERDELL IB_HL-2ed (243) 243 MATHEMATICAL INDUCTION (Chapter 9) 13 Use the principle of mathematical induction to prove the following propositions: µ ¶µ ¶µ ¶µ ¶ µ ¶ 1 1 1 a 1¡ 1¡ 1¡ 1¡ ::::: ¡ = , n Z + n+1 n+1 ¶ n µ X Note: The LHS 6= 1¡ since it is a product not a sum i+1 i=1 b If n straight lines are drawn on a plane such that each line intersects every other line and no three lines have a common point of intersection, then the plane is divided n(n + 1) into + regions c If n points are placed inside a triangle and non-intersecting lines are drawn connecting the vertices of the triangle and the points within it to partition the triangle into smaller triangles, then the number of triangles resulting is 2n + ¶µ ¶µ ¶ µ ¶ µ 1 n+1 1¡ ¡ ::::: ¡ = for all integers n > 2: d 1¡ 2 n 2n 14 Prove the following propositions to be true using the principle of mathematical induction: a 3n > + 2n for all n Z , n > b n! > 2n for all n Z , n > for all n Z + c >n n d (1 ¡ h)n 1 + nh for h and all n Z + INVESTIGATION SEQUENCES, SERIES AND INDUCTION This investigation involves the principle of mathematical induction as well as concepts from sequences, series, and counting What to do: The sequence of numbers fun g is defined by u1 = £ 1!, u2 = £ 2!, u3 = £ 3!, etc What is the nth term of the sequence? Let Sn = u1 + u2 + u3 + ::::: + un Investigate Sn for several different values of n Based on your results from 2, conjecture an expression for Sn Prove your conjecture to be true using the principle of mathematical induction Show that un can be written as (n + 1)! ¡ n! and devise an alternative direct proof of your conjecture for Sn Let Cn = un + un+1 Write an expression for Cn in factorial notation and simplify it Let Tn = C1 + C2 + C3 + ::::: + Cn and find Tn for n = 1, 2, 3, and Conjecture an expression for Tn cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Prove your conjecture for Tn by any method black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_09\243IB_HL-2_09.CDR Friday, 12 December 2008 12:01:33 PM TROY IB_HL-2ed (244) 244 MATHEMATICAL INDUCTION (Chapter 9) INDIRECT PROOF (EXTENSION) C Some propositions may be proven to be true by using an indirect proof such as proof by contradiction In such proofs we suppose the opposite of the statement to be true and, by using correct argument, hope to obtain a contradiction Example Prove that the sum of any positive real number and its reciprocal is at least Proof: (by contradiction) Suppose that x + < for some x > x µ ¶ ) x x+ < 2x fmultiplying both sides by x where x > 0g x ) x2 + < 2x ) x2 ¡ 2x + < ) (x ¡ 1)2 < which is a contradiction as no perfect square of a real number can be negative So, the supposition is false and its opposite x + > 2, x > must be true x Example Prove that the solution of 2x = is an irrational number Proof: (by contradiction) Suppose that if 2x = then x is rational ) p q = for some positive integers p and q, q 6= p ) (2 q )q = 3q ) 2p = 3q which is clearly a contradiction, as the LHS = 2p is even and the RHS = 3q is odd ) the supposition is false and its opposite is true i.e., if 2x = then x is irrational p where p and q are q integers, q = and p, q have no common factors magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 25 95 100 50 75 25 cyan 75 A rational number can be written in the form Reminder: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_09\244IB_HL-2_09.CDR Friday, 26 October 2007 3:18:54 PM PETERDELL IB_HL-2ed (245) MATHEMATICAL INDUCTION (Chapter 9) 245 EXERCISE 9C Use proof by contradiction to prove that: a the sum of a positive number and nine times its reciprocal is at least b the solution of 3x = is an irrational number c log2 is irrational p Challenge: Prove by contradiction that is irrational REVIEW SET 9A Prove the following propositions using the principle of mathematical induction: n X (2i ¡ 1) = n2 , n Z + i=1 7n + is divisible by 3, n Z + n X i(i + 1)(i + 2) = i=1 n(n + 1)(n + 2)(n + 3) , n Z + 4 + r + r2 + r3 + r4 + ::::: + rn¡1 = ¡ rn , n Z + , provided that r 6= 1¡r 52n ¡ is divisible by 24, n Z + : 5n > + 4n, n Z + : If u1 = and un+1 = 3un + 2n , then un = 3n ¡ 2n , n Z + REVIEW SET 9B Prove the following propositions using the principle of mathematical induction: n X (2i ¡ 1)2 = i=1 n(2n + 1)(2n ¡ 1) n Z + , n > 1: 32n+2 ¡ 8n ¡ is divisible by 64 for all positive integers n n X (2i + 1)2i¡1 = + (2n ¡ 1) £ 2n for all positive integers n i=1 5n + is divisible by for all integers n > 0: n X i(i + 1)2 = i=1 n(n + 1)(n + 2)(3n + 5) 12 for all positive integers n 5n + 3n > 22n+1 , n Z + : cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 If u1 = and un+1 = 2un + 3(5n ), then un = 2n+1 + 5n , n Z + black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_09\245IB_HL-2_09.cdr Friday, November 2007 10:27:41 AM PETERDELL IB_HL-2ed (246) 246 MATHEMATICAL INDUCTION (Chapter 9) REVIEW SET 9C Prove the following propositions using the principle of mathematical induction: n X i(i + 2) = i=1 n(n + 1)(2n + 7) , n Z +: 7n ¡ is divisible by 6, n Z + : n X (2i ¡ 1)3 = n2 (2n2 ¡ 1) for all positive integers n > 1: i=1 3n ¡ ¡ 2n is divisible by for all non-negative integers n n X i=1 n = for all positive integers n (2i ¡ 1)(2i + 1) 2n + If u1 = and un+1 = 2un ¡ 3(¡1)n , then un = 3(2n ) + (¡1)n, n Z + cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 p n+1 n n! , n Z +: 100 50 75 25 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_09\246IB_HL-2_09.cdr Thursday, 11 March 2010 10:23:47 AM PETER IB_HL-2ed (247) Chapter 10 The unit circle and radian measure Contents: A B C D Radian measure Arc length and sector area The unit circle and the basic trigonometric ratios Areas of triangles cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 10A Review set 10B Review set 10C black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\247IB_HL-2_10.CDR Tuesday, 30 October 2007 2:44:25 PM PETERDELL IB_HL-2ed (248) 248 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) Before starting this chapter you should make sure that you have a good understanding of the necessary background knowledge in trigonometry and Pythagoras Click on the icon alongside to obtain a printable set of exercises and answers on this background knowledge BACKGROUND KNOWLEDGE OPENING PROBLEM Consider an equilateral triangle with sides 10 cm long All its angles are of size 60o Altitude AN bisects side BC and the vertical angle BAC Can you see from this figure that sin 30o = 12 ? Use your calculator to find the values of sin 30o , sin 150o , sin 390o , sin 1110o and sin(¡330o ) What you notice? Can you explain why this result occurs even though the angles are not between 0o and 90o ? ² ² A A 30° 30° 60° B N 10 cm 60° C cm RADIAN MEASURE DEGREE MEASUREMENT OF ANGLES Recall that one full revolution makes an angle of 360o and a straight angle is 180o Hence, th of one full revolution This measure of angle is one degree, 1o , can be defined as 360 probably most useful for surveyors and architects, and is the one you have probably used in earlier years th of one degree and one second, 100 , For greater accuracy we define one minute, 10 , as 60 th of one minute Obviously a minute and a second are very small angles as 60 Most graphics calculators have the capacity to convert fractions of angles measured in degrees into minutes and seconds This is also useful for converting fractions of hours into minutes 1 and seconds for time measurement, as one minute is 60 th of one hour, and one second is 60 th of one minute RADIAN MEASUREMENT OF ANGLES An angle is said to have a measure of radian (1c ) if it is subtended at the centre of a circle by an arc equal in length to the radius The symbol ‘c’ is used for radian measure but is usually omitted, whilst the degree symbol is always used r 60° when the measure of an angle is r given in degrees c arc length =¡r 1c radius¡=¡r From the diagram to the right, it can be seen that 1c is slightly smaller than 60o In fact, 1c ¼ 57:3o cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The word “radian” is an abbreviation of “radial angle” black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\248IB_HL-2_10.CDR Friday, November 2007 9:36:33 AM PETERDELL IB_HL-2ed (249) 249 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) DEGREE-RADIAN CONVERSIONS If the radius of a circle is r, then an arc of length 2r will subtend an angle of radians at the centre An arc of length ¼r (half the circumference) will subtend an angle of ¼ radians ¼ radians ´ 180o Therefore, So, 1c = ¡ 180 ¢o ¼ ¼ 57:3o and 1o = ¡ ¼ 180 ¢c ¼ 0:0175c To convert from degrees to radians, multiply by ¼ 180 : To convert from radians to degrees, multiply by 180 ¼ : We can summarise these results in the conversion diagram: If degrees are used we indicate this with a small o To indicate radians, we can use a small c or else use no symbol at all £ ¼ 180 Degrees Radians £ 180 ¼ Example Convert 45o to radians in terms of ¼ 45o = (45 £ = ¼ ¼ 180 ) radians 180o = ¼ radians ¡ 180 ¢o = ¼4 radians or ) radians ) 45o = ¼ radians Example Convert 126:5o to radians 126:5o ¼ = (126:5 £ 180 ) radians ¼ 2:21 radians (3 s.f.) Example Convert degrees 5¼ 5¼ to = ¡ 5¼ £ ¢ 180 o ¼ Notice that angles in radians are expressed either in terms of p or as decimals = 150o Example cyan magenta yellow 95 100 50 75 25 95 0:638 radians o = (0:638 £ 180 ¼ ) o ¼ 36:6 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Convert 0:638 radians to degrees black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\249IB_HL-2_10.CDR Tuesday, 30 October 2007 2:44:35 PM PETERDELL IB_HL-2ed (250) 250 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) EXERCISE 10A Convert to radians, in b a 90o f 135o g o k 315 l terms of ¼: 60o c 225o h o 540 m 30o 270o 36o Convert to radians (correct to s.f.): b 137:2o c 317:9o a 36:7o d Convert the following radian measure to degrees: b 3¼ c 3¼ a ¼5 d 7¼ f ¼ 10 g 3¼ 20 h 18o 360o 80o d i n 219:6o ¼ 18 5¼ i Convert the following radians to degrees (to decimal places): a b 1:53 c 0:867 d 3:179 e j o e e j e 9o 720o 230o 396:7o ¼ ¼ 5:267 Copy and complete: a Degrees Radians 45 90 135 180 225 270 315 360 b Degrees Radians 30 60 90 B 120 150 180 210 240 270 300 330 360 ARC LENGTH AND SECTOR AREA ARC LENGTH In the diagram, µ is measured in radians A l q O arc length µ = circumference 2¼ l µ ) = 2¼r 2¼ B r ) l = µr AREA OF SECTOR µ is measured in radians X area of minor sector XOY µ = area of circle 2¼ A µ ) = ¼r2 2¼ r O q Y r ) A = 12 µr2 cyan magenta yellow 95 µ £ ¼r2 360 100 50 75 and A = 25 95 100 50 µ £ 2¼r 360 75 25 95 100 50 75 25 95 100 50 75 25 If µ is in degrees, l = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\250IB_HL-2_10.CDR Wednesday, February 2008 4:42:29 PM PETERDELL IB_HL-2ed (251) 251 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) Example A sector has radius 12 cm and angle 65o Use radians to find its: a arc length b area a area = 12 µr2 65¼ = 12 £ £ 122 180 ¼ 81:7 cm2 b arc length = µr 65¼ = £ 12 180 ¼ 13:6 cm Example A sector has radius 8:2 cm and arc length 13:3 cm Find its angle in radians and degrees fµ in radiansg l = µr 13:3 l = ¼ 1:62c r 8:2 13:3 180 and µ = £ ¼ 92:9o 8:2 ¼ ) µ= EXERCISE 10B Use radians to find: i the arc length a radius cm and angle 41:6o ii b the area of a sector of a circle of radius 4:93 cm and angle 122o A sector has an angle of 107:9o and an arc length of 5:92 m Find: a its radius b its area A sector has an angle of 68:2o and an area of 20:8 cm2 Find: a its radius b its perimeter Find the angle of a sector of: a radius 4:3 m and arc length 2:95 m radius 10 cm and area 30 cm2 b Find µ (in radians) for each of the following, and hence find the area of each figure: a b c 31.7 cm q cm cm cm q cm 8.4 cm q Find the arc length and area of a sector of radius cm and angle radians cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 If a sector has radius 10 cm and arc length 13 cm, find its area black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\251IB_HL-2_10.CDR Tuesday, 30 October 2007 2:44:44 PM PETERDELL IB_HL-2ed (252) 252 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) This cone is made from this sector: 10 cm s cm q° 12 cm Find correct to significant figures: a the slant length s cm b c the arc length of the sector d A the value of r the sector angle (µ o ) The end wall of a building has the shape illustrated, where the centre of arc AB is at C Find: a ® to significant figures b µ to significant figures c the area of the wall B 30 m r cm 5m q° a° C nautical mile (nmi) 10 A nautical mile (nmi) is the distance on the Earth’s surface that subtends an angle of minute (where 1 degree) of the Great Circle arc measured minute = 60 from the centre of the Earth A knot is a speed of nautical mile per hour a Given that the radius of the Earth is 6370 km, show that nmi is approximately equal to 1:853 km b Calculate how long it would take a plane to fly from Perth to Adelaide (a distance of 2130 km) if the plane can fly at 480 knots N 1' P Q C S 11 A sheep is tethered to a post which is m from a long fence The length of rope is m Find the area which is available for the sheep to feed on fence 6m post S 12 A belt fits tightly around two pulleys of radii cm and cm respectively which have a distance of 20 cm between their centres a° f° q° Find, correct to significant figures: cyan magenta yellow 95 100 50 75 25 95 100 50 d the length of the belt 75 25 95 c Á 100 50 75 25 b µ 95 100 50 75 25 a ® black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\252IB_HL-2_10.CDR Friday, November 2007 9:37:45 AM PETERDELL IB_HL-2ed (253) 253 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) C THE UNIT CIRCLE AND THE BASIC TRIGONOMETRIC RATIOS y The unit circle is the circle with centre (0, 0) and radius unit x the unit circle CIRCLES WITH CENTRE (0, 0) y Consider a circle with centre (0, 0) and radius r units, and suppose P(x, y) is any point on this circle P(x, y) r Since OP = r, then p (x ¡ 0)2 + (y ¡ 0)2 = r fdistance formulag x (0,¡0) x2 + y = r2 ) x2 + y2 = r2 is the equation of a circle with centre (0, 0) and radius r The equation of the unit circle is x2 + y2 = So, ANGLE MEASUREMENT y Positive direction Suppose P lies anywhere on the unit circle and A is (1, 0) Let µ be the angle measured from [OA] on the positive x-axis q µ is positive for anticlockwise rotations and negative for clockwise rotations x A P For example, µ = 210o and Á = ¡150o y Negative direction q x 30° f DEFINITION OF SINE, COSINE AND TANGENT So, as point P moves anywhere on the unit circle, cos µ is the x-coordinate of P sin µ is the y-coordinate of P 1 -1 yellow 95 100 50 75 25 95 100 50 75 25 95 50 75 25 95 100 50 75 25 100 magenta black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\253IB_HL-2_10.CDR Friday, January 2008 9:14:22 AM DAVID3 P(cos¡q, sin¡q) q -1 where µ is the angle made by [OP] with the positive x-axis cyan y x -1 q -1 y x P(cos¡q, sin¡q) IB_HL-2ed (254) 254 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) For all points on the unit circle: ¡1 x and ¡1 y So, ¡1 cos µ and ¡1 sin µ for all µ Also, we have already seen that the equation of the unit circle is x2 + y = cos2 µ + sin2 µ = for all µ This leads to the identity tan µ = The tangent ratio is defined as sin µ cos µ By considering the first quadrant, we can easily see that the right angled triangle definitions of sine, cosine and tangent are consistent with the unit circle definition, but are restricted to acute angles only The unit circle definition applies to angles of any value From the definition, cos ¼2 = and sin ¼2 = If any integer multiple of 2¼ is added to µ, P will still be at (0, 1) ¡ ¢ ¡ ¢ Therefore, cos ¼2 + 2k¼ = and sin ¼2 + 2k¼ = 1, k Z : More generally, adding integer multiples of 2¼ to any value of µ will not change the position of P So for all k Z and angles µ, cos (µ + 2k¼) = cos µ and sin (µ + 2k¼) = sin µ This periodic feature is an important property of the trigonometric functions Example Use the unit circle to show that cos ¡¼ ¢ + µ = ¡ sin µ Consider point B(a, b) on the unit circle such that [OB] makes an angle of µ with the positive x-axis Thus a = cos µ and b = sin µ ¢ ¡ Now consider point P on the unit circle such that [OP] makes an angle of ¼2 + µ with the positive x-axis y b = ¼ Now OB = OP = unit and BOP P so P is obtained from B by an anticlockwise B(a,¡b) ¼ o rotation of or 90 about the origin b ) from transformation geometry, P has q a coordinates (¡b, a) or (¡ sin µ, cos µ) x But the coordinates of P are (cos( ¼2 + µ), sin( ¼2 + µ)) ) cos( ¼2 + µ) = ¡ sin µ Example y (0'\1) cyan magenta yellow 95 100 50 75 25 95 100 50 ) cos(¡270o ) = and sin(¡270o ) = 75 25 95 x -270° 100 50 75 25 95 100 50 75 25 Use a unit circle diagram to find the values of cos(¡270o ) and sin(¡270o ): black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\254IB_HL-2_10.CDR Friday, January 2008 9:15:58 AM DAVID3 fthe x-coordinateg fthe y-coordinateg IB_HL-2ed (255) 255 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) Example Find the possible values of cos µ for sin µ = 23 Illustrate cos2 µ + sin2 µ = ¡ ¢2 ) cos2 µ + 23 = cos2 µ = ) ) cos µ = y We_ p § 35 ~`5 -\ _ x ~`5 _ Helpful hint: We have seen previously how the quadrants of the Cartesian Plane are labelled in anticlockwise order from 1st to 4th We can also use a letter to show which trigonometric ratios are positive in each quadrant You might like to remember them using y 2nd 1st S A T C 3rd All Silly Turtles Crawl x 4th Example 10 If sin µ = ¡ 34 and ¼ < µ < 3¼ , find cos µ without using a calculator Since ¼ < µ < µ is a quad angle and ) cos µ is negative y S Now cos2 µ + sin2 µ = =1 ) cos2 µ + 16 A q T 3¼ , x ) C -\Er_ ) cos2 µ = cos µ = and since cos µ is negative, cos µ = 16 p § 47 p ¡ 47 : or we can use a working angle ® in quadrant 1, where ® is the acute angle symmetric with µ In this case sin ® = 4 2 so n = ¡ = fPythagorasg p p ) n = and so cos ® = 47 a S But µ is in quad where cos µ is negative cyan magenta yellow 95 100 50 75 25 95 100 50 T 75 25 p 95 100 50 75 25 95 100 50 75 25 so cos µ = ¡ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\255IB_HL-2_10.CDR Friday, November 2007 9:45:42 AM PETERDELL n q A a C IB_HL-2ed (256) 256 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) Example 11 If sin x = ¡ 13 and ¼ < x < Consider sin X = X is acute find the value of tan x, without finding x where So X p 8: This side is p1 ) tan X = fPythagorasg y p1 ) tan x = 3¼ , S fsince x lies in quad where tan x > 0g A x x T C Example 12 If tan x = and ¼ < x < 3¼ , find sin x and cos x x is in quadrant ) sin x < and cos x < y S A x T Consider tan X = X x C ) sin X = where X is acute and cos X = ) sin x = ¡ 35 and cos x = ¡ 45 EXERCISE 10C.1 Sketch the graph of the curve with equation: b x2 + y = a x2 + y = c x2 + y = 1, y > For each angle illustrated: i write down the actual coordinates of points A, B and C in terms of sine or cosine ii use your calculator to give the coordinates of A, B and C correct to significant figures a b y y B -1 A 146° A 26° 199° 123° -1 x 251° C x -35° C cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 B -1 100 50 75 25 95 100 50 75 25 -1 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\256IB_HL-2_10.CDR Wednesday, November 2007 4:12:31 PM PETERDELL IB_HL-2ed (257) 257 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) With the aid of a unit circle, complete the following table: 0o µ (degrees) µ (radians) sine cosine tangent 90o 180o 270o 360o 450o Use a unit circle to show the following: a sin(¼ + µ) = ¡ sin µ b sin Without using a sin 137o c cos 143o e sin 115o b d f sin 59o if sin 121o ¼ 0:8572 cos 24o if cos 156o ¼ ¡0:9135 cos 132o if cos 48o ¼ 0:6691 your calculator find: if sin 43o ¼ 0:6820 if cos 37o ¼ 0:7986 if sin 65o ¼ 0:9063 ¡ 3¼ ¢ + µ = ¡ cos µ Find the possible exact values of cos µ for: b sin µ = ¡ 13 a sin µ = 12 c sin µ = d sin µ = ¡1 Find the possible exact values of sin µ for: b cos µ = ¡ 34 a cos µ = 45 c cos µ = d cos µ = y The diagram alongside shows the quadrants They are numbered anticlockwise 2nd 1st 3rd 4th x a Copy and complete: Quadrant Degree measure < µ < 90 Radian measure 0<µ< ¼ cos µ sin µ positive positive tan µ b In which quadrants are the following true? i cos µ is positive ii cos µ is negative iii cos µ and sin µ are both negative iv cos µ is negative and sin µ is positive Without using a calculator, find: a sin µ if cos µ = 23 , < µ < magenta yellow 2nd S A 1st 3rd T C 4th All Silly Turtles Crawl 95 sin µ if cos µ = ¡ 13 , ¼<µ< 100 d 50 < µ < 2¼ 75 cos µ if sin µ = 25 , 25 b ¼ 95 3¼ 100 50 75 25 95 100 50 25 95 100 50 75 25 cyan 75 cos µ if sin µ = ¡ 35 , c Remember: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\257IB_HL-2_10.CDR Tuesday, 30 October 2007 2:45:08 PM PETERDELL ¼ <µ<¼ 3¼ IB_HL-2ed (258) 258 10 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) a If sin x = and ¼ b If cos x = and 3¼ c If sin x = ¡ p13 d If cos x = ¡ 34 < x < ¼, find tan x in radical (surd) form < x < 2¼, find tan x in radical (surd) form 3¼ , and ¼ < x < ¼ and find tan x in radical (surd) form < x < ¼, find tan x in radical (surd) form 11 Find sin x and cos x given that: a tan x = c tan x = p and < x < ¼ and ¼ < x < 3¼ INVESTIGATION b tan x = ¡ 43 d tan x = ¡ 12 and and ¼ <x<¼ 3¼ < x < 2¼ NEGATIVE AND COMPLEMENTARY ANGLE FORMULAE The purpose of this investigation is to discover relationships (if they exist) between: ² cos(¡µ), sin(¡µ), cos µ and sin µ ² cos( ¼2 ¡ µ), sin( ¼2 ¡ µ), cos µ and sin µ Note: ¡µ is the negative of µ ¼ and ¡µ is the complement of µ What to do: Copy and complete, adding angles of your choice to the table: µ sin µ cos µ sin(¡µ) sin( ¼2 ¡ µ) cos(¡µ) cos( ¼2 ¡ µ) 2:67 0:642 ¼ From your table in make a prediction on how to simplify sin(¡µ), cos(¡µ), sin( ¼2 ¡ µ) and cos( ¼2 ¡ µ) NEGATIVE ANGLE FORMULAE P and P0 have the same x-coordinate, but their y-coordinates are negatives y P(cos q,sin ¡ q) x ) tan(¡µ) = P'(cos (-q),sin ¡ (-q)) yellow 95 100 50 75 25 95 100 50 75 25 95 50 75 25 95 100 50 75 25 100 magenta cos(¡µ) = cos µ sin(¡µ) = ¡ sin µ tan(¡µ) = ¡ tan µ So, GRAPHING PACKAGE cyan ¡ sin µ sin(¡µ) = cos(¡µ) cos µ ) tan(¡µ) = ¡ tan µ q -q Hence cos(¡µ) = cos µ and sin(¡µ) = ¡ sin µ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\258IB_HL-2_10.CDR Tuesday, 30 October 2007 2:45:12 PM PETERDELL IB_HL-2ed (259) 259 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) COMPLEMENTARY ANGLE FORMULAE y Consider P0 on the unit circle which corresponds to the angle ¼2 ¡ µ ) P is (cos( ¼2 ¡ µ), sin( ¼2 q P' y=x P(cos¡q,¡sin¡q) q ¡ µ)) (1) But P is the image of P under a reflection in the line y = x: p -q ¡ is shaded GRAPHING PACKAGE ) P0 is (sin µ, cos µ) (2) Comparing (1) and (2) gives cos( ¼2 ¡ µ) = sin µ and sin( ¼2 ¡ µ) = cos µ x cos( ¼2 ¡ µ) = sin µ sin( ¼2 ¡ µ) = cos µ Example 13 a Simplify: a sin(¡µ) + sin µ b cos µ + cos(¡µ) b sin(¡µ) + sin µ = ¡2 sin µ + sin µ = sin µ cos µ + cos(¡µ) = cos µ + cos µ = cos µ Example 14 sin( ¼2 ¡ µ) + cos µ = cos µ + cos µ = cos µ Simplify: ¡ ¢ sin ¼2 ¡ µ + cos µ EXERCISE 10C.2 Simplify: a sin µ + sin(¡µ) b tan(¡µ) ¡ tan µ sin µ ¡ sin(¡µ) e cos (¡®) cos(¡®) cos ® ¡ sin(¡®) sin ® d g Simplify: a sin µ ¡ cos(90o ¡ µ) cos(¡µ) ¡ sin( ¼2 ¡ µ) d c cos µ + cos(¡µ) f sin2 (¡®) b sin(¡µ)¡cos(90o ¡µ) c sin(90o ¡ µ) ¡ cos µ e cos µ + sin( ¼2 ¡ µ) f cos( ¼2 ¡ µ) + sin µ Explain why sin(µ ¡ Á) = ¡ sin(Á ¡ µ), cos(µ ¡ Á) = cos(Á ¡ µ): cyan magenta yellow 95 100 50 75 25 cos( ¼2 ¡ µ) cos µ 95 f 100 cos( ¼2 ¡ µ) sin( ¼2 ¡ µ) 50 e 75 ¡ sin(¡µ) cos µ 25 d sin( ¼2 ¡ µ) cos µ c 95 sin(¡µ) cos(¡µ) 100 b 50 sin µ cos µ 75 a 25 95 100 50 75 25 Simplify: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\259IB_HL-2_10.CDR Tuesday, 30 October 2007 2:45:17 PM PETERDELL IB_HL-2ed (260) 260 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) INVESTIGATION PARAMETRIC EQUATIONS Usually we write functions in the form y = f(x) For example: y = 3x + 7, y = x2 ¡ 6x + 8, y = sin x However, sometimes it is useful to express both x and y in terms of another variable, t say, called the parameter In this case we say we have parametric equations GRAPHING PACKAGE TI C What to do: Either click on the icon or use your graphics calculator (with the same scale on both axes) to plot f(x, y): x = cos t, y = sin t, 0o t 360o g Note: Your calculator will need to be set to degrees Describe the resulting graph What is the equation of this graph? (There are two possible answers.) If using a graphics calculator, use the trace key to move along the curve What you notice? MULTIPLES OF ¼ AND ¼ The following diagrams may be helpful when finding exact trigonometric ratios 30° ~`2 45° 60° 1 Consider µ = ¼ y ¼ MULTIPLES OF ~`3 P(a, a) = 45o : 45° a x 45° a B Triangle OBP is isosceles as angle OPB measures 45o also ) OB = BP = a, say and a2 + a2 = 12 ) 2a2 = p1 ) p1 where (-1, 0) ³ yellow 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 So, we have: magenta ³ p1 ; p1 2 ´ p 5p Consequently we can find the coordinates 5¼ 7¼ corresponding to angles of 3¼ , and using suitable rotations and reflections cyan y (0, 1) ´ 3p ¼ 0:7 : ¡ p12 ; ¡ p12 95 P is ( p12 , as a > ¡ p12 ; p12 (1, 0) 7p ³ ´ 100 Hence, a= ³ 50 ) p1 75 a2 = ) fPythagorasg black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\260IB_HL-2_10.CDR Friday, November 2007 10:09:19 AM PETERDELL p1 ; ¡ p1 2 x ´ (0,-1) IB_HL-2ed (261) 261 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) ¼ MULTIPLES OF ¼ Consider µ = Triangle OAP is isosceles with vertical angle 60o The remaining angles are therefore 60o and so triangle AOP is equilateral The altitude [PN] bisects base [OA], = 60o : y P(1–2 , k) ) ON = If P is ( 12 , k), then ( 12 )2 + k2 = ) k2 = 34 k 60° x A(1'\\0) Qw_ N ) ) P is ( 12 , Consequently, we can find the coordinates of all points on the unit circle corresponding to multiples of ¼6 using rotations and reflections p fas k > 0g p ) ´ ³ p ¡ 23 ; 12 p where y (0,¡1) ³ p ´ ¡ 12 ; 23 So we have: k= ³ p ´ 2; ³p 7p ; 2p ¡ 12 ; ¡ p ´ x (1,¡0) ´ ³p ; ¡2 (-1,¡0) ³ p ´ ¡ 23 ; ¡ 12 ³ ¼ 0:9 ´ ³ (0,-1) p ´ 2;¡ Summary: ² If µ is a multiple of and §1 ¼ , the coordinates of the points on the unit circle involve ² If µ is a multiple of ¼ , (but not a multiple of ¼ ), the coordinates involve § p12 : ² If µ is a multiple of ¼ , (but not a multiple of ¼ ), the coordinates involve § 12 and p § 23 You should not try to memorise the coordinates on the above circles for every multiple of and ¼4 , but rather use the summary to work out the correct result Consider 225o = For example: 5¼ y y 5p 5p x ¡ p12 ; ¡ p12 ´ »-0.9 95 100 50 75 25 95 50 75 25 95 100 50 75 25 95 100 50 75 25 100 yellow x ³ p ´ 2;¡ is in quad 4, so signs are (+, ¡) and from the diagram the x-value is 12 is in quad 3, so signs are both negative and both have p12 size magenta Qw_ 5¼ 5¼ cyan 5¼ of ¼6 Consider 300o = i.e., a multiple ³ ¼ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\261IB_HL-2_10.CDR Friday, January 2008 9:21:42 AM DAVID3 IB_HL-2ed (262) 262 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) Example 15 Use a unit circle to find the exact values of sin ®, cos ® and tan ® for ® = ³ ¡ p12 ; p12 3¼ : y ´ » 0.7 3p ) x 3¼ p1 cos( 3¼ ) = ¡ , sin( ) = ¡ ¢ = ¡1 tan 3¼ p1 »-0.7 Example 16 y Use a unit circle diagram to find the exact values of sin A, cos A and tan A for A = 4¼ : x p ´ ¡ 2; p p ¡ 23 ¡ 12 and sin( 4¼ ) = ¡ 4p -1–2 ³ ) cos( 4¼ ) = ¡2 ) tan( 4¼ ) = »-0.9 ) tan( 4¼ ) = p Example 17 Without using a calculator, find the value of sin( ¼3 ) cos( 5¼ ): ³ y ³ ´ p ¡ ;2 p ´ 2; sin( ¼3 ) = 5p ) x p p p p p 8( 23 )(¡ 23 ) and cos( 5¼ )=¡ sin( ¼3 ) cos( 5¼ ) = = 2(¡3) = ¡6 Example 18 Use a unit circle diagram to find all angles in [0, 2¼] with a cosine of 12 As the cosine is 12 , we draw the vertical line x = 12 : y 5p p Because 12 is involved we know the required angles are multiples of ¼6 They are ¼3 and 5¼ x Qw_ cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 !=\Qw_ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\262IB_HL-2_10.CDR Friday, November 2007 10:17:36 AM PETERDELL IB_HL-2ed (263) 263 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) EXERCISE 10C.3 Use a unit circle diagram to find sin µ, cos µ and tan µ for µ equal to: b 5¼ c 7¼ d ¼ e a ¼4 4 ¡3¼ Use a unit circle diagram to find sin ¯, cos ¯ and tan ¯ for ¯ equal to: ¼ a 2¼ b 7¼ c d 5¼ 11¼ e Without using a calculator, evaluate: a sin2 60o d ¡ cos2 g j c sin 60o cos 30o e sin 30o cos 60o ¡ ¢ sin2 2¼ ¡1 f cos2 ( ¼4 ) ¡ sin( 7¼ ) 5¼ sin( 3¼ ) ¡ cos( ) h ¡ sin2 ( 7¼ ) i 5¼ cos2 ( 5¼ ) ¡ sin ( ) tan2 ( ¼3 ) ¡ sin2 ( ¼4 ) k 3¼ tan(¡ 5¼ ) ¡ sin( ) l tan 150o ¡ tan2 150o b ¡¼¢ Check all answers using your calculator Use a unit circle diagram to find all angles between 0o and 360o with: p a a sine of d a cosine of ¡ 12 b a sine of e a cosine of ¡ p12 p1 p ¡ c a cosine of f a sine of Use a unit circle diagram to find all angles between and 2¼ which have: p a a tangent of b a tangent of ¡1 c a tangent of p f a tangent of ¡ d a tangent of e a tangent of p13 Use a unit circle diagram to find all angles between and 4¼ with: a a cosine of p b a sine of ¡ 12 Find µ in radians if µ 2¼ and: a cos µ = e i p b sin µ = cos µ = ¡ p12 f sin2 µ = tan µ = ¡ p13 j tan2 µ = D c a sine of ¡1 c cos µ = ¡1 d sin µ = g cos2 µ = h cos2 µ = AREAS OF TRIANGLES DEMO height base base base If we know the base and height measurements of a triangle we can calculate the area using cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 base £ height 100 50 75 25 95 100 50 75 25 area = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\263IB_HL-2_10.CDR Tuesday, 30 October 2007 2:45:35 PM PETERDELL IB_HL-2ed (264) 264 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) However, cases arise where we not know the height but we can still calculate the area These cases are: ² knowing two sides and the included angle between them For example: ² knowing all three sides cm cm cm 49° 10 cm 10 cm USING THE INCLUDED ANGLE If triangle ABC has angles of size A, B and C, the sides opposite these angles are labelled a, b and c respectively Using trigonometry, we can develop an alternative formula that does not depend on a perpendicular height A A c B B Any triangle that is not right angled must be either acute or obtuse We will consider both cases: b C a A A c c b h B C D h b B C a a (180°-C) D C In both triangles a perpendicular is constructed from A to D on [BC] (extended if necessary) h b h = b sin C ) So, area = 12 ah gives h b ) h = b sin(180o ¡ C) but sin(180o ¡ C) = sin C ) h = b sin C sin(180o ¡ C) = sin C = area = 12 ab sin C bc sin A Using different altitudes we can show that the area is also or ac sin B: Summary: Given the lengths of two sides of a triangle and the included angle between them, the area of the triangle is a half of the product of two sides and the sine of the included angle side included angle side Example 19 Area = 12 ac sin B Find the area of triangle ABC: A = 11 cm magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 C 15 cm 95 100 50 75 25 cyan £ 15 £ 11 £ sin 28o ¼ 38:7 cm2 28° B black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\264IB_HL-2_10.CDR Tuesday, 30 October 2007 2:45:39 PM PETERDELL IB_HL-2ed (265) 265 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) Example 20 A triangle has sides of length 10 cm and 11 cm and an area of 50 cm2 Show that the included angle may have two possible sizes If the included angle measures µo , then Now arcsin ¡ 50 ¢ 55 £ 10 £ 11 £ sin µ = 50 ) sin µ = 50 55 ¼ 65:4 10 ) µ ¼ 65:4 or 180 ¡ 65:4 i.e., µ ¼ 65:4 or 114:6 10 11 65.4° 11 114.6° So, the two different possible angles are 65:4o and 114:6o EXERCISE 10D Find the area of: a b cm c 31 km 82° 40° 10.2 cm 25 km 2p 10 cm 6.4 cm A 2 If triangle ABC has area 150 cm , find the value of x: 17 cm B 68° C x cm A parallelogram has two adjacent sides of length cm and cm respectively If the included angle measures 52o , find the area of the parallelogram A rhombus has sides of length 12 cm and an angle of 72o Find its area Find the area of a regular hexagon with sides of length 12 cm A rhombus has an area of 50 cm2 and an internal angle of size 63o Find the length of its sides A regular pentagonal garden plot has centre of symmetry O and an area of 338 m2 Find the distance OA A cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find the possible values of the included angle of a triangle with: a sides cm and cm, and area 15 cm2 b sides 45 km and 53 km, and area 800 km2 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\265IB_HL-2_10.CDR Tuesday, 30 October 2007 2:45:43 PM PETERDELL IB_HL-2ed (266) 266 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) The Australian 50 cent coin has the shape of a regular dodecagon (12 sides) Eight of these 50 cent coins will fit exactly on an Australian $10 note as shown What fraction of the $10 note is not covered? 10 Find the shaded area in: a b A 12 cm 12 cm O B 18 cm P 0.66 c 1.5 HERON’S FORMULA In the first century A.D., Heron of Alexandria showed that if a triangle has sides of length a, b and c, then its area can be calculated using A= p s(s ¡ a)(s ¡ b)(s ¡ c) s= where a+b+c a Find the area of the right angled triangle with sides cm, cm and cm: i without using Heron’s formula ii using Heron’s formula b Find the area of a triangle with sides of length: ii 7:2 cm, 8:9 cm and 9:7 cm i cm, cm and 12 cm 11 REVIEW SET 10A Determine the area of: 7.3 km 38° 9.4 km Determine the area of: a a sector of angle 80o and radius 13 cm b a triangle with sides 11 cm, cm and included angle 65o y 10° Find the coordinates of the points M, N and P on the unit circle N M 73° -53° x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 P black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\266IB_HL-2_10.CDR Wednesday, November 2007 4:24:12 PM PETERDELL IB_HL-2ed (267) 267 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) Find the angle [OA] makes with the positive x-axis if the x-coordinate of the point A on the unit circle is ¡0:222 Find the acute angles that would have the same: b sine as 165o a sine as 2¼ c cosine as 276o Find the obtuse angles which have the same: ¼ b sine as 15 a sine as 47o c cosine as 186o Without using your calculator, find: a sin 159o if sin 21o ¼ 0:358 c cos 75o if cos 105o ¼ ¡0:259 b d cos 92o if cos 88o ¼ 0:035 sin 227o if sin 47o ¼ 0:731 Use a unit circle diagram to find: a cos 360o and sin 360o b cos(¡¼) and sin(¡¼) Explain how to use the unit circle to find µ when cos µ = ¡ sin µ 10 If sin 74o ¼ 0:961, find without using a calculator the value of: b sin 254o c sin 286o d sin 646o a sin 106o ¡ ¢ ¡ ¢ ¡ 3¼ ¢ b cos 3¼ 11 Without a calculator, evaluate: a tan2 2¼ ¡ sin 12 If sin x = ¡ 14 and ¼ < x < 3¼ , find tan x in radical (surd) form REVIEW SET 10B a 120o Convert these to radians in terms of ¼: Convert to radians (to sig figs.): a 71o Convert these radian measure to degrees: b 225o b 124:6o 2¼ a c ¡142o 5¼ b c 150o c Convert these radian measure to degrees (to decimal places): a b 1:46 c 0:435 d 540o d ¡25:3o 7¼ d 11¼ d ¡5:271 Find the perimeter and area of a sector of radius 11 cm and angle 63o Find the radius and area of a sector of perimeter 36 cm with an angle of 2¼ A triangle has sides of length cm and 13 cm and its area is 42 cm2 Find the size of its included angle B Anke and Lucas are considering buying a block of land The land agent supplies them with the given accurate sketch Find the area of the property, giving your answer in: b hectares a m2 C 30° 90 m 125 m A 75° D 120 m If cos 42 ¼ 0:743, without using a calculator, find the value of: b cos 222o c cos 318o d a cos 138o o cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 find the possible values of sin µ 95 100 50 75 25 95 100 50 75 25 10 If cos µ = cos(¡222o ) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\267IB_HL-2_10.CDR Friday, January 2008 9:23:26 AM DAVID3 IB_HL-2ed (268) 268 THE UNIT CIRCLE AND RADIAN MEASURE (Chapter 10) 11 Without a calculator evaluate: ¡ ¢ ¡ ¢ b a sin ¼3 cos ¼3 12 Given tan x = ¡ 32 tan2 3¼ and ¡¼ ¢ ¡1 c cos2 ¡¼¢ a sin x < x < 2¼, find: ¡ sin2 ¡¼¢ b cos x REVIEW SET 10C Use your calculator to determine the coordinates of the point on the unit circle corresponding to an angle of: a 320o b 163o Illustrate the regions where sin µ and cos µ have the same sign Use a unit circle diagram to find exact values for sin µ and cos µ for µ equal to: b 8¼ a 2¼ 3 Use a unit circle diagram to find: ¡ ¢ ¡ ¢ and sin 3¼ a cos 3¼ 2 If cos µ = ¡ 34 , ¼ <µ<¼ ¡ ¢ ¡ ¢ cos ¡ ¼2 and sin ¡ ¼2 b a sin µ find: b tan µ Use a unit circle diagram to find all angles between 0o and 360o which have: p p a a cosine of ¡ 23 b a sine of p12 c a tangent of ¡ Find µ in radians if: cos µ = ¡1 a sin2 µ = b Without using a calculator, evaluate: b cos2 ( ¼4 ) + sin( ¼2 ) a tan2 60o ¡ sin2 45o a cos( ¼2 ¡ µ) ¡ sin µ Simplify: c 5¼ cos( 5¼ ) ¡ tan( ) b cos µ tan µ 10 Find the value of x if the area is 80 cm2 Hence, find the length of [AC] 11 Determine the yellow shaded area: A 19.2 cm x° 11.3 cm B magenta yellow B C 95 100 50 A 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 12 Three circles with radius r are drawn as shown, each with its centre on the circumference of the other two circles A, B and C are the centres of the three circles Prove that an expression for the area of the shaded region is: p r2 A = (¼ ¡ 3) cyan cm 13p 18 C black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_10\268IB_HL-2_10.CDR Friday, January 2008 9:23:46 AM DAVID3 IB_HL-2ed (269) 11 Chapter Non-right angled triangle trigonometry Contents: A B C The cosine rule The sine rule Using the sine and cosine rules cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 11A Review set 11B black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_11\269IB_HL-2_11.CDR Monday, 29 October 2007 12:40:04 PM PETERDELL IB_HL-2ed (270) 270 NON-RIGHT ANGLED TRIANGLE TRIGONOMETRY (Chapter 11) A THE COSINE RULE A The cosine rule involves the sides and angles of a triangle In any ¢ABC: A a2 = b2 + c2 ¡ 2bc cos A or b2 = a2 + c2 ¡ 2ac cos B or c2 = a2 + b2 ¡ 2ab cos C c b C B B a C We will develop the first formula for both an acute and an obtuse triangle Proof: C C a b a h A B c-x B A x D c b h A x D (180°-A) In both triangles drop a perpendicular from C to meet [AB] (extended if necessary) at D Let AD = x and let CD = h Apply the theorem of Pythagoras in ¢BCD: a2 = h2 + (c ¡ x)2 a2 = h2 + c2 ¡ 2cx + x2 ) ) a2 = h2 + (c + x)2 a2 = h2 + c2 + 2cx + x2 In both cases, applying Pythagoras to ¢ADC gives h2 + x2 = b2 ) h2 = b2 ¡ x2 , and we substitute this into the equations above a2 = b2 + c2 ¡ 2cx x cos A = b b cos A = x ) a2 = b2 + c2 ¡ 2bc cos A a2 = b2 + c2 + 2cx x cos(180o ¡ A) = b o ) b cos(180 ¡ A) = x But cos(180o ¡ A) = ¡ cos A ) ¡b cos A = x ) a2 = b2 + c2 ¡ 2bc cos A ) In ADC: ) ) The other variations of the cosine rule could be developed by rearranging the vertices of ¢ABC Note that if A = 90o then cos A = and a2 = b2 + c2 ¡ 2bc cos A reduces to a2 = b2 + c2 , which is the Pythagorean Rule The cosine rule can be used to solve triangles given: ² two sides and an included angle ² three sides cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 If we are given two sides and a non-included angle, then when we try to find the third side we will end up with a quadratic equation This is an ambiguous case where there may be two plausible solutions black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_11\270IB_HL-2_11.CDR Wednesday, 31 October 2007 9:06:36 AM PETERDELL IB_HL-2ed (271) 271 NON-RIGHT ANGLED TRIANGLE TRIGONOMETRY (Chapter 11) Example Find, correct to decimal places, the length of [BC] By the cosine rule: BC2 = 112 + 132 ¡ £ 11 £ 13 £ cos 42o p ) BC ¼ (112 + 132 ¡ £ 11 £ 13 £ cos 42o ) ) BC ¼ 8:801 B 11 cm 42° A ) BC is 8:80 cm in length C 13 cm Rearrangement of the original cosine rule formulae can be used for finding angles if we know all three sides The formulae for finding the angles are: b2 + c2 ¡ a2 2bc cos A = cos B = c2 + a2 ¡ b2 2ca cos C = a2 + b2 ¡ c2 2ab Example In triangle ABC, if AB = cm, BC = cm and CA = cm, find the measure of angle BCA By the cosine rule: (52 + 82 ¡ 72 ) cos C = (2 £ £ 8) µ ¶ + 82 ¡ 72 ¡1 ) C = cos 2£5£8 ) C = 60o A cm cm C B cm So, angle BCA measures 60o : EXERCISE 11A Find the length of the remaining side in the given triangle: a b c A Q 15 cm B 105° K 4.8 km 21 cm R 32° 6.2 m 6.3 km C L P Find the measure of all angles of: 72° 14.8 m M Find the measure of obtuse angle PQR P C 12 cm 10 cm 11 cm cm cyan yellow 95 100 50 75 25 Q 95 100 50 75 25 95 50 100 magenta R B 13 cm 75 25 95 100 50 75 25 A black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_11\271IB_HL-2_11.CDR Monday, 29 October 2007 12:48:00 PM PETERDELL cm IB_HL-2ed (272) 272 NON-RIGHT ANGLED TRIANGLE TRIGONOMETRY (Chapter 11) Find: a the smallest angle of a triangle with sides 11 cm, 13 cm and 17 cm b the largest angle of a triangle with sides cm, cm and cm Find: a cos µ but not µ b the value of x q cm The smallest angle is opposite the shortest side cm cm cm x cm Find the exact value of x in each of the following diagrams: a b c cm cm cm cm x cm cm 60° 60° 120° x cm x cm 2x cm Find x in each of the following diagrams: a b 11 cm x cm 13 cm cm 130° 70° cm x cm In the diagram alongside, cos µ = ¡ 15 a Find x b Hence find the exact value of the area of the triangle (3x¡+¡1) cm (x¡+¡2) cm q (x¡+¡3) cm B THE SINE RULE The sine rule is a set of equations which connects the lengths of the sides of any triangle with the sines of the angles of the triangle The triangle does not have to be right angled for the sine rule to be used A In any triangle ABC with sides a, b and c units in length, and opposite angles A, B and C respectively, cyan magenta yellow 95 100 50 75 25 95 100 50 a b c = = sin A sin B sin C 75 25 or 95 100 50 75 25 95 100 50 75 25 sin A sin B sin C = = a b c b c black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_11\272IB_HL-2_11.CDR Monday, 29 October 2007 12:51:43 PM PETERDELL C B a IB_HL-2ed (273) 273 NON-RIGHT ANGLED TRIANGLE TRIGONOMETRY (Chapter 11) The area of any triangle ABC is given by Proof: Dividing each expression by abc gives bc sin A = ac sin B = ab sin C: sin A sin B sin C = = : a b c The sine rule is used to solve problems involving triangles given either: ² two angles and one side ² two sides and a non-included angle FINDING SIDES Example Find the length of [AC] correct to two decimal places By the sine rule: b 12 = o sin 58 sin 39o A A b cm 12 cm 39° 58° 12 cm C b= ) b ¼ 16:170 74 B 39° C 58° B 12 £ sin 58o sin 39o ) ) AC is about 16:17 cm long EXERCISE 11B.1 Find the value of x: a b 11 cm x cm 51° 115° 23 cm c 4.8 km 37° 48° x cm 48° 80° x km In triangle ABC find: a a if A = 63o , B = 49o and b = 18 cm b b if A = 82o , C = 25o and c = 34 cm c c if B = 21o , C = 48o and a = 6:4 cm FINDING ANGLES The problem of finding angles using the sine rule is more complicated because there may be two possible answers We call this situation the ambiguous case INVESTIGATION THE AMBIGUOUS CASE cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 You will need a blank sheet of paper, a ruler, a protractor and a compass for the tasks that follow In each task you will be required to construct triangles from given information You could also this using a computer package such as “The Geometer’s Sketchpad” black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_11\273IB_HL-2_11.CDR Monday, 10 December 2007 2:59:02 PM PETERDELL IB_HL-2ed (274) 274 NON-RIGHT ANGLED TRIANGLE TRIGONOMETRY (Chapter 11) Task 1: Draw AB = 10 cm At A construct an angle of 30o Using B as centre, draw an arc of a circle of radius cm Let the arc intersect the ray from A at C How many different positions may C have and therefore how many different triangles ABC may be constructed? Task 2: As before, draw AB = 10 cm and construct a 30o angle at A This time draw an arc of radius cm centred at B How many different triangles are possible? Task 3: Repeat, but this time draw an arc of radius cm centred at B How many different triangles are possible? Task 4: Repeat with an arc of radius 12 cm from B How many triangles are possible now? You should have discovered that when you are given two sides and a non-included angle there are a number of different possibilities You could get two triangles, one triangle or it may be impossible to draw any triangles from the given data Now consider the calculations involved in each of the cases of the investigation Task 1: C1 Given: c = 10 cm, a = cm, A = 30o sin C sin A = c a c sin A sin C = a 10 £ sin 30o sin C = ¼ 0:8333 Finding C: ) ) C2 cm cm 30° 10 cm A B Because sin µ = sin(180o ¡ µ) there are two possible angles: C = 56:44o or 180o ¡ 56:44o = 123:56o Given: c = 10 cm, a = cm, A = 30o Task 2: C sin C sin A = c a c sin A sin C = a 10 £ sin 30o sin C = =1 Finding: C ) ) cm 30° 10 cm A B There is only one possible solution for C in the range from 0o to 180o and that is C = 90o Only one triangle is therefore possible Complete the solution of the triangle yourself Given: c = 10 cm, a = cm, A = 30o cyan magenta yellow cm 30° 10 cm 95 100 50 75 A 25 95 100 50 75 25 95 25 95 100 50 75 25 ) 100 ) 50 sin C sin A = c a c sin A sin C = a 10 £ sin 30o sin C = ¼ 1:6667 Finding C: 75 Task 3: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_11\274IB_HL-2_11.CDR Monday, 29 October 2007 1:44:31 PM PETERDELL B IB_HL-2ed (275) 275 NON-RIGHT ANGLED TRIANGLE TRIGONOMETRY (Chapter 11) There is no angle that has a sine ratio > Therefore there is no solution for this given data, and no possible triangle can be drawn Given: c = 10 cm, a = 12 cm, A = 30o Task 4: Finding C: sin C sin A = c a ) sin C = ) sin C = C c sin A a 10 £ sin 30o 12 sin C = 0:4167 ) 12 cm Two angles have a sine ratio of 0:4167 : C ¼ 24:62o or 180o ¡ 24:62o C ¼ 24:62o or 155:38o 30° 10 cm A B However, in this case only one of these two angles is valid If A = 30o then C cannot possibly equal 155:38o because 30o + 155:38o > 180o Therefore, there is only one possible solution, C ¼ 24:62o Once again, you may wish to carry on and complete the solution Conclusion: Each situation using the sine rule with two sides and a non-included angle must be examined very carefully Example Find the measure of angle C in triangle ABC if AC is cm, AB is 11 cm and angle B measures 25o sin C sin B = fBy the sine ruleg c b sin 25o sin C = ) 11 7 cm 11 £ sin 25o ) sin C = C µ ¶ 11 £ sin 25o ¡1 ) C = sin or its supplement A 11 cm 25° B ) ) C ¼ 41:6o or 180o ¡ 41:6o fas C may be obtuseg o C ¼ 41:6 or 138:4o ) C measures 41:6 if angle C is acute, or 138:4o if angle C is obtuse In this example there is insufficient information to determine the actual shape of the triangle cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 o black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_11\275IB_HL-2_11.CDR Monday, 29 October 2007 1:47:31 PM PETERDELL IB_HL-2ed (276) 276 NON-RIGHT ANGLED TRIANGLE TRIGONOMETRY (Chapter 11) Sometimes there is information in the question which enables us to reject one of the answers Example Find the measure of angle L in triangle KLM given that angle LKM measures 56o , LM = 16:8 m and KM = 13:5 m sin L sin 56o = fby the sine ruleg 13:5 16:8 13:5 £ sin 56o ) sin L = 16:8 µ ¶ 13:5 £ sin 56o ¡1 ) L = sin or its supplement 16:8 L 16.8 m 56° K We reject L = 138:2o ) L ¼ 41:8o or 180o ¡ 41:8o or 138:2o ) L ¼ 41:8o ) L ¼ 41:8o M 13.5 m as 138:2o + 56o > 180o which is impossible EXERCISE 11B.2 Triangle ABC has angle B = 40o , b = cm and c = 11 cm Find the two possible values for angle C In triangle ABC, find the measure of: b = 65o a angle A if a = 14:6 cm, b = 17:4 cm and ABC b = 43o b angle B if b = 43:8 cm, c = 31:4 cm and ACB b = 71o c angle C if a = 6:5 km, c = 4:8 km and BAC Is it possible to have a triangle with measurements as shown? Explain! Find the magnitude of the angle ABC and hence the length BD B 85° 12° 9.8 cm 68° 20 cm C A 11.4 cm Find x and y in the given figure D 78° ym 95° xm 30° 118° 22 m b = 58o , AB = 10 cm and AC = 5:1 cm Find: Triangle ABC has A b correct to the nearest tenth of a degree using the sine rule a C b correct to the nearest tenth of a degree using the cosine rule b C cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 c Copy and complete: “When faced with using either the sine rule or the cosine rule it is better to use the as it avoids ” black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_11\276IB_HL-2_11.CDR Friday, 14 December 2007 10:40:14 AM PETERDELL IB_HL-2ed (277) 277 NON-RIGHT ANGLED TRIANGLE TRIGONOMETRY (Chapter 11) b = 30o , AC = cm and AB = cm Find the area of the In triangle ABC, ABC triangle In the diagram alongside, find the exact value of x Express your answer in the p form a + b where a, b Q C 45° 30° (x¡+¡3) cm (2x¡-¡5) cm USING THE SINE AND COSINE RULES If we are given a problem involving a triangle, we must first decide which rule to use If the triangle is right angled then the trigonometric ratios or Pythagoras’ Theorem can be used For some problems we can add an extra line or two to the diagram to create a right angled triangle However, if we not have a right angled triangle and we have to choose between the sine and cosine rules, the following checklist may be helpful: ² ² Use the cosine rule when given: three sides two sides and an included angle Use the sine rule when given: ² one side and two angles ² two sides and a non-included angle, but beware of the ambiguous case which can occur when the smaller of the two given sides is opposite the given angle Example T The angles of elevation to the top of a mountain are measured from two beacons A and B at sea These angles are as shown on the diagram If the beacons are 1473 m apart, how high is the mountain? A 29.7° 41.2° B b = 41:2o ¡ 29:7o ATB = 11:5o T cyan magenta yellow 95 100 50 95 100 50 75 N 25 95 100 50 75 25 95 100 50 75 25 29.7° 41.2° 1473 m B A 1473 x = o sin 29:7 sin 11:5o 1473 ) x= £ sin 29:7o sin 11:5o ¼ 3660:62 hm 75 xm fexterior angle of ¢g We find x in ¢ABT using the sine rule: 25 11.5° N black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_11\277IB_HL-2_11.CDR Friday, 14 December 2007 10:41:44 AM PETERDELL IB_HL-2ed (278) 278 NON-RIGHT ANGLED TRIANGLE TRIGONOMETRY (Chapter 11) Now, in ¢BNT, sin 41:2o = h h ¼ x 3660:62 h ¼ sin 41:2o £ 3660:62 h ¼ 2410 ) ) So, the mountain is about 2410 m high Example S R Find the measure of angle RPV P W cm T U p p 52 + 32 = 34 cm p p PV = 62 + 32 = 45 cm p p PR = 62 + 52 = 61 cm V cm In ¢RVW, RV = fPythagorasg In ¢PUV, fPythagorasg In ¢PQR, ~`6`1 cm P fPythagorasg p p p ( 61)2 + ( 45)2 ¡ ( 34)2 p p cos µ = 61 45 R q = ~`4`5 cm cm Q ~`3`4 cm V ) 61 + 45 ¡ 34 p p 61 45 72 = p p 61 45 ¶ µ 36 µ = cos¡1 p p ¼ 46:6o 61 45 ) angle RPV measures about 46:6o EXERCISE 11C Rodrigo wishes to determine the height of a flagpole He takes a sighting of the top of the flagpole from point P He then moves further away from the flagpole by 20 metres to point Q and takes a second sighting The information is shown in the diagram alongside How high is the flagpole? 63 m 112° P magenta yellow 95 100 50 75 25 95 50 75 25 95 100 50 75 25 95 100 50 75 25 P 53° R lake cyan 28° 20 m To get from P to R, a park ranger had to walk along a path to Q and then to R as shown What is the distance in a straight line from P to R? Q 175 m 100 Q black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_11\278IB_HL-2_11.CDR Friday, November 2007 9:31:58 AM PETERDELL IB_HL-2ed (279) NON-RIGHT ANGLED TRIANGLE TRIGONOMETRY (Chapter 11) A golfer played his tee shot a distance of 220 m to a point A He then played a 165 m six iron to the green If the distance from tee to green is 340 m, determine the number of degrees the golfer was off line with his tee shot 279 A 165 m G 220 m 340 m T A Communications Tower is constructed on top of a building as shown Find the height of the tower tower 15.9° building 23.6° 200 m A football goal is metres wide When a player is 26 metres from one goal post and 23 metres from the other, he shoots for goal What is the angle of view of the goals that the player sees? angle of view goal posts player A tower 42 metres high stands on top of a hill From a point some distance from the base of the hill, the angle of elevation to the top of the tower is 13:2o and the angle of elevation to the bottom of the tower is 8:3o Find the height of the hill From the foot of a building I have to look upwards at an angle of 22o to sight the top of a tree From the top of the building, 150 metres above ground level, I have to look down at an angle of 50o below the horizontal to sight the tree top a How high is the tree? b How far from the building is this tree? cm Q Find the measure of angle PQR in the rectangular box shown cm P cm R Two observation posts are 12 km apart at A and B A third observation post C is located such that angle CAB is 42o and angle CBA is 67o Find the distance of C from both A and B cyan magenta yellow Q 10 km R 30° km 95 P 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 10 Stan and Olga are considering buying a sheep farm A surveyor has supplied them with the given accurate sketch Find the area of the property, giving your answer in: a km2 b hectares black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_11\279IB_HL-2_11.CDR Wednesday, November 2007 4:40:33 PM PETERDELL 70° 12 km S IB_HL-2ed (280) 280 NON-RIGHT ANGLED TRIANGLE TRIGONOMETRY (Chapter 11) 11 Thabo and Palesa start at point A They each walk in a straight line at an angle of 120o to each other Thabo walks at km h¡1 and Palesa walks at km h¡1 How far apart are they after 45 minutes? A 12 The cross-section design of the kerbing for a driverless-bus roadway is shown opposite The metal strip is inlaid into the concrete and is used to control the direction and speed of the bus Find the width of the metal strip 30° 38° 200 mm 110° B C D metal strip 13 An orienteer runs for 12 km, then turns through an angle of 32o and runs for another km How far is she from her starting point? 14 Sam and Markus are standing on level ground 100 metres apart A large tree is due North of Markus and on a bearing of 065o from Sam The top of the tree appears at an angle of elevation of 25o to Sam and 15o to Markus Find the height of the tree 15 A helicopter A observes two ships B and C B is 23:8 km from the helicopter and C is 31:9 km from it The angle of view from the helicopter to B and C (angle BAC) is 83:6o How far are the ships apart? REVIEW SET 11A Determine the value of x: a 11 cm b 15 km 72° 13 cm 17 km 19 cm x° x km Find the value of x: a 13 cm x cm b 14 cm 11 cm 47° x° 21 cm 19 cm Find the unknown side and angles: Find the area of quadrilateral ABCD: C A 110° cm A 9.8 cm 11 cm B 40° D 74° 11 cm 16 cm cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 C 95 100 50 75 25 95 100 50 75 25 B black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_11\280IB_HL-2_11.CDR Thursday, 11 March 2010 10:24:11 AM PETER IB_HL-2ed (281) NON-RIGHT ANGLED TRIANGLE TRIGONOMETRY (Chapter 11) 281 A vertical tree is growing on the side of a hill with slope of 10o to the horizontal From a point 50 m downhill from the tree, the angle of elevation to the top of the tree is 18o Find the height of the tree From point A, the angle of elevation to the top of a tall building is 20o On walking 80 m towards the building the angle of elevation is now 23o How tall is the building? Peter, Sue and Alix are sea-kayaking Peter is 430 m from Sue on a bearing of 113o while Alix is on a bearing of 203o and a distance 310 m from Sue Find the distance and bearing of Peter from Alix A family in Germany drives at 140 km h¡1 for 45 minutes on a bearing of 032o and then 180 km h¡1 for 40 minutes on a bearing 317o Find the distance and bearing of the car from its starting point You are given details of a triangle such that you could use either the cosine rule or the sine rule to find an unknown Which rule should you use? Explain your answer 10 Kady was asked to draw the illustrated triangle exactly a Use the cosine rule to find x b What should Kady’s response be? cm cm 60° x cm 11 Soil contractor Frank was given the following dimensions over the telephone: The triangular garden plot ABC has angle CAB measuring 44o , [AC] is m long and [BC] is m long Soil to a depth of 10 cm is required a Explain why Frank needs extra information from his client b What is the maximum volume of soil needed if his client is unable to supply the necessary information? REVIEW SET 11B Find the measure of angle EDG: F G H E 3m B C 4m A km A D 40° P B km D Hikers Andrew and Brett take separate trails from their starting point P to get to their destination at D They walk at an angle of 40o apart from each other as shown, and camp overnight at positions A and B respectively How far does Brett have to walk the next day to reach the destination? 100° km 6m Consider the kite ABCD alongside: b = ABC b a Use the cosine rule to show that ADC b = BAC b b Use the sine rule to show that DAC D A C cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 B black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_11\281IB_HL-2_11.CDR Monday, 29 October 2007 2:38:36 PM PETERDELL IB_HL-2ed (282) 282 NON-RIGHT ANGLED TRIANGLE TRIGONOMETRY (Chapter 11) A boat sailing from A to B travels in a straight line until the captain realises he is off course The boat is turned through an angle of 60o, then travels another 10 km to B The trip would have been km shorter if the boat had gone straight from A to B How far did the boat travel? 60° B C 10 km A At pm, runner A runs at 14 km h¡1 on a bearing of 25o , while runner B starting at the same point as A runs at 12 km h¡1 on a bearing of 97o a At what time will A and B be 20 km apart? b What will be the bearing of B from A at this time? Triangle ABC has [AB] of length m, [AC] of length d m, C [BC] of length x m, and angle ABC measures 20o a Find an expression for d2 in terms of x dm b By using the fact that, for d > 0, d is minimised when d2 is minimised, find the exact value of x which A minimises d b is a right angle c Hence, show that d is minimised when BCA B 20° 5m 12 m a Explain why cos(180o ¡ µ) = ¡ cos µ b For the given quadrilateral: i Show that 300 cos ¡ 192 cos bo = 117 ii If b + d = 180, find the values of b and d iii Hence, find the values of a and c xm a° b° 10 m 8m d° c° 15 m a For the quadratic function y = ¡x2 +12x¡20, find the maximum or minimum value and the corresponding value of x: B b In triangle ABC, AB = y, BC = x, AC = 8, and the perimeter of the triangle is 20 y x i Write y in terms of x: ii Use the cosine rule to write y in terms of x and q A cos µ: C 3x ¡ 10 iii Hence show that cos µ = 2x c If the area of the triangle is A, show that A2 = 20(¡x2 + 12x ¡ 20): d Hence, find the maximum area of the triangle and the triangle’s shape when this occurs Q The given figure shows quadrilateral PQRS which has been divided into two triangles by the P f diagonal [QS] PQ = 3, QR = 7, PS = 6, 32° b = 32o Qb PS = Á and QRS a Find QS in terms of cos Á S b If Á = 50o : b i find the possible values of RSQ ii given that Rb SQ is acute, find the perimeter of the quadrilateral R cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 c Find the area of quadrilateral PQRS if Á = 50o black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_11\282IB_HL-2_11.CDR Wednesday, November 2007 4:52:25 PM PETERDELL IB_HL-2ed (283) 12 Chapter Advanced trigonometry Contents: A B C D E F G H I J K L M Observing periodic behaviour The sine function Modelling using sine functions The cosine function The tangent function Trigonometric equations Using trigonometric models Reciprocal trigonometric functions Trigonometric relationships Compound angle formulae Double angle formulae Trigonometric equations in quadratic form Trigonometric series and products cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review Set 12A Review Set 12B Review Set 12C Review Set 12D black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\283IB_HL-2_12.CDR Monday, 29 October 2007 2:58:40 PM PETERDELL IB_HL-2ed (284) 284 ADVANCED TRIGONOMETRY (Chapter 12) INTRODUCTION Periodic phenomena occur all the time in the physical world For example, in: ² ² ² ² ² ² seasonal variations in our climate variations in average maximum and minimum monthly temperatures the number of daylight hours at a particular location variations in the depth of water in a harbour due to tidal movement the phases of the moon animal populations These phenomena illustrate variable behaviour which is repeated over time The repetition may be called periodic, oscillatory or cyclic in different situations In this topic we will consider how trigonometric functions can be used to model periodic phenomena We will then extend our knowledge of the trigonometric functions by considering formulae that connect them OPENING PROBLEM A Ferris wheel rotates at a constant speed The wheel’s radius is 10 m and the bottom of the wheel is m above ground level From a point in front of the wheel, Andrew is watching a green light on the perimeter of the wheel Andrew notices that the green light moves in a circle He estimates how high the light is above ground level at two second intervals and draws a scatterplot of his results ² What does his scatterplot look like? ² Could a known function be used to model the data? ² How could this function be used to find the light’s position at any point in time? ² How could this function be used to find the times when the light is at its maximum and minimum heights? ² What part of the function would indicate the time interval over which one complete cycle occurs? Click on the icon to visit a simulation of the Ferris wheel You will be able to view the light on the Ferris wheel: ² from a position in front of the wheel ² from a side-on position ² from above the wheel DEMO cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Observe the graph of height above (or below) the wheel’s axis as the wheel rotates at a constant rate black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\284IB_HL-2_12.CDR Wednesday, 31 October 2007 9:07:58 AM PETERDELL IB_HL-2ed (285) ADVANCED TRIGONOMETRY (Chapter 12) A 285 OBSERVING PERIODIC BEHAVIOUR Consider the table below which shows the mean monthly maximum temperature (o C) for Cape Town, South Africa Month Jan Temp 28 Feb Mar 27 25 12 Apr May 22 18 12 Jun 16 Jul Aug 15 16 Sep Oct Nov Dec 18 21 12 24 26 T, Temp (°C) The data is graphed alongside using a scatterplot, assigning January = 1, February = etc., for the 12 months of the year 30 20 Note: The points are not joined as interpolation has no meaning here (10,¡21\Qw_\) 10 t (months) 12 JAN JAN The temperature shows a variation from an average of 28o C in January through a range of values across the months and the cycle will repeat itself for the next 12 months It is worthwhile noting that later we will be able to establish a function which approximately fits this set of points T, Temp (°C) 30 20 10 t (months) 12 15 JAN JAN 18 21 24 HISTORICAL NOTE direction of rotation + voltage lines of force 90° - 180° 270° 360° In 1831 Michael Faraday discovered that an electric current was generated by rotating a coil of wire in a magnetic field The electric current produced showed a voltage which varied between positive and negative values as the coil rotated through 360 o cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Graphs with this basic shape where the cycle is repeated over and over are called sine waves black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\285IB_HL-2_12.CDR Wednesday, 31 October 2007 9:08:12 AM PETERDELL IB_HL-2ed (286) 286 ADVANCED TRIGONOMETRY (Chapter 12) GATHERING PERIODIC DATA Data on a number of periodic phenomena can be found online or in other publications For example: ² Maximum and minimum monthly temperatures can be found at http://www.bom.gov.au/silo/ ² Tidal details can be obtained from daily newspapers or http://tidesandcurrents.noaa.gov or http://www.bom.gov.au/oceanography TERMINOLOGY USED TO DESCRIBE PERIODICITY A periodic function is one which repeats itself over and over in a horizontal direction The period of a periodic function is the length of one repetition or cycle f(x) is a periodic function with period p , f (x + p) = f (x) for all x, and p is the smallest positive value for this to be true Use a graphing package to examine the function: f : x 7! x ¡ [x] where [x] is “the largest integer less than or equal to x” GRAPHING PACKAGE Is f(x) periodic? What is its period? A cycloid is another example of a periodic function It is the curve traced out by a point on a circle as the circle moves along a flat surface However, the cycloid function cannot be written as y = :::::: or f (x) = :::::: DEMO horizontal flat surface In this course we are mainly concerned with periodic phenomena which show a wave pattern when graphed the wave principal axis cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The wave oscillates about a horizontal line called the principal axis or mean line A maximum point occurs at the top of a crest and a minimum point at the bottom of a trough The amplitude is the distance between a maximum (or minimum) point and the principal axis black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\286IB_HL-2_12.CDR Wednesday, 31 October 2007 9:08:18 AM PETERDELL IB_HL-2ed (287) 287 ADVANCED TRIGONOMETRY (Chapter 12) maximum point amplitude principal axis period minimum point EXERCISE 12A For each set of data below, draw a scatterplot and decide whether or not the data exhibits approximately periodic behaviour a x y b x y c x y d x y 0 1 1 0 0 1:4 0:5 1:9 4:7 ¡1 1:5 4:5 2:0 4:7 ¡1:4 ¡1 3:0 3:4 3:5 2:4 10 1:4 11 12 4 1:0 3:5 3:4 1:7 2:1 2:5 4:3 5:2 8:9 10:9 10:2 10 8:4 12 10:4 The following tabled values show the height above the ground of a point on a bicycle wheel as the bicycle is wheeled along a flat surface Distance travelled (cm) Height above ground (cm) 0 20 40 23 60 42 80 57 100 120 140 160 180 200 64 59 43 23 Distance travelled (cm) 220 240 260 280 300 320 340 360 380 400 Height above ground (cm) 27 40 55 63 60 44 24 a Plot the graph of height against distance b Is the data periodic? If so, estimate: i the equation of the principal axis ii the maximum value iii the period iv the amplitude c Is it reasonable to fit a curve to this data, or should we leave it as discrete points? Which of these graphs show periodic behaviour? y a b x y c y 12 15 18 x y d cyan magenta yellow 95 100 50 x 75 25 x 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\287IB_HL-2_12.CDR Monday, November 2007 12:13:55 PM PETERDELL IB_HL-2ed (288) 288 ADVANCED TRIGONOMETRY (Chapter 12) e f y 10 15 y 20 x B x THE SINE FUNCTION In previous studies of trigonometry we have only considered right angled triangles, or static situations where an angle µ is fixed However, when an object moves in a circle, the situation is dynamic with the angle between the radius OP and the horizontal axis continually changing y Consider again the Opening Problem in which a Ferris wheel of radius 10 m revolves at constant speed The height of P, the point representing the green light on the wheel relative to the principal axis O, can be determined using right angled triangle trigonometry As sin µ = P 10 h q x 10 h , then h = 10 sin µ: 10 As time goes by µ changes and so does h So, h is a function of µ, but more importantly h is a function of time t Suppose the Ferris wheel observed by Andrew takes 100 seconds for a full revolution The graph below shows the height of the light above or below the principal axis against the time in seconds 10 height (metres) DEMO 50 100 time (seconds) -10 We observe that the amplitude is 10 metres and the period is 100 seconds The family of sine curves can have different amplitudes and different periods We will examine such families in this section THE BASIC SINE CURVE If we project the values of sin µ from the unit circle to the set of axes on the right we obtain the graph of y = sin x y y y = sin x 90° 180° p p 270° 3p 360° 2p x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 -1 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\288IB_HL-2_12.CDR Wednesday, 31 October 2007 9:09:43 AM PETERDELL IB_HL-2ed (289) 289 ADVANCED TRIGONOMETRY (Chapter 12) The wave of course can be continued beyond x 2¼ y 90° 180° p p 270° 360° 2p 3p 540° x 3p y = sin x -1 We expect the period to be 2¼, since the Ferris wheel repeats its positioning after one full revolution The maximum value is and the minimum is ¡1 as ¡1 y on the unit circle The amplitude is GRAPHING TI PACKAGE Use your graphics calculator or graphing package to obtain the graph of y = sin x to check these features C When patterns of variation can be identified and quantified in terms of a formula or equation, predictions may be made about behaviour in the future Examples of this include tidal movement which can be predicted many months ahead, and the date of a future full moon THE FAMILY y = Asin¡x INVESTIGATION What to do: GRAPHING Use technology to graph on the same set of axes: PACKAGE a y = sin x and y = sin x b y = sin x and y = 0:5 sin x c y = sin x and y = ¡ sin x (A = ¡1) If using a graphics calculator, make sure that the mode is set in radians and that your viewing window is appropriate For each of y = sin x, y = sin x, y = 0:5 sin x, y = ¡ sin x record the maximum and minimum values and state the period and amplitude How does A affect the function y = A sin x? State the amplitude of: a y = sin x b y= p sin x c y = ¡2 sin x THE FAMILY y = sin¡Bx, B > INVESTIGATION What to do: Use technology to graph on the same set of axes: a y = sin x and y = sin 2x b y = sin x and y = sin( 12 x) For each of y = sin x, y = sin 2x, y = sin( 12 x) record the maximum and minimum values and state the period and amplitude How does B affect the function y = sin Bx? cyan magenta yellow 95 c y = sin(1:2x) 100 50 75 25 95 50 100 y = sin( 13 x) b 75 25 95 100 50 75 25 95 100 50 75 25 State the period of: a y = sin 3x black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\289IB_HL-2_12.CDR Wednesday, 31 October 2007 9:10:32 AM PETERDELL GRAPHING PACKAGE d y = sin Bx IB_HL-2ed (290) 290 ADVANCED TRIGONOMETRY (Chapter 12) From the previous investigations you should have observed that: ² in y = A sin x, j A j determines the amplitude ² in y = sin Bx, B > 0, B affects the period and the period is Recall j x j is the modulus of x, or size of x ignoring its sign 2¼ : B The modulus sign ensures that the final answer is non-negative This must be so for amplitudes Example Without using technology sketch the graphs of: a y = sin x b y = ¡2 sin x for x 2¼ a The amplitude is and the period is 2¼: y 3p p 2p x p -2 We place the points as shown and fit the sine wave to them b The amplitude is 2, the period is 2¼, and it is the reflection of y = sin x in the x-axis y y¡=¡-2sin¡¡x 3p p 2p x p y¡=¡2sin¡¡x -2 Example Without using technology sketch the graph of y = sin 2x, x 2¼ 2¼ The period is = ¼: As sin 2x has half the period of sin x, the first maximum is at ¼4 not ¼2 So, for example, the maximum values are ¼ units apart y 3p 2p p p magenta yellow 95 100 50 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 cyan x y¡=¡sin¡¡2x -1 75 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_12\290IB_HL-2_12.CDR Thursday, 11 March 2010 10:24:34 AM PETER IB_HL-2ed (291) 291 ADVANCED TRIGONOMETRY (Chapter 12) EXERCISE 12B.1 Without using technology sketch the graphs of the following for x 2¼: a y = ¡3 sin x b y = sin x c y= sin x d y = ¡ 32 sin x Without using technology sketch the graphs of the following for x 3¼: ¡ ¢ c y = sin(¡2x) a y = sin 3x b y = sin x2 State the period of: a y = sin 4x b c y = sin(¡4x) y = sin ¡x¢ d Find B given that the function y = sin Bx, B > has period: c 12¼ d e a 5¼ b 2¼ y = sin(0:6x) 100 Use a graphics calculator or graphing package to help you graph, for 0o x 720o : b y = sin x + sin 2x + sin 3x c y= a y = sin x + sin 2x sin x Use a graphing package or graphics calculator to graph: a f (x) = sin x + sin 3x sin 5x + b f (x) = sin x + sin 3x sin 5x sin 7x sin 9x sin 11x + + + + 11 Predict the graph of f (x) = sin x + GRAPHING PACKAGE sin 3x sin 5x sin 7x sin 1001x + + + :::::: + 1001 INVESTIGATION THE FAMILIES y = sin(x ¡ C) AND y = sin¡x + D What to do: Use a b c technology to graph on the same set of axes: y = sin x and y = sin(x ¡ 2) y = sin x and y = sin(x + 2) y = sin x and y = sin(x ¡ ¼3 ) GRAPHING PACKAGE For each of y = sin x, y = sin(x ¡ 2), y = sin(x + 2), y = sin(x ¡ ¼3 ) record the maximum and minimum values and state the period and amplitude What transformation moves y = sin x to y = sin(x ¡ C)? Use technology to graph on the same set of axes: a y = sin x and y = sin x + b y = sin x and y = sin x ¡ For each of y = sin x, y = sin x + and y = sin x ¡ record the maximum and minimum values and state the period and amplitude What transformation moves y = sin x to y = sin x + D? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 What transformation moves y = sin x to y = sin(x ¡ C) + D? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\291IB_HL-2_12.CDR Monday, November 2007 12:14:42 PM PETERDELL IB_HL-2ed (292) 292 ADVANCED TRIGONOMETRY (Chapter 12) From Investigation we observe that: ² ² y = sin(x ¡ C) is a horizontal translation of y = sin x through C units y = sin x + D is a vertical translation of y = sin x through D units h i C y = sin(x ¡ C) + D is a translation of y = sin x through vector D ² Example On the same set of axes graph for x 4¼: a y = sin x and y = sin(x ¡ 1) b y = sin x and y = sin x ¡ a y 1 p 2p 4p 3p -1 b x y¡=¡sin¡x y¡=¡sin(¡ x¡-¡1) y -1 -1 p 2p -1 4p 3p -1 -1 -2 x y¡=¡sin¡x y¡=¡sin¡x¡-¡1 THE GENERAL SINE FUNCTION y = A sin B(x ¡ C) + D affects amplitude affects period is called the general sine function affects horizontal translation affects vertical translation The principal axis of the general sine function is y = D The period of the general sine function is 2¼ B Consider y = sin 3(x ¡ ¼4 ) + 1: It is a translation of y = sin 3x under ·¼¸ So, starting with y = sin x we would: cyan magenta yellow 95 100 50 75 25 y = sin x, then y = sin 3x, then ¢ ¡ y = sin x ¡ ¼4 + 1: 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ² double the amplitude to produce divide the period by to produce ·¼¸ to produce translate by ² ² black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\292IB_HL-2_12.CDR Wednesday, 31 October 2007 9:38:39 AM PETERDELL IB_HL-2ed (293) 293 ADVANCED TRIGONOMETRY (Chapter 12) EXERCISE 12B.2 Draw sketch graphs of: a y = sin x ¡ d y = sin(x ¡ 2) b e y = sin x + ¼ 4) y = sin(x + c y = sin(x + 2) f y = sin(x ¡ ¼6 ) + Check your answers to using technology State the period of: a y = sin 5t b y = sin GRAPHING PACKAGE ¡t¢ c y = sin(¡2t) Find B in y = sin Bx if B > and the period is: ¼ c 100¼ a 3¼ b 10 State the transformation(s) which maps: a y = sin x onto y = sin x ¡ c y = sin x onto y = sin x d 50 b y = sin x onto y = sin(x ¡ ¼4 ) d e y = sin x onto y = sin x f y = sin x onto y = sin 4x ¡ ¢ y = sin x onto y = sin x4 g y = sin x onto y = ¡ sin x h y = sin x onto y = ¡3 + sin(x + 2) i y = sin x onto y = sin 3x j y = sin x onto y = sin(x ¡ ¼3 ) + C MODELLING USING SINE FUNCTIONS Sine functions can be useful for modelling certain biological and physical phenomena in nature which are approximately periodic MEAN MONTHLY TEMPERATURE Consider again the mean monthly maximum temperature (o C) for Cape Town: Month Jan Temp 28 Feb Mar 27 25 12 Apr May 22 18 12 Jun Jul 16 15 Aug 16 Sep Oct Nov Dec 18 21 12 24 26 The graph over a two year period is shown below: 40 30 T, temperature (°C) A 20 10 cyan t (months) magenta yellow 95 Mar 100 50 Jan 75 25 95 100 50 Nov Sep Jul 75 25 May 95 Mar 100 50 Jan 75 25 95 100 50 75 25 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\293IB_HL-2_12.CDR Wednesday, 31 October 2007 9:45:26 AM PETERDELL May Jul Sep Nov IB_HL-2ed (294) 294 ADVANCED TRIGONOMETRY (Chapter 12) We attempt to model this data using the general sine function y = A sin B(x ¡ C) + D, or in this case T = A sin B(t ¡ C) + D 2¼ = 12 and ) B = ¼6 : The period is 12 months, so B The amplitude = max ¡ 28 ¡ 15 ¼ ¼ 6:5, so A ¼ 6:5 : 2 The principal axis is midway between max and min., so D ¼ 28 + 15 ¼ 21:5 So, the model is T ¼ 6:5 sin ¼6 (t ¡ C) + 21:5 for some constant C We notice that point A on the original graph lies on the principle axis and is a point at which we are starting a new period ) since A is at (10, 21:5), C = 10 The model is therefore T ¼ 6:5 sin ¼6 (t ¡ 10) + 21:5 and we can superimpose it on the original data as follows 40 T, temperature (°C) 30 20 10 t (months) Jan Mar May Nov Sep Jul Jan Mar May Nov Sep Jul TIDAL MODELS The tides at Juneau, Alaska were recorded over a two day period The results are shown in the table opposite: Day Day high tide low tide high tide low tide 1.18 pm 6.46 am, 7.13 pm 1.31 am, 2.09 pm 7.30 am, 7.57 pm Suppose high tide corresponds to height and low tide to height ¡1 Plotting these times with t being the time after midnight before the first low tide, we get: 12 hrs 13 Tide height, H 12 hrs 38 X t am 12 noon pm 12 pm am 12 noon pm midnight -1 12 hrs 27 12 hrs 17 12 hrs 27 We attempt to model this periodic data using H = A sin B(t ¡ C) + D cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The principal axis is H = 0, so D = The amplitude is 1, so A = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\294IB_HL-2_12.CDR Monday, November 2007 12:15:41 PM PETERDELL IB_HL-2ed (295) 295 ADVANCED TRIGONOMETRY (Chapter 12) The graph shows that the ‘average’ period is about 12 hours 24 ¼ 12:4 hours 2¼ 2¼ 2¼ But the period is ) ¼ 12:4 and so B ¼ ¼ 0:507 B B 12:4 The model is now H ¼ sin 0:507(t ¡ C) for some constant C We find point X which is midway between the first minimum and the following maximum, 6:77 + 13:3 ¼ 10:0 ) C¼ So, the model is H ¼ sin 0:507(t ¡ 10:04): Below is our original graph of seven plotted points and our model which attempts to fit them H¡¼¡sin\\0"507(t-10"04) H pm am 12 noon pm 12 pm am t 12 noon -1 Use your graphics calculator to check this result Times must be given in hours after midnight, i.e., (6:77, ¡1), (13:3, 1), (19:22, ¡1), etc MODELLING TI C EXERCISE 12C Below is a table which shows the mean monthly maximum temperatures (o C) for a city in Greece Month Jan Feb Mar Apr May Jun July Aug Sept Oct Nov Dec o Temperature ( C) 15 14 15 18 21 25 27 26 24 20 18 16 a Use a sine function of the form T ¼ A sin B(t ¡ C) + D to model the data Find good estimates of the constants A, B, C and D without using technology Use Jan ´ 1, Feb ´ 2, etc b Use technology to check your answer to a How well does your model fit? The data in the table shows the mean monthly temperatures for Christchurch Month Jan Feb Mar Apr May Jun July Aug Sept Oct Nov Dec o Temperature ( C) 15 16 14 12 12 10 12 7 12 12 10 12 12 12 14 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a Find a sine model for this data in the form T ¼ A sin B(t ¡ C) + D Assume Jan ´ 1, Feb ´ 2, etc Do not use technology b Use technology to check your answer to a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\295IB_HL-2_12.CDR Wednesday, 31 October 2007 10:03:18 AM PETERDELL IB_HL-2ed (296) 296 ADVANCED TRIGONOMETRY (Chapter 12) At the Mawson base in Antarctica, the mean monthly temperatures for the last 30 years are as follows: Month Jan Feb Mar Apr May Jun July Aug Sept Oct Nov Dec Temperature (o C) ¡4 ¡10 ¡15 ¡16 ¡17 ¡18 ¡19 ¡17 ¡13 ¡6 ¡1 Find a sine model for this data using your calculator Use Jan ´ 1, Feb ´ 2, etc How appropriate is the model? Some of the largest tides in the world are observed in Canada’s Bay of Fundy The difference between high and low tides is 14 metres and the average time difference between high tides is about 12:4 hours a Find a sine model for the height of the tide H in terms of the time t b Sketch the graph of the model over one period Revisit the Opening Problem on page 284 The wheel takes 100 seconds to complete one revolution Find the sine model which gives the height of the light above the ground at any point in time Assume that at time t = 0, the light is at its lowest point D 10 m 2m THE COSINE FUNCTION We return to the Ferris wheel to see the cosine function being generated DEMO y Click on the icon to inspect a simulation of the view from above the wheel 10 q The graph being generated over time is a cosine function d This is no surprise as cos µ = 10 and so d = 10 cos µ x d y Now view the relationship between the sine and cosine functions Notice that the functions are identical in shape, but the cosine function is ¼2 units left of the sine function under a horizontal translation y¡=¡sin¡x -p p p ¡ cos x = sin x + This suggests that -p ¼ 3p -p ¢ cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Use your graphing package or graphics calculator to check this by graphing ¢ ¡ y = cos x and y = sin x + ¼2 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\296IB_HL-2_12.CDR Monday, November 2007 12:29:45 PM PETERDELL 2p x y¡=¡cos¡x GRAPHING PACKAGE IB_HL-2ed (297) 297 ADVANCED TRIGONOMETRY (Chapter 12) Example ¢ ¡ On the same set of axes graph y = cos x and y = cos x ¡ ¼3 ¢ ¡ y = cos x ¡ ¼3 comes from y = cos x under a horizontal translation through y y = cos x -p p -2p p Qw_ p x 2p p -1 p ¼ y = cos&x - p3* y EXERCISE 12D Given the graph of y = cos x, sketch the graphs of: x -p p 2p -1 a d g j y y y y b e h k = cos x + = cos(x + ¼6 ) = ¡ cos x = cos 2x y = cos x ¡ y = 23 cos x y = cos(x ¡ ¼6 ) + ¡ ¢ y = cos x2 Without graphing them, state the periods of: ¡ ¢ a y = cos 3x b y = cos x3 = cos(x ¡ ¼4 ) = 32 cos x = cos(x + ¼4 ) ¡ = cos 2x c f i l y y y y c y = cos ¡¼ ¢ 50 x The general cosine function is y = A cos B(x ¡ C) + D State the geometrical significance of A, B, C and D For the following graphs, find the cosine function representing them: a b c y y y x 2p -2 2p E 4p magenta yellow 95 y Q(1, tan¡q) P black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\297IB_HL-2_12.CDR Thursday, November 2007 9:14:40 AM PETERDELL sin¡q tan¡q x q -1 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 -5 THE TANGENT FUNCTION Consider the unit circle diagram given P(cos µ, sin µ) is a point which is free to move around the circle In the first quadrant we extend [OP] to meet the tangent at A(1, 0) so the intersection occurs at Q As P moves, so does Q The position of Q relative to A is defined as the tangent function Notice that ¢s ONP and OAQ are equiangular and therefore similar cyan x x cos¡q -1 N A(1, 0) tangent IB_HL-2ed (298) 298 ADVANCED TRIGONOMETRY (Chapter 12) AQ NP = OA ON Consequently AQ sin µ = cos µ i.e., which suggests that tan µ = sin µ cos µ y P The question arises: “If P does not lie in the first quadrant, how is tan µ defined?” q A(1, 0) x For µ obtuse, since sin µ is positive and cos µ is sin µ negative, tan µ = is negative cos µ As before, [PO] is extended to meet the tangent at A at Q(1, tan µ) Q(1, tan¡q) y q Q(1, tan¡q) A(1, 0) For µ in quadrant 3, sin µ and cos µ are both negative and so tan µ is positive This is clearly demonstrated as Q is back above the x-axis x P y For µ in quadrant 4, sin µ is negative and cos µ is positive So, tan µ is negative q A(1, 0) x P Q(1, tan¡q) DISCUSSION ² What is tan µ when P is at (0, 1)? ² What is tan µ when P is at (0, ¡1)? EXERCISE 12E.1 Use your calculator to find the value of: b tan 15o a tan 0o e tan 35o f tan 45o tan 20o tan 50o c g d h tan 25o tan 55o cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Explain why tan 45o = exactly black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\298IB_HL-2_12.CDR Wednesday, 31 October 2007 10:51:07 AM PETERDELL IB_HL-2ed (299) 299 ADVANCED TRIGONOMETRY (Chapter 12) THE GRAPH OF y = tan¡x y The graph of y = tan x is x DEMO -p - p2 p 3p p 2p 5p -3 DISCUSSION ² ² Is the tangent function periodic? If so, what is its period? For what values of x does the graph not exist?¡ What physical characteristics are shown near these values?¡ Explain why these values must occur when cos x = 0: ² ² ² Discuss how to find the x-intercepts of y = tan x What must tan(x ¡ ¼) simplify to? How many solutions does the equation tan x = have? EXERCISE 12E.2 a Use a transformation approach to sketch the graphs of these functions, x [0, 3¼]: ii y = ¡ tan x iii y = tan 2x i y = tan(x ¡ ¼2 ) b Use technology to check your answers to a Look in particular for: ² asymptotes ² x-axis intercepts GRAPHING PACKAGE Use the graphing package to graph, on the same set of axes: a y = tan x and y = tan(x ¡ 1) b y = tan x and y = ¡ tan x ¡x¢ c y = tan x and y = tan GRAPHING PACKAGE Describe the transformation which moves the first curve to the second in each case What is the period of: a y = tan x F b y = tan 2x c y = tan nx? TRIGONOMETRIC EQUATIONS cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Linear equations such as 2x + = 11 have exactly one solution Quadratic equations of the form ax2 + bx + c = 0, a 6= have at most two real solutions Trigonometric equations generally have infinitely many solutions unless a restrictive domain such as x 3¼ is given We will examine solving trigonometric equations using: ² preprepared graphs ² technology ² algebraic methods black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\299IB_HL-2_12.CDR Wednesday, 31 October 2007 10:57:41 AM PETERDELL IB_HL-2ed (300) 300 ADVANCED TRIGONOMETRY (Chapter 12) ¼ For the Ferris wheel Opening Problem the model is H = 10 sin 50 (t ¡ 25) + 12 t = 50 We can easily check this by substituting t = 0, 25, 50, 75 ¡ ¢ H(0) = 10 sin ¡ ¼2 + 12 = ¡10 + 12 = X 10 m t = 75 H(25) = 10 sin + 12 = 12 X ¡ ¢ H(50) = 10 sin ¼2 + 12 = 22 X etc t = 25 10 m 2m However, we may be interested in the times when the light is some other height above the ground, for example 16 m We would then need to solve the equation ¼ (t ¡ 25) + 12 = 16 which is called a sine equation 10 sin 50 GRAPHICAL SOLUTION OF TRIGONOMETRIC EQUATIONS Sometimes simple sine or cosine graphs are available on grid paper In such cases we can estimate solutions straight from the graph For example, we could use a graph to find approximate solutions for trigonometric equations such as cos µ = 0:4 for µ 10 radians y A B p p -1 C 2p 3p 3p y¡¡=¡0.4 q 10 5p y = cos¡q y = 0:4 meets y = cos µ at A, B and C and hence µ ¼ 1:2, 5:1 or 7:4 So, the solutions of cos µ = 0:4 for µ 10 radians are 1:2, 5:1 and 7:4 EXERCISE 12F.1 y y¡=¡sin¡x 0.5 x 10 11 12 13 14 15 -0.5 -1 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Use the graph of y = sin x to find, correct to decimal place, the solutions of: a sin x = 0:3 for x 15 b sin x = ¡0:4 for x 15 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\300IB_HL-2_12.CDR Wednesday, 31 October 2007 11:00:53 AM PETERDELL IB_HL-2ed (301) 301 ADVANCED TRIGONOMETRY (Chapter 12) y y=cos¡x 0.5 x 10 11 12 13 14 15 -0.5 -1 Use the graph of y = cos x to find, to decimal place, approximate solutions of: a cos x = 0:4, x [ 0, 10 ] b cos x = ¡0:3, x [ 4, 12 ] y y¡=¡sin2 ¡ x 0.5 x 10 11 12 13 14 15 -0.5 -1 Use the graph of y = sin 2x to find, correct to decimal place, the solutions of: a sin 2x = 0:7, x [ 0, 16 ] b sin 2x = ¡0:3, x [ 0, 16 ] The graph of y = tan x is illustrated y magenta yellow x -1 -2 95 100 50 75 25 -3 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a Use the graph to find estimates of: i tan ii tan 2:3 b Check your answers with a calculator c Find, correct to decimal place, the solutions of: i tan x = for 06x68 ii tan x = ¡1:4 for 26x67 cyan y¡=¡tan¡x black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\301IB_HL-2_12.CDR Thursday, November 2007 9:16:59 AM PETERDELL IB_HL-2ed (302) 302 ADVANCED TRIGONOMETRY (Chapter 12) SOLVING TRIGONOMETRIC EQUATIONS USING TECHNOLOGY Trigonometric equations may be solved using either a graphing package or a graphics display calculator (gdc) GRAPHING PACKAGE When using a graphics calculator make sure that the mode is set to radians TI C Example Solve sin x ¡ cos x = ¡ x for x 2¼ We need to use window settings just larger than the domain Xmin = ¡ ¼6 In this case, Xmax = 13¼ Xscale = ¼ The grid facility on the gdc can also be helpful, particularly when a sketch is required Using the appropriate function on the gdc gives the following solutions: x = 1:82, 3:28, 5:81 (3 s.f.) EXERCISE 12F.2 Solve each of the following for x 2¼: a c sin(x + 2) = 0:0652 µ 2¶ x x tan = x2 ¡ 6x + 10 b sin2 x + sin x ¡ = d sin(2x) cos x = ln x cos(x ¡ 1) + sin(x + 1) = 6x + 5x2 ¡ x3 Solve for x, ¡2 x 6: SOLVING TRIGONOMETRIC EQUATIONS ALGEBRAICALLY Using a calculator we get approximate decimal or numerical solutions to trigonometric equations Sometimes exact solutions are needed in terms of ¼, and these arise when the solutions are multiples of ¼6 or ¼4 Exact solutions obtained using algebra are called analytical solutions y ³ ´ ¡ p12 ; p12 (0, 1) ³ p1 ; p1 2 ³ p ´ ¡ 12 ; 23 ´ x (1, 0) (-1, 0) ³ ´ ¡ p12 ; ¡ p12 ³ p1 ; ¡ p1 2 yellow ´ ³ p ´ ¡ 12 ; ¡ 23 95 50 75 25 95 100 50 75 25 95 50 75 25 95 100 50 75 25 100 magenta (0, 1) ³ p ´ 2; ³p ;2 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\302IB_HL-2_12.CDR Wednesday, 31 October 2007 11:37:49 AM PETERDELL ´ x (1, 0) ´ ³p ; ¡2 (-1, 0) ³ p ´ ¡ 23 ; ¡ 12 (0,-1) cyan y ³ p ´ ¡ 23 ; 12 100 Reminder: ³ (0,-1) p ´ 2;¡ IB_HL-2ed (303) 303 ADVANCED TRIGONOMETRY (Chapter 12) Example Use the unit circle to find the exact solutions of x, x 3¼ for: ¢ ¡ b sin 2x = ¡ 12 c sin x ¡ ¼6 = ¡ 12 a sin x = ¡ 12 sin x = ¡ 12 , so from the unit circle 7¼ ¾ x = 11¼ + k2¼, k an integer a y x ) 7¼ , 11¼ , 19¼ k=0 k=0 k=1 x= is too big 7p 11p @=-\Qw_ Substituting k = 1, 2, 3, gives answers outside the required domain Likewise k = ¡1, ¡2, gives answers outside the required domain 11¼ ) there are two solutions: x = 7¼ or sin 2x = ¡ 12 b is solved in exactly the same way: ¾ 7¼ 2x = 11¼ + k2¼, k an integer ) x= ) x= 7¼ 11¼ 19¼ 23¼ 31¼ 35¼ 12 , 12 , 12 , 12 , 12 , 12 k=0 k=1 or 2¼ which is three solutions yellow Don’t forget to try k¡=¡¡1, ¡2, etc as sometimes we get solutions from them 95 k = ¡1 100 50 is too big, 75 4¼ 25 k=0 10¼ to both sidesg 2¼, 95 50 75 25 95 100 50 75 25 100 magenta ¼ 4¼ , 95 x= 100 ) fadding + k2¼ 2¼ 50 x= ¾ 8¼ 75 ) ¼ So, x = 0, fobtained by letting k = 0, 1, 2g is also solved in the same way: ¾ 7¼ = 11¼ + k2¼ 25 x¡ fdivide each term by 2g + k¼ 11¼ 12 sin(x ¡ ¼6 ) = ¡ 12 c cyan ¾ 7¼ 12 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\303IB_HL-2_12.CDR Wednesday, 31 October 2007 11:52:45 AM PETERDELL IB_HL-2ed (304) 304 ADVANCED TRIGONOMETRY (Chapter 12) Example p ¡ cos x ¡ Find exact solutions of p ¡ cos x ¡ Rearranging We recognise p1 ) 3¼ x¡ ) If k = ¡1, If k = 1, 3¼ ¢ 3¼ ¢ + = for x [ 0, 6¼ ] ¡ + = 0, we find cos x ¡ x= y x + k2¼ - x = ¡ ¼2 or 0: x= = ¡ p12 ¾ 2¼ 7¼ ¢ ¼ 4) as a special fraction (for multiples of 3¼ ¾ = 5¼ + k2¼, k an integer 3¼ 3¼ or 4¼: If k = 0, x= If k = 2, x= » -0.7 3¼ 11¼ or 2¼: or 6¼: If k = 3, the answers are greater than 6¼ 3¼ , So, the solutions are: x = 0, 2¼, 7¼ , 11¼ 4¼, or 6¼: Since the tangent function is periodic with period ¼ we see that tan(x + ¼) = tan x for all values of x This means that equal tan values are ¼ units apart Example ¢ ¡ Find exact solutions of tan 2x ¡ ¼3 = for x [¡¼, ¼ ] 2x ¡ ) ¼ x= x= ¡ 17¼ 24 , ¡ 5¼ 24 , 7¼ 24 , 19¼ 24 k = ¡2 k = ¡1 k=0 k=1 2x = ) ) fsince tan ¼4 = 1g ¼ + k¼ 7¼ 12 + k¼ 7¼ ¼ 24 + k = y ( , ) x p Example p sin x = cos x for x 2¼: p sin x = cos x p sin x ) fdividing both sides by cos xg = p cos x cyan tan x = 95 100 50 75 25 95 50 75 100 magenta yellow p1 x= ¼ or 7¼ 95 ) ) 25 95 , - 12 ) 100 50 75 25 - ( 100 x p 50 75 25 ( ,) y Find the exact solutions of black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\304IB_HL-2_12.CDR Wednesday, 31 October 2007 12:09:01 PM PETERDELL IB_HL-2ed (305) 305 ADVANCED TRIGONOMETRY (Chapter 12) EXERCISE 12F.3 List the possible solutions for x if k is an integer and: b x = ¡ ¼3 + k2¼, ¡2¼ x 2¼ a x = ¼6 + k2¼, x 6¼ ¡¼¢ c x = ¡ ¼2 + k¼, ¡4¼ x 4¼ d x = 5¼ + k , x 4¼ Find the exact solutions of: a cos x = ¡ 12 , x [ 0, 5¼ ] p c cos x + = 0, x 3¼ ¡ ¢ e sin x + ¼3 = 1, x [¡3¼, 3¼ ] sin x ¡ = 0, ¡2¼ x 2¼ ¢ ¡ = , x [¡2¼, 2¼ ] cos x ¡ 2¼ p ¢ ¡ sin x ¡ ¼4 + = 0, x [ 0, 3¼ ] b d f p Find the exact solutions of tan X = in terms of ¼ only Hence solve the equations: p ¢ p ¡ b tan 4x = c tan2 x = a tan x ¡ ¼6 = Find the zeros of: a y = sin 2x between and ¼ (inclusive) b y = sin(x ¡ ¼4 ) between and 3¼ (inclusive) a Use your graphics calculator to sketch the graphs of y = sin x and y = cos x on the same set of axes on the domain x [ 0, 2¼ ] b Find the x values of the points of intersection of the two graphs c Confirm that these values are the solutions of sin x = cos x on x [ 0, 2¼ ] Find the exact solutions to these equations for x 2¼ : a sin x = ¡ cos x b sin(3x) = cos(3x) c sin(2x) = p cos(2x) Check your answers to question using a graphics calculator Find the points of intersection of the appropriate graphs G USING TRIGONOMETRIC MODELS Example 10 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The height h(t) metres of the tide above mean sea level on January 24th at Cape ¡ ¢ where t is the number Town is modelled approximately by h(t) = sin ¼t of hours after midnight a Graph y = h(t) for t 24: b When is high tide and what is the maximum height? c What is the height at pm? d If a ship can cross the harbour provided the tide is at least m above mean sea level, when is crossing possible on January 24? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\305IB_HL-2_12.CDR Wednesday, 31 October 2007 12:10:38 PM PETERDELL IB_HL-2ed (306) 306 ADVANCED TRIGONOMETRY (Chapter 12) a h(t) = sin h(t) ¡ ¼t ¢ has period = 2¼ ¼ A 12 = 12 hours and h(0) = 15 18 21 24 t noon -3 c ¼ B b = 2¼ £ High tide is at am and pm The maximum height is m above the mean as seen at points A and B ¢ ¡ ¼ 2:60 (3 sig figs) At pm, t = 14 and h(14) = sin 14¼ So the tide is 2:6 m above the mean d h(t) A t1 (14, 2.60) B t2 12 t3 15 -3 We need to solve h(t) = h¡=¡2 i.e., sin t4 18 21 24 t ¡ ¼t ¢ = 2: ¡ ¼X ¢ Using a graphics calculator with Y1 = sin and Y2 = we obtain t1 = 1:39, t2 = 4:61, t3 = 13:39, t4 = 16:61 or you could trace across the graph to find these values Now 1:39 hours = hour 23 minutes, etc ) can cross between 1:23 am and 4:37 am or 1:23 pm and 4:37 pm EXERCISE 12G The population of grasshoppers after t weeks where t 12 is estimated by ¡ ¢ P (t) = 7500 + 3000 sin ¼t : cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 What is: i the initial estimate ii the estimate after weeks? What is the greatest population size over this interval and when does it occur? When is the population i 9000 ii 6000? During what time interval(s) does the population size exceed 10 000? ¡ ¢ The model for the height of a light on a Ferris wheel is H(t) = 20 ¡ 19 sin 2¼t , where H is the height in metres above the ground, and t is in minutes a Where is the light at time t = 0? b At what time is the light at its lowest in the first revolution of the wheel? c How long does the wheel take to complete one revolution? d Sketch the graph of the function H(t) over one revolution a b c d black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\306IB_HL-2_12.CDR Wednesday, 31 October 2007 12:13:21 PM PETERDELL IB_HL-2ed (307) ADVANCED TRIGONOMETRY (Chapter 12) 307 The population of water buffalo is given by ¡ ¢ where t is the number of P (t) = 400 + 250 sin ¼t years since the first estimate was made a What was the initial estimate? b What was the population size after: i months ii two years? c Find P (1) What is the significance of this value? d Find the smallest population size and when it first occurs e Find the first time when the herd exceeded 500 A paint spot X lies on the outer rim of the wheel of a paddle-steamer The wheel has radius m and as it rotates at a constant rate, X is seen entering the water every seconds H is the distance of X above the Hm bottom of the boat At time t = 0, X is at its highest point a Find a cosine model for H in the form H(t) = A cos B(t ¡ C) + D b At what time does X first enter the water? X 3m water level 1m 1m bottom of the boat Over a 28 day period, the cost per litre of petrol was modelled by C(t) = 9:2 sin ¼7 (t ¡ 4) + 107:8 cents L¡1 a True or false? i “The cost per litre oscillates about 107:8 cents with maximum price $1:17:” ii “Every 14 days, the cycle repeats itself.” b What was the cost of petrol at day 7? c On what days was the petrol priced at $1:10 per litre? d What was the minimum cost per litre and when did it occur? H RECIPROCAL TRIGONOMETRIC FUNCTIONS We define the reciprocal trigonometric functions cosec x, secant x and cotangent x as: csc x = , sin x sec x = cos x and cot x = cos x = tan x sin x Note the important identities: tan2 x + = sec2 x and + cot2 x = csc2 x Start with sin2 x + cos2 x = Proof: sin2 x cos2 x + = 2 cos x cos x cos2 x ) fdividing each term by cos2 xg ) tan2 x + = sec2 x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 To prove the second identity we instead divide each term by sin2 x black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\307IB_HL-2_12.CDR Wednesday, 31 October 2007 12:17:06 PM PETERDELL IB_HL-2ed (308) 308 ADVANCED TRIGONOMETRY (Chapter 12) Example 11 y¡=¡csc¡x y Use sketching techniques from y¡=¡sin¡x Chapter to sketch the graph 1 = csc x from sin x the graph of y = sin x for x [¡2¼, 2¼ ] of y = x -2p -p p 2p -1 Check your answer using your calculator EXERCISE 12H Without using a calculator, find csc x, sec x and cot x for: b cos x = 23 , 3¼ a sin x = 35 , x ¼2 < x < 2¼ Without using a calculator, find: ¡ ¢ ¡ ¢ b cot 2¼ a csc ¼3 c sec ¡ 5¼ ¢ d cot (¼) Find the other five trigonometric ratios if: a cos x = c sec x = 12 e tan ¯ = 3¼ and < x < 2¼ and < x < and ¼ < ¯ < Simplify: a tan x cot x d sin x cot x ¼ 3¼ b sin x = ¡ 23 d csc x = and f cot µ = and ¼ < x < ¼ 3¼ <x<¼ and ¼ < µ < 3¼ b sin x csc x c csc x cot x e cot x csc x f sin x cot x + cos x cot x Use technology to help sketch graphs on [¡2¼, 2¼ ] of: a y = sec x b y = cot x cyan magenta yellow 95 100 50 75 25 f csc 3x = ¡ 23 c 95 50 75 25 95 e 100 50 75 25 sec 2x = 95 100 50 75 25 d p csc x = ¡ 100 Solve for x when x [ 0, 2¼ ] : a sec x = b black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\308IB_HL-2_12.CDR Monday, November 2007 12:41:25 PM PETERDELL cot x = ¢ ¡ cot 2x ¡ ¼4 + = IB_HL-2ed (309) 309 ADVANCED TRIGONOMETRY (Chapter 12) I TRIGONOMETRIC RELATIONSHIPS There are a vast number of trigonometric relationships However, we only need to remember a few because we can obtain the rest by rearrangement or substitution SIMPLIFYING TRIGONOMETRIC EXPRESSIONS For any given angle µ, sin µ and cos µ are real numbers, so the algebra of trigonometry is identical to the algebra of real numbers Consequently, expressions like sin µ + sin µ compare with 2x + 3x when we wish to simplification So, sin µ + sin µ = sin µ To simplify complicated trigonometric expressions, we often use the Pythagorean identities: sin2 µ + cos2 µ = tan2 µ + = sec2 µ + cot2 µ = csc2 µ Notice that we can also use the rearrangements: sin2 µ = ¡ cos2 µ tan2 µ = sec2 µ ¡ cot2 µ = csc2 µ ¡ cos2 µ = ¡ sin2 µ EXERCISE 12I Factorise: a ¡ sin2 µ d cot2 x ¡ cot x + b e tan2 ® ¡ tan ® c sec2 ¯ ¡ csc2 ¯ sin2 x + sin x cos x + cos2 x Simplify: a sin2 µ + cos2 µ c ¡ cos2 µ b d ¡ sec2 ¯ cos2 ® ¡ e tan2 µ(cot2 µ + 1) tan2 µ + f cos2 ®(sec2 ® ¡ 1) g (2 sin µ + cos µ)2 + (3 sin µ ¡ cos µ)2 h i sec A ¡ sin A tan A ¡ cos A j (1 + csc µ)(sin µ ¡ sin2 µ) cos2 µ 1¡ + sin µ k + cot µ sec µ ¡ csc µ tan µ + cot µ l m tan2 µ sec µ ¡ cos2 ¯ ¡ sin2 ¯ cos ¯ ¡ sin ¯ cyan magenta yellow 95 100 50 75 25 95 100 50 75 sin µ + cos µ + = csc µ + cos µ sin µ 25 d cos ® sin ® + = sin ® + cos ® ¡ tan ® ¡ cot ® c 95 cos µ = sec µ + tan µ ¡ sin µ 100 b 50 sec A ¡ cos A = tan A sin A 75 a 25 95 100 50 75 25 Prove: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\309IB_HL-2_12.CDR Wednesday, 31 October 2007 12:27:58 PM PETERDELL IB_HL-2ed (310) 310 ADVANCED TRIGONOMETRY (Chapter 12) J COMPOUND ANGLE FORMULAE INVESTIGATION COMPOUND ANGLE FORMULAE What to do: Copy and complete for angles A and B in radians or degrees: A B cos A cos B cos(A ¡ B) cos A ¡ cos B cos A cos B + sin A sin B o 47 24o 138o 49o 3c 2c Make sure you include some angles of your choosing What you suspect from the results of this table? Make another table with columns A, B, sin A, sin B, sin(A + B), sin A + sin B, sin A cos B + cos A sin B and complete it for four sets of angles of your choosing What is your conclusion? cos (A § B) = cos A cos B ¨ sin A sin B sin (A § B) = sin A cos B § cos A sin B tan A § tan B tan (A § B) = ¨ tan A tan B If A and B are any two angles then: These are known as the compound angle formulae There are many ways of establishing them, but many are unsatisfactory as the arguments limit the angles A and B to being acute Proof: Consider P(cos A, sin A) and Q(cos B, sin B) as any two points on the unit circle, as shown Angle POQ is A ¡ B Using the distance formula: p PQ = (cos A ¡ cos B)2 + (sin A ¡ sin B)2 y P(cosA, sinA) 1 Q(cos B, sinB) A B 1 x ) (PQ)2 = cos2 A ¡ cos A cos B + cos2 B + sin2 A ¡ sin A sin B + sin2 B = (cos2 A + sin2 A) + (cos2 B + sin2 B) ¡ 2(cos A cos B + sin A sin B) = ¡ 2(cos A cos B + sin A sin B) (1) But, by the cosine rule in ¢POQ, (PQ)2 = 12 + 12 ¡ 2(1)(1) cos(A ¡ B) = ¡ cos(A ¡ B) (2) ) cos(A ¡ B) = cos A cos B + sin A sin B fcomparing (1) and (2)g cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 From this formula the other formulae can be established: cos(A + B) = cos(A ¡ (¡B)) = cos A cos(¡B) + sin A sin(¡B) = cos A cos B + sin A(¡ sin B) fcos(¡µ) = cos µ and sin(¡µ) = ¡ sin µg = cos A cos B ¡ sin A sin B black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\310IB_HL-2_12.CDR Wednesday, 31 October 2007 12:33:19 PM PETERDELL IB_HL-2ed (311) 311 ADVANCED TRIGONOMETRY (Chapter 12) Also sin(A ¡ B) = cos( ¼2 ¡ (A ¡ B)) = cos(( ¼2 ¡ A) + B) = cos( ¼2 ¡ A) cos B ¡ sin( ¼2 ¡ A) sin B = sin A cos B ¡ cos A sin B sin(A + B) = sin(A ¡ (¡B)) = sin A cos(¡B) ¡ cos A sin(¡B) = sin A cos B ¡ cos A(¡ sin B) = sin A cos B + cos A sin B tan(A ¡ B) = tan(A + (¡B)) tan(A + B) = = sin(A + B) cos(A + B) sin A cos B + cos A sin B cos A cos B ¡ sin A sin B sin A cos B cos A sin B + cos A cos B cos A cos B = = tan A + tan(¡B) ¡ tan A tan(¡B) tan A ¡ tan B = + tan A tan B = ftan(¡B) = ¡ tan Bg cos A cos B sin A sin B ¡ cos A cos B cos A cos B tan A + tan B ¡ tan A tan B Example 12 y o sin(270 + ®) = sin 270o cos ® + cos 270o sin ® = ¡1 £ cos ® + £ sin ® = ¡ cos ® Expand and simplify sin(270o + ®) 270° x (0,-1) Example 13 cos 3µ cos µ ¡ sin 3µ sin µ = cos(3µ + µ) fcompound formula in reverseg = cos 4µ Simplify: cos 3µ cos µ ¡ sin 3µ sin µ Example 14 sin 75o = sin(45o + 30o ) = sin 45o cos 30o + cos 45o sin 30o Without using your calculator, show that sin 75o = p p 6+ : p = ( p12 )( 23 ) + ( p12 )( 12 ) ´p ³p p2 p = 23+1 2 cyan magenta yellow 95 p p 6+ 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\311IB_HL-2_12.CDR Wednesday, 31 October 2007 12:36:09 PM PETERDELL IB_HL-2ed (312) 312 ADVANCED TRIGONOMETRY (Chapter 12) EXERCISE 12J Expand and simplify: a sin(90o + µ) b cos(90o + µ) c sin(180o ¡ ®) d cos(¼ + ®) ¡ ¢ tan ¼4 + µ e sin(2¼ ¡ A) ¢ ¡ tan µ ¡ 3¼ f cos( 3¼ ¡ µ) i tan(¼ + µ) g h Expand, then simplify and write your answer in the form A sin µ + B cos µ: a sin(µ + ¼3 ) b cos( 2¼ ¡ µ) cos(µ + ¼4 ) c sin( ¼6 ¡ µ) d Simplify using appropriate compound formulae in reverse: b sin 2A cos A + cos 2A sin A a cos 2µ cos µ + sin 2µ sin µ c cos A sin B ¡ sin A cos B d sin ® sin ¯ + cos ® cos ¯ e sin Á sin µ ¡ cos Á cos µ f sin ® cos ¯ ¡ cos ® sin ¯ tan 2A + tan A tan 2µ ¡ tan µ h g + tan 2µ tan µ ¡ tan 2A tan A Prove: a sin 2µ ¡ tan µ = + cos 2µ c b csc 2µ = tan µ + cot 2µ sin 2µ cos 2µ ¡ = sec µ sin µ cos µ Simplify using compound formulae: a cos(® + ¯) cos(® ¡ ¯) ¡ sin(® + ¯) sin(® ¡ ¯) b sin(µ ¡ 2Á) cos(µ + Á) ¡ cos(µ ¡ 2Á) sin(µ + Á) c cos ® cos(¯ ¡ ®) ¡ sin ® sin(¯ ¡ ®) Without using your calculator, show that the following are true: a cos 75o = p p 6¡ sin 105o = b p p 6+ Find the exact value of each of the following: If tan A = a c tan ¡ 5¼ ¢ 12 cos ¡ 13¼ ¢ 12 b = p p ¡ 6¡ tan 105o and tan B = ¡ 15 , find the exact value of tan(A + B): ¢ ¡ If tan A = 34 , evaluate tan A + ¼4 10 Simplify each of the following: ¡ ¢ ¡ ¢ a tan A + ¼4 tan A ¡ ¼4 tan(A + B) + tan(A ¡ B) ¡ tan(A + B) tan(A ¡ B) b tan 80o ¡ tan 20o + tan 80o tan 20o 11 Simplify, giving your answer as an exact value: 12 If tan(A + B) = and tan B = 23 , find the exact value of tan A: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 13 Find tan A if tan(A ¡ B) tan(A + B) = 1: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\312IB_HL-2_12.CDR Wednesday, 31 October 2007 12:40:57 PM PETERDELL IB_HL-2ed (313) ADVANCED TRIGONOMETRY (Chapter 12) 14 Find the exact value of tan ® in the diagram: 313 C cm a A cm B 10 cm 15 Find the exact value of the tangent of the acute angle between two lines if their gradients are 12 and 23 16 Express tan(A + B + C) in terms of tan A, tan B and tan C: Hence show that, if A, B and C are the angles of a triangle, tan A + tan B + tan C = tan A tan B tan C: 17 Show that p ¢ ¡ a cos µ + ¼4 = cos µ ¡ sin µ: p ¡ ¢ b cos µ ¡ ¼3 = cos µ + sin µ: c cos(® + ¯) ¡ cos(® ¡ ¯) = ¡2 sin ® sin ¯ d cos(® + ¯) cos(® ¡ ¯) = cos2 ® ¡ sin2 ¯: 18 Prove that, in the given figure, ® + ¯ = ¼4 E A 19 p b a B C D a Show that: sin(A + B) + sin(A ¡ B) = sin A cos B b From a we notice that sin A cos B = 12 sin(A + B) + 12 sin(A ¡ B) and this formula enables us to convert a product into a sum Use the formula to write the following as sums: i sin 3µ cos µ ii sin 6® cos ® iii sin 5¯ cos ¯ vi 13 cos 5A sin 3A v cos 4® sin 3® iv cos µ sin 4µ 20 a Show that cos(A + B) + cos(A ¡ B) = cos A cos B b From a we notice that cos A cos B = 12 cos (A + B) + 12 cos (A ¡ B) Use this formula to convert the following to a sum of cosines: i cos 4µ cos µ ii cos 7® cos ® iii cos 3¯ cos ¯ vi 14 cos 4x cos 2x v cos P cos 4P iv cos x cos 7x 21 a Show that cos(A ¡ B) ¡ cos(A + B) = sin A sin B: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b From a we notice that sin A sin B = 12 cos(A ¡ B) ¡ 12 cos(A + B) Use this formula to convert the following to a difference of cosines: i sin 3µ sin µ ii sin 6® sin ® iii sin 5¯ sin ¯ vi 15 sin 3M sin 7M v 10 sin 2A sin 8A iv sin µ sin 4µ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\313IB_HL-2_12.CDR Wednesday, 31 October 2007 12:42:32 PM PETERDELL IB_HL-2ed (314) 314 ADVANCED TRIGONOMETRY (Chapter 12) 22 sin A cos B = cos A cos B = sin A sin B = sin(A + B) + sin(A ¡ B) (1) cos(A + B) + cos(A ¡ B) (2) cos(A ¡ B) ¡ cos(A + B) (3) are called products to sums formulae What formulae result if we replace B by A in each of these formulae? and A ¡ B = D: 23 Suppose A + B = S a Show that A = S+D and B = S¡D : b For the substitution A + B = S and A ¡ B = D, show that equation (1) in ¢ ¡ ¢ ¡ cos S¡D (4) question 22 becomes sin S + sin D = sin S+D 2 ¢ ¡ ¢ ¡ c In (4) replace D by (¡D) and show that sin S ¡ sin D = cos S+D sin S¡D : 2 d What results when the substitution A = (2) of question 22? S+D and B = S¡D is made into e What results when the substitution A = (3) of question 22? S+D and B = S¡D is made into 24 From question 23 we obtain the formulae: ¡ ¢ ¡ ¢ ¢ ¡ ¢ ¡ cos S¡D cos S¡D cos S + cos D = cos S+D sin S + sin D = sin S+D 2 2 ¡ ¡ ¢ ¡ ¢ ¢ ¡ ¢ sin S ¡ sin D = cos S+D cos S ¡ cos D = ¡2 sin S+D sin S¡D sin S¡D 2 2 These are called the factor formulae as they convert sums and differences into factorised forms Use these formulae to convert the following to products: a d g b e h sin 5x + sin x sin 5µ ¡ sin 3µ cos 2B ¡ cos 4B K c f i cos 8A + cos 2A cos 7® ¡ cos ® sin(x + h) ¡ sin x cos 3® ¡ cos ® sin 3® + sin 7® cos(x + h) ¡ cos x DOUBLE ANGLE FORMULAE By replacing B by A in each of the addition compound angle formulae (on page 310), we obtain the double angle formulae Replacing B by A in the formula sin(A + B) = sin A cos B + cos A sin B we obtain sin(A + A) = sin A cos A + cos A sin A ) sin 2A = sin A cos A Replacing B by A in the formula cos(A + B) = cos A cos B ¡ sin A sin B we obtain cos(A + A) = cos A cos A ¡ sin A sin A magenta yellow 95 100 50 75 25 95 50 75 25 95 100 50 75 25 95 100 50 75 25 cyan 100 cos 2A = cos2 A ¡ sin2 A ) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\314IB_HL-2_12.CDR Wednesday, 31 October 2007 12:46:52 PM PETERDELL IB_HL-2ed (315) ADVANCED TRIGONOMETRY (Chapter 12) 315 Two other forms of this formula can be obtained by respectively replacing cos2 A by ¡ sin2 A and sin2 A by ¡ cos2 A We obtain cos 2A = (1 ¡ sin2 A) ¡ sin2 A = ¡ sin2 A and cos 2A = cos2 A ¡ (1 ¡ cos2 A) = cos2 A ¡ tan A + tan B , ¡ tan A tan B tan A + tan A we obtain tan(A + A) = ¡ tan A tan A tan A ) tan 2A = ¡ tan2 A Replacing B by A in the formula tan(A + B) = The double angle formulae are: sin 2A = sin A cos A cos 2A = cos2 A ¡ sin2 A GRAPHING PACKAGE = ¡ sin2 A = cos2 A ¡ tan 2A = tan A ¡ tan2 A Example 15 Given that sin ® = and cos ® = ¡ 45 a find: a sin 2® b cos 2® sin 2® = sin ® cos ® = 2( 35 )(¡ 45 ) = b ¡ 24 25 cos 2® = cos2 ® ¡ sin2 ® = (¡ 45 )2 ¡ ( 35 )2 = 25 Example 16 Given that sin ® = ¡ 13 tan ® = and cos ® = 12 13 , find tan 2®: sin ® = ¡ 12 cos ® cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ¡ 5¢ ¡ 56 ¡ 12 tan ® ¡120 ) tan 2® = = = = = ¡ 120 ¢ ¡ 2 25 119 144 ¡ 25 ¡ tan ® ¡ ¡ ¡ 12 144 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\315IB_HL-2_12.CDR Wednesday, 31 October 2007 12:53:12 PM PETERDELL IB_HL-2ed (316) 316 ADVANCED TRIGONOMETRY (Chapter 12) Example 17 If ® is acute and cos 2® = find the values of a cos ® b sin ® cos 2® = cos2 ® ¡ ) 34 = cos2 ® ¡ a cos2 ® = ) b cos ® = § 2p72 ) cos ® = p ¡ cos2 ® fas ® is acute, sin ® > 0g q ) sin ® = ¡ 78 q ) sin ® = 18 p ) sin ® = p p7 2 ) fas ® is acute, cos ® > 0g sin ® = p 2 Example 18 Use an appropriate ‘double angle formula’ to simplify: a sin µ cos µ b cos2 2B ¡ a cos2 2B ¡ = 2(2 cos2 2B ¡ 1) = cos 2(2B) = cos 4B b sin µ cos µ = 32 (2 sin µ cos µ) = sin 2µ EXERCISE 12K If sin A = 5 and cos A = a If cos A = 13 , find cos 2A a If sin ® = ¡ 23 where ¼ < ® < value of sin 2®: b If cos ¯ = 25 where value of sin 2¯ 3¼ a find the values of: sin 2A b cos 2A b If sin Á = ¡ 23 , find cos 2Á 3¼ find the value of cos ® and hence the < ¯ < 2¼, find the value of sin ¯ and hence the If ® is acute and cos 2® = ¡ 79 , find without a calculator: Find the exact value of tan A if tan 2A = 21 20 a cos ® b sin ® and A is obtuse Find the exact value of tan A if tan 2A = ¡ 12 and A is acute ¡¼¢ Find the exact value of tan If sin A = ¡ 13 , ¼ A and cos B = p1 , B ¼2 , find cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 tan(A + B) 95 100 50 75 25 b tan 2A £ ¡¼¢ â Ị đô Find the exact value of cos 12 + sin 12 a 3¼ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\316IB_HL-2_12.CDR Wednesday, 31 October 2007 12:54:49 PM PETERDELL IB_HL-2ed (317) ADVANCED TRIGONOMETRY (Chapter 12) 10 Use an appropriate ‘double angle’ formula to simplify: a sin ® cos ® b cos ® sin ® c sin ® cos ® 317 d cos ¯ ¡ e ¡ cos Á f ¡ sin2 N g j sin2 M ¡ sin 2A cos 2A h k cos2 ® ¡ sin2 ® cos 3® sin 3® i l sin2 ® ¡ cos2 ® cos2 4µ ¡ m ¡ cos2 3¯ n ¡ sin2 5® o sin2 3D ¡ p cos2 2A ¡ sin2 2A q cos2 ( ®2 ) ¡ sin2 ( ®2 ) r sin2 3P ¡ cos2 3P 2 11 Show that: a (sin µ + cos µ)2 = + sin 2µ 4 cos µ ¡ sin µ = cos 2µ b GRAPHING PACKAGE 12 Find the exact value of cos A in the diagram: a b A 2A cm cm A 2A cm cm 13 Find the domain and range of: a x 7! tan x b a sin2 µ = 14 Prove that: ¡ x 7! sec 2x c b cos2 µ = cos 2µ + x 7! cot(3x) cos 2µ 15 Prove the identities: sin 2µ sin µ + sin 2µ a = cot µ b = tan µ ¡ cos 2µ + cos µ + cos 2µ p 16 If sin x + cos x = k sin(x + b) for k > and < b < 2¼, find k and b 17 Use the identity sin A ¡ sin B = cos 12 (A + B) sin 12 (A ¡ B) to show that if sin A = sin B, then either A = B + k2¼ or A + B = ¼ + k2¼, for some k Z : 18 a Prove that cos 3µ = cos3 µ ¡ cos µ by replacing 3µ by (2µ + µ) b Hence, solve the equation cos3 µ ¡ cos µ + = for µ [ ¡¼, ¼ ]: 19 a Write sin 3µ in the form a sin3 µ + b sin µ where a, b Z b Hence, solve the equation sin 3µ = sin µ for µ [ 0, 3¼ ]: 20 Use the basic definition of periodicity to show algebraically that the period of 2¼ f(x) = sin(nx) is , for all n > n 21 a Write cos x ¡ sin x in the form k cos(x + b) for k > 0, < b < 2¼ cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b Use a to solve the equation cos x ¡ sin x = ¡2 for x ¼ 2t ¡ t2 c Given that t = tan( x2 ), prove that sin x = and cos x = + t2 + t2 d Solve cos x ¡ sin x = ¡2 for x ¼ using c black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_12\317IB_HL-2_12.CDR Friday, 12 December 2008 12:04:27 PM TROY IB_HL-2ed (318) 318 ADVANCED TRIGONOMETRY (Chapter 12) L TRIGONOMETRIC EQUATIONS IN QUADRATIC FORM Sometimes we may be given trigonometric equations in quadratic form For example, sin2 x + sin x = and cos2 x + cos x ¡ = are clearly quadratic equations where the variables are sin x and cos x respectively These equations can be factorised and then solved: sin2 x + sin x = sin x(2 sin x + 1) = ) sin x = or ¡ 12 etc ) cos2 x + cos x + = (2 cos x ¡ 1)(cos x + 1) = ) cos x = 12 or ¡1 etc and ) The use of the quadratic formula is often necessary EXERCISE 12L Solve for x [ 0, 2¼ ] giving your answers as exact values: b cos2 x = cos x c cos2 x + cos x ¡ = a sin2 x + sin x = d sin2 x + sin x + = g sin 4x = sin 2x e sin2 x = ¡ cos x f p h sin x + cos x = tan x = cot x Solve for x [¡¼, ¼ ] giving your answers as exact values: a sin x + csc x = b sin 2x + cos x ¡ sin x ¡ = c tan x ¡ tan x ¡ = Solve for x [ 0, 2¼ ] : a cos2 x = sin x b cos 2x + sin x = tan2 x + sec2 x = c M TRIGONOMETRIC SERIES AND PRODUCTS EXERCISE 12M a Simplify + sin x + sin2 x + sin3 x + sin4 x + ::::: + sinn¡1 x b What is the sum of the infinite series + sin x + sin2 x + sin3 x + ::::: ? c If the series in b has sum 23 , find x such that x [ 0, 2¼ ] We know that sin x cos x = sin 2x a Show that: i sin x(cos x + cos 3x) = sin 4x ii sin x(cos x + cos 3x + cos 5x) = sin 6x b What you suspect the following would simplify to? i sin x(cos x + cos 3x + cos 5x + cos 7x) ii cos x + cos 3x + cos 5x + ::::: + cos 19x (i.e., 10 terms) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 c Write down the possible generalisation of b ii to n terms black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\318IB_HL-2_12.CDR Thursday, November 2007 9:17:54 AM PETERDELL IB_HL-2ed (319) 319 ADVANCED TRIGONOMETRY (Chapter 12) From sin 2x = sin x cos x we observe that sin x cos x = sin 2x sin(21 x) = 21 a Prove that: sin(23 x) sin(22 x) ii sin x cos x cos 2x cos 4x = i sin x cos x cos 2x = 22 23 b If the pattern observed in a continues, what would: i sin x cos x cos 2x cos 4x cos 8x ii sin x cos x cos 2x:::: cos 32x simplify to? c What is the generalisation of the results in a and b? a Use the principle of mathematical induction to prove that: sin 2nµ cos µ + cos 3µ + cos 5µ + :::::: + cos(2n ¡ 1)µ = , n Z + sin µ b What does cos µ + cos 3µ + cos 5µ + :::::: + cos 31µ simplify to? Use the principle of mathematical induction to prove that: sin µ + sin 3µ + sin 5µ + ::::: + sin(2n ¡ 1)µ = for all positive integers n, and hence find the value of 5¼ 9¼ 11¼ 13¼ sin ¼7 + sin 3¼ + sin + sin ¼ + sin + sin + sin ¡ cos 2nµ sin µ Use the principle of mathematical induction to prove that: sin(2n x) cos x £ cos 2x £ cos 4x £ cos 8x::::::: cos(2n¡1 x) = n £ sin x for all n Z + Use the principle of mathematical induction to prove that: · ¸ cos(n + 1)µ sin nµ cos2 µ + cos2 2µ + cos2 3µ + cos2 4µ::::::: cos2 (nµ) = 12 n + sin µ for all n Z + REVIEW SET 12A Without using technology draw the graph of y = sin x for x 2¼: Without using technology draw the graph of y = sin 3x for x 2¼ ¡ ¢ State the period of: a y = sin x3 b y = ¡2 tan 4x ¡ ¢ Without using technology draw a sketch graph of y = sin x ¡ ¼3 + 2: The table below gives the mean monthly maximum temperature (o C) for Perth Airport in Western Australia Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Temp 31:5 31:8 29:5 25:4 21:5 18:8 17:7 18:3 20:1 22:4 25:5 28:8 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a A sine function of the form T ¼ A sin B(t¡C)+D is used to model the data Find good estimates of the constants A, B, C and D without using technology Use Jan ´ 1, Feb ´ 2, etc b Check your answer to a using your technology How well does your model fit? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\319IB_HL-2_12.CDR Wednesday, 31 October 2007 2:08:33 PM PETERDELL IB_HL-2ed (320) 320 ADVANCED TRIGONOMETRY (Chapter 12) Use technology to solve for x [ 0, ] : a sin x = 0:382 b tan Use technology to solve for x [ 0, ] : a sin(x ¡ 2:4) = 0:754 b ¢ ¡ sin x + ¼3 = 0:6049 Solve algebraically in terms of ¼: a sin x = ¡1 for x [ 0, 4¼ ] p sin x ¡ = for x [¡2¼, 2¼] b ¡x¢ = ¡0:458 Solve algebraically in terms of ¼: p a sin 3x + = for x [ 0, 2¼] b sec2 x = tan x + for x [ 0, 2¼] 10 An ecologist studying a species of water beetle estimates the population of a colony over an eight week period If t is the number of weeks after the initial estimate is made, then the population can ¡ ¢ where be modelled by P (t) = + sin ¼t t a What was the initial population? b What were the smallest and largest populations? c During what time interval(s) did the population exceed 6000? REVIEW SET 12B Solve algebraically, giving answers in terms of ¼: b sin2 x = a sin2 x ¡ sin x ¡ = a On the same set of axes, sketch y = cos x and y = cos x ¡ 3: ¡ ¢ b On the same set of axes, sketch y = cos x and y = cos x ¡ ¼4 c On the same set of axes, sketch y = cos x and y = cos 2x: ¡ ¢ d On the same set of axes, sketch y = cos x and y = cos x ¡ ¼3 + 3: In an industrial city, the amount of pollution in the air becomes greater during the working week when factories are operating, and lessens over the weekend The number of milligrams of pollutants in a cubic metre of air is given by ¡ ¢ 37 P (t) = 40 + 12 sin 2¼ t ¡ 12 where t is the number of days after midnight on Saturday night cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a What is the minimum level of pollution? b At what time during the week does this minimum level occur? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\320IB_HL-2_12.CDR Wednesday, 31 October 2007 2:10:58 PM PETERDELL IB_HL-2ed (321) 321 ADVANCED TRIGONOMETRY (Chapter 12) Find the cosine function represented in the following graphs: y y a b 2p x x -4 Use technology to solve: a cos x = 0:4379 for x 10 b cos(x ¡ 2:4) = ¡0:6014 for x 6 ¡¼¢ b tan 12 Find the exact value of: a cos(165o ) Find the exact solutions of: ¢ ¡ a tan x ¡ ¼3 = p13 , x [ 0, 4¼ ] ¡ b cos x + 2¼ ¢ = 12 , x [¡2¼, 2¼ ] Find the exact solutions of: p p ¢ ¡ a cos x + ¼4 ¡ = 0, x [ 0, 4¼ ] b tan 2x ¡ = 0, x [ 0, 2¼ ] Simplify: a cos3 µ + sin2 µ cos µ b d sin2 µ ¡ cos µ cos2 µ ¡ sin µ e cos2 µ (tan µ + 1)2 ¡ 10 Expand and simplify if possible: (2 sin ® ¡ 1)2 a ¡ sin2 µ c (cos ® ¡ sin ®)2 b REVIEW SET 12C Simplify: ¡ cos2 µ + cos µ a sin ® ¡ cos ® sin2 ® ¡ cos2 ® b Show that: cos µ + sin µ a + = sec µ + sin µ cos µ If sin A = 13 and cos A = If sin ® = ¡ 34 , ® [ ¼, sin 2® If cos x = ¡ 34 3¼ 12 13 µ b 1+ cos µ c ¶ find the values of: sin2 ® ¡ cos ® ¡ ¢ cos µ ¡ cos2 µ = sin2 µ: a sin 2A b cos 2A ] find the value of cos ® and hence the value of and ¼ < x < 3¼ find the exact value of sin ii tan a Solve algebraically: i tan x = ¡x¢ =4 b Find the exact solutions in terms of ¼ only for: p p ¢ ¡ i tan x + ¼6 = ¡ ii tan 2x = ¡ ¡x¢ iii tan (x ¡ 1:5) = iii tan2 x ¡ = c Use technology to solve tan(x ¡ 1:2) = ¡2: cyan magenta yellow 95 100 50 75 25 95 100 50 < µ < ¼, find sin µ and cos µ without using a calculator 75 ¼ 25 95 100 50 75 25 95 100 50 75 25 If tan µ = ¡ 23 , black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\321IB_HL-2_12.CDR Wednesday, 31 October 2007 2:14:18 PM PETERDELL IB_HL-2ed (322) 322 ADVANCED TRIGONOMETRY (Chapter 12) sin 2® ¡ sin ® simplifies to tan ®: cos 2® ¡ cos ® + ¢ ¡ ¢ ¡ Simplify: a cos 3¼ b sin µ + ¼2 ¡µ Show that D 10 Find the length of BC: 4m C q q A B 3m REVIEW SET 12D Show, in the simplest possible way, that: p ¢ ¡ cos µ + ¼4 = cos µ ¡ sin µ a b cos ® cos(¯ ¡ ®) ¡ sin ® sin(¯ ¡ ®) = cos ¯ for x [ ¼2 , ¼ ], find without using a calculator the exact values of: b sin 2x c cos 2x d tan 2x p p ¡ ¢ Show that sin ¼8 = 12 ¡ using a suitable double angle formula If sin x = a cos x If ® and ¯ are the other angles of a right angled triangle, show that sin 2® = sin 2¯ Prove that: (sin µ + cos µ)2 = + sin 2µ a Solve for x in [ 0, 2¼ ] : q a cos 2x + = c a Prove that: b If tan 2® = for ® ] 0, A 2a ¼ csc 2x + cot 2x = cot x p b sin 2x = ¡ cos 2x a sin 2µ = 2ab c2 b cos 2µ = a2 ¡ b2 c2 [, find sin ® without using a calculator Without using a calculator: cm a Show that cos ® = 56 C x cm a b b Show that x is a solution of 3x2 ¡ 25x + 48 = c Find x by solving the equation in b cm B cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 10 From ground level, a shooter is aiming at targets on a vertical brick wall At the current angle of elevation of his rifle, he will hit a target 20 m above ground level If he doubles the angle of elevation of the rifle, he will hit a target 45 m above ground level How far is the shooter from the wall? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_12\322IB_HL-2_12.CDR Monday, November 2007 12:42:25 PM PETERDELL IB_HL-2ed (323) Chapter Matrices Contents: A B C D E F G 13 Matrix structure Matrix operations and definitions The inverse of a 2×2 ¡ ¡ matrix 3¡×3 and larger matrices Solving systems of linear equations Solving systems using row operations Induction with matrices cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 13A Review set 13B Review set 13C Review set 13D Review set 13E black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\323IB_HL-2_13.CDR Wednesday, 31 October 2007 2:36:06 PM PETERDELL IB_HL-2ed (324) 324 MATRICES (Chapter 13) Matrices are rectangular arrays of numbers which are used to organise information of a numeric nature They are used in a wide array of fields extending far beyond mathematics, including: ² Solving systems of equations in business, physics, engineering, etc ² Linear programming where we may wish to optimise a linear expression subject to linear constraints For example, optimising profits of a business ² Business inventories involving stock control, cost, revenue and profit calculations Matrices form the basis of business computer software ² Markov chains for predicting long term probabilities such as in weather ² Strategies in games where we wish to maximise our chance of winning ² Economic modelling where the input from various suppliers is needed to help a business be successful Graph (network) theory used to determine routes for trucks and airlines to minimise distance travelled and therefore costs Assignment problems to direct resources in industrial situations in the most cost effective way ² ² ² Forestry and fisheries management where we need to select an appropriate sustainable harvesting policy ² Cubic spline interpolation used to construct curves and fonts Each font is stored in matrix form in the memory of a computer ² Computer graphics, flight simulation, Computer Aided Tomography (CAT scanning) and Magnetic Resonance Imaging (MRI), Fractals, Chaos, Genetics, Cryptography (coding, code breaking, computer confidentiality), etc A MATRIX STRUCTURE A matrix is a rectangular array of numbers arranged in rows and columns In general the numbers within a matrix represent specific quantities You have been using matrices for many years without realising it For example: Arsenal Liverpool Chelsea Leeds Won Lost Drew Points 24 76 23 73 21 68 20 5 65 Ingredients sugar flour milk salt Amount tspn cup 200 mL pinch Consider these two items of information: magenta yellow 95 100 50 75 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 cyan 25 Furniture inventory chairs tables beds Flat Unit House 10 Shopping list Bread loaves Juice carton Eggs Cheese black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\324IB_HL-2_13.CDR Wednesday, 31 October 2007 2:48:28 PM PETERDELL IB_HL-2ed (325) MATRICES (Chapter 13) 325 We can write these tables as matrices by extracting the numbers and placing them in square or round brackets: number B J 6 E C C T B F " 2# or simply 6 U H 10 and and " 10 3 # Notice how the organisation of the data is maintained in matrix form 2 has rows and column and we say 7 that this is a 4¡£¡1 column matrix or column vector column 2 10 row has rows and columns and is called a 3¡£¡3 square matrix this element, 3, is in row 3, column £ Note: ² ¡1 ¤ has row and columns and is called a 1¡£¡4 row matrix or row vector An m £ n matrix has m rows and n columns rows columns ² m £ n specifies the order of a matrix In business, a matrix can be used to represent numbers of items to be purchased, prices of items to be purchased, and so on Example Lisa goes shopping at store A to buy loaves of bread at $2:65 each, litres of milk at $1:55 per litre, and one 500 g tub of butter at $2:35 a Represent the quantities purchased in a row matrix Q and the costs in a column matrix A b Lisa goes to a different supermarket (store B) and finds that the prices for the same items are $2:25 for bread, $1:50 for milk, and $2:20 for butter Write the costs for both stores in a single costs matrix C The quantities matrix is Q = £ bread milk 2:65 The costs matrix is A = 1:55 2:35 cyan magenta yellow 95 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ¤ butter bread milk butter 100 a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\325IB_HL-2_13.CDR Thursday, November 2007 1:55:42 PM PETERDELL IB_HL-2ed (326) 326 MATRICES (Chapter 13) b We write the costs for each store in separate columns 2:65 2:25 bread milk The new costs matrix is C = 1:55 1:50 2:35 2:20 butter store A store B EXERCISE 13A Write down the order of: a £ b ¤ · ¸ c · ¡1 ¸ d 3 42 45 A grocery list consists of loaves of bread, kg of butter, eggs and carton of cream The cost of each grocery item is $1:95, $2:35, $0:15 and $0:95 respectively a Construct a row matrix showing quantities b Construct a column matrix showing prices c What is the significance of (2 £ 1:95) + (1 £ 2:35) + (6 £ 0:15) + (1 £ 0:95)? Big Bart’s Baked Beans factory produces cans of baked beans in sizes: 200 g, 300 g and 500 g In February they produced respectively: 1000, 1500 and 1250 cans of each in week 1; 1500, 1000 and 1000 of each in week 2; 800, 2300 and 1300 cans of each in week 3; 1200 cans of each in week Construct a matrix to show February’s production levels Over a long weekend holiday, a baker produced the following food items: On Friday he baked 40 dozen pies, 50 dozen pasties, 55 dozen rolls and 40 dozen buns On Saturday, 25 dozen pies, 65 dozen pasties, 30 dozen buns and 44 dozen rolls were made On Sunday 40 dozen pasties, 40 dozen rolls, 35 dozen of each of pies and buns were made On Monday the totals were 40 dozen pasties, 50 dozen buns and 35 dozen of each of pies and rolls Represent this information as a matrix B MATRIX OPERATIONS AND DEFINITIONS MATRIX NOTATION Consider a matrix A which has order m £ n A = (aij ) where i = 1, 2, 3, , m j = 1, 2, 3, , n and aij is the element in the ith row, jth column We can write cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 For example, a23 is the number in row and column of matrix A black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\326IB_HL-2_13.CDR Thursday, November 2007 9:13:02 AM PETERDELL IB_HL-2ed (327) MATRICES (Chapter 13) 327 EQUALITY Two matrices are equal if they have exactly the same shape (order) and the elements in corresponding positions are equal · For example, if ¸ · a b w = c d y x z ¸ then a = w, b = x, c = y and d = z A = B , (aij ) = (bij ) for all i, j We can write: MATRIX ADDITION Store A B C 23 41 68 dresses Thao has three stores (A, B and C) Her stock levels for 28 39 79 skirts dresses, skirts and blouses are given by the matrix: 46 17 62 blouses Some newly ordered stock has just arrived For each 20 20 20 store 20 dresses, 30 skirts and 50 blouses must be added 30 30 30 to stock levels Her stock order is given by the matrix: 50 50 50 23 + 20 41 + 20 68 + 20 Clearly the new levels are shown as: 28 + 30 39 + 30 79 + 30 46 + 50 17 + 50 62 + 50 3 23 41 68 20 20 20 43 61 88 or 28 39 79 + 30 30 30 = 58 69 109 46 17 62 50 50 50 96 67 112 to add two matrices they must be of the same order and then we add corresponding elements So, MATRIX SUBTRACTION 29 51 19 31 28 32 40 17 29 If Thao’s stock levels were and her sales matrix for the week was 15 12 20 16 19 what are the current stock levels? 19 14 What Thao has left is her original stock levels less what she has sold Clearly, we need to subtract corresponding elements: 3 29 51 19 15 12 14 39 13 31 28 32 ¡ 20 16 19 = 11 12 13 40 17 29 19 14 21 15 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 to subtract matrices they must be of the same order and then we subtract corresponding elements 25 95 100 50 75 25 So, black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\327IB_HL-2_13.CDR Thursday, November 2007 9:25:30 AM PETERDELL IB_HL-2ed (328) 328 MATRICES (Chapter 13) Summary: ² A § B = (aij ) § (bij ) = (aij § bij ) ² We can only add or subtract matrices of the same order ² We add or subtract corresponding elements ² The result of addition or subtraction is another matrix of same order Example · ¸ · ¸ If A = , B= 5 a A+B a A+B = · and C = ¸ find: b A+C · ¸ ¸ · + · ¸ 1+2 2+1 3+6 = 6+0 5+3 4+5 · ¸ 3 = b A + C cannot be found as A and C are not the same sized matrices i.e., they have different orders Example 3 If A = and B = 4 5 A¡B find A ¡ B 3 = 42 05 ¡ 43 45 3 3¡2 4¡0 8¡6 = 42 ¡ ¡ 0 ¡ 45 1¡5 4¡2 7¡3 = ¡1 ¡4 ¡4 EXERCISE 13B.1 · If A = a ¸ · ¸ ¡3 , B= ¡2 b A+B · and C = c A+B+C ¸ ¡3 , find: ¡4 ¡2 B+C d C+B¡A 3 17 ¡4 3 ¡11 and Q = ¡2 ¡8 5, find: If P = 10 ¡2 ¡1 ¡4 11 cyan yellow 95 100 50 Q¡P 75 25 95 c 100 50 75 P¡Q 25 95 50 75 25 100 magenta b P+Q 95 100 50 75 25 a black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_13\328IB_HL-2_13.CDR Thursday, 11 March 2010 10:25:01 AM PETER IB_HL-2ed (329) MATRICES (Chapter 13) 329 A restaurant served 85 men, 92 women and 52 children on Friday night On Saturday night they served 102 men, 137 women and 49 children a Express this information in two column matrices b Use the matrices to find the totals of men, women and children served over the Friday-Saturday period On Monday David bought shares in five companies and on Friday he sold them The details are: a Write down David’s column matrix for: i cost price ii selling price b What matrix operation is needed to find David’s profit or loss matrix? A B C D E Cost price per share $1:72 $27:85 $0:92 $2:53 $3:56 Selling price per share $1:79 $28:75 $1:33 $2:25 $3:51 c Find David’s profit or loss matrix In November, Lou E Gee sold 23 fridges, 17 stoves and 31 microwave ovens His partner Rose A Lee sold 19 fridges, 29 stoves and 24 microwave ovens In December, Lou’s sales were: 18 fridges, stoves and 36 microwaves while Rose’s sales were: 25 fridges, 13 stoves and 19 microwaves a Write their sales for November as a £ matrix b Write their sales for December as a £ matrix c Write their total sales for November and December as a £ matrix ¸ · ¸ · ¸ · ¸ b y x y ¡y x x x2 = = ¡1 y+1 y x x ¡y · ¸ · ¸ ¡1 a If A = and B = find A + B and B + A ¡1 Find x and y if: · a b Explain why A + B = B + A for all £ matrices A and B · ¸ · ¸ · ¸ ¡1 4 ¡1 a For A = , B= and C = ¡1 ¡2 ¡1 find (A + B) + C and A + (B + C) b Prove that, if A, B and C are any £ matrices then (A + B) + C = A + (B + C) · ¸ · ¸ · a b p q w Hint: Let A = , B= and C = c d r s y ¸ x z MULTIPLES OF MATRICES cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 In the pantry there are cans of peaches, cans of apricots and cans of pears This information could be represented by the column vector C = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\329IB_HL-2_13.CDR Thursday, November 2007 9:49:52 AM PETERDELL IB_HL-2ed (330) 330 MATRICES (Chapter 13) 12 Doubling these cans in the pantry we would have which is C + C or 2C 16 Notice that to get 2C from C we simply multiply all matrix elements by 2 3 3 Likewise, trebling and halving 3£6 18 £6 61 the fruit cans in them gives: 3C = 43 £ 45 = 12 £ 47 = C = 42 the pantry gives: 3£8 24 £ If A = (aij ) is of order m £ n and k is a scalar, then kA = (kaij ) So, to find kA, we multiply each element in A by k The result is another matrix of order m £ n Note: The notation we use is capital letters for matrices and lower-case letters for scalars Example · If A is · find a ¸ ¸ 3A = ¸ · 15 = a 2A 3A b b 2A = · "1 = 2 12 ¸ # EXERCISE 13B.2 · 12 If B = 24 · ¸ If A = a a 2B find: ¸ · and B = b A+B 3B b A¡B c 12 B d ¡ 12 B ¸ find: c 2A + B d 3A ¡ B Isabelle sells clothing made by four different companies which we will call A, B, C and D Her usual monthly order is: A B C D 30 40 40 60 skirt dress 6 50 40 30 75 evening 40 40 50 50 10 20 20 15 suit cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find her order, to the nearest whole number, if: a she increases her total order by 15% b she decreases her total order by 15% black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\330IB_HL-2_13.CDR Thursday, November 2007 9:26:16 AM PETERDELL IB_HL-2ed (331) 331 MATRICES (Chapter 13) During weekdays a video store finds that its average hirings are: 75 movies (DVD), 27 movies (VHS) and 102 video/computer games On the weekends the average figures are: 43 VHS movies, 136 DVD movies and 129 games a Represent the data using two column matrices DVD VHS b Find the sum of the matrices in a games c What does the sum matrix in b represent? A builder builds a block of 12 identical flats Each flat is to contain table, chairs, beds and wardrobe 647 is the matrix representing the furniture in one flat, If F = 425 what, in terms of F, is the matrix representing the furniture in all flats? ZERO OR NULL MATRIX For real numbers, it is true that a + = + a = a for all values of a So, is there a matrix O such that A + O = O + A = A for any matrix A? · ¸ · ¸ · ¸ 0 Simple examples like: + = suggest that O consists of all ¡1 0 ¡1 zeros A zero matrix is a matrix in which all elements are zero · For example, the £ zero matrix is ¸ · ¸ 0 0 ; the £ zero matrix is 0 0 Zero matrices have the property that: If A is a matrix of any order and O is the corresponding zero matrix, then A + O = O + A = A NEGATIVE MATRICES The negative matrix A, denoted ¡A is actually ¡1A · ¸ · ¸ · ¸ ¡1 ¡1 £ ¡1 £ ¡1 ¡3 So, if A = , then ¡A = = ¡1 £ ¡1 £ ¡2 ¡4 ¡A is obtained from A by reversing the sign of each element of A Thus The addition of a matrix and its negative always produces a zero matrix For example: · ¸ · ¸ · ¸ ¡1 ¡3 0 + = ¡2 ¡4 0 magenta yellow 95 100 50 75 25 95 50 75 25 95 100 50 75 25 95 100 50 75 25 cyan 100 A + (¡A) = (¡A) + A = O Thus, in general, black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\331IB_HL-2_13.CDR Thursday, November 2007 10:06:08 AM PETERDELL IB_HL-2ed (332) 332 MATRICES (Chapter 13) MATRIX ALGEBRA Compare our discoveries about matrices so far with ordinary algebra We will assume that A and B are matrices of the same order Ordinary algebra ² Matrix algebra If a and b are real numbers then a + b is also a real number a+b=b+a (a + b) + c = a + (b + c) a+0=0+a=a a + (¡a) = (¡a) + a = a a half of a is ² ² ² ² ² ² If A and B are matrices then A¡+¡B is a matrix of the same order ² ² ² ² A+B=B+A (A + B) + C = A + (B + C) A+O=O+A=A A + (¡A) = (¡A) + A = O µ ¶ A not a half of A is 12 A ² Example Explain why it is true that: a if X + A = B then X = B ¡ A b if 3X = A then X = 13 A a b If 3X = A then 13 (3X) = 13 A If X + A = B then X + A + (¡A) = B + (¡A) ) X+O=B¡A ) X=B¡A ) 1X = 13 A ) X = 13 A EXERCISE 13B.3 Simplify: a A + 2A d ¡B + B g ¡(2A ¡ C) 3B ¡ 3B 2(A + B) 3A ¡ (B ¡ A) b e h Find X in terms of A, B and C if: a X+B=A b B+X=C d 2X = A e 3X = B 2X g c f 4B + X = 2C A¡X=B i A ¡ 4X = C ¸ , find X if 13 X = M · ¸ ¡1 b If N = , find X if 4X = N · ¸ · ¸ 1 c If A = and B = , find X if A ¡ 2X = 3B ¡1 ¡1 cyan · magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 2(X + A) = B C ¡ 2C ¡(A + B) A + 2B ¡ (A ¡ B) a If M = h =C c f i black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\332IB_HL-2_13.CDR Thursday, November 2007 10:10:25 AM PETERDELL IB_HL-2ed (333) MATRICES (Chapter 13) 333 MATRIX MULTIPLICATION Suppose you go to a shop and purchase soft drink cans, chocolate bars and icecreams £ ¤ We can represent this by the quantities matrix A = soft drink cans $1:30 ice creams $1:20 1:30 then we can represent these using the costs matrix B = 0:90 1:20 If the prices are: chocolate bars $0:90 We can find the total cost of the items by multiplying the number of each item by its respective cost, and then adding the results The total cost is thus £ $1:30 + £ $0:90 + £ $1:20 = $9:90 We can also determine the total cost by the matrix multiplication: £ ¤ 1:30 AB = 4 0:90 1:20 = (3 £ 1:30) + (4 £ 0:90) + (2 £ 1:20) = 9:90 Notice that we write the row matrix first and the column matrix second £ In general, p a b c q = ap + bq + cr r ¤ EXERCISE 13B.4 £ ¤6 7 ¡1 b c ¡1 £ ¤6 7 w x y z Show that the sum of w, x, y and z is given by 415: Represent the average of w, x, y and z in the same way £ ¤ 415 Determine: · ¸ £ ¤ ¡1 a Lucy buys shirts, skirts and blouses costing $27, $35 and $39 respectively a Write down a quantities matrix Q and a price matrix P b Show how to use P and Q to determine the total cost cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 In the interschool public speaking competition a first place is awarded 10 points, second place points, third place points and fourth place point One school won first places, seconds, thirds and fourths a Write down this information in terms of points matrix P and a numbers matrix N b Show how to use P and N to find the total number of points awarded to the school black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\333IB_HL-2_13.CDR Thursday, November 2007 10:15:30 AM PETERDELL IB_HL-2ed (334) 334 MATRICES (Chapter 13) MORE COMPLICATED MULTIPLICATIONS Consider again Example on page 325 where Lisa needed loaves of bread, litres of milk and tub of butter £ ¤ We represented this by the quantities matrix Q = 2:65 2:25 The prices for each store were summarised in the costs matrix C = 1:55 1:50 2:35 2:20 To find the total cost of the items in each store, Lisa needs to multiply the number of items by their respective cost In Store A a loaf of bread is $2:65, a litre of milk is $1:55 and a tub of butter is $2:35, so the total cost is £ $2:65 + £ $1:55 + £ $2:35 = $12:30 In Store B a loaf of bread is $2:25, a litre of milk is $1:50 and a tub of butter is $2:20, so the total cost is £ $2:25 + £ $1:50 + £ $2:20 = $11:20 To this using matrices notice that: £ ¤ £ QC = 1£3 orders: 2:65 2:25 1:55 1:50 2:35 2:20 = £ 12:30 11:20 3£2 the same ¤ 1£2 resultant matrix Now suppose Lisa’s friend Olu needs loaf of bread, litres of milk and tubs of butter · ¸ Lisa The quantities matrix for both Lisa and Olu would be 2 Olu bread milk butter Lisa’s total cost at Store A is $12:30 and at store B is $11:20 Olu’s total cost at £ $2:65 + £ $1:55 + £ $2:35 = $10:45 £ $2:25 + £ $1:50 + £ $2:20 = $9:65 Store A is Store B is So, using matrices we require that row ´ column · 1 2 ¸ £ 2:65 2:25 1:55 1:50 2:35 2:20 · = row ´ column 12:30 11:20 10:45 9:65 row ´ column 2£3 3£2 the same ¸ row ´ column 2£2 resultant matrix cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Having observed the usefulness of multiplying matrices in the contextual examples above, we are now in a position to define matrix multiplication more formally black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\334IB_HL-2_13.CDR Thursday, November 2007 9:27:00 AM PETERDELL IB_HL-2ed (335) MATRICES (Chapter 13) 335 The product of an m¡£¡n matrix A with an n¡£¡p matrix B, is the m¡£¡p matrix AB in which the element in the rth row and cth column is the sum of the products of the elements in the rth row of A with the corresponding elements in the cth column of B n P If C = AB then (cij ) = air brj = ai1 b1j + ai2 b2j + :::::: + ain bnj r=1 for each pair i and j with i m and j p Note: The product AB exists only if the number of columns of A equals the number of rows of B For example: · ¸ a b If A = c d · ¸ · ¸ p q ap + br aq + bs and B = , then AB = r s cp + dr cq + ds · ¸ · ¸ x ax + by + cz a b c If C = and D = y , then CD = dx + ey + fz d e f z 2£1 2£3 3£1 To get the matrix AB you multiply rows by columns To get the element in the 5th row and 3rd column of AB (if it exists) multiply the 5th row of A by the 3rd column of B Example £ · ¤ If A = , B = ¸ A is £ and C is £ a ü AC = £ , and C = find: a AC b BC ) AC is £ 2 ¤ £ ¤ = 1£1+3£2+5£1 1£0+3£3+5£4 £ ¤ = 12 29 b B is £ and C is £ ü · BC = ¸ · ) BC is £ 2 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 1£1+3£2+5£1 1£0+3£3+5£4 = 2£1+1£2+3£1 2£0+1£3+3£4 · ¸ 12 29 = 15 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_13\335IB_HL-2_13.CDR Thursday, 11 March 2010 10:25:35 AM PETER ¸ IB_HL-2ed (336) 336 MATRICES (Chapter 13) EXERCISE 13B.5 £ Explain why AB cannot be found for A = ¤ · ¸ and B = If A is £ n and B is m £ 3: a When can we find AB? b If AB can be found, what is its order? c Why can BA never be found? · ¸ £ ¤ and B = find i AB ii BA a For A = ¤ £ and B = find i AB ii BA b For A = Find: £ a 3 1 40 05 ¤ 2 b 32 ¡1 ¡1 35 ¡1 At a fair, tickets for the Ferris wheel are $12:50 per adult and $9:50 per child On the first day of the fair, 2375 adults and 5156 children ride this wheel On the second day the figures are 2502 adults and 3612 children a Write the costs matrix C as a £ matrix and the numbers matrix N as a £ matrix b Find NC and interpret the resulting matrix c Find the total income for the two days You and your friend each go to your local hardware stores A and B to price items you wish to purchase You want to buy hammer, screwdriver and cans of white paint and your friend wants hammer, screwdrivers and cans of white paint The prices of these goods are: Hammer E7 E6 Store A Store B a b c d e Screwdriver E3 E2 Can of paint E19 E22 Write the requirements matrix R as a £ matrix Write the prices matrix P as a £ matrix Find PR What are your costs at store A and your friend’s costs at store B? Should you buy from store A or store B? USING TECHNOLOGY FOR MATRIX OPERATIONS Click on the appropriate icon to obtain graphics calculator instructions on how to perform operations with matrices cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Alternatively, click on the Matrix Operations icon to obtain computer software for these tasks black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\336IB_HL-2_13.CDR Thursday, November 2007 10:50:15 AM PETERDELL TI MATRIX OPERATIONS C IB_HL-2ed (337) 337 MATRICES (Chapter 13) EXERCISE 13B.6 Use technology to find: a 13 12 3 11 11 12 + 9 13 17 b 22 5 4 c d 13 11 2 63 41 3 12 11 12 ¡ 9 13 17 32 657 67 76 254 11 Use technology to assist in solving the following problems: For their holiday, Lars and Simke are planning to spend time at a popular tourist resort They will need accommodation at one of the local motels and they are not certain how long they will stay Their initial planning is for three nights and includes three breakfasts and two dinners They have gathered prices from three different motels The Bay View has rooms at $125 per night A full breakfast costs $22 per person (and therefore $44 for them both) An evening meal for two usually costs $75 including drinks By contrast, ‘The Terrace’ has rooms at $150 per night, breakfast at $40 per double and dinner costs on average $80 Things seem to be a little better at the Staunton Star Motel Accommodation is $140 per night, full breakfast (for two) is $40, while an evening meal for two usually costs $65 a Write down a ‘numbers’ matrix as a £ row matrix b Write down a ‘prices’ matrix in £ form c Use matrix multiplication to establish total prices for each venue d Instead of the couple staying three nights, the alternative is to spend just two nights In that event Lars and Simke decide on having breakfast just once and one evening meal before moving on Recalculate prices for each venue e Now remake the ‘numbers’ matrix (2 £ 3) so that it includes both scenarios Calculate the product with the ‘prices’ matrix to check your answers to c and d A bus company runs four tours Tour A costs $125, Tour B costs $315, Tour C costs $405, and Tour D costs $375 The numbers of clients they had over the summer period are shown in the table below 2Tour A Tour B Tour C Tour D3 Use the information and matrix 50 42 18 65 November methods to find the total income 65 37 25 82 December for the tour company 29 23 75 January 120 42 36 19 72 February cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The Oregon Motel has three types of suites for guests Standard suites cost $125 per night They have 20 suites Deluxe suites cost $195 per night They have 15 suites Executive suites cost $225 per night They have suites black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\337IB_HL-2_13.CDR Thursday, November 2007 10:51:25 AM PETERDELL IB_HL-2ed (338) 338 MATRICES (Chapter 13) The rooms which are occupied also have a maintenance cost: Standard suites cost $85 per day to maintain Deluxe suites cost $120 per day to maintain Executive suites cost $130 per day to maintain The hotel has confirmed room bookings for the next week: M T W Th F S Su3 Standard 15 12 13 11 14 16 4 75 Deluxe Executive 4 a The profit per day is given by (income from room) £ (bookings per day) ¡ (maintenance cost per room) £ (bookings per day) Create the matrices required to show how the profit per week can be found b How would the results alter if the hotel maintained (cleaned) all rooms every day? Show calculations c Produce a profit per room matrix and show how a could be done with a single matrix product PROPERTIES OF MATRIX MULTIPLICATION In the following exercise we should discover the properties of £ matrix multiplication which are like those of ordinary number multiplication, and those which are not EXERCISE 13B.7 For ordinary arithmetic For matrices, does AB · Hint: Try A = · a b If A = c d ¸ £ = £ and in algebra ab = ba always equal BA? ¸ · ¸ ¡1 and B = · 0 and O = 0 ¸ find AO and OA For all real numbers a, b and c it is true that a(b + c) = ab + ac This is known as the distributive law a Use any three £ matrices A, B and C to verify that A(B + C) = AB + AC · ¸ · ¸ · ¸ a b p q w x , B= and C = b Now let A = c d r s y z Prove that in general, A(B + C) = AB + AC cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 c Use the matrices you ‘made up’ in a to verify that (AB)C = A(BC) d Prove that (AB)C = A(BC) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\338IB_HL-2_13.CDR Thursday, November 2007 10:54:26 AM PETERDELL IB_HL-2ed (339) 339 MATRICES (Chapter 13) · a If a b c d ¸· ¸ w y · x a b = z c d ¸ i.e., AX = A, show that w = z = and x = y = is a solution b For any real number a, it is true that a £ = £ a = a Is there a matrix I such that AI = IA = A for all £ matrices A? Suppose A2 = AA or A £ A and that A3 = AAA · ¸ · ¸ ¡1 b Find A if A = : a Find A if A = ¡2 a If A = 4 try to find A2 6 b When can A2 be found, i.e., under what conditions can we square a matrix? · Show that if I = ¸ then I2 = I and I3 = I · I= You should have discovered from the above exercise that: 0 ¸ is called the identity matrix Ordinary algebra Matrix algebra ² If a and b are real numbers then so is ab ² If A and B are matrices that can be multiplied then AB is also a matrix fclosureg ² ab = ba for all a, b ² In general AB 6= BA ² a0 = 0a = for all a ² If O is a zero matrix then AO = OA = O for all A ² a(b + c) = ab + ac ² A(B + C) = AB + AC fnon-commutativeg fdistributive lawg · ¸ ² If I is the identity matrix then AI = IA = A for all £ matrices A fidentity lawg ² a£1=1£a= a ² An exists provided A is square and n Z + ² an exists for all a > and n R cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Note: In general, A(kB) = k(AB) 6= kBA We can change the order in which we multiply by a scalar, but in general we cannot reverse the order in which we multiply matrices black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_13\339IB_HL-2_13.CDR Thursday, 11 March 2010 10:26:53 AM PETER IB_HL-2ed (340) 340 MATRICES (Chapter 13) Example a (A + 2I)2 Expand and simplify where possible: a b b (A ¡ B)2 (A + 2I)2 = (A + 2I)(A + 2I) = (A + 2I)A + (A + 2I)2I = A2 + 2IA + 2AI + 4I2 = A2 + 2A + 2A + 4I = A2 + 4A + 4I fX2 = XX by definitiong fB(C + D) = BC + BDg fB(C + D) = BC + BD again, twiceg fAI = IA = A and I2 = Ig (A ¡ B)2 = (A ¡ B)(A ¡ B) = (A ¡ B)A ¡ (A ¡ B)B = A2 ¡ BA ¡ AB + B2 fX2 = XX by definitiong fC(D ¡ E) = CD ¡ CE twiceg Note: b cannot be simplified further since, in general, AB 6= BA Example If A2 = 2A + 3I, find A3 and A4 in the form kA + lI where k and l are scalars A3 = = = = = A £ A2 A(2A + 3I) 2A2 + 3AI 2(2A + 3I) + 3AI 7A + 6I A4 = A £ A3 = A(7A + 6I) = 7A2 + 6AI = 7(2A + 3I) + 6A = 20A + 21I Example · Find constants a and b such that A2 = aA + bI for A equal to ¸ · ¸ · 2 =a +b 4 · ¸ · ¸ · 1+6 2+8 a 2a b ) = + + 12 + 16 3a 4a · ¸ · ¸ 10 a+b 2a ) = 15 22 3a 4a + b magenta yellow 50 4a + b = 4(5) + (2) = 22 X 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Checking for consistency: 3a = 3(5) = 15 X cyan ¸ ¸ b 2a = 10 b=2 95 Thus a + b = and ) a = and ¸· 100 · Since A2 = aA + bI, ¸ : black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\340IB_HL-2_13.CDR Thursday, November 2007 10:59:41 AM PETERDELL IB_HL-2ed (341) 341 MATRICES (Chapter 13) EXERCISE 13B.8 Given that all matrices are £ a A(A + I) b e d A(A + A ¡ 2I) g (A + B)(A ¡ B) h and I is the identity matrix, expand and simplify: (B + 2I)B c A(A2 ¡ 2A + I) (A + B)(C + D) f (A + B)2 (A + I)2 i (3I ¡ B)2 a If A2 = 2A ¡ I, find A3 and A4 in the linear form kA + lI where k and l are scalars b If B2 = 2I ¡ B, find B3 , B4 and B5 in linear form c If C2 = 4C ¡ 3I, find C3 and C5 in linear form a If A2 = I, simplify: i A(A + 2I) ii (A ¡ I)2 iii A(A + 3I)2 iii A(A + I)3 b If A3 = I, simplify A2 (A + I)2 c If A2 = O, simplify: i A(2A ¡ 3I) ii A(A + 2I)(A ¡ I) The result “if ab = then a = or b = 0” for real numbers does not have an equivalent result for matrices · ¸ · ¸ 0 a If A = and B = find AB 0 This example provides us with evidence that “If AB = O then A = O or B = O” is a false statement "1 1# 2 b If A = 2 determine A2 c Comment on the following argument for a £ matrix A: It is known that A2 = A, so A2 ¡ A = O ) A(A ¡ I) = O ) A = O or A ¡ I = O ) A = O or I d Find all £ matrices A for which A2 = A Hint: Let A = · ¸ a b c d Give one example which shows that “if A2 = O then A = O” is a false statement Find constants a and b such that A2 = aA + bI for A equal to: · ¸ · ¸ a b ¡1 2 ¡2 · ¸ If A = , find constants p and q such that A2 = pA + qI ¡1 ¡3 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a Hence, write A3 in the linear form rA + sI where r and s are scalars b Write A4 in linear form black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\341IB_HL-2_13.CDR Thursday, November 2007 11:01:34 AM PETERDELL IB_HL-2ed (342) 342 MATRICES (Chapter 13) C THE INVERSE OF A 2×2 ¡ ¡ MATRIX ½ 2x + 3y = 5x + 4y = 17 We can solve algebraically to get x = 5, y = ¡2 · Notice that this system can be written as a matrix equation The solution x = 5, · y = ¡2 is easily checked as ¸· ¸ · ¸ · ¸ 2(5) + 3(¡2) = = ¡2 5(5) + 4(¡2) 17 ¸· ¸ · ¸ x = y 17 X In general, a system of linear equations can be written in the form AX = B where A is the matrix of coefficients, X is the unknown column matrix, and B is the column matrix of constants The question arises: If AX = B, how can we find X using matrices? To answer this question, suppose there exists a matrix C such that CA = I If we premultiply each side of AX = B by C we get C(AX) = CB ) (CA)X = CB ) IX = CB and so X = CB Premultiply means multiply on the left of each side If C exists such that CA = I then C is said to be the multiplicative inverse of A, and we denote C by A¡1 The multiplicative inverse of A, denoted A¡1 , satisfies A¡1 A = AA¡1 = I · Suppose A = · so AA¡1 = a b c d a b c d · ¸ and A¡1 = ¸· · ¸ x =I z ¸ · ¸ ax + bz = cx + dz x z ¸ w y aw + by cw + dy ½ aw + by = (1) ) cw + dy = (2) ) w y ½ and ax + bz = (3) cx + dz = (4) d ad ¡ bc Solving (1) and (2) simultaneously for w and y gives: w = Solving (3) and (4) simultaneously for x and z gives: x = · a b So, if A = c d ¸ ¡1 then A · ¸ d ¡b = ad ¡ bc ¡c a ¡b ad ¡ bc and y = and z = ¡c ad ¡ bc a ad ¡ bc cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 If ad ¡ bc 6= then A¡1 exists and we say that A is invertible or non-singular If ad ¡ bc = then A¡1 does not exist and we say that A is singular black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\342IB_HL-2_13.CDR Thursday, November 2007 11:31:34 AM PETERDELL IB_HL-2ed (343) 343 MATRICES (Chapter 13) · ¸ a b If A = , the value ad ¡ bc is called the determinant of A, c d denoted j A j or detA If j A j = then A¡1 does not exist and A is singular · ¸ d ¡b ¡1 If j A j 6= then A is invertible and A = j A j ¡c a EXERCISE 13C.1 · ¸ ¡6 a Find ¡2 · ¸ · ¸ ¡4 b Find ¡1 ¸ · · ¸ and hence find the inverse of · ¸ ¡4 and hence find the inverse of 2 Find j A j for A equal to: · ¸ · ¸ ¡1 a b ¡2 · c Find detB for B equal to: · ¸ · ¸ ¡2 a b · ¡1 For A = ¡1 ¡1 · c ¸ a find: jA j 0 0 1 ¸ · d ¸ · d j A j2 b c 0 ¸ a ¡a a ¸ j 2A j Prove that if A is any £ matrix and k is a constant, then j kA j = k2 j A j · ¸ · ¸ a b w x By letting A = and B = c d y z a find j A j and j B j b find AB and j AB j c Hence show that j AB j = j A j j B j for all £ matrices A and B · A= ¸ · and B = ¸ ¡1 a Using the results of and above and the calculated values of jAj and jBj, find: i jAj ii j 2A j iii j¡A j iv j¡3B j v j AB j b Check your answers without using the results of and above cyan magenta yellow c 95 · · 100 50 g 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find, if it exists, the inverse matrix of: · ¸ · ¸ a b ¡1 ¡1 · ¸ · ¸ e f ¡1 ¡6 ¡10 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\343IB_HL-2_13.CDR Thursday, November 2007 1:56:20 PM PETERDELL ¸ ¡1 d ¸ h · · 0 ¸ ¡1 ¡1 ¸ IB_HL-2ed (344) 344 MATRICES (Chapter 13) · a If A = ¡1 ¸ and B = , find AB ¡1 ¡4 ¡1 b Does your result in a imply that A and B are inverses? Hint: Find BA The above example illustrates that only square matrices can have inverses Why? SOLVING A PAIR OF LINEAR EQUATIONS We have already seen how a system of linear equations can be written in matrix form We can solve the system using an inverse matrix if such an inverse exists If two lines are parallel then the resulting matrix will be singular and no inverse exists This indicates that either the lines are coincident and there are infinitely many solutions, or the lines never meet and there are no solutions If the lines are not parallel then the resulting matrix will be invertible We premultiply by the inverse to find the unique solution which is the point of intersection never meet (no solution) unique solution coincident (infinitely many solutions) Example 10 ( · a a ¸ If A = find j A j : j A j = 2(4) ¡ 1(3) =8¡3 =5 b Does b 2x + y = 3x + 4y = ¡1 · The system in matrix form is: have a unique solution? ¸· ¸ · ¸ x = y ¡1 Now as j A j = 6= 0, A¡1 exists and there is a unique solution Example 11 · ¸ ½ 2x + 3y = ¡1 If A = , find A and hence solve : 5x + 4y = 17 · In matrix form the system is: ¸ · ¸ · ¸ x = y 17 Notice that we premultiply with the inverse matrix on both sides i.e., AX = B where j A j = ¡ 15 = ¡7 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Now A¡1 AX = A¡1 B black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\344IB_HL-2_13.CDR Friday, November 2007 9:18:12 AM PETERDELL IB_HL-2ed (345) 345 MATRICES (Chapter 13) ¸ · ¸ ¡3 ¡5 17 · ¸ · ¸ ¡35 = and so x = 5, y = ¡2 ) X = ¡7 14 ¡2 Check this answer! ¡7 ) IX = · Example 12 · Find A¡1 when A = k ¡1 ¸ and state k when A¡1 exists A¡1 If jAj¡=0 ¡ , the matrix A is singular and is not invertible · ¸ 2k +4 ¡1 ¡k = =6 ¡4 ¡ 2k ¡2 2k + k 2k + 7 ¡4 2k + So A¡1 exists provided that 2k + 6= 0, i.e., k 6= ¡2: EXERCISE 13C.2 Convert into matrix equations: a 3x ¡ y = b 2x + 3y = 4x ¡ 3y = 11 3x + 2y = ¡5 Use matrix algebra to solve the system: b 5x ¡ 4y = a 2x ¡ y = 2x + 3y = ¡13 x + 3y = 14 d 4x ¡ 7y = 3x ¡ 5y = e 3x + 5y = 2x ¡ y = 11 c 3a ¡ b = 2a + 7b = ¡4 c x ¡ 2y = 5x + 3y = ¡2 f 7x + 11y = 18 11x ¡ 7y = ¡11 a Show that if AX = B then X = A¡1 B whereas if XA = B then X = BA¡1 : b Find X if: · ¸ · ¸ 14 ¡5 i X = ¡1 22 · For a A= k ¡6 · ii ¸ · b A= ¡1 k ¸ · ¸ ¡3 X= ¡1 ¸ · c A= k+1 k ¸ i ii find the values of k for which the matrix A is singular find A¡1 when A is non-singular ½ 2x ¡ 3y = a Consider the system 4x ¡ y = 11 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 i Write the equations in the form AX = B and find j A j ii Does the system have a unique solution? If so, find it black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\345IB_HL-2_13.CDR Thursday, November 2007 9:39:34 AM PETERDELL IB_HL-2ed (346) 346 MATRICES (Chapter 13) ½ b Consider the system 2x + ky = 4x ¡ y = 11 i Write the system in the form AX = B and find j A j ii For what value(s) of k does the system have a unique solution? Find the unique solution iii Find k when the system does not have a unique solution How many solutions does it have in this case? FURTHER MATRIX ALGEBRA The following exercise requires matrix algebra with inverse matrices Be careful that you use multiplication correctly In particular, remember that: ² We can only perform matrix multiplication if the orders of the matrices allow it ² If we premultiply on one side then we must premultiply on the other This is important because, in general, AB 6= BA The same applies if we postmultiply Example 13 If A2 = 2A + 3I, find A¡1 in linear form rA + sI, where r and s are scalars A2 = 2A + 3I ) A¡1 A2 = A¡1 (2A + 3I) ) A¡1 AA = 2A¡1 A + 3A¡1 I ) IA = 2I + 3A¡1 ) A ¡ 2I = 3A¡1 ) A¡1 = 13 (A ¡ 2I) fpremultiply both sides by A¡1 g i.e., A¡1 = 13 A ¡ 23 I EXERCISE 13C.3 · ¸ · ¸ · ¸ 1 Given A = , B= and C = , find X if AXB = C ¡1 2 If a matrix A is its own inverse, then A = A¡1 · ¸ · ¸ · ¸ 1 ¡1 ¡1 ¡1 For example, if A = then A = = = A ¡1 ¡1 ¡1 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a Show that, if A = A¡1 , then A2 = I · ¸ a b b If is its own inverse, show that there are exactly matrices of this form b a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\346IB_HL-2_13.CDR Thursday, November 2007 11:54:04 AM PETERDELL IB_HL-2ed (347) 347 MATRICES (Chapter 13) · a If A = ¡1 ¸ find A¡1 and (A¡1 )¡1 b If A is any square matrix which has inverse A¡1 , simplify (A¡1 )¡1 (A¡1 ) and (A¡1 )(A¡1 )¡1 by replacing A¡1 by B c What can be deduced from b? · ¸ · ¸ 1 a If A = and B = ¡1 ¡3 A¡1 (BA)¡1 i iv find in simplest form: B¡1 A¡1 B¡1 ii v iii vi (AB)¡1 B¡1 A¡1 b Choose any two invertible matrices and repeat question a c What the results of a and b suggest? d Simplify (AB)(B¡1 A¡1 ) and (B¡1 A¡1 )(AB) given that A¡1 and B¡1 exist Conclusion? 1 If k is a non-zero number and A¡1 exists, simplify (kA)( A¡1 ) and ( A¡1 )(kA) k k What conclusion follows from your results? Suppose X, Y and Z are £ matrices and A, B are £ matrices If X = AY and Y = BZ where A and B are invertible, find: b Z in terms of X a X in terms of Z · ¸ If A = , write A2 in the form pA + qI where p and q are scalars ¡2 ¡1 Hence write A¡1 in the form rA + sI where r and s are scalars Find A¡1 in linear form given that b 5A = I ¡ A2 a A2 = 4A ¡ I c 2I = 3A2 ¡ 4A It is known that AB = A and BA = B where the matrices A and B are not necessarily invertible Prove that A2 = A (Note: From AB = A, you cannot deduce that B = I Why?) 10 Under what condition is it true that “if AB = AC then B = C”? 11 If X = P¡1 AP and A3 = I, prove that X3 = I 12 If aA2 + bA + cI = O and X = P¡1 AP, prove that aX2 + bX + cI = O Summary: During this exercise you should have discovered that: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ² (A¡1 )¡1 = A ² (AB)¡1 = B¡1 A¡1 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\347IB_HL-2_13.CDR Thursday, November 2007 1:56:58 PM PETERDELL IB_HL-2ed (348) 348 MATRICES (Chapter 13) 3×3 ¡ ¡ AND LARGER MATRICES D The principles of determinants and inverses are equally applicable to £ and other larger square matrices THE DETERMINANT OF A 3×3 ¡ ¡ MATRIX a1 b1 c1 The determinant of A = a2 b2 c2 is defined as a3 b3 c3 ¯ ¯ ¯ a1 b1 c1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ b2 c2 ¯ ¯ a2 c2 ¯ ¯a ¯ ¯ ¯ ¯ ¯ ¯ ¡ b1 ¯ + c1 ¯¯ j A j = ¯ a2 b2 c2 ¯ = a1 ¯ ¯ ¯ b c a c a3 3 3 ¯ a3 b3 c3 ¯ ¯ b2 ¯¯ b3 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1¯ ¯ ¡ 2¯2 1¯ + 4¯2 ¯ j A j = ¯¯ ¯ ¯ ¯ ¯ ¡1 ¯ ¡1 Example 14 Find j A j for A = 42 15 ¡1 same same same = 1(0 ¡ ¡1) ¡ 2(4 ¡ 3) + 4(¡2 ¡ 0) =1¡2¡8 = ¡9 Just like £ systems, a £ system of linear equations in matrix form AX = B will have a unique solution if j A j 6= Remember that you can use your graphics calculator or matrix software to find the value of a determinant MATRIX OPERATIONS TI C EXERCISE 13D.1 ¯ ¯ ¯ ¡1 ¡3 ¯ ¯ ¯ ¯ 0 ¯ ¯ ¯ ¯ ¡1 ¯ ¯ ¯ ¯0 2¯ ¯ ¯ ¯0 0¯ ¯ ¯ ¯3 0¯ a Find the values of x for which the matrix b What does your answer to a mean? cyan yellow 25 95 100 50 75 25 95 magenta ¯ ¯ ¯ ¯ ¡1 ¯ ¯ ¯ ¯ ¯ ¯ ¡1 ¯ ¯ ¡1 ¯ ¯¯ ¯¯ 3¯ ¯ ¯¯ ¯¯ 1¯ x is singular 25 ¡1 x ¯ y ¯¯ z ¯¯ 0¯ ¯ ¯ x ¯ ¯ ¡x ¯ ¯ ¡y ¡z b 100 50 75 25 95 100 50 75 25 Evaluate: ¯ ¯ a ¯a 0¯ ¯ ¯ ¯0 b 0¯ ¯ ¯ ¯0 c¯ f c 95 c 100 e 50 b 75 Evaluate: ¯ ¯ a ¯ 0¯ ¯ ¯ ¯ ¡1 ¯ ¯ ¯ ¯ 5¯ ¯ ¯ d ¯¯ 0 ¯¯ ¯0 0¯ ¯ ¯ ¯0 3¯ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\348IB_HL-2_13.CDR Thursday, November 2007 12:17:13 PM PETERDELL ¯ ¯ ¯a b c ¯ ¯ ¯ ¯ b c a¯ ¯ ¯ ¯c a b¯ IB_HL-2ed (349) 349 MATRICES (Chapter 13) < x + 2y ¡ 3z 2x ¡ y ¡ z For what values of k does : kx + y + 2z < 2x ¡ y ¡ 4z 3x ¡ ky + z For what values of k does : 5x ¡ y + kz Find k given that: ¯ ¯1 ¯ ¯k ¯ ¯3 a k =5 =8 = 14 have a unique solution? =8 =1 = ¡2 have a unique solution? ¯ ¯¯ ¡1 ¯¯ = ¯ b ¯ ¯k ¯ ¯2 ¯ ¯1 k ¯ ¯¯ ¯¯ = k¯ Use technology to find the determinant and inverse of: 3 a b 62 07 62 27 6 61 47 7 43 05 42 1 55 If Jan bought one orange, two apples, a pear, a cabbage and a lettuce the total cost would be $6:30 Two oranges, one apple, two pears, one cabbage and one lettuce would cost a total of $6:70 One orange, two apples, three pears, one cabbage and one lettuce would cost a total of $7:70 Two oranges, two apples, one pear, one cabbage and three lettuces would cost a total of $9:80 Three oranges, three apples, five pears, two cabbages and two lettuces would cost a total of $10:90 a Write this information in the form AX = B where A is the quantities matrix, X is the cost per item column matrix, and B is the total costs column matrix b Explain why X cannot be found from the given information c If the last lot of information is deleted and in its place “three oranges, one apple, two pears, two cabbages and one lettuce cost a total of $9:20” is substituted, can the system be solved now, and if so, what is the solution? THE INVERSE OF A 3×3 ¡ ¡ MATRIX cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 There is no simple rule for finding the inverse of a £ matrix like there is for a £ matrix Hence we use technology For example, if A = what is A¡1 ? ¡1 2 ¡0:111 0:888 ¡0:222 We obtain A¡1 = 0:111 1:111 ¡0:7775 which converts to 0:222 ¡0:777 0:444 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\349IB_HL-2_13.CDR Thursday, November 2007 12:27:20 PM PETERDELL MATRIX OPERATIONS TI C 6 ¡ 19 9 10 ¡ 79 ¡ 29 ¡ 79 IB_HL-2ed (350) 350 MATRICES (Chapter 13) EXERCISE 13D.2 32 3 ¡11 15 and hence the inverse of 5 Find 54 ¡1 1 ¡3 ¡6 ¡10 ¡3 3 A = ¡1 Use technology to find A¡1 for: a Find B¡1 for: 13 43 ¡11 27 B = 16 ¡8 31 ¡13 a b 3 A=4 5 ¡3 b 1:61 4:32 6:18 B = 0:37 6:02 9:41 7:12 5:31 2:88 Check that your answers to and are correct Note: ² In general, we can only find the determinants of square matrices i.e., for £ 2, £ 3, £ 4, etc matrices ² If A is square and j A j 6= 0, then A¡1 exists and A is called an invertible or non-singular matrix ² If A is square and j A j = 0, then A¡1 does not exist and A is called a singular matrix ² det(AB) = detA detB or j AB j = j A j j B j for all square matrices A, B of equal size E SOLVING SYSTEMS OF LINEAR EQUATIONS Example 15 Solve the system x¡y¡z = x + y + 3z = 9x ¡ y ¡ 3z = ¡1 using matrix methods and a graphics calculator 32 3 ¡1 ¡1 x 1 54 y = ¡1 ¡3 z ¡1 In matrix form AX = B the system is: We enter A and B into our GDC and calculate [ A ]¡1 [ B ] 3 3¡1 2 0:6 x ¡1 ¡1 ) y = 1 = ¡5:3 ¡1 3:9 z ¡1 ¡3 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 So, x = 0:6, y = ¡5:3, z = 3:9 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\350IB_HL-2_13.CDR Thursday, November 2007 9:41:59 AM PETERDELL IB_HL-2ed (351) 351 MATRICES (Chapter 13) Example 16 Rent-a-car has three different makes of vehicles, P, Q and R, for hire These cars are located at yards A and B on either side of a city Some cars are out (being rented) In total they have 150 cars At yard A they have 20% of P, 40% of Q and 30% of R which is 46 cars in total At yard B they have 40% of P, 20% of Q and 50% of R which is 54 cars in total How many of each car type does Rent-a-car have? Suppose Rent-a-car has x of P, y of Q and z of R Then as it has 150 cars in total, x + y + z = 150 (1) But yard A has 20% of P + 40% of Q + 30% of R and this is 46 10 x ) 10 y + + 10 z = 46 ) 2x + 4y + 3z = 460 (2) Yard B has 40% of P + 20% of Q + 50% of R and this is 54 10 x ) 10 y + + 10 z = 54 ) 4x + 2y + 5z = 540 (3) We need to solve the system: x + y + z = 150 2x + 4y + 3z = 460 4x + 2y + 5z = 540 32 3 1 x 150 i.e., 4 y = 460 5 z 540 3 3¡1 150 45 x 1 4 5 460 = 55 fusing techg y = So, 540 50 z Thus, Rent-a-car has 45 of P, 55 of Q and 50 of R A £ system of linear equations in the form AX = B has a unique solution if j A j 6= EXERCISE 13E Write as a matrix equation: x¡y¡z = x + y + 3z = 9x ¡ y ¡ 3z = ¡1 a b 2x + y ¡ z = y + 2z = x ¡ y + z = 13 c a+b¡c = a¡b+c = 2a + b ¡ 3c = ¡2 3 2 ¡1 ¡3 For A = ¡1 and B = ¡1 ¡2 , 6 12 ¡5 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 calculate AB and hence solve the system of equations black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\351IB_HL-2_13.CDR Thursday, November 2007 1:46:23 PM PETERDELL 4a + 7b ¡ 3c = ¡8 ¡a ¡ 2b + c = 6a + 12b ¡ 5c = ¡15 IB_HL-2ed (352) 352 MATRICES (Chapter 13) 3 ¡7 3 For M = ¡1 ¡3 and N = ¡1 , ¡3 ¡1 calculate MN and hence solve the system 3u + 2v + 3w = 18 u ¡ v + 2w = 2u + v + 3w = 16 Use matrix methods and technology to solve: a 3x + 2y ¡ z = 14 x ¡ y + 2z = ¡8 2x + 3y ¡ z = 13 x ¡ y ¡ 2z = 5x + y + 2z = ¡6 3x ¡ 4y ¡ z = 17 b c x + 3y ¡ z = 15 2x + y + z = x ¡ y ¡ 2z = c 2x ¡ y + 3z = 17 2x ¡ 2y ¡ 5z = 3x + 2y + 2z = 10 Use your graphics calculator to solve: a x+y+z = 2x + 4y + z = 2x + 3y + z = b d x + 2y ¡ z = 23 x ¡ y + 3z = ¡23 7x + y ¡ 4z = 62 x + 4y + 11z = x + 6y + 17z = x + 4y + 8z = f 10x ¡ y + 4z = ¡9 7x + 3y ¡ 5z = 89 13x ¡ 17y + 23z = ¡309 e 1:3x + 2:7y ¡ 3:1z = 8:2 2:8x ¡ 0:9y + 5:6z = 17:3 6:1x + 1:4y ¡ 3:2z = ¡0:6 Westwood School bought two footballs, one baseball and three basketballs for a total cost of $90 Sequoia School bought three footballs, two baseballs and a basketball for $81 Lamar School bought five footballs and two basketballs for $104 a State clearly what the variables x, y and z must represent if this situation is to be described by the set of equations: 2x + y + 3z = 90, 3x + 2y + z = 81, 5x + 2z = 104 b If Kato International School needs footballs and baseballs, and wishes to order as many basketballs as they can afford, how many basketballs will they be able to purchase if there is a total of $315 to be spent? Managers, clerks and labourers are paid according to an industry award Xenon employs managers, clerks and labourers with a total salary bill of E352 000 Xanda employs manager, clerks and labourers with a total salary bill of E274 000 Xylon employs manager, clerks and 11 labourers with a total salary bill of E351 000 a If x, y and z represent the salaries (in thousands of euros) for managers, clerks and labourers respectively, show that the above information can be represented by a system of three equations b Solve the above system of equations c Determine the total salary bill for Xulu company which employs managers, clerks and 37 labourers cyan magenta yellow 95 100 50 Cashews Macas Brazils 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A mixed nut company uses cashews, macadamias and Brazil nuts to make three gourmet mixes The table alongside indicates the weight in hundreds of grams of each kind of nut required to make a kilogram of mix black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_13\352IB_HL-2_13.CDR Thursday, 11 March 2010 10:27:24 AM PETER Mix A Mix B Mix C 4 IB_HL-2ed (353) MATRICES (Chapter 13) 353 If kg of mix A costs $12:50 to produce, kg of mix B costs $12:40 and kg of mix C costs $11:70, determine the cost per kilogram of each of the different kinds of nuts Hence, find the cost per kilogram to produce a mix containing 400 grams of cashews, 200 grams of macadamias and 400 grams of Brazil nuts Klondike High has 76 students in total in classes P, Q and R There are p students in P, q in Q and r in R One-third of P, one-third of Q and two-fifths of R study Chemistry One-half of P, two-thirds of Q and one-fifth of R study Mathematics One-quarter of P, one-third of Q and three-fifths of R study Geography Given that 27 students study Chemistry, 35 study Mathematics and 30 study Geography: a find a system of equations which contains this information, making sure that the coefficients of p, q and r are integers b Solve for p, q and r 10 Susan and James opened a new business in 2001 Their annual profit was $160 000 in 2004, $198 000 in 2005 and $240 000 in 2006 Based on the information from these three years they believe that their annual profit could be predicted by the model c P (t) = at + b + pounds t+4 where t is the number of years after 2004, i.e., t = gives the 2004 profit a Determine the values of a, b and c which fit the profits for 2004, 2005 and 2006 b If the profit in 2003 was $130¡000, does this profit fit the model in a? c Susan and James believe their profit will continue to grow according to this model Predict their profit in 2007 and 2009 The system of equations in each of the examples above could be represented as AX = B where we are looking for matrix X In every case above we could find X by pre-multiplying each side by A¡1 to get X = A¡1 B This pre-supposes that A is non-singular which may not be the case Hence, in the next section we will investigate another technique for solving the system of equations that will still work if A is singular, i.e., detA = and A¡1 does not exist INVESTIGATION USING MATRICES IN CRYPTOGRAPHY Cryptography is the study of encoding and decoding messages Cryptography was first developed for the military to send secret messages However, today it is used to maintain privacy when information is being transmitted via public communication services such as the internet cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Messages are sent in code or cipher form The method of converting text to ciphertext is called enciphering and the reverse process is called deciphering The operations of matrix addition and multiplication can be used to create codes and the coded messages are transmitted Decoding using additive or multiplicative PRINTABLE inverses is required by the receiver in order to read the message INVESTIGATION Click on the icon for a printable investigation on cryptography black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_13\353IB_HL-2_13.CDR Thursday, 11 March 2010 10:27:37 AM PETER IB_HL-2ed (354) 354 MATRICES (Chapter 13) F SOLVING SYSTEMS USING ROW OPERATIONS The system of equations ½ 2x + y = ¡1 x ¡ 3y = 17 is called a £ system, because there are equations in unknowns In the method of ‘elimination’ used to solve these equations, we observe that the following operations produce equations with the same solutions as the original pair ² The equations can be interchanged without affecting the solutions ½ ½ 2x + y = ¡1 x ¡ 3y = 17 For example, has the same solutions as x ¡ 3y = 17 2x + y = ¡1 ² An equation can be replaced by a non-zero multiple of itself For example, 2x + y = ¡1 could be replaced by ¡6x ¡ 3y = (obtained by multiplying each term by ¡3) ² Any equation can be replaced by a multiple of itself plus (or minus) a multiple of another equation For example, suppose we replace the second equation by “twice the second equation minus the first equation” In this case: ½ ½ 2x + y = ¡1 x ¡ 3y = 17 becomes 2x + y = ¡1 ¡7y = 35 using: 2x ¡ 6y = 34 ¡ (2x + y = ¡1) ¡7y = 35 We can use these principles to develop row operations for matrices AUGMENTED MATRICES We will now solve systems of equations using row operations on an augmented matrix ½ 2x + y = ¡1 Instead of writing we detach the coefficients and write the system in x ¡ 3y = 17 · ¸ ¡1 augmented matrix form ¡3 17 In this form we can use elementary row operations equivalent to the three legitimate operations with equations We can hence: ² ² ² interchange rows replace any row by a non-zero multiple of itself replace any row by itself plus (or minus) a multiple of another row cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Interchanging rows is equivalent to writing the equations in a different order It is often desirable to have in the top left hand corner black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_13\354IB_HL-2_13.CDR Thursday, 11 March 2010 10:28:04 AM PETER IB_HL-2ed (355) MATRICES (Chapter 13) 355 We can this by swapping rows and We can write this as R1 $ R2 · ¸ · ¸ 1 ¡3 17 ¡1 So, becomes ¡1 ¡3 17 We now attempt to eliminate one of the variables in the second equation by obtaining a in its place To this we replace R2 by R2 ¡ 2R1 , i.e., row by (row ¡ £ row 1) ¸ · ¸ · ¡1 R2 ¡3 17 ¡3 17 becomes So, ¡2 ¡34 ¡2R1 ¡1 ¡35 7 ¡35 adding The second row of the matrix is now really 7y = ¡35 and so y = ¡5 Substituting y = ¡5 into the first equation, x ¡ 3(¡5) = 17 and so x = So, the solution is x = 2, y = ¡5 This process of solving a system using elementary row operations is known as row reduction When we have rewritten the augmented matrix so there are all zeros in the bottom left hand corner, the system is said to be in echelon form We may not see the benefit of this method right now but we will certainly appreciate it when solving £ or higher order systems Example 17 ½ Use elementary row operations to solve: 2x + 3y = 5x + 4y = 17 In augmented matrix form the system is: · ¸ 17 · ¸ » R2 ! 5R1 ¡ 2R2 ¡14 » is read as “which has the same solution as” 10 ¡10 Re-introducing the variables we have 7y = ¡14 ) y = ¡2 and on ‘back substituting’ into the first equation we have 2x + 3(¡2) = ) 2x ¡ = ) 2x = 10 ) x=5 So the solution is x = 5, y = ¡2 15 ¡8 20 ¡34 ¡14 Don’t forget to check your solution cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Reminder: In two dimensional geometry ax¡+¡by¡=¡c where a, b and c are constants, are the equations of straight lines If we are given two such equations then there are three different cases which could occur: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\355IB_HL-2_13.CDR Thursday, November 2007 10:55:36 AM PETERDELL IB_HL-2ed (356) 356 MATRICES (Chapter 13) intersecting lines parallel lines l1 l1 2x + 3y = x ¡ 2y = e.g one point of intersection ) a unique solution Example 18 l1 or l2 l2 e.g coincident lines l2 2x + 3y = 2x + 3y = e.g no points of intersection ) no solution ½ Find all solutions to x + 3y = 4x + 12y = k 2x + 3y = 4x + 6y = infinitely many points of intersection ) infinitely many solutions where k is a constant, by using elementary row operations In augmented matrix form, the system is: · ¸ 12 k ¸ · » R2 ! R2 ¡ 4R1 0 k ¡ 20 ¡4 12 ¡12 The second equation actually reads 0x + 0y = k ¡ 20 If k 6= 20 we have = a non-zero number, which is absurd ) no solution exists and the lines are parallel but not coincident k ¡20 k ¡ 20 If k = 20 we have = 0, which is true for any x and y This means that all solutions come from x + 3y = alone Letting y = t, x = ¡ 3t for all values of t ) there are infinitely many solutions of the form x = ¡ 3t, y = t, t R In this case the lines are coincident EXERCISE 13F.1 Solve by row reduction: a x ¡ 2y = 4x + y = b c 4x + 5y = 21 5x ¡ 3y = ¡20 3x + y = ¡10 2x + 5y = ¡24 By inspection, classify the following pairs of equations as either intersecting, parallel or coincident lines: a x ¡ 3y = b x+y = c 4x ¡ y = 3x + y = 3x + 3y = y=2 cyan magenta yellow 95 100 50 f 75 25 95 5x ¡ 11y = 6x + y = 100 50 75 25 95 e 100 50 75 25 x ¡ 2y = 2x ¡ 4y = 95 100 50 75 25 d black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\356IB_HL-2_13.CDR Thursday, November 2007 10:58:09 AM PETERDELL 3x ¡ 4y = ¡3x + 4y = IB_HL-2ed (357) MATRICES (Chapter 13) ½ Consider the equation pair 357 x + 2y = 2x + 4y = a Explain why there are infinitely many solutions, giving geometric evidence b Explain why the second equation can be ignored when finding all solutions c Give all solutions in the form: i x = t, y = :::::: ii y = s, x = :::::: ½ 2x + 3y = to show that it a Use elementary row operations on the system 2x + 3y = 11 · ¸ reduces to What does the second row indicate? What is the 0 geometrical significance of your result? ½ 2x + 3y = to show that it b Use elementary row operations on the system 4x + 6y = 10 · ¸ reduces to Explain this result geometrically 0 ½ 3x ¡ y = a By using augmented matrices show that has infinitely many 6x ¡ 2y = solutions of the form x = t, y = 3t ¡ ½ 3x ¡ y = b Discuss the solutions to where k can take any real value 6x ¡ 2y = k ½ 3x ¡ y = Consider where k is any real number 6x ¡ 2y = k · ¸ ¡1 a Use elementary row operations to reduce the system to: 0 b For what value of k is there infinitely many solutions? c What form the infinite number of solutions have? d When does the system have no solutions? ½ 4x + 8y = Consider 2x ¡ ay = 11 · ¸ a Use elementary row operations to reduce the system to: b For what values of a does the system have a unique solution? a + 88 ¡21 c Show that the unique solution is x = , y= for these a values 4a + 16 2a + d What is the solution in all other cases? ½ mx + 2y = Use elementary row operations to find the values of m when the system 2x + my = has a unique solution a Find the unique solution b Discuss the solutions in the other two cases USING A GRAPHICS CALCULATOR cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Click on the appropriate icon to obtain instructions on how to enter an augmented matrix You can then obtain the reduced row-echelon form which has row reduction performed so that there are 0s in the bottom left corner and in the higher rows wherever possible black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\357IB_HL-2_13.CDR Thursday, November 2007 2:32:40 PM PETERDELL TI C IB_HL-2ed (358) 358 MATRICES (Chapter 13) ½ For the example · 2x + y = ¡1 x ¡ 3y = 17 the reduced row-echelon form is 0 ¡5 ¸ Consequently x = and y = ¡5 Try solving other £ systems using your calculator: ½ ½ 0:83x + 1:72y = 13:76 3x + 5y = b a 6x ¡ y = ¡11 1:65x ¡ 2:77y = 3:49 USING ROW OPERATIONS TO SOLVE A 3×3 ¡ ¡ SYSTEM A general £ system in variables x, y and z has the form where a1 a2 a3 the coefficients of x, y and z are constants a b1 c1 d1 is the system’s augmented b2 c2 d2 matrix form which we need b3 c3 d3 to reduce to echelon form: b e < a1 x + b1 y + c1 z = d1 a2 x + b2 y + c2 z = d2 : a3 x + b3 y + c3 z = d3 c d using f g elementary h i row operations Notice the creation of a triangle of zeros in the bottom left hand corner In this form we can easily solve the system because the last row is really hz = i i , and h likewise y and x from the other two rows Thus we arrive at a unique solution ² If h 6= (i may or may not be 0) we can determine z uniquely using z = ² If h = and i 6= 0, the last row reads £ z = i where i 6= which is absurd Hence, there is no solution and we say that the system is inconsistent ² If h = and i = 0, the last row is all zeros Consequently, there are infinitely many solutions of the form x = p + kt, y = q + lt and z = t where t R Note: ² The parametric representation of infinite solutions in terms of the parameter t is not unique This particular form assumes you are eliminating x and y to get z from Row It may be easier to eliminate, for example, x and z to get y ² A geometric interpretation of the different cases for three equations in three unknowns will be given later in Chapter 16 Example 19 < x + 3y ¡ z = 15 2x + y + z = : x ¡ y ¡ 2z = Solve the system: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 In augmented matrix form, the system is 3 ¡1 15 1 ¡1 ¡2 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\358IB_HL-2_13.CDR Thursday, November 2007 2:44:49 PM PETERDELL IB_HL-2ed (359) 359 MATRICES (Chapter 13) » 0 » 0 15 ¡23 ¡15 ¡5 ¡4 ¡1 ¡1 ¡5 ¡1 ¡17 The last row gives 15 ¡23 17 R2 ! R2 ¡ 2R1 ¡2 ¡6 ¡5 R3 ! R3 ¡ R1 ¡1 ¡2 ¡1 ¡3 ¡4 ¡1 R3 ! 5R3 ¡ 4R2 ¡20 ¡5 ¡75 20 ¡12 92 0 ¡17 17 ¡17z = 17 ) z = ¡1 ¡5y + 3z = ¡23 ¡5y ¡ = ¡23 ) ¡5y = ¡20 ) y=4 Using the final row we get ¡30 ¡23 ¡15 ¡15 A typical graphics calculator solution: x + 3y ¡ z = 15 x + 12 + = 15 ) x=2 Using the final row we get Thus we have a unique solution x = 2, y = 4, z = ¡1 Remember: You can use matrix algebra to find unique solutions: 3 3¡1 15 x ¡1 y 5=4 1 = 4 ¡1 z ¡1 ¡2 CASES WITH NON-UNIQUE SOLUTIONS As with £ systems of linear equations, £ systems may have a unique solution where a single value of each variable satisfies all three equations simultaneously Alternatively, it could have no solutions or infinitely many solutions We will now consider examples which show each of these situations Example 20 < x + 2y + z = 2x ¡ y + z = : 3x ¡ 4y + z = 18 Solve the system: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 In augmented matrix form, the system is: 2 ¡1 R2 ! R2 ¡ 2R1 ¡4 18 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\359IB_HL-2_13.CDR Thursday, November 2007 11:43:42 AM PETERDELL ¡2 ¡1 ¡4 ¡5 ¡2 ¡1 ¡6 IB_HL-2ed (360) 360 MATRICES (Chapter 13) »4 0 » ¡5 ¡10 3 ¡1 ¡2 ¡5 3 5 ¡1 R2 ! R2 ¡ 2R1 R3 ! R3 ¡ 3R1 ¡3 R3 ! R3 ¡ 2R2 0 ¡4 ¡6 ¡10 ¡10 10 ¡3 ¡2 18 ¡9 ¡2 ¡4 The last equation is really 0x + 0y + 0z = i.e., = 5, which is absurd, ) the system has no solution Use a graphics calculator to confirm there is no solution Example 21 < 2x ¡ y + z = x+y¡z = : 3x ¡ 3y + 3z = Solve the system: In augmented matrix form, the system is: »4 0 »4 0 3¾ ¡1 ¡3 ¡1 ¡3 ¡6 ¡1 ¡3 ¡1 5 5 Notice the swapping R1 $ R2 R2 ! R2 ¡ 2R1 R3 ! R3 ¡ 3R1 ¡2 ¡1 ¡2 ¡3 ¡4 ¡3 ¡3 ¡3 ¡6 3 ¡6 ¡6 ¡2 0 0 R3 ! R3 ¡ 2R2 ¡6 The row of zeros indicates infinitely many solutions If we let z = t in row 2, ¡3y + 3t = ) ¡3y = ¡ 3t ¡ 3t = ¡ 13 + t ) y= ¡3 Thus in equation 1, x + (¡ 13 + t) ¡ t = ) x ¡ 13 = and so x = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ) the solutions have the form: x = 73 , y = ¡ 13 + t, z = t, where t R black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\360IB_HL-2_13.CDR Thursday, November 2007 3:12:12 PM PETERDELL IB_HL-2ed (361) 361 MATRICES (Chapter 13) Example 22 < x ¡ 2y ¡ z = ¡1 2x + y + 3z = 13 : x + 8y + 9z = a where a takes all real values Consider the system a b c Use elementary row operations to reduce the system to echelon form When does the system have no solutions? When does the system have infinitely many solutions? What are the solutions? a In augmented matrix form, the ¡2 ¡1 ¡1 13 a ¡2 ¡1 ¡1 15 5 » 0 10 10 a + ¡2 ¡1 ¡1 15 5 » a ¡ 29 0 system is: ¡2 R2 ! R2 ¡ 2R1 R3 ! R3 ¡ R1 ¡1 10 R3 ! R3 ¡ 2R2 0 10 ¡10 13 15 10 a a+1 10 ¡10 a+1 ¡30 a ¡ 29 b Now if a 6= 29, we have an inconsistent system as zero = non-zero, and ) no solutions c If a = 29, the last row is all zeros indicating infinitely many solutions Letting z = t, in equation gives 5y + 5t = 15 ) y = 3¡t Using the first equation, gives x ¡ 2y ¡ z = ¡1 x ¡ 2(3 ¡ t) ¡ t = ¡1 ) x ¡ + 2t ¡ t = ¡1 ) x = 5¡t Thus we have infinitely many solutions, in the form: x = ¡ t, y = ¡ t, z = t, where t R EXERCISE 13F.2 Solve the following systems using row reduction: a x ¡ 2y + 5z = 2x ¡ 4y + 8z = ¡3x + 6y + 7z = ¡3 b x + 4y + 11z = x + 6y + 17z = x + 4y + 8z = c 2x ¡ y + 3z = 17 2x ¡ 2y ¡ 5z = 3x + 2y + 2z = 10 d 2x + 3y + 4z = 5x + 6y + 7z = 8x + 9y + 10z = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Use technology to check your answers to b and c black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\361IB_HL-2_13.CDR Thursday, November 2007 3:14:30 PM PETERDELL IB_HL-2ed (362) 362 MATRICES (Chapter 13) Without using technology, solve using row operations on the augmented matrix: a x+y+z = b x + 2y ¡ z = c 2x + 4y + z = 2x + 4y + z = 3x + 2y + z = 3x + 5y = 2x + 3y + z = 5x + 2y + 3z = 11 5x + 13y + 7z = Write the system of equations x + 2y + z = 2x ¡ y + 4z = x + 7y ¡ z = k in augmented matrix a Use elementary row operations to reduce the system to ² form as shown: echelon b Show that the system has either no solutions or infinitely many solutions and write down these solutions c Why does the system not have a unique solution? form ² ² ² ² ² ² ² ² x + 2y ¡ 2z = x ¡ y + 3z = ¡1 x ¡ 7y + kz = ¡k Consider the system of equations a Reduce the system to echelon form b Show that for one value of k the system has infinitely many solutions and find the solutions in this case c Show that a unique solution exists for all other k Find this solution x + 3y + 3z = a ¡ 2x ¡ y + z = 3x ¡ 5y + az = 16 A system of equations is a Reduce the system to echelon form using elementary row operations b Show that if a = ¡1 the system has infinitely many solutions, and find their form c If a 6= ¡1, find the unique solution in terms of a 2x + y ¡ z = mx ¡ 2y + z = x + 2y + mz = ¡1 Reduce the system of equations: to a form in which the solutions may be determined for all real values of m a Show that the system has no solutions for one value of m (m = m1 , say) b Show that there are infinitely many solutions for another value of m (m = m2 , say) c For what values of m does the system have a unique solution? 3(m ¡ 2) ¡7 Show that the unique solution is x = , y= , z= m+5 m+5 m+5 Consider the system of equations x + 3y + kz = kx ¡ 2y + 3z = k 4x ¡ 3y + 10z = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a Write the system in augmented matrix form and reduce it by elementary row operations to the form: b Show that for one value of k the system has infinitely many solutions and find the form of these solutions c For what value(s) of k does the system have no solutions? d For what values of k does the system have a unique solution? (There is no need to find the unique solution.) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\362IB_HL-2_13.CDR Thursday, November 2007 3:15:25 PM PETERDELL ² k ² ² ² ² IB_HL-2ed (363) 363 MATRICES (Chapter 13) NOT ENOUGH OR TOO MANY EQUATIONS ½ is an example of a £ system of two linear equations in three unknowns which has infinitely many solutions x + y + 2z = 2x + y ¡ z = It requires a further equation if a unique solution is to be obtained underspecified system We call this an If AX = B where A is m £ n, m < n, the system is underspecified If AX = B where A is m £ n, m > n, the system is overspecified and any solutions must fit all of the equations You need to check they Example 23 ½ x + y + 2z = 2x + y ¡ z = a Solve: b What can be deduced if the following equation is added to the system: i 3x ¡ y ¡ 4z = 18 ii 3x + y ¡ 4z = 18? a The augmented matrix is: ¸ · 1 2 ¡1 · ¸ 1 2 » ¡1 ¡5 R2 ! R2 ¡ 2R1 ¡2 ¡2 ¡1 ¡1 ¡4 ¡5 ¡4 So, ¡y ¡ 5z = and if z = t, y = ¡5t and as x + y + 2z = 2, x ¡ 5t + 2t = ) x = + 3t So, x = + 3t, y = ¡5t, z = t for all t R b i ii cyan If in addition 3x + y ¡ 4z = 18 then 3(2 + 3t) + (¡5t) ¡ 4t = 18 ) = 18 which is absurd, so no solution exists magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 Underspecified systems may also have no solutions This will be evident by any inconsistency 75 25 95 100 50 75 25 Note: If in addition 3x ¡ y ¡ 4z = 18 then 3(2 + 3t) ¡ (¡5t) ¡ 4t = 18 ) + 9t + 5t ¡ 4t = 18 ) 10t = 12 ) t = 1:2 When t = 1:2, x = 5:6, y = ¡6, z = 1:2 is a unique solution black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\363IB_HL-2_13.CDR Thursday, November 2007 11:44:45 AM PETERDELL IB_HL-2ed (364) 364 MATRICES (Chapter 13) EXERCISE 13F.3 Solve the systems: a 2x + y + z = x¡y+z = ½ Solve ½ Solve b c 3x + y + 2z = 10 x ¡ 2y + z = ¡4 x + 2y + z = 2x + 4y + 2z = 16 < x ¡ 3y + z = 2x + y ¡ 2z = and hence solve the system : 3x ¡ y + z = 18 x ¡ 3y + z = 2x + y ¡ 2z = < 2x + 3y + z = x ¡ y + 2z = and hence solve : ax + y ¡ z = 2x + 3y + z = x ¡ y + 2z = for all a R An economist producing x thousand items attempts to model a profit function as a quadratic model P (x) = ax2 + bx + c thousand dollars She knows that when producing 1000 items the profit is $8000, and when producing 4000 items the profit is $17 000 ½ a+b+c = a Using the supplied information show that 16a + 4b + c = 17 b Show that a = t, b = ¡ 5t, c = + 4t represents the possible solutions for the system c If she discovers that the profit for producing 2500 items is $19 750, find the actual profit function d What is the maximum profit to be made and what level of production is needed to achieve it? G INDUCTION WITH MATRICES Example 24 · Let the matrix B = a b b Find B2 , B3 and B4 i Give a proposition (conjecture) for Bn , n Z + ii Prove your proposition is true using mathematical induction · a ¸ ¡1 B = i Pn ¸ · ¸ · ¸ 16 = Also B = and B = ¡3 ¡7 ¡15 · ¸ · ¸ 2n n is that “if B = then B = ” for all n Z + ¡1 1 ¡ 2n ¡1 ¸2 · · ii For n = 1, B = 21 ¡ 21 ¸ · = ¡1 ¸ cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ) P1 is true black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\364IB_HL-2_13.CDR Thursday, November 2007 3:27:49 PM PETERDELL IB_HL-2ed (365) MATRICES (Chapter 13) · + Assume that Pk is true for some k Z , so B = k 2k ¡ 2k 365 ¸ Now Bk+1 = B Bk · ¸· ¸ 2k = fby assumptiong ¡1 1 ¡ 2k · ¸ 2(2k ) = ¡2k + ¡ 2k · ¸ 2k+1 = ¡ 2(2k ) · ¸ 2k+1 = as required ¡ 2k+1 So, P1 is true and Pn+1 is true whenever Pn is true · ¸ 2n n ) B = for all n Z + by mathematical induction ¡ 2n EXERCISE 13G · Let M = ¸ a Find M2 , M3 and M4 b State a proposition for Mn c Prove your conjecture is true using mathematical induction · a Given A = 2 ¸ , find A2 , A3 , A4 and A5 b Conjecture a value for An in terms of n where n Z + c Use mathematical induction to prove your conjecture is true d Is the result true when n = ¡1? · ¸ Let the matrix P = ¡1 a Find P2 , P3 and P4 b Give a proposition (conjecture) for Pn where n Z + c Prove your proposition is true using mathematical induction cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 For the sequence 1, 1, 2, 3, 5, 8, 13, where u1 = u2 = and un+2 = un +un+1 , · ¸ 1 + , prove by the principle of mathematical n Z , and the matrix A = · ¸ un+2 un+1 n+1 induction that A = for all integers n > un+1 un black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\365IB_HL-2_13.CDR Thursday, November 2007 3:45:42 PM PETERDELL IB_HL-2ed (366) 366 MATRICES (Chapter 13) REVIEW SET 13A · If A = a e i ¡1 ¸ · and B = b f j A+B B ¡ 2A A¡1 ¸ ¡2 find: Find a, b, c and d if: · ¸ · ¸ a b¡2 ¡a a = c d ¡ c ¡4 Make Y the subject of: a B¡Y=A d YB = C b e ¡2B AB ABA c g k 3A 3A ¡ 2B A2 · 2a b ¡2 b ¸ · + c f 2Y + C = D C ¡ AY = B Solve using an inverse matrix: 3x ¡ 4y = a 5x + 2y = ¡1 · ¸ · ¸ c X = 1 ¡2 · ¸ · ¸ 1 e X= ¡2 b d f b ¡a c d a · = a 2 ¸ 4x ¡ y = Why is this possible? 2x + 3y = · ¸ · ¸ ¡1 X= ¡1 · ¸ · ¸ · ¸ 1 X = ¡1 1 ¡1 ¤ Q¡P b P+Q ¸ AY = B AY¡1 = B If A is and B is find, if possible: a 2B b 2B c AB 3 For P = and Q = 4 find: 1 £ A¡B BA (AB)¡1 d h l 2P c d BA ¡Q x + 4y = have a unique solution? kx + 3y = ¡6 Comment on the solutions for the non-unique cases When does the system Solve the system 3x ¡ y + 2z = 2x + 3y ¡ z = ¡3 x ¡ 2y + 3z = The two points (¡2, 4) and (1, 3) lie on a circle x2 + y + ax + by + c = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a Find two equations in a, b and c and solve the system of equations b Explain why infinitely many solutions are obtained in a c If (2, 2) is also on the circle, find the equation of the circle black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\366IB_HL-2_13.CDR Thursday, November 2007 11:45:06 AM PETERDELL IB_HL-2ed (367) 367 MATRICES (Chapter 13) 10 2x + 3y ¡ 4z = 13 x ¡ y + 3z = ¡1 3x + 7y ¡ 11z = k 11 Solve the system: Solve the system using elementary row operations, and describe the solution set as k takes all real values 3x + y ¡ z = 0, x + y + 2z = 0: REVIEW SET 13B · Determine the £ matrix which when multiplied by · ¸ · ¸ a b of Hint: Let the matrix be 1 c d 1 £ ¤ A = , B = and C = 2 a AB b BA c AC d 1 ¸ gives an answer 3 Find, if possible: CA e CB Find, if they exist, the inverse matrices of each of the following: · ¸ · ¸ · ¸ ¡3 11 a b c ¡6 ¡6 ¡3 a If A = 2A¡1 , show that A2 = 2I b If A = 2A¡1 , simplify (A ¡ I)(A + 3I) giving your answer in the form rA + sI where r and s are real numbers A café sells two types of cola drinks The drinks each come in three sizes: small, medium and large At the beginning and end of the day the stock in the fridge was counted The results are shown below: Start of the day: End of the day: Brand C 42 36 34 small medium large The profit matrix is: Brand P 54 27 30 small medium large Brand C 27 28 28 Brand P 31 15 22 £ small medium large ¤ $0:75 $0:55 $1:20 Use matrix methods to calculate the profit made for the day from the sale of these drinks 3 3 ¡2 ¡1 and B = ¡4 ¡1 , find AB and BA and If A = ¡2 ¡4 ¡5 cyan magenta yellow 95 100 50 75 2x + y + z = 4x ¡ 7y + 3z = 10 : 3x ¡ 2y ¡ z = 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Solve the system of equations: < hence find A¡1 in terms of B black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\367IB_HL-2_13.CDR Thursday, November 2007 4:05:45 PM PETERDELL IB_HL-2ed (368) 368 MATRICES (Chapter 13) x + 2y ¡ 3z = 6x + 3y + 2z = ½ 2x ¡ 3y = a Show that the system mx ¡ 7y = n · elementary row operations of 14 ¡ 3m Solve the system: has an augmented matrix after ¡3 63 ¡ 3n ¸ b Under what conditions does the system have a unique solution? REVIEW SET 13C kx + 3y = ¡6 x + (k + 2)y = When does the system have a unique solution? Comment on the solutions for the non-unique cases ¯ ¯ ¯ ¯ x ¯ ¯ ¯ Find x if ¯ x + ¡2 ¯¯ = 0, given that x is real ¯ ¡2 x + ¯ · If A = ¸ · ¸ · ¸ ¡2 ¡7 ¡1 , B= , C= , evaluate if possible: ¡1 ¡3 2A ¡ 2B a b AC c · Find X if AX = B, A = ¡1 ¸ d D, given that DA = B · ¸ and B = ¡1 13 18 CB Write 5A2 ¡ 6A = 3I in the form AB = I and hence find A¡1 in terms of A and I a If A and B are square matrices, under what conditions are the following true? i If AB = B then A = I ii (A + B)2 = A2 + 2AB + B2 · ¸ · ¸ k k ¡ ¡2 b If M = has an inverse M¡1 , what values can k k ¡3 k have? < x ¡ y ¡ 2z = ¡3 tx + y ¡ z = 3t Find the values of t for which the system of equations does : x + 3y + tz = 13 not have a unique solution for x, y and z Show that no solution exists for one of these values of t, and find the solution set for the other values of t cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A rock was thrown from the top of a cliff such that its distance above sea level was given by s(t) = at2 + bt + c, where t is the time in seconds after the rock was released After second the rock was 63 m above sea level, after seconds 72 m, and after seconds 27 m a Find a, b and c and hence an expression for s(t) b Find the height of the cliff c Find the time taken for the rock to reach sea level black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\368IB_HL-2_13.CDR Thursday, November 2007 4:41:26 PM PETERDELL IB_HL-2ed (369) 369 MATRICES (Chapter 13) REVIEW SET 13D 3x ¡ y + 2z = 2x + 3y ¡ z = ¡3 x ¡ 2y + 3z = ¯ a+b c c ¯¯ a b+c a ¯¯ = 4abc b b c+a ¯ Solve the system: ¯ ¯ ¯ Prove that ¯¯ ¯ If A2 = 5A + 2I, find A3 , A4 , A5 and A6 in the form rA + sI The cost of producing x hundred bottles of correcting fluid per day is given by the function C(x) = ax3 + bx2 + cx + d dollars where a, b, c and d are constants a If it costs $80 before any bottles are produced, find d b It costs $100 to produce 100 bottles, $148 to produce 200 bottles and $376 to produce 400 bottles per day Determine a, b and c · If A = ¸ · ¸ · ¸ ¡3 ¡1 ¡12 ¡11 , B= , C= , find X if AXB = C ¡10 ¡1 Find the solution set of the following: 2x + y ¡ z = 3x + 2y + 5z = 19 x + y ¡ 3z = a ½ Solve the system of equations ½ Solve the system 2x + y ¡ z = 3x + 2y + z = x ¡ 3y = b kx + 2y = 2x + ky = ¡2 2x ¡ 3y + z = 10 4x ¡ 6y + kz = m Consider the system of equations as k takes all real values for all possible values of k and m x + 5y ¡ 6z = kx + y ¡ z = 5x ¡ ky + 3z = as k takes all real values a Write the system in augmented matrix form Show using elementary row operations that it reduces to: ¡6 ¡ 5k ¡ 2k 6k ¡ 0 (k ¡ 2)(3k ¡ 2) ¡(k ¡ 2)(k + 18) b For what values of k does the system have a unique solution? c For what value of k does the system have infinitely many solutions? Find the solutions cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 d For what value of k is the system inconsistent? How many solutions does the system have in this case? black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_13\369IB_HL-2_13.CDR Friday, 12 December 2008 12:13:25 PM TROY IB_HL-2ed (370) 370 MATRICES (Chapter 13) REVIEW SET 13E Hung, Quan and Ariel bought tickets for three separate performances The table below shows the number of tickets bought by each person: Hung Quan Ariel Opera Play Concert a If the total cost for Hung was E267, for Quan E145 and for Ariel E230, represent this information in the form of three equations b Find the cost per ticket for each of the performances c Determine how much it would cost Phuong to purchase opera, play and concert tickets Solve the system of equations: · If A = a 2x + y + z = 4x ¡ 7y + 3z = 10 3x ¡ 2y ¡ z = ¸ · ¸ ¡3 2 ¡2 5 , B = ¡3 and C = find, if possible: ¡1 b 3A c AB d BA AC e BC 3 2 ¡1 ¡3 If A = 4 and B = ¡1 show by calculation that 3 det(AB) = detA £ detB = 80 A matrix A has the property that A2 = A ¡ I Find expressions for An for n = 3, 4, , in terms of A and I Hence: a deduce simple expressions for A6n+3 and A6n+5 b express A¡1 in terms of A and I c Prove your result for A6n+5 is true by mathematical induction 1 a Find the values of a and b for which the matrix 1 has an inverse 2 a b of the form b a b cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b Use your answer to a above to solve the system of equations: 2x + y + z = x+y+z = 2x + 2y + z = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_13\370IB_HL-2_13.CDR Thursday, November 2007 11:53:28 AM PETERDELL IB_HL-2ed (371) Chapter 14 Vectors in and dimensions Contents: A B C D E F G H I J Vectors Operations with vectors 2-D vectors in component form 3-D coordinate geometry 3-D vectors in component form Algebraic operations with vectors Parallelism Unit vectors The scalar product of two vectors The vector product of two vectors cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 14A Review set 14B Review set 14C Review set 14D Review set 14E black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\371IB_HL-2_14.CDR Friday, November 2007 10:41:02 AM PETERDELL IB_HL-2ed (372) 372 VECTORS IN AND DIMENSIONS (Chapter 14) A VECTORS OPENING PROBLEM An aeroplane in calm conditions is flying due east A cold wind suddenly blows in from the south west The aeroplane, cruising at 800 km h¡1 , is blown slightly off course by the 35 km h¡1 wind ² ² ² What effect does the wind have on the speed and direction of the aeroplane? How can we accurately determine the new speed and direction using mathematics? How much of the force of the wind operates in the direction of the aeroplane? How does this affect fuel consumption and the time of the flight? VECTORS AND SCALARS To solve questions like those in the Opening Problem, we need to examine the size or magnitude of the quantities under consideration as well as the direction in which they are acting For example, the effect of the wind on an aeroplane would be different if the wind was blowing from behind the plane rather than against it Consider the problem of forces acting at a point force A If three equal forces act on point A and they are from directions 120o apart then clearly A would not move 120° 120° 120° For example, imagine three people 120o apart around a statue, pushing with equal force force force force Now suppose three forces act on the point A as shown What is the resultant force acting on A? In what direction would A move under these three forces? force 3N 5N 30° A 30° 60° x 4N force To handle these situations we need to consider quantities called vectors which have both size (magnitude) and direction Quantities which have only magnitude are called scalars Quantities which have both size (magnitude) and direction are called vectors cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Velocity is a vector since it describes speed (a scalar) in a particular direction black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\372IB_HL-2_14.CDR Thursday, November 2007 4:23:39 PM PETERDELL IB_HL-2ed (373) 373 VECTORS IN AND DIMENSIONS (Chapter 14) ² ² ² ² ² Other examples of vector quantities are: acceleration force displacement momentum weight DIRECTED LINE SEGMENT REPRESENTATION Consider a car that is travelling at 80 km h¡1 in a NE direction N W E S One good way of representing this is to use an arrow on a scale diagram N 45° Scale: cm represents 40 km h¡1 The length of the arrow represents the size (magnitude) of the quantity and the arrowhead shows the direction of travel Consider the vector represented by the line segment from O to A ² This vector could be represented by ¡! ! OA or a or a ~ or ¡ a " " bold used used by in text books students A a O ² The magnitude (length) could be represented by ¡! ! j OA j or OA or j a j or j a ~ j or j ¡ a j For ¡! AB is the vector which emanates from A and terminates at B, ¡! AB is the position vector of B relative to A we say that B A and that Example Scale: cm ´ 10 m s¡1 cyan magenta yellow 95 100 50 20 ms-1 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 On a scale diagram, sketch the vector which represents “a force of 20 m s¡1 in a southerly direction” black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\373IB_HL-2_14.CDR Tuesday, 29 January 2008 10:26:30 AM PETERDELL IB_HL-2ed (374) 374 VECTORS IN AND DIMENSIONS (Chapter 14) Example Scale: cm ´ 10 N N Draw a scaled arrow diagram representing “ a force of 40 Newtons on a bearing 115o ” 115° 25° 40 N EXERCISE 14A.1 Using a scale of cm represents 10 units, sketch a vector to represent: a 30 Newtons in a SE direction b 25 m s¡1 in a northerly direction c a displacement of 35 m in the direction 070o d an aeroplane taking off at an angle of 10o to the runway with a speed of 50 m s¡1 represents a velocity of 50 ms¡ ¡1 due east, draw a directed If line segment representing a velocity of: b 75 m s¡1 north east a 100 m s¡1 due west Draw a scaled arrow diagram representing the following vectors: a a force of 30 Newtons in the NW direction b a velocity of 40 m s¡1 in the direction 146o c a displacement of 25 km in the direction S32o E d an aeroplane taking off at an angle of 8o to the runway at a speed of 150 km h¡1 VECTOR EQUALITY Two vectors are equal if they have the same magnitude and direction So, if arrows are used to represent vectors, then equal vectors are parallel and equal in length a The arrows that represent them are translations of one another a a As we can draw a vector with given magnitude and direction from any point, we consider vectors to be free They are sometimes referred to as free vectors NEGATIVE VECTORS ¡! ¡! Notice that AB and BA have the same length but have opposite directions ¡! ¡! We say that BA is the negative of AB ¡! ¡! and write BA = ¡AB B AB BA A magenta yellow 95 100 50 75 25 95 100 50 75 25 since a and ¡a must be parallel and equal in length, but opposite in direction -a 95 50 75 25 95 cyan 100 then a 100 50 75 25 If black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\374IB_HL-2_14.CDR Friday, November 2007 11:38:47 AM PETERDELL IB_HL-2ed (375) 375 VECTORS IN AND DIMENSIONS (Chapter 14) Example b Q PQRS is a parallelogram in which ¡ ! ¡! PQ = a and QR = b Find vector expressions for: ¡ ! ¡! ¡ ! ¡ ! a QP b RQ c SR d SP R a P a b c d S ¡! fthe negative vector of PQg ¡! fthe negative vector of QRg ¡ ! QP = ¡a ¡! RQ = ¡b ¡ ! SR = a ¡ ! SP = ¡b ¡ ! fparallel to and the same length as PQg ¡! fparallel to and the same length as RQg EXERCISE 14A.2 State the vectors which are: a equal in magnitude b c in the same direction d e negatives of one another parallel equal p q r s t The figure alongside consists of equilateral triangles A, B and C lie on a straight ¡! ¡! ¡! E D line AB = p, AE = q and DC = r Which of the following statements is true? ¡! ¡! q r a EB = r b jpj = jqj c BC = r ¡! ¡! d DB = q e ED = p f p=q p A B C DISCUSSION Could we have a zero vector? What would its length be? What would its direction be? B OPERATIONS WITH VECTORS We have already been operating with vectors without realising it Bearing problems are an example of this The vectors in these cases are displacements km cyan magenta yellow 95 q° N W km E x km S 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A typical problem could be: A runner runs in an easterly direction for km and then in a southerly direction for km How far is she from her starting point and in what direction? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\375IB_HL-2_14.CDR Friday, November 2007 11:49:07 AM PETERDELL IB_HL-2ed (376) 376 VECTORS IN AND DIMENSIONS (Chapter 14) Trigonometry and Pythagoras’ theorem are used to answer such problems as we need to find µ and x VECTOR ADDITION P Suppose we have three towns P, Q and R A trip from P to Q followed by a trip from Q to R is equivalent to a trip from P to R This can be expressed in a vector form as ¡ ! ¡! ¡ ! the sum PQ + QR = PR This triangular diagram could take all sorts of shapes, but in each case the sum will be true For example: R Q Q R Q P Q R R P P After considering diagrams like those above, we can now define vector addition geometrically: To add a and b: Step 1: Step 2: Step 3: Draw a At the arrowhead end of a, draw b Join the beginning of a to the arrowhead end of b This is vector a + b b b a So, given a¡+¡b we have DEMO a Example b Given a and b as shown, construct a + b a a b a+b EXERCISE 14B.1 Copy the given vectors p and q and hence show how to find p + q: a b c q p q p q cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 p black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\376IB_HL-2_14.CDR Friday, January 2008 9:24:41 AM DAVID3 IB_HL-2ed (377) 377 VECTORS IN AND DIMENSIONS (Chapter 14) d e f p p q p q q Example A Find a single vector which is equal to: ¡! ¡! a BC + CA ¡! ¡! ¡ ! b BA + AE + EC ¡! ¡! ¡! c AB + BC + CA ¡! ¡! ¡! ¡! d AB + BC + CD + DE ¡! ¡! BC + CA ¡! ¡! BA + AE ¡! ¡! AB + BC ¡! ¡! AB + BC a b c d B E C D ¡! = BA fas showng ¡ ! ¡! + EC = BC ¡! ¡! + CA = AA ¡! ¡! ¡! + CD + DE = AE A B E C D Find a single vector which is equal to: ¡! ¡! ¡! ¡! a AB + BC b BC + CD c a Use vector diagrams to find i p is ¡! ¡! ¡! AB + BC + CD ii p+q d ¡! ¡! ¡! AC + CB + BD q + p given that: and q is b For any two vectors p and q, is p + q = q + p? Q Consider: b ¡ ! One way of finding PS is: ! ¡ ! ! ¡ ¡ PS = PR + RS = (a + b) + c R a c P Use the diagram to show that (a + b) + c = a + (b + c) S THE ZERO VECTOR Having defined vector addition, we are now able to state that: The zero vector is a vector of length For any vector a: a+0=0+a=a a + (¡a) = (¡a) + a = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ¡ ! When we write the zero vector by hand, we usually write black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\377IB_HL-2_14.CDR Friday, November 2007 12:08:04 PM PETERDELL IB_HL-2ed (378) 378 VECTORS IN AND DIMENSIONS (Chapter 14) VECTOR SUBTRACTION To subtract one vector from another, we simply add its negative, i.e., a ¡ b = a + (¡b) For example: for and then b a b -b a a-b Example For r, s and t as shown find geometrically: a r¡s b s¡t¡r r a s t b -r s -t r s-t-r -s r-s s EXERCISE 14B.2 For the following vectors p and q, show how to construct p ¡ q: a b c d p p q p q p q q For the following vectors: q magenta yellow 95 p¡q¡r 100 50 25 b 95 100 50 p+q¡r 75 95 100 50 75 25 95 100 50 75 25 cyan 25 a show how to construct: r 75 p black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\378IB_HL-2_14.CDR Friday, November 2007 12:13:13 PM PETERDELL c r¡q¡p IB_HL-2ed (379) VECTORS IN AND DIMENSIONS (Chapter 14) 379 Example For points A, B, C and D, simplify the following vector expressions: ¡! ¡! ¡! ¡! ¡! a AB ¡ CB b AC ¡ BC ¡ DB ¡! ¡! a AB ¡ CB B ¡! ¡! ¡! ¡! = AB + BC fas BC = ¡CBg ¡! A = AC b C C ¡! ¡! ¡! AC ¡ BC ¡ DB ¡! ¡! ¡! = AC + CB + BD ¡! = AD B A D Example Construct vector equations for: a b c d r s p q t a b c e g r f t=r+s r = ¡p + q f = ¡g + d + e We select any vector for the LHS and then take another path from its starting point to its finishing point Example S R Find, in terms of r, s and t: ¡ ! ¡ ! ¡ ! a RS b SR c ST s r t O a ¡ ! RS ¡! ¡! = RO + OS ¡! ¡ ! = ¡OR + OS = ¡r + s =s¡r b ¡ ! SR ¡! ¡! = SO + OR ¡! ¡! = ¡OS + OR = ¡s + r =r¡s c T ¡ ! ST ¡! ¡! = SO + OT ¡! ¡! = ¡OS + OT = ¡s + t =t¡s cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 For points A, B, C and D, simplify the following vector expressions: ¡! ¡! ¡! ¡! ¡! ¡! a AC + CB b AD ¡ BD c AC + CA ¡! ¡! ¡! ¡! ¡! ¡! ¡! ¡! ¡! d AB + BC + CD e BA ¡ CA + CB f AB ¡ CB ¡ DC black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\379IB_HL-2_14.CDR Friday, November 2007 12:27:02 PM PETERDELL IB_HL-2ed (380) 380 VECTORS IN AND DIMENSIONS (Chapter 14) Construct vector equations for: a b r c s p s r s q t t r d e p q s r f q p q p r t u s a Find, in terms of p, q and r: ¡! ¡! ¡! i AD ii BC iii AC B p B s s t b Find, in terms of r, s and t: ¡! ¡! ¡! i OB ii CA iii OC A r A t r q O C D r C AN APPLICATION OF VECTOR SUBTRACTION Vector subtraction is used to solve problems involving displacement, velocity and force Consider the following velocity application: An aeroplane needs to fly due east from one city to another at a speed of 400 km h¡1 However, a 50 km h¡1 wind blows constantly from the north-east In what direction must the aeroplane head to compensate for the wind, and how does the wind affect its speed? On this occasion we are given: f w E final vector SIMULATION wind vector We also know that the aeroplane will have to head a little north of its final destination so the north-easterly wind will blow it back to the correct direction p w f cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 In order to move in the f direction, the aeroplane must actually head in the p direction black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\380IB_HL-2_14.CDR Wednesday, 14 November 2007 9:06:52 AM PETERDELL IB_HL-2ed (381) VECTORS IN AND DIMENSIONS (Chapter 14) p Notice that p+w=f ) p + w + (¡w) = f + (¡w) ) p+0=f¡w ) p=f¡w 381 -w f By the cosine rule, x2 = 502 + 4002 ¡ £ 50 £ 400 cos 135o ) x ¼ 437 By the sine rule, x q° 135° sin µ sin 135o = 50 436:8 ) µ ¼ 4:64 50 400 Consequently, the aeroplane must head 4:64o north of east It needs to fly so that its speed in still air would be 437 km h¡1 The wind slows the aeroplane down to 400 km h¡1 EXERCISE 14B.3 A boat needs to travel south at a speed of 20 km h¡1 However a constant current of km h¡1 is flowing from the south-east Use vector subtraction to find: a the equivalent speed in still water for the boat to achieve the actual speed of 20 km h¡1 b the direction in which the boat must head to compensate for the current As part of an endurance race, Stephanie needs to swim from X to Y across a wide river Stephanie swims at 1:8 m s¡1 in still water 20 m Y If the river flows with a consistent current of 0:3 m s¡1 as shown, find: current 0.3 m¡s-1 80 m a the distance from X to Y b the direction in which Stephanie should head c the time Stephanie will take to cross the river X SCALAR MULTIPLICATION A scalar is a non-vector quantity It has a size but no direction We can multiply vectors by scalars such as and ¡3 If a is a vector, what 2a and ¡3a mean? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 We define 2a = a + a and 3a = a + a + a so ¡3a = 3(¡a) = (¡a) + (¡a) + (¡a) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\381IB_HL-2_14.CDR Wednesday, 14 November 2007 9:08:15 AM PETERDELL IB_HL-2ed (382) 382 VECTORS IN AND DIMENSIONS (Chapter 14) If a is then a a a 2a a So, -a a 3a -a -3a -a a 2a is in the same direction as a but is twice as long as a 3a is in the same direction as a but is three times longer than a ¡3a has the opposite direction to a and is three times longer than a If a is a vector and k is a scalar, then: ka is also a vector and we are performing scalar multiplication If k > 0, ka and a have the same direction If k < 0, ka and a have opposite directions In general: Example 10 r Given vectors s and b r ¡ 3s a 2r + s show how to find a b -s s r r geometrically -s r¡-¡3s 2r¡+¡s -s r Example 11 b p = ¡ 12 q Draw sketches of vectors p and q if a p = 3q q Let q be a b q q p = ¡ 12 q p = 3q EXERCISE 14B.4 s ¡r b 2s c e 2r ¡ s f 2r + 3s g magenta yellow 50 25 + 2s p = 13 q d p = 2q 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 cyan c 2r 2r 95 a Draw sketches of p and q if: a p=q b p = ¡q , show how to find geometrically: 100 and 75 r Given vectors black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\382IB_HL-2_14.CDR Thursday, November 2007 4:35:34 PM PETERDELL d ¡ 32 s h (r e + 3s) p = ¡3q IB_HL-2ed (383) 383 VECTORS IN AND DIMENSIONS (Chapter 14) a Copy this diagram and on it mark the points: ¡! ¡! ¡! i X such that MX = MN + MP ¡! ¡! ¡! ii Y such that MY = MN ¡ MP ¡ ! ¡! iii Z such that PZ = 2PM P N M b What type of figure is MNYZ? C 2-D VECTORS IN COMPONENT FORM So far we have examined vectors from their geometric representation N We have used arrows where: ² ² the length of the arrow represents size or magnitude the arrowhead indicates direction 45° 80 a a Consider a car travelling at 80 km h¡1 in a NE direction The velocity vector could be represented using the x and y steps which are necessary to go from the start to the finish ´ ³ 56:6 In this case the column vector 56:6 gives the x and y steps a2 + a2 = 802 ) 2a2 = 6400 ) a2 = 3200 ) a ¼ 56:6 ³ ´ x y y-component is the component form of a vector x-component ³ ´ For example, given we could draw is the horizontal step and is the vertical step EXERCISE 14C.1 Draw arrow diagrams to represent the vectors: ³ ´ ³ ´ b c a Write the illustrated vectors in component form: a b cyan magenta yellow 95 100 50 75 25 ¡5 ´ ³ d ¡1 ¡3 ´ c f 95 100 50 75 25 95 100 50 e 75 25 95 100 50 75 25 d ³ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\383IB_HL-2_14.CDR Friday, January 2008 10:06:18 AM DAVID3 IB_HL-2ed (384) 384 VECTORS IN AND DIMENSIONS (Chapter 14) VECTOR ADDITION ³ Consider adding vectors a = a1 a2 ´ ³ and b = b1 b2 ´ a+b b2 a2+b2 b Notice that: the horizontal step for a + b is a1 + b1 a the vertical step for a + b is a2 + b2 ³ So, if a = a1 a2 If a = ¡3 ´ ´ ³ and b = ³ ´ b1 b2 µ ´ and b = a1 + b1 a2 + b2 ³ Graphical check: µ find a + b = NEGATIVE VECTORS Consider the vector a = Notice that ¡a = ³ ´ 1+4 ¡3 + b The zero vector is = a a -a -3 if a = ZERO VECTOR a+b -2 ´ ¡2 ¡3 ³ In general, ³ ´ + ¶ = ³ ´ ¡3 ³ ´ Check graphically ¶ then a + b = a+b= b1 a1 a1+ b1 Example 12 ³ a2 a1 a2 ´ then ¡a = ³ ´ ³ ¡a1 ¡a2 ´ For any vector a : a+0=0+a=a a + (¡a) = (¡a) + a = VECTOR SUBTRACTION To subtract one vector from another, we simply add its negative, i.e., a ¡ b = a + (¡b) ³ ´ ³ ´ a b and b = b1 Notice that, if a = a1 2 ´ µa ¡ b ¶ ³ ´ ³ a1 ¡b1 then a ¡ b = a + (¡b) = a + ¡b = a1 ¡ b1 magenta yellow µ 95 100 50 then a ¡ b = 75 ´ 25 95 50 75 25 b1 b2 ³ and b = ´ 95 50 75 25 95 100 50 75 25 cyan a1 a2 100 ³ If a = 100 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\384IB_HL-2_14.CDR Friday, January 2008 9:32:09 AM DAVID3 a1 ¡ b1 a2 ¡ b2 ¶ IB_HL-2ed (385) 385 VECTORS IN AND DIMENSIONS (Chapter 14) Example 13 ³ ¡2 Given p = ´ , a ³ ´ q= ³ ¡2 ¡5 and r = q¡p ³ ´ ³ ´ = ¡ ¡2 ´ ³ find: = ³ a q¡p b p¡q¡r = 1¡3 4+2 ¡2 b p¡q¡r ³ ´ ³ ´ ³ ´ ¡2 = ¡2 ¡ ¡ ¡5 ´ ³ = ´ ³ = 3¡1+2 ¡2 ¡ + ¡1 ´ ´ VECTORS BETWEEN TWO POINTS y B(b1,¡b2) b2 a2 The position vector of B(b1 , b2 ) relative to µ ¶ ¡! b ¡a A(a1 , a2 ) is: AB = b1 ¡ a1 A(a1,¡a2) x b1 a1 Notice that this result could be found using: ¡! ¡! ¡! AB = AO + OB = b ¡ a µ ¶ ¡! ³ b1 ´ ³ a1 ´ b1 ¡ a1 ) AB = b ¡ a = b ¡ a B A a b 2 2 EXERCISE 14C.2 ³ If a = a e ¡3 ´ b= ³ ´ ³ ´ ¡2 , c = find: ¡5 b f a+b a+c ³ Given p = ¡4 d h b+c a+a c+b b+a+c ´ ³ ´ ³ ´ ¡1 , q = ¡5 and r = ¡2 find: p¡q p¡q¡r a d c g b+a c+a b e q¡r q¡r¡p c f p+q¡r r+q¡p ¡! ³ ´ ¡! ³ ¡3 ´ ¡! a Given BA = ¡3 and BC = find AC ¡! ³ ¡1 ´ ¡! ³ ´ ¡! b If AB = and CA = ¡1 , find CB ¡! ³ ¡1 ´ ¡! ³ ´ ¡ ! ³ ¡3 ´ ¡ ! c If PQ = , RQ = and RS = , find SP ¡! Find AB given: a A(2, 3) and B(4, 7) b A(3, ¡1) and B(1, 4) c A(¡2, 7) and B(1, 4) cyan yellow 95 100 50 75 25 95 100 50 75 25 95 50 100 magenta B(6, ¡1) and A(0, 4) e B(3, 0) and A(2, 5) 75 25 95 100 50 75 25 d black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\385IB_HL-2_14.CDR Friday, January 2008 12:03:12 PM DAVID3 f B(0, 0) and A(¡1, ¡3) IB_HL-2ed (386) 386 VECTORS IN AND DIMENSIONS (Chapter 14) SCALAR MULTIPLICATION Recall the geometric approach for scalar multiplication For example: a a ³ ´ ³ ´ Consider a = 2a a 2a = a + a = ³ ´ ³ ´ 3a = a + a + a = ² Notice that: (¡1)a = ¶ (0)a = (0) a1 ¶ and ³ ´ + ³ ´ = ³ If k is a scalar, then ka = µ = (¡1) a2 µ ² (¡1) a1 = ³ ´ + Examples like these suggest the following definition for scalar multiplication: µ ³ ´ + ¡a1 ¡a2 ka1 ka2 ´ ¶ = ¡a µ ¶ 0 = (0) a2 =0 Example 14 ³ ´ ³ ´ , q = find: ¡3 For p = a b 3q ³ ´ = ¡3 ³ = ¡9 a 3q b p + 2q c p + 2q ³ ´ ³ ´ = + ¡3 µ ´ = + 2(2) ³ ¡5 2p ¡ 3q 2p ¡ 3q ³ ´ ³ ´ = 12 ¡ ¡3 ¶ µ = + 2(¡3) = c µ ´ = (4) ¡ 3(2) (1) ¡ 3(¡3) ¡4 ¶ ¶ 12 EXERCISE 14C.3 2q c 2p + q d p ¡ 2q f 2p + 3r g 2q ¡ 3r h 2p ¡ q + 13 r cyan find by diagram and then comment on the results: b yellow 95 100 50 75 p+q+p+q+q 25 95 50 75 25 100 magenta ´ ¡1 and q = p+p+q+q+q 95 100 50 75 25 a ³ 1 95 ³ ´ If p = 100 p¡ 2r b 50 e ¡3p 25 a ³ ´ ³ ´ ³ ´ ¡2 ¡3 , q = and r = find: ¡1 75 For p = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\386IB_HL-2_14.CDR Monday, 12 November 2007 4:12:32 PM PETERDELL c q+p+q+p+q IB_HL-2ed (387) 387 VECTORS IN AND DIMENSIONS (Chapter 14) LENGTH OF A VECTOR ³ ´ Consider vector a = as illustrated Recall that j a j represents the length of a a By Pythagoras, j a j2 = 22 + 32 = + = 13 p ) j a j = 13 units fsince j a j > 0g ³ if a = In general, a1 a2 ´ p , then j a j = a12 + a22 Example 15 ³ ¡5 If p = ³ a p= ³ b q= ¡5 ¡1 ¡2 ´ ³ and q = ´ ´ ¡1 ¡2 ´ a jpj find: p + 25 p = 34 units ) jpj = b jqj p ¡ 2q = c ³ ¡5 ´ ¡2 c ³ j p ¡ 2q j ¡1 ¡2 ´ ³ ¡1 = ´ p 52 + (¡1)2 p = 26 units ) j p ¡ 2q j = p 1+4 p = units ) jqj = EXERCISE 14C.4 ³ ´ For r = jrj a ³ and s = jsj b ³ ´ ´ find: jr + sj c ³ If p = ¡1 and q = ¡2 d jr ¡ sj j s ¡ 2r j e ´ find: a jpj b j 2p j c j¡2p j d f jqj g j 4q j h j ¡4q j i j 3p j ¯1 ¯ ¯ q¯ e j j ¡3p j ¯ ¯ ¯¡ q¯ From your answers in 2, you should have noticed that j ka j = j k j j a j So, (the length of ka) = (the modulus of k) £ (the length of a) ³ ´ a By letting a = a1 , prove that j ka j = j k j j a j : ¯¡!¯ ¯ ¯ The length of the vector between A and B is denoted ¯AB¯ or simply AB cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Given A(2, ¡1), B(3, 5), C(¡1, 4) and D(¡4, ¡3), find: ¡! ¡! a AB and AB b BA and BA c ¡! ¡! d DC and DC e CA and CA f black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\387IB_HL-2_14.CDR Friday, January 2008 9:45:32 AM DAVID3 ¡! BC and BC ¡! DA and DA IB_HL-2ed (388) 388 VECTORS IN AND DIMENSIONS (Chapter 14) D 3-D COORDINATE GEOMETRY To specify points in 3-dimensional space we need a point of reference, O, called the origin Z Through O we draw mutually perpendicular lines and call them the X, Y and Z-axes We often think of the Y Z-plane as the plane of the page, with the X-axis coming directly out of the page However, we cannot of course draw this Once the positive directions of the X and Y -axes are determined, convention requires the positive Z-axis is given by the right hand rule (See page 411.) Y X In the diagram alongside the coordinate planes divide space into regions, each pair of planes intersecting on the axes Z The positive direction of each axis is a solid line whereas the negative direction is ‘dashed’ P(x, y, z) z x Any point P in space can be specified by an ordered triple of numbers (x, y, z) where x, y and z are the steps in the X, Y and Z directions from the origin O, to P à ! x ¡ ! The position vector of P is OP = y Y y X Z z To help us visualise the 3-D position of a point on our 2-D paper, it is useful to complete a rectangular prism or box with the 3-D POINT origin O as one vertex, the axes PLOTTER as sides adjacent to it, and P being the vertex opposite O P(x, y, z) z x Y y X Example 16 a A(0, 2, 0) Illustrate the points: a b B(3, 0, 2) b Z c c Z Z yellow Y 95 100 50 75 25 X 95 100 50 75 25 95 50 75 25 100 magenta X 95 100 50 75 25 -1 Y Y X cyan C B A C(¡1, 2, 3) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\388IB_HL-2_14.CDR Monday, 12 November 2007 4:13:53 PM PETERDELL IB_HL-2ed (389) 389 VECTORS IN AND DIMENSIONS (Chapter 14) DISTANCE AND MIDPOINTS Z Triangle OAB is right angled at A ) OB2 = a2 + b2 (1) fPythagorasg P(a, b, c) Triangle OBP is right angled at B ) OP2 = OB2 + c2 ) OP2 = a2 + b2 + c2 p ) OP = a2 + b2 + c2 c Y a A B b X For two general points A(x1 , y1 , z1 ) and B(x2 , y2 , z2 ) B(x2, y2, z2) A(x1, y1, z1) a, the x-step from A to B = x2 ¡ x1 = ¢x b, the y-step from A to B = y2 ¡ y1 = ¢y c, the z-step from A to B = z2 ¡ z1 = ¢z c a P Q b fPythagorasg ffrom (1)g p (x2 ¡ x1 )2 + (y2 ¡ y1 )2 + (z2 ¡ z1 )2 So, AB = A simple extension from 2-D to 3-D geometry also gives µ the midpoint of [AB] is ¶ x1 + x2 y1 + y2 z1 + z2 , , 2 Note: As with the 2-D case, a proof of this rule can be made using similar triangles Example 17 If A(¡1, 2, 4) and B(1, 0, ¡1) are two points in space, find: a the distance from A to B b the coordinates of the midpoint of [AB] a b AB p = (1 ¡ ¡1)2 + (0 ¡ 2)2 + (¡1 ¡ 4)2 p = + + 25 p = 33 units The midpoint is ¶ µ ¡1 + + + (¡1) , , 2 which is (0, 1, 32 ) EXERCISE 14D Illustrate P and find its distance from the origin O if P is: a (0, 0, ¡3) b (0, ¡1, 2) c (3, 1, 4) yellow 95 50 b d 75 25 95 100 50 75 25 95 50 75 25 95 100 50 75 25 100 magenta (¡1, ¡2, 3) find the midpoint of [AB] A(¡1, 2, 3) and B(0, ¡1, 1) A(3, ¡1, ¡1) and B(¡1, 0, 1) a c cyan ii 100 For each of the following: i find the distance AB d black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\389IB_HL-2_14.CDR Friday, November 2007 2:43:02 PM PETERDELL A(0, 0, 0) and B(2, ¡1, 3) A(2, 0, ¡3) and B(0, 1, 0) IB_HL-2ed (390) 390 VECTORS IN AND DIMENSIONS (Chapter 14) Show that P(0, 4, 4), Q(2, 6, 5) and R(1, 4, 3) are vertices of an isosceles triangle Determine the nature of triangle ABC using distances: a b c d A(2, A(0, A(5, A(1, ¡1, 7), B(3, 1, 4) and C(5, 4, 5) 0, 3), B(2, 8, 1) and C(¡9, 6, 18) 6, ¡2), B(6, 12, 9) and C(2, 4, 2) 0, ¡3), B(2, 2, 0) and C(4, 6, 6) A sphere has centre C(¡1, 2, 4) and diameter [AB] where A is (¡2, 1, 3) Find the coordinates of B and the radius of the circle Z a State the coordinates of any point on the Y -axis b Use a and the diagram opposite to find the coordinates of two points on the Y -axis which p are 14 units from B(¡1, ¡1, 2) B ~`1`4 -1 -1 Y X E 3-D VECTORS IN COMPONENT FORM Consider a point P(x1 , y1 , z1 ) The x, y and z-steps from the origin to P are x1 , y1 and z1 respectively à ! x1 ¡ ! is the vector which emanates from O So, OP = y1 P(xz, yz, zz) Z zz Y xz z1 and terminates at P A yz B X In general, if A(x1 , y1 , z1 ) and B(x2 , y2 , z2 ) are two points in space then: à ! x2 ¡ x1 x-step p ¡! ¡! y-step AB = y2 ¡ y1 and j AB j = (x2 ¡ x1 )2 + (y2 ¡ y1 )2 + (z2 ¡ z1 )2 z2 ¡ z1 z-step ¡! AB is called ‘vector AB’ or the ‘position vector of B relative to A’ ¡ ! OP, the position vector of P relative to O, is called the position vector of the point P in both 2-D and 3-D Its usefulness is marked by the fact that its components are exactly the same as the coordinates of the point P Example 18 ¡! ¡! If A is (3, ¡1, 2) and B is (1, 0, ¡2) find: a OA b AB ! à ! à ! à ! à 3¡0 1¡3 ¡2 ¡! ¡! b AB = ¡ (¡1) = a OA = ¡1 ¡ = ¡1 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 ¡2 ¡ 2 100 50 75 25 95 100 50 75 25 2¡0 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\390IB_HL-2_14.CDR Monday, 12 November 2007 4:15:25 PM PETERDELL ¡4 IB_HL-2ed (391) 391 VECTORS IN AND DIMENSIONS (Chapter 14) Example 19 ¡ ! If P is (¡3, 1, 2) and Q is (1, ¡1, 3), find j PQ j ¡! PQ = à ¡ (¡3) ¡1 ¡ 3¡2 ! à = ¡2 ! p ¡! ) j PQ j = 42 + (¡2)2 + 12 p = 21 units Example 20 If A is (¡1, 3, 2) and B is (2, 1, ¡4), find: a the position vector of A from B b the distance between A and B ! à ! à ¡1 ¡ ¡3 ¡! 3¡1 = a The position vector of A from B is BA = ¡ (¡4) ¡! b AB = j BA j p ¡! ¡! = + + 36 jABj¡=¡jBAj is the = units distance between A and B EXERCISE 14E.1 Consider the point T(3, ¡1, 4) a Draw a diagram to locate the position of T in space ¡! c How far is it from O to T? b Find OT Given A(¡3, 1, 2) and B(1, 0, ¡1) find: ¡! ¡! ¡! ¡! a AB and BA b the length of AB and BA ¡! ¡! ¡! Given A(3, 1, 0) and B(¡1, 1, 2) find OA, OB, and AB Given M(4, ¡2, ¡1) and N(¡1, 2, 0) find: a the position vector of M from N b the position vector of N from M c the distance between M and N For A(¡1, 2, 5), B(2, 0, 3) and C(¡3, 1, 0) find the position vector of: a A from O and the distance from O to A b C from A and the distance from A to C c B from C and the distance from C to B Find the distance from Q(3, 1, ¡2) to: a the Y -axis b the origin c the Y OZ plane GEOMETRIC REPRESENTATION cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 As for 2-D vectors, 3-D vectors are represented by directed line segments or arrows Consider the vector represented by the directed line segment from O to A black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\391IB_HL-2_14.CDR Monday, 12 November 2007 4:19:15 PM PETERDELL IB_HL-2ed (392) 392 VECTORS IN AND DIMENSIONS (Chapter 14) ² A This vector could be represented by ¡! ! OA or a or a or ¡ a a e bold used used by O in text books students ¡! ! The magnitude could be represented by j OA j, OA, j a j, j e a j or j ¡ a j: à ! a1 p If a = a2 then j a j = a12 + a22 + a32 : ² a3 VECTOR EQUALITY Two vectors are equal if they have the same magnitude and direction So, if arrows are used to represent vectors, then equal vectors are parallel and equal in length This means that equal vector arrows are translations of one another, but in space (free vectors) à ! à ! a1 a2 a3 If a = b1 b2 b3 and b = a a a , then a = b , a1 = b1 , a2 = b2 , a3 = b3 a = b implies that vector a is parallel to vector b, and j a j = j b j : a b Consequently, a and b are opposite sides of a parallelogram and lie in the same plane DISCUSSION ² Do any three points in space define a plane? What about four points? Illustrate ² What simple tests on four points in space enable us to deduce that the points are vertices of a parallelogram? Consider using vectors and not using vectors à Example 21 Find a, b, and c if Equating components, a¡3 b¡2 c¡1 a ¡ = ¡ a, ) 2a = 4, ) a = 2, ! à = 1¡a ¡b ¡3 ¡ c b ¡ = ¡b 2b = b=1 ! and and and c ¡ = ¡3 ¡ c 2c = ¡2 c = ¡1 Example 22 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ABCD is a parallelogram A is (¡1, 2, 1), B is (2, 0, ¡1) and D is (3, 1, 4) Find the coordinates of C black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\392IB_HL-2_14.CDR Friday, November 2007 3:17:37 PM PETERDELL IB_HL-2ed (393) 393 VECTORS IN AND DIMENSIONS (Chapter 14) First we sketch the points as shown: Let C be (a, b, c) Now [AB] is parallel to [DC] and has the same ¡! ¡! length, so DC = AB ! à ! à a¡3 b¡1 c¡4 ) ) ¡2 ¡2 = a ¡ = 3, ) a = 6, b ¡ = ¡2, b = ¡1, c ¡ = ¡2 c=2 So, C is (6, ¡1, 2) midpoint of [AC] is ³ ´ midpoint of [DB] is ´ ¢ ¡ + + + ¡1 or 52 , 12 , 32 , , C(a,¡b,¡c) D(3,¡1,¡4) Check: ³ B(2,¡0,-1) A(-1,¡2,¡1) ¡1 + + ¡1 + , , 2 2 or ¡5 2, 2, ¢ What property of a parallelogram are we checking? EXERCISE 14E.2 à Find a, b and c if: a a¡4 b¡3 c+2 ! à = Find scalars a, b and c if: ! à ! à ! à a 3a = b c¡1 2 a b à = ¡4 ! b a2 a+b à b ! a¡5 b¡2 c+3 à ! 1 c a ! à = à +b ¡1 3¡a 2¡b 5¡c ! ! à ! +c 1 à = ¡1 3 ! A(¡1, 3, 4), B(2, 5, ¡1), C(¡1, 2, ¡2) and D (r, s, t) are four points in space ¡! ¡! ¡! ¡! Find r, s and t if: a AC = BD b AB = DC A quadrilateral has vertices A(1, 2, 3), B(3, ¡3, 2), C(7, ¡4, 5) and D(5, 1, 6) ¡! ¡! b What can be deduced about the quadrilateral ABCD? a Find AB and DC PQRS is a parallelogram P is (¡1, 2, 3), Q(1, ¡2, 5) and R(0, 4, ¡1) a Use vectors to find the coordinates of S b Use midpoints of diagonals to check your answer F ALGEBRAIC OPERATIONS WITH VECTORS The rules for algebra with vectors readily extend from 2-D to 3-D: ! à and b = b1 b2 b3 ! à then a + b = à cyan magenta yellow 50 ka = 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 and 95 If a = a1 a2 a3 100 à black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\393IB_HL-2_14.CDR Thursday, November 2007 4:45:50 PM PETERDELL a1 + b1 a2 + b2 a3 + b3 ka1 ka2 ka3 ! à , a¡b= a1 ¡ b1 a2 ¡ b2 a3 ¡ b3 ! ! for some scalar k IB_HL-2ed (394) 394 VECTORS IN AND DIMENSIONS (Chapter 14) SOME PROPERTIES OF VECTORS ² a+b=b+a ² (a + b) + c = a + (b + c) ² a+0=0+a=a ² a + (¡a) = (¡a) + a = ² j ka j = j k j j a j where ka is parallel to a length of ka length of a modulus of k The rules for solving vector equations are similar to those for solving real number equations However, there is no such thing as dividing a vector by a scalar Instead, we multiply by reciprocals a For example, if 2x = a then x = 12 a and not a has no meaning in vector algebra Two useful rules are: To establish these notice that: ² if x + a = b then x = b ¡ a ² if kx = a then x = ka (k 6= 0) if x + a = b then x + a + (¡a) = b + (¡a) ) x+0=b¡a ) x=b¡a and if then kx = a k (kx) = k1 a ) 1x = k1 a ) x = k1 a Another useful property is that: ¡! ¡! if OA = a and OB = b where O is the origin ¡! ¡! then AB = b ¡ a and BA = a ¡ b B A b a O Example 23 3x ¡ r = s ) 3x = s + r ) x = 13 (s + r) a Solve for x: a 3x ¡ r = s b c ¡ 2x = d b c ¡ 2x = d ) c ¡ d = 2x ) 12 (c ¡ d) = x Example 24 cyan magenta yellow 95 100 50 25 95 100 50 75 25 95 , find j 2a j 75 j 2a j = j a j p = (¡1)2 + 32 + 22 p = 1+9+4 p = 14 units ! 100 50 25 95 100 50 75 25 If a = ¡1 75 à black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\394IB_HL-2_14.CDR Friday, November 2007 3:35:41 PM PETERDELL IB_HL-2ed (395) 395 VECTORS IN AND DIMENSIONS (Chapter 14) EXERCISE 14F.1 Solve the following vector equations for x: a 2x = q b 12 x = n d ³ If r = a ´ ¡2 4s ¡ 5x = t e q + 2x = r ¡3x = p f 4m ¡ 13 x = n ³ ´ , find y if: and s = 2y b 2y = r c c =s µ Show by equating components, that if x = x = k1 a x1 x2 ¶ , µ a= 3s ¡ 4y = r d r + 2y = s ¶ a1 a2 and kx = a, then Find B if C is the centre of a circle with diameter [AB]: A is (3, ¡2) and C(1, 4) A is (¡1, ¡4) and C(3, 0) a c b Find: a the coordinates of M ¡! ¡! ¡! b vectors CA, CM and CB: ¡! ¡! ¡! c Verify that CM = 12 CA + 12 CB A(3'\\6) M C(-4'\\1) B(-1'\\2) à If a = ¡! If OA = ¡1 à ! ¡2 ¡1 à and b = ! ¡2 A is (0, 5) and C(¡1, ¡2) ! ¡! and OB = find x if: à ¡1 ! a c 2a + x = b b 3x ¡ a = 2b 2b ¡ 2x = ¡a ¡! find AB and hence the distance from A to B à The position vectors of A, B, C and D from O are à ! ¡2 ¡! ¡! ¡3 respectively Deduce that BD = 2AC ¡2 ! à , ¡4 ! à , ¡2 ! and Find the coordinates of C, D and E B(2'\\3'-3) A(-1'\\5' 2) C D E 10 Use vectors to find whether or not ABCD is a parallelogram: a A(3, ¡1), B(4, 2), C(¡1, 4) and D(¡2, 1) b A(5, 0, 3), B(¡1, 2, 4), C(4, ¡3, 6) and D(10, ¡5, 5) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 c A(2, ¡3, 2), B(1, 4, ¡1), C(¡2, 6, ¡2) and D(¡1, ¡1, 2) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\395IB_HL-2_14.CDR Friday, November 2007 3:54:29 PM PETERDELL IB_HL-2ed (396) 396 VECTORS IN AND DIMENSIONS (Chapter 14) 11 Use vector methods to find the remaining vertex of: a b B(2,-1) A(3, 0) P(-1, 4, 3) A b O à ¡! 13 If AB = a ! ¡1 ! à , b= à ¡1 ! à , b= ! ¡! and BD = ¡3 ! à and c = ¡3 ¡3 c g ! à and c = ¡2 à ¡3 ! find: ¡! CD c a¡b c ¡ 12 a b f a+b a+b+c ¡1 ¡! CB b 14 For a = 15 If a = à ¡! , AC = ¡! AD à a e ¡1 Y(3,-2,-2) In the given figure BD is parallel to OA and half its length Find, in terms of a and b, vector expressions for: ¡! ¡! ¡! a BD b AB c BA ¡! ¡! ¡! d OD e AD f DA B a X R S(4, 0, 7) D 12 W(-1, 5, 8) Z(0, 4, 6) C(8,-2) D c Q(-2, 5, 2) ! , find: a ¡ 3c 2b ¡ c + a b + 2c a¡b¡c d h find: a j a j b jbj ! c jb + cj d j a ¡ cj e j a j b 16 Find scalars r and s such that: µ ¶ µ ¶ µ ¶ ¡8 a r ¡1 + s = ¡27 à b r ¡3 ! à ! +s à = ¡19 f a jaj ! THE DIVISION OF A LINE SEGMENT Consider these points which are equally spaced on a number line: A B C D Notice that B divides [AD] in the ratio : But in what ratio does B divide [DA] and F divide [AD] ? E F Because confusion can arise we adopt the following convention to interpret the ratio of the division of a line segment: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ¡! ¡! X divides [AB] in the ratio a : b means AX : XB = a : b black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_14\396IB_HL-2_14.CDR Thursday, 11 March 2010 10:28:40 AM PETER IB_HL-2ed (397) VECTORS IN AND DIMENSIONS (Chapter 14) ¡! ¡! So, to find the ratio in which B divides [DA], we find DB : BA ¡! ¡! Notice that both DB and BA have the same direction B lies in between A and D, so this is internal division A 397 This is clearly : B D A D ¡! ¡! To find the ratio in which F divides [AD], we find AF : FD This is : ¡2 or ¡5 : 2 F The minus sign indicates that these two vectors are opposite in direction F is outside the interval [AD], so this is external division We say that F divides [AD] externally in the ratio : Example 25 If A is (¡1, 4, 7) and B is (3, 0, 5) find: a P if P divides [AB] in the ratio : b Q if Q divides [BA] externally in the ratio : ¡ ! ¡! ¡ ! OP = OA + AP ¡! ¡! = OA + 34 AB à ! à ¡ ! ¡ ! AP : PB = : ¡! ¡! ) AP = 34 AB a P A B = à = b ¡1 12 + ! ¡4 ¡2 ! ) P is (2, 1, 12 ) Since Q divides [BA] externally in the ratio : 1, we need to use a minus sign ¡! ¡! ¡! ¡! ¡! OQ = OA + AQ BQ : QA = ¡3 : ¡! ¡! ¡! ¡! = OA + 12 BA ) AQ = BA à ! à ! B Q A = à = ¡1 ¡3 + ! ¡4 ) Q is (¡3, 6, 8) EXERCISE 14F.2 The points A to F are equally spaced on a number line Find the ratio in which: cyan magenta yellow 95 50 75 25 100 a B divides [AC] c C divides [FB] e C divides [AB] F 95 E 100 50 75 D 25 95 C 100 50 B 75 25 95 100 50 75 25 A black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\397IB_HL-2_14.CDR Monday, 12 November 2007 4:33:39 PM PETERDELL b C divides [BF] d D divides [EA] f F divides [DA] IB_HL-2ed (398) 398 VECTORS IN AND DIMENSIONS (Chapter 14) For A(3, 1, 1), B(¡1, 2, 0), C(1, ¡1, 4) and D(3, ¡2, 4) find: a b c d e f Q if Q divides [BC] internally in the ratio : R if R divides [CA] externally in the ratio : S if S divides [BA] internally in the ratio : T if T divides [CB] externally in the ratio : X if X divides [AD] internally in the ratio : Y if Y divides [DB] externally in the ratio : 3 A P P lies on the line passing through points A and B Find the position vector p of point P if: a P divides [AB] internally in the ratio : b P divides [AB] externally in the ratio : c P divides [AB] in the ratio m : n B p a b O G PARALLELISM If two non-zero vectors are parallel, then one is a scalar multiple of the other and vice versa Note: ² If a is parallel to b, then there exists a scalar k such that a = kb ² If a = kb for some scalar k, then I a is parallel to b, and I j a j = j k jj b j: a b à Notice that a = à a= ¡4 ¡4 ! à is parallel to b = ¡2 ! ! à and c = à à Example 26 Find r and s given that a = ! as a = 2b and a = 12 c ! ¡3 ¡9 is also parallel to d = 12 ¡8 as a = ¡ 23 d ¡1 r ! à is parallel to b = s ¡3 ! Since a and b are parallel, then a = kb for some scalar k à ! à ! = ks, ¡1 = 2k Consequently, k = ¡ 12 magenta yellow r= 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ) cyan and r = ¡3k ¡ ¢ = ¡ 12 s and r = ¡3 ¡ 12 and ) and s = ¡4 95 ) 100 s ¡3 50 =k 75 ¡1 r ) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\398IB_HL-2_14.CDR Monday, 12 November 2007 4:36:07 PM PETERDELL IB_HL-2ed (399) 399 VECTORS IN AND DIMENSIONS (Chapter 14) Three or more points are said to be collinear if they lie on the same straight line ¡! ¡! A, B and C are collinear if AB = kBC for some scalar k C B A Example 27 Prove that A(¡1, 2, 3), B(4, 0, ¡1) and C(14, ¡4, ¡9) are collinear and hence find the ratio in which B divides [CA] ¡! AB = à ¡2 ¡4 ! ¡! BC = à 10 ¡4 ¡8 ! à =2 ¡2 ¡4 ! ¡! ¡! ) BC = 2AB ) [BC] is parallel to [AB] and since B is common to both, A, B and C are collinear To find the ratio in which B divides [CA], we find à ! à ! 5 ¡! ¡! CB : BA = ¡2 ¡2 : ¡ ¡2 = : ¡4 ¡4 ) B divides [CA] internally in the ratio : EXERCISE 14G à a= ¡1 ! à and b = ¡6 r s ! are parallel Find r and s à Find scalars a and b given that ¡1 ! à ! a b and are parallel à a Find a vector of length unit which is parallel to a = Hint: Let the vector be ka à b Find a vector of length units which is parallel to b = What can be deduced from the following? ¡! ¡! ¡ ! ¡! a AB = 3CD b RS = ¡ 12 KL ¡1 ¡2 ! ¡2 ¡1 ¡! ¡! AB = 2BC c à The position vectors of P, Q, R and S from O are respectively ¡1 ! à , ¡3 ! d ! à , ¡1 ¡! ¡! BC = 13 AC ! à and ¡1 ¡2 ! a Deduce that [PR] and [QS] are parallel b What is the relationship between the lengths of [PR] and [QS]? magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 95 100 50 75 25 cyan 25 a Prove that A(¡2, 1, 4), B(4, 3, 0) and C(19, 8, ¡10) are collinear and hence find the ratio in which A divides [CB] b Prove that P(2, 1, 1), Q(5, ¡5, ¡2) and R(¡1, 7, 4) are collinear and hence find the ratio in which Q divides [PR] black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\399IB_HL-2_14.CDR Friday, November 2007 4:44:09 PM PETERDELL IB_HL-2ed (400) 400 VECTORS IN AND DIMENSIONS (Chapter 14) a A(2, ¡3, 4), B(11, ¡9, 7) and C(¡13, a, b) are collinear Find a and b b K(1, ¡1, 0), L(4, ¡3, 7) and M(a, 2, b) are collinear Find a and b Triangle inequality In any triangle, the sum of any two sides must always be greater than the third side This is based on the well known result: “the shortest distance between two points is a straight line” a Prove that j a + b j j a j + j b j using a geometrical argument Hint: Consider ² ² b c a is not parallel to b and use the triangle inequality a and b parallel ² any other cases H UNIT VECTORS A unit vector is any vector which has a length of one unit à ! 0 For example: ² is a unit vector as its length is 1 p ² @ p 12 + 02 + 02 = A is a unit vector as its length is r³ ¡ p1 p1 ´2 ³ ´2 + 02 + ¡ p12 = à ! 0 i= à ! , j= à Notice that a = a1 a2 a3 à ! and k = Thus, a = ¡5 are special unit vectors in the directions of the positive X, Y and Z-axes respectively ! , a = a1 i + a2 j + a3 k component form à 0 unit vector form ! can be written as a = 2i + 3j ¡ 5k and vice versa We call i, j and k the base vectors as any vector can be written as a linear combination of the vectors i, j and k magenta yellow 95 ¶ p 22 + (¡5)2 , its length is p = 29 units 100 50 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 cyan ¡5 25 µ As 2i ¡ 5j = Find the length of the 2-D vector 2i ¡ 5j 75 Example 28 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\400IB_HL-2_14.CDR Friday, November 2007 4:49:43 PM PETERDELL IB_HL-2ed (401) 401 VECTORS IN AND DIMENSIONS (Chapter 14) Example 29 à Find a vector b of length in the opposite direction to the vector a = p 4+1+1 The unit vector in the direction of a is à ¡1 ! à = p1 ¡1 ¡1 ! ! We now multiply this unit vector by ¡7 The negative reverses the direction and the gives the required length à ! ¡1 Thus b = ¡ p76 Check that j b j = EXERCISE 14H Express the following vectors in component form and find their length: a i¡j+k b 3i ¡ j + k c i ¡ 5k d 12 (j + k) Find k for the unit vectors: µ ¶ µ ¶ k a b k 0 µ ¶ k c d @ ¡ 12 k A @ e Find the lengths of the vectors: a 3i + 4j b 2i ¡ j + k i + 2j ¡ 2k c d µ ¡1 a it has the same direction as µ b it has the opposite direction to à c it has the same direction as ¡1 à d it has the opposite direction to ¡2i ¡ 5j ¡ 2k and has length units ¡1 ¡4 ¶ and has length units ! and has length units ¡1 ¡2 ¡2 ! and has length units k a jaj ² A vector b of length k in the opposite direction to a is b = ¡ magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ² A vector b of length k which is parallel to a could be b = § cyan A ¶ ² A vector b of length k in the same direction as a is b = Note: ¡ 13 ¡2:36i + 5:65j Find the unit vector in the direction of: a i + 2j b 2i ¡ 3k c Find a vector b if: k black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\401IB_HL-2_14.CDR Friday, November 2007 4:53:53 PM PETERDELL k a jaj k a jaj IB_HL-2ed (402) 402 VECTORS IN AND DIMENSIONS (Chapter 14) I THE SCALAR PRODUCT OF TWO VECTORS We have learned how to add, subtract and multiply vectors by a scalar These operations have all been demonstrated to have practical uses For example, scalar multiplication is used in the concept of parallelism and finding unit vectors VECTOR PRODUCTS For ordinary numbers numbers a and b we can write the product of a and b as ab or a £ b There is only one interpretation for this product, and so we have developed power notation as a shorthand a2 = a £ a, a3 = a £ a £ a, and so on However, there are two different types of product involving two vectors These are: ² The scalar product of vectors, which results in a scalar answer and has the notation a ² b (read “a dot b”) ² The vector product of vectors, which results in a vector answer and has the notation a £ b (read “a cross b”) Consequently, for vector a, a2 or (a)2 has no meaning, as it not clear which of the vector products we mean So, we should never write an or (a)n SCALAR DOT PRODUCT The scalar product of two vectors is also known as the dot product or inner product à ! à ! a1 a2 a3 If a = Definition: b1 b2 b3 and b = , the scalar product of a and b is defined as a ² b = a1 b1 + a2 b2 + a3 b3 ANGLE BETWEEN VECTORS à Consider vectors: a1 a2 a3 a= ! à and b= b1 b2 b3 ! We translate one of the vectors so that they both emanate from the same point This vector is ¡a + b = b ¡ a and has length j b ¡ a j : a q b j b ¡ a j2 = j a j2 + j b j2 ¡ j a j j b j cos µ à ! à ! à ! magenta yellow 95 = b1 ¡ a1 b2 ¡ a2 b3 ¡ a3 100 50 25 95 50 75 25 95 100 50 75 25 95 100 50 75 25 cyan a1 a2 a3 ¡ 75 b1 b2 b3 But b ¡ a = 100 Using the cosine rule, black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\402IB_HL-2_14.CDR Monday, November 2007 9:16:18 AM PETERDELL IB_HL-2ed (403) 403 VECTORS IN AND DIMENSIONS (Chapter 14) ) (b1 ¡ a1 )2 + (b2 ¡ a2 )2 + (b3 ¡ a3 )2 = a12 + a22 + a32 + b12 + b22 + b32 ¡ j a j j b j cos µ which simplifies to a1 b1 + a2 b2 + a3 b3 = j a j j b j cos µ ) a ² b = j a j j b j cos µ So, cos µ = a²b jajjbj can be used to find the angle between two vectors a and b ALGEBRAIC PROPERTIES OF THE SCALAR PRODUCT The scalar product has the following algebraic properties for both 2-D and 3-D vectors: I I I a²b=b²a and a ² a = j a j2 a ² (b + c) = a ² b + a ² c (a + b) ² (c + d) = a ² c + a ² d + b ² c + b ² d à ! à ! These properties are proven in general by using vectors such as: a1 a2 a3 a= and b = b1 b2 b3 Be careful not to confuse the scalar product, which is the product of two vectors to give a scalar answer, with scalar multiplication, which is the product of a scalar and a vector to give a parallel vector They are quite different GEOMETRIC PROPERTIES OF THE SCALAR PRODUCT I If µ is the angle between vectors a and b then: a ² b = j a j j b j cos µ So, if µ is acute, cos µ > and ) a ² b > if µ is obtuse, cos µ < and ) a ² b < For non-zero vectors a and b: a ² b = , a and b are perpendicular a ² b = § j a j j b j , a and b are non-zero parallel vectors I I Two vectors form two angles µ and ® as in the diagram drawn The angle between two vectors is always taken as the smaller angle, so we take µ to be the angle between the two vectors with µ 180o a q a b Example 30 ! à and q = p²q à ! a ¡1 = à ² ¡1 ¡1 ! , find: a p ² q b ! = yellow 50 75 25 ) 95 100 50 75 25 95 50 75 25 95 100 50 75 25 100 magenta the angle between p and q p ² q jpjjqj ¡4 p = p 4+9+1 1+0+4 = 2(¡1) + 3(0) + (¡1)2 = ¡2 + ¡ = ¡4 cyan b cos µ = 95 If p = ¡1 ¡4 p 70 µ = arccos 100 à black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\403IB_HL-2_14.CDR Tuesday, 13 November 2007 9:34:00 AM PETERDELL ³ ¡4 p 70 ´ ¼ 119o IB_HL-2ed (404) 404 VECTORS IN AND DIMENSIONS (Chapter 14) Example 31 Since a and b are perpendicular, a ² b = µ ¶ µ ¶ ¡1 ² t =0 ) Find t such that µ ¶ ¡1 a= and µ ¶ ) t b= (¡1)(2) + 5t = ) ¡2 + 5t = ) 5t = and so t = are perpendicular If two vectors are perpendicular then their scalar product is zero Example 32 Find the measure of the angle between the lines 2x + y = and 3x ¡ 2y = 2x + y = has slope ¡ 21 and ) direction vector ¡2 ¶ which we call a µ ¶ 3x ¡ 2y = has slope µ and ) direction vector which we call b If the angle between the lines is µ, then cos µ = If a line has slope ab , it has direction ¡ ¢ vector ab a ²b ¡4 (1 £ 2) + (¡2 £ 3) p =p p = p jajjbj 1+4 4+9 13 ´ ³ ¼ 119:7o ) µ = arccos p¡4 65 When finding the angle between two lines we choose the acute angle, in this case 180o ¡ µ ) the angle is about 60:3o EXERCISE 14I µ ¶ µ ¶ µ ¶ ¡1 ¡2 For p = , q = and r = , find: q²p 2p ² 2p à ! yellow 3r ² q i²i c 95 50 j a j2 a²b+a²c b the angle between p and q i²i 75 95 100 50 95 100 magenta c f , find: a p ² q b d h find: i²j 100 ¡2 (i + j ¡ k) ² (2j + k) 50 ¡1 q ² (p + r) q²j b b²a e a ² (b + c) à ! and q = 75 25 a 95 100 50 75 25 Find: and c = c g ! 25 ¡1 If p = à a²b a²a à ! a d cyan , b= 75 For a = ¡1 1 q²r i²p ! b f à 25 a e black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\404IB_HL-2_14.CDR Monday, 12 November 2007 4:38:14 PM PETERDELL IB_HL-2ed (405) 405 VECTORS IN AND DIMENSIONS (Chapter 14) à Using a = ! a1 a2 a3 à , b= b1 b2 b3 ! à c1 c2 c3 and c = ! prove that a ² (b + c) = a ² b + a ² c Hence, prove that (a + b) ² (c + d) = a ² c + a ² d + b ² c + b ² d Find t given that these vectors are perpendicular: µ ¶ µ ¶ ¡2 a p= t and q = b µ c a= t t+2 ¶ µ ¡ 3t t and b = µ r= t t+2 à ¶ d a= ¡1 t ¶ µ and s = ! à and b = ¡4 2t ¡3 ¡4 ¶ ! For question find, where possible, the value(s) of t for which the given vectors are parallel Explain why in some cases the vectors can never be parallel à ! à ! à ! Show that a = à ! 1 a Show that à b Find t if t ¡2 ¡1 1 , b= à and ! ¡1 and c = ¡4 are mutually perpendicular ! are perpendicular à is perpendicular to 1¡t ¡3 ! 10 Consider triangle ABC in which A is (5, 1, 2), B(6, ¡1, 0) and C(3, 2, 0) Using scalar product only, show that the triangle is right angled 11 A(2, 4, 2), B(¡1, 2, 3), C(¡3, 3, 6) and D(0, 5, 5) are vertices of a quadrilateral a Prove that ABCD is a parallelogram ¡! ¡! b Find j AB j and j BC j What can be said about ABCD? ¡! ¡! c Find AC ² BD What property of figure ABCD has been found to be valid? 12 Find the measure of the angle between the lines: a x ¡ y = and 3x + 2y = 11 b y = x + and y = ¡ 3x d y = ¡ x and x ¡ 2y = c y + x = and x ¡ 3y + = 13 Find p ² q if: a j p j = 2, j q j = 5, µ = 60o Example 33 ³ ´ Find the form of all vectors which are ³ ´ perpendicular to ² ³ So, ³ ¡4 ¡4 ´ b j p j = 6, j q j = 3, µ = 120o = ¡12 + 12 = ´ is one such vector cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ) required vectors have form k black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\405IB_HL-2_14.CDR Friday, November 2007 9:17:05 AM PETERDELL ³ ¡4 ´ , k 6= IB_HL-2ed (406) 406 VECTORS IN AND DIMENSIONS (Chapter 14) 14 Find the form of all vectors which are perpendicular to: ³ ´ ³ ´ ³ ´ ³ ´ ¡1 ¡4 a b c d ¡2 ¡1 15 Find the angle ABC of triangle ABC for A(3, 0, 1), B(¡3, 1, 2) and C(¡2, 1, ¡1) ¡! ¡! Hint: To find the angle at B, use BA and BC ¡! ¡! What angle is found if BA and CB are used? ³ ´ e A q B C Example 34 B cm Use vector methods to determine the measure of angle ABC cm Notice that vectors used must both be away from B (or towards B) If this is not done you will be finding the exterior angle at B C A cm Placing the coordinate axes as illustrated, A is (2, 0, 0), B is (0, 4, 3) and C is (1, 4, 0) à ! à ! ¡! ¡! ¡4 and BC is ) BA is ¡3 ¡3 ¡! ¡! BA ² BC b and cos ABC = ¡! ¡! j BA j j BC j Z B = 2(1) + (¡4)(0) + (¡3)(¡3) p p + 16 + + + = p11 290 Y C A b = arccos ) ABC X ³ p11 290 ´ ¼ 49:8o 16 For the cube alongside with sides of length cm, find using vector methods: a the measure of angle ABS b the measure of angle RBP c the measure of angle PBS Q P S B C A D 17 KL, LM and LX are 8, and units long respectively P is the midpoint of KL Find, using vector methods: W a the measure of angle YNX b the measure of angle YNP magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 K cyan R black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\406IB_HL-2_14.CDR Monday, November 2007 9:54:34 AM PETERDELL Z Y X M N P L IB_HL-2ed (407) 407 VECTORS IN AND DIMENSIONS (Chapter 14) D(3,¡2,¡0) 18 For the tetrahedron ABCD: a find the coordinates of M b find the measure of angle DMA C(2,¡2,¡2) M A(2,¡1,¡1) 19 B(1,¡3,¡1) a Find t if 2i + tj + (t ¡ 2)k and ti + 3j + tk are perpendicular à ! à ! à ! b Find r, s and t if a = perpendicular 2 r , b = and c = s t are mutually 20 Find the angle made by: à ! à a the X-axis and b a line parallel to the Y -axis and ¡1 ! 21 Find three vectors a, b and c such that a 6= and a ² b = a ² c but b 6= c 22 Show, using j x j = x ² x, that: a j a + b j2 + j a ¡ b j2 = j a j2 + j b j2 b j a + b j2 ¡ j a ¡ b j2 = a ² b 23 a and b are the position vectors of two distinct points A and B, neither of which is the origin Show that if j a + b j = j a ¡ b j then a is perpendicular to b using: a a vector algebraic method b a geometric argument 24 If j a j = and j b j = 4, find (a + b) ² (a ¡ b) 25 Explain why a ² b ² c is meaningless J THE VECTOR PRODUCT OF TWO VECTORS We have now seen how the scalar product of two vectors results in a scalar The second form of product between two vectors is the vector product or vector cross product, and this results in a vector The vector product arises when we attempt to find a vector which is perpendicular to two other known vectors Following is such an attempt: à ! à ! à ! x y z Suppose x = is perpendicular to both a = ½ ) ½ ) a1 x + a2 y + a3 z = b1 x + b2 y + b3 z = a1 a2 a3 and b = b1 b2 b3 fas dot products are zerog a1 x + a2 y = ¡a3 z (1) b1 x + b2 y = ¡b3 z (2) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 We will now try to solve these two equations to get expressions for x and y in terms of z To eliminate x, we multiply (1) by ¡b1 and (2) by a1 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\407IB_HL-2_14.CDR Friday, November 2007 9:19:38 AM PETERDELL IB_HL-2ed (408) 408 VECTORS IN AND DIMENSIONS (Chapter 14) ¡a1 b1 x ¡ a2 b1 y = a3 b1 z a1 b1 x + a1 b2 y = ¡a1 b3 z (a1 b2 ¡ a2 b1 )y = (a3 b1 ¡ a1 b3 )z Adding these gives y a3 b1 ¡ a1 b3 = z a1 b2 ¡ a2 b1 and so ) y = (a3 b1 ¡ a1 b3 )t and z = (a1 b2 ¡ a2 b1 )t for any non-zero t a1 x = ¡a3 (a1 b2 ¡ a2 b1 )t ¡ a2 (a3 b1 ¡ a1 b3 )t ) a1 x = (¡a1 a3 b2 + a2 a3 b1 ¡ a2 a3 b1 + a1 a2 b3 )t ) a1 x = a1 (a2 b3 ¡ a3 b2 )t ) x = (a2 b3 ¡ a3 b2 )t Substituting into (1), So, the simplest vector perpendicular to both a and b is obtained by letting t = à ! In this case this gives x = a2 b3 ¡ a3 b2 a3 b1 ¡ a1 b3 a1 b2 ¡ a2 b1 We call this vector the cross product of a and b, and it is written as a £ b à a£b= a2 b3 ¡ a3 b2 a3 b1 ¡ a1 b3 a1 b2 ¡ a2 b1 ¯ ¯ i ¯ Notice also that: a £ b = ¯ a1 ¯ b1 ! j a2 b2 ¯ ¯a = ¯ b2 ¯ a3 ¯ b3 ¯ ¯ ¯a i ¡ ¯ b1 ¯ a3 ¯ b3 ¯ ¯ ¯a j + ¯ b1 ¯ a2 ¯ b2 ¯ ¯ k ¯ ¯ a3 ¯ b3 ¯ This form is known as a £ determinant After finding a¡£¡b, check that your answer is perpendicular to both a and b Example 35 à If a = ¡1 à and b = ! ¡1 a£b ¯ ¯ ¯ i j k ¯ ¯ ¡1 ¯ =¯ ¯ ¯ ¡1 ¯ ¯ ¯ ¯ ¯ ¡1 ¯ ¯ = ¯ ¯ i ¡ ¯ ¡1 ! , k ¯ ¡1 ¯ ¯ ¯ ¯ j + ¯ ¡1 ¯ 3¯ 2¯ k = 14i ¡ 7j + 7k find a £ b Example 36 ! à ! , b= b£c ¯ ¯ ¯ i j k¯ ¯1 3¯ =¯ ¯ ¯2 4¯ ¯ ¯ ¯ ¯2 3¯ ¯1 = ¯0 4¯ i ¡ ¯2 a à ! and c = ¯ 3¯ 4¯ ¯ ¯1 j + ¯2 ¯ 2¯ 0¯ k yellow 50 75 25 ² ¡4 ! = 16 + + = 22 95 100 50 75 25 ¡1 = 95 50 75 25 95 100 50 75 25 100 magenta a ² (b £ c) à ! à b = 8i + 2j ¡ 4k cyan a b£c b a ² (b £ c) find: 95 For a = ¡1 100 à black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\408IB_HL-2_14.CDR Monday, November 2007 10:06:40 AM PETERDELL IB_HL-2ed (409) 409 VECTORS IN AND DIMENSIONS (Chapter 14) Example 37 à ¡1 Find all vectors perpendicular to both a = ¯ ¯i ¯ a £ b = ¯1 ¯1 ¯ ¯ ¯2 = ¯0 k ¯ ¯ ¡1 ¯ ¡3 ¯ j ¯ ¡1 ¯ ¡3 ¯ ¯ ¯1 i ¡ ¯1 ! à and b = ¯ ¡1 ¯ ¡3 ¯ ¯ ¯1 j + ¯1 ¡3 ! ¯ 2¯ 0¯ k = ¡6i + 2j ¡ 2k which is ¡2(3i ¡ j + k) ) the vectors have form k(3i ¡ j + k), where k is any non-zero real number Example 38 Find a direction vector which is perpendicular to the plane passing through the points A(1, ¡1, 2), B(3, 1, 0) and C(¡1, 2, ¡3) à ! à ! ¡2 ¡! ¡! , AC = AB = v ¡2 B The vector v must be perpendicular to ¡! ¡! both AB and AC C A ¯ ¯ i ¯ Thus v = ¯ ¯ ¡2 ¯ k ¯ ¯ ¡2 ¯ ¡5 ¯ j ¡5 ¯ ¯2 = ¯3 ¯ ¯ i ¡ ¯ ¡2 ¯ ¡2 ¯ ¡5 ¯ ¯ ¡2 ¯ ¡5 ¯ ¯ ¯ j + ¯ ¡2 ¯ 2¯ 3¯ k = ¡4i + 14j + 10k which is ¡2(2i ¡ 7j ¡ 5k) Thus any non-zero multiple of (2i ¡ 7j ¡ 5k) will EXERCISE 14J.1 Calculate: à ! à ¡3 a £ ¡2 ! à b à ! Given a = à and b = ¡1 ¡1 ¡1 ! à £ ¡1 ¡2 ! c d (i + j ¡2k) £ (i ¡ k) (2i ¡ k) £ (j + 3k) ! , find a £ b and hence determine a ² (a £ b) and b ² (a £ b) What has been verified from these results? If i, j and k are the unit vectors parallel to the coordinate axes: a find i £ i, j £ j, and k £ k b find i £ j and j £ i, j £ k and k £ j, and i £ k and k £ i cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 What you suspect a £ a equals for any vector a? What you suspect is the relationship between a £ b and b £ a? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\409IB_HL-2_14.CDR Monday, November 2007 10:13:49 AM PETERDELL IB_HL-2ed (410) 410 VECTORS IN AND DIMENSIONS (Chapter 14) à Using a = a1 a2 a3 ! à and b = b1 b2 b3 ! , prove that: a a £ a = for all space vectors a b a £ b = ¡b £ a for all space vectors a and b à ! For a = a à , b= ¡1 b£c ! à and c = ¡2 ! find: a ² (b £ c) b ¯ ¯1 ¯2 ¯ ¯0 c ¡1 Explain why the answers to b and c are the same ¯ ¯ ¯ ¯ ¡2 ¯ Repeat for vectors of your choosing If a = i + 2k, b = ¡j + k and c = 2i ¡ k, determine: a a£b b a£c c (a £ b) + (a £ c) a £ (b + c) d What you suspect to be true from 7? Check with vectors a, b and c of your choosing à Prove that a £ (b + c) = a £ b + a £ c using a = à ! c1 c2 c3 c= a1 a2 a3 ! à , b= b1 b2 b3 ! and 10 Use a £ (b + c) = (a £ b) + (a £ c) to prove that (a + b) £ (c + d) = (a £ c) + (a £ d) + (b £ c) + (b £ d) Notice that the order of the vectors must be maintained as x £ y = ¡y £ x 11 Use the properties found in and to simplify: a c a £ (a + b) (a + b) £ (a ¡ b) (a + b) £ (a + b) 2a ² (a £ b) b d 12 Find all vectors perpendicular to both: à ! à ! ¡1 a c à 1 and b i + j and i ¡ j ¡ k ¡1 ! à ! and i ¡ j ¡ k and 2i + 2j ¡ 3k d à 13 Find all vectors perpendicular to both a = ¡1 ! à and b = ¡2 ! Hence find a vector of length units which is perpendicular to both a and b 14 Find a direction vector which is perpendicular to the plane passing through the points: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a A(1, 3, 2), B(0, 2, ¡5) and C(3, 1, ¡4) b P(2, 0, ¡1), Q(0, 1, 3) and R(1, ¡1, 1) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\410IB_HL-2_14.CDR Friday, November 2007 9:20:41 AM PETERDELL IB_HL-2ed (411) VECTORS IN AND DIMENSIONS (Chapter 14) 411 In the above exercise you should have observed the following properties of vector cross products: I I I a£ a£ a£ i.e., b is a vector which is perpendicular to both a and b a = for all space vectors a b = ¡b £ a for all space vectors a and b, a £ b and b £ a have the same length but opposite direction ¯ ¯ ¯ a1 a2 a3 ¯ ¯b b b ¯ a ² (b £ c) = ¯ ¯ and is called the scalar triple product ¯ c1 c2 c3 ¯ I I a £ (b + c) = (a £ b) + (a £ c) and hence (a + b) £ (c + d) = (a £ c) + (a £ d) + (b £ c) + (b £ d) DIRECTION OF a¡×¡b Z We have already observed that as a £ b = ¡b £ a then a £ b and b £ a are oppositely directed Z k However, what is the direction of each? Consider i £ j and j £ i In the last Exercise, we saw that i £ j = k and j £ i = ¡k X i j Y i -k j Y X In general, the direction of x £ y is determined by the right hand rule: To determine the direction of x £ y use your right hand If your fingers turn from x to y then your thumb points in the direction of x £ y x´y x y THE LENGTH OF a¡×¡b a2 b3 ¡ a3 b2 p (a2 b3 ¡ a3 b2 )2 + (a3 b1 ¡ a1 b3 )2 + (a1 b2 ¡ a2 b1 )2 As a £ b = @ a3 b1 ¡ a1 b3 A, j a £ b j = a1 b2 ¡ a2 b1 However, another very useful form of the length of a £ b exists This is: j a £ b j = j a j j b j sin µ Proof: where µ is the angle between a and b j a j2 j b j2 sin2 µ We start with 2 = j a j j b j (1 ¡ cos2 µ) = j a j2 j b j2 ¡ j a j2 j b j2 cos2 µ 2 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 = j a j j b j ¡ (a ² b)2 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\411IB_HL-2_14.CDR Monday, November 2007 10:19:24 AM PETERDELL IB_HL-2ed (412) 412 VECTORS IN AND DIMENSIONS (Chapter 14) = (a12 + a22 + a32 )(b12 + b22 + b32 ) ¡ (a1 b1 + a2 b2 + a3 b3 )2 which on expanding and then factorising gives = (a2 b3 ¡ a3 b2 )2 + (a3 b1 ¡ a1 b3 )2 + (a1 b2 ¡ a2 b1 )2 = j a £ b j2 and so j a £ b j = j a j j b j sin µ fas sin µ > 0g Immediate consequences are: ² If u is a unit vector in the direction of a £ b then a £ b = j a j j b j sin µ u In some texts this is the geometric definition of a £ b ² If a and b are non-zero vectors, then a £ b = , a is parallel to b EXERCISE 14J.2 a Find i £ k and k £ i using the original definition of a £ b b Check that the right-hand rule correctly gives the direction of i £ k and k £ i c Check that a £ b = j a j j b j sin µ u could be used to find i £ k and k £ i X à ! à ! ¡1 Consider a = ¡1 and b = Z k i j Y a Find a ² b and a £ b b Find cos µ using a ² b = j a j j b j cos µ 2 c Find sin µ using sin µ + cos µ = d Find sin µ using j a £ b j = j a j j b j sin µ Prove the property: “If a and b are non-zero vectors then a £ b = , a is parallel to b.” A(2,¡3,-1) O is the origin Find: ¡! ¡! ¡! ¡! ¡! ¡! a OA and OB b OA £ OB and j OA £ OB j : O ¡! ¡! c Explain why the area of triangle OAB is 12 j OA £ OB j : B(-1,¡1,¡2) A, B and C are distinct points with non-zero position vectors a, b and c respectively ¡! ¡! a If a £ c = b £ c, what can be deduced about OC and AB? b If a + b + c = 0, what relationship exists between a £ b and b £ c? c If c 6= and b £ c = c £ a, prove that a + b = kc for some scalar k AREAS AND VOLUMES TRIANGLES a If a triangle has defining vectors a and b then its area is 12 j a £ b j units2 Proof: Area = q b £ product of two sides £ sine of included angle = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\412IB_HL-2_14.CDR Friday, November 2007 9:21:17 AM PETERDELL 2 j a j j b j sin µ ja £ bj IB_HL-2ed (413) 413 VECTORS IN AND DIMENSIONS (Chapter 14) Example 39 Find the area of ¢ABC given A(¡1, 2, 3), B(2, 1, 4) and C(0, 5, ¡1) ¯ ¯i ¡! ¡! ¯ AB £ AC = ¯ ¯1 ¯ ¯ ¡1 =¯ j ¡1 ¯ k ¯ ¯ ¯ ¡4 ¯ ¯ ¯ ¡4 ¯ ¯ ¯3 i ¡ ¯1 ¯ ¯ ¡4 ¯ ¯ ¯3 j + ¯1 ¯ ¡1 ¯ ¯ k = i + 13j + 10k ) area = p ¡! ¡! j AB £ AC j = 12 + 169 + 100 p = 12 270 units2 PARALLELOGRAMS If a parallelogram has defining vectors a and b then its area is j a £ b j units2 a b The proof follows directly from that of a triangle as the parallelogram consists of two congruent triangles with defining vectors a and b PARALLELEPIPED (EXTENSION) If a parallelepiped has defining vectors a, b and c then its volume is determinant ¯¯ ¯¯ ¯¯ a1 a2 a3 ¯¯ ¯¯ ¯¯ j a ² (b £ c ) j = ¯¯¯¯ b1 b2 b3 ¯¯¯¯ units3 ¯¯ c1 c2 c3 ¯¯ modulus a c b modulus Volume = (area of base) £ (perp height) = j b £ c j £ AN Proof: f c N magenta yellow 95 50 75 25 95 100 50 25 95 100 50 75 25 95 100 50 75 25 B b fas sin µ = AN g jaj = j a j j b £ c j sin µ = j a j j b £ c j cos Á where Á is the angle between a and b £ c = j a ² (b £ c ) j as cos Á > C O cyan = j b £ c j £ j a j sin µ 75 a q a 100 b´ c f A black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\413IB_HL-2_14.CDR Tuesday, 29 January 2008 10:56:04 AM PETERDELL IB_HL-2ed (414) 414 VECTORS IN AND DIMENSIONS (Chapter 14) TETRAHEDRON (EXTENSION) If a tetrahedron has defining vectors a, b and c then its volume is ¯¯ ¯¯ ¯¯ a1 a2 a3 ¯¯ ¯¯ ¯¯ 1 ¯¯ b b2 b3 ¯¯¯¯ units3 : j a ² (b £ c) j = ¯¯ ¯¯ c1 c2 c3 ¯¯ a c b Volume = 13 (area of base) £ (perp height) Proof: b¡´¡c f a A = £ = j b £ c j j a j sin µ j a j j b £ c j cos Á C j b £ c j £ AN fas sin µ = AN g jaj a f c = q N where Á is the angle between a and b £ c O b = B j a ² (b £ c ) j as cos Á > Example 40 Find the volume of the tetrahedron with vertices P(0, 0, 1), Q(2, 3, 0), R(¡1, 2, 1) and S(1, ¡2, 4) ¡ ! PQ = à ¡1 ! ¡ ! , PR = à ¡1 ! ¡ ! , PS = à ¡2 ! are the defining vectors from P Q ¯¯ ¯¯ ¯¯ ¡1 ¯¯ ¯¯ ¯¯ ¡1 ¯¯ ) volume = ¯¯ ¯¯ ¡2 ¯¯ ¯ ¯ ¯ ¯ ¯ ¡1 ¯ ¯ 0¯ = 16 ¯ ¯ ¡2 ¯ ¡ ¯ = R P ¯ ¯ ¯ ¡1 0¯ ¡ ¯ ¯ S ¯¯ ¯¯ ¡2 ¯¯ j12 + ¡ 0j = 12 units3 EXERCISE 14J.3 Calculate the area of triangle ABC for: a A(2, 1, 1), B(4, 3, 0), C(1, 3, ¡2) b A(0, 0, 0), B(¡1, 2, 3) and C(1, 2, 6) c A(1, 3, 2), B(2, ¡1, 0) and C(1, 10, 6) Calculate the area of parallelogram ABCD for A(¡1, 2, 2), B(2, ¡1, 4) and C(0, 1, 0) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ABCD is a parallelogram where A is (¡1, 3, 2), B(2, 0, 4) and C(¡1, ¡2, 5) Find the a coordinates of D b area of ABCD black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\414IB_HL-2_14.CDR Tuesday, 29 January 2008 10:58:03 AM PETERDELL IB_HL-2ed (415) 415 VECTORS IN AND DIMENSIONS (Chapter 14) ABCD is a tetrahedron with A(1, ¡1, 0), B(2, 1, ¡1), C(0, 1, ¡3) and D(¡1, 1, 2) Find the: a volume of the tetrahedron b total surface area of the tetrahedron A(3, 0, 0), B(0, 1, 0) and C(1, 2, 3) are the vertices of a parallelepiped which are adjacent to another vertex at O(0, 0, 0) Find the: a coordinates of the other four vertices b measure of ]ABC c volume of the parallelepiped If A(¡1, 1, 2), B(2, 0, 1) and C(k, 2, ¡1) are three points in space, find k if the p area of triangle ABC is 88 units2 A, B and C are three points with position vectors a, b and c respectively Find a formula for S, the total surface area of the tetrahedron OABC Three distinct points A, B and C have position vectors a, b and c respectively Prove that A, B and C are collinear , (b ¡ a) £ (c ¡ b) = TEST FOR COPLANAR POINTS Four points in space are either coplanar or form the vertices of a tetrahedron If they are coplanar, the volume of the tetrahedron is zero So: If four points A, B, C and D have position vectors a, b, c and d respectively then A, B, C and D are coplanar , (b ¡ a) ² (c ¡ a) £ (d ¡ a) = Example 41 Are the points A(1, 2, ¡4), B(3, 2, 0), C(2, 5, 1) and D(5, ¡3, ¡1) coplanar? à ¡! b ¡ a = AB = ¡! d ¡ a = AD = à 3¡1 2¡2 ¡ ¡4 ! à ! = 5¡1 ¡3 ¡ ¡1 ¡ ¡4 ! à = ¡! c ¡ a = AC = ¡5 à 2¡1 5¡2 ¡ ¡4 ! à ! = ! ¯ ¯2 ¯ and (b ¡ a) ² (c ¡ a) £ (d ¡ a) = ¯ ¯4 ¡5 ¯ 4¯ ¯ 5¯ 3¯ = 2(9 + 25) + 4(¡5 ¡ 12) =0 ) A, B, C and D are coplanar Are these points coplanar? a A(1, 1, 2), B(2, 4, 0), C(3, 1, 1) and D(4, 0, 1) b P(2, 0, 5), Q(0, ¡1, 4), R(2, 1, 0), S(1, 1, 1) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 10 Find k given that A(2, 1, 3), B(4, 0, 1), C(0, k, 2), D(1, 2, ¡1) are coplanar black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\415IB_HL-2_14.CDR Monday, November 2007 10:44:22 AM PETERDELL IB_HL-2ed (416) 416 VECTORS IN AND DIMENSIONS (Chapter 14) REVIEW SET 14A (Mainly 2-D) Using a scale of cm represents 10 units, sketch a vector to represent: a an aeroplane taking off at an angle of 8o to the runway with a speed of 60 m s¡1 b a displacement of 45 m in a direction of 060o Copy the given vectors and find geometrically: a x+y b y ¡ 2x ¡ ! ¡! a PR + RQ Find a single vector which is equal to: y x ¡! ¡! ¡! b PS + SQ + QR Dino walks for km in the direction 246o and then for km in the direction 096o Find his displacement from his starting point ¡! ¡! ¡! ¡! ¡! Simplify a AB ¡ CB b AB + BC ¡ DC What geometrical facts can be deduced from the equations: ¡! ¡! ¡! ¡! b AB = 2AC ? a AB = 12 CD Construct vector equations for: a b r p l m k Q P n j q ¡! ¡! ¡! In the figure alongside OP = p, OR = r and RQ = q M p If M and N are midpoints of the sides as shown, find in terms of p, q and r: O ¡! ¡ ! ¡! ¡! r a OQ b PQ c ON d MN R ³ ´ ³ ´ ³ ´ Draw arrow diagrams to represent: a b c ¡5 ¡4 N q ³ 10 If p = a ¡3 ´ ³ ´ ³ ´ , q = ¡4 and r = find: q ¡ 3r b 2p + q c p¡q+r ¡ ! ³ ¡4 ´ ¡! ³ ¡1 ´ ¡ ! ³ ´ ¡ ! 11 If PQ = , RQ = and RS = ¡3 , find SP ³ ´ 12 If r = p O ¡3 ´ find: a j r j A M q magenta yellow 95 100 50 75 25 95 50 75 25 95 100 50 25 95 100 50 75 25 14 If p = ´ ³ ´ ³ ´ ¡3 , q = and r = , find x if: ¡4 75 ³ c jr + sj d j 2s ¡ r j C B cyan b jsj BC is parallel to OA and is twice its length Find, in terms of p and q, vector expressions ¡! ¡! for a AC b OM 100 13 ³ and s = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\416IB_HL-2_14.CDR Monday, November 2007 10:48:56 AM PETERDELL a b p ¡ 3x = 2q ¡ x = r IB_HL-2ed (417) 417 VECTORS IN AND DIMENSIONS (Chapter 14) 15 Use vectors to show that WYZX is a parallelogram if X is (¡2, 5), Y(3, 4), W(¡3, ¡1), and Z(4, 10) ³ ´ ³ ´ ³ ´ ¡2 13 16 Find scalars r and s such that r + s ¡4 = ¡24 ¡! ¡! 17 [AB] and [CD] are diameters of a circle centre O If OC = q and OB = r, find: ¡! ¡! a DB in terms of q and r b AC in terms of q and r What can be deduced about [DB] and [AC]? REVIEW SET 14B (Mainly 3-D) Given P(2, ¡5, 6) and Q(¡1, 7, 9), find: a c b the position vector of Q from P the distance from P to the x-axis à ! à ! For m = ¡3 ¡4 , n= the distance from P to Q à and p = ¡1 ! , find: m¡n+p b 2n ¡ 3p à ! à ! ¡6 ¡! ¡! ¡! and AC = , find CB If AB = ¡7 a jm + pj c ¡3 à Find m and n if m n ! à and ¡12 ¡20 ! are parallel vectors Prove that P(¡6, 8, 2), Q(4, 6, 8) and R(19, 3, 17) are collinear Hence find the ratio in which Q divides [PR] à Find t if ¡4 t+2 t ! à and t 1+t ¡3 ! are perpendicular vectors à Determine the angle between ¡4 ! à and ¡1 ! E F H G cm Find the measure of angle GAC in the rectangular box alongside Use vector methods D A For P(2, 3, ¡1) and Q(¡4, 4, 2) find: ¡! a PQ b the distance between P and Q à ! à ! à ! 10 For p = ¡1 ¡1 , q= a p²q 1 and r = b p + 2q ¡ r c cm c B C cm the midpoint of [PQ] find: the angle between p and r cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 11 Find all angles of the triangle with vertices K(3, 1, 4), L(¡2, 1, 3) and M(4, 1, 3) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\417IB_HL-2_14.CDR Monday, November 2007 10:54:55 AM PETERDELL IB_HL-2ed (418) 418 VECTORS IN AND DIMENSIONS (Chapter 14) à 12 Find the angle between ¡2 ! à ! and 13 If A(4, 2, ¡1), B(¡1, 5, 2), C(3, ¡3, c) are vertices of triangle ABC which is right angled at B, find the possible values of c 14 Explain why: a a ² b ² c is meaningless b you not need brackets for a ² b £ c µ4¶ µ ¶ k 15 Find k if the following are unit vectors: a b k k REVIEW SET 14C (Mainly 2-D) ³ If p = ¡2 ´ ³ ´ ³ ´ ¡1 ¡3 , q= and r = find: a p ² q ³ Using p = ¡2 b q ² (p ¡ r) ´ ³ ´ ³ ´ ¡2 , q= and r = ¡3 verify that: p ² (q ¡ r) = p ² q ¡ p ² r ¶ µ ¶ µ t2 + t and are perpendicular Determine the value of t if ¡ 2t ¡2 b is a right angle Given A(2, 3), B(¡1, 4) and C(3, k), find k if BAC ³ ´ ¡4 Find all vectors which are perpendicular to the vector Find the measure of all angles of triangle KLM for K(¡2, 1), L(3, 2) and M(1, ¡3) Find the angle between the two lines with equations 4x ¡ 5y = 11 and 2x +3y = In this question you may not assume any diagonal properties of parallelograms ¡! OABC is a parallelogram with OA = p ¡! and OC = q M is the midpoint of AC A B M a Find in terms of p and q: ¡! ¡! O C i OB ii OM b Show using a only that O, M and B are collinear and M is the midpoint of OB cyan magenta yellow C Q P O 95 A 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 AP and BQ are altitudes of triangle ABC ¡! ¡! ¡! Let OA = p, OB = q and OC = r ¡! ¡! a Find vector expressions for AC and BC in terms of p, q and r b Deduce that q ² r = p ² q = p ² r c Hence prove that OC is perpendicular to AB black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\418IB_HL-2_14.CDR Monday, November 2007 10:58:41 AM PETERDELL B IB_HL-2ed (419) 419 VECTORS IN AND DIMENSIONS (Chapter 14) à 10 If a = ¡3 ! à and b = ¡1 ! a b , find: 2a ¡ 3b x if a ¡ 3x = b p ¡! ¡! 11 If OA = a, OB = b, j a j = 3, j b j = and a £ b = i + 2j ¡ 3k find: a a²b b the area of triangle OAB c the volume of tetrahedron OABC if C is the point (1, ¡1, 2) REVIEW SET 14D (Mainly 3-D) If a = 3i ¡ j + 2k and b = i ¡ 2k find: a 3a ¡ 2b b jaj If k is a scalar and a is any 3-dimensional vector, prove that j ka j = j k j j a j : P(¡1, 2, 3) and Q(4, 0, ¡1) are two points in space Find: ¡ ! ¡! a PQ b the angle that PQ makes with the X-axis c the coordinates of R if R divides QP in the ratio : The triangle with vertices P(¡1, 2, 1), Q(0, 1, 4) and R(a, ¡1, ¡2) has an area p of 118 units2 Find a M is (¡1, 3, 4) and N is (2, 0, 1) Find: p a the coordinates of two points on MN such that their distance from N is units ¡! b a vector in the direction of MN with length units Show that A(1, ¡3, 2), B(2, 0, 1) and C(¡1, ¡9, 4) are collinear and hence find the ratio in which C divides BA Find the coordinates of the point which divides the line segment joining A(¡2, 3, 5) to B(3, ¡1, 1) externally in the ratio : Find the volume of tetrahedron ABCD given A(3, 1, 2), B(¡1, 2, 1), C(¡2, 0, 3) and D(4, 3, ¡1) S divides AB externally in the ratio : and T divides CS internally in the ratio : ¡! If A, B and C have position vectors a, b and c respectively, find t = OT 10 Find a unit vector parallel to i + rj + 2k and perpendicular to 2i + 2j ¡ k à ! 11 Given j u j = and ¡3 ¡4 j v j = and u £ v = of u ² v find the possible values 12 If v = i + 2j ¡ 3k, u = 2i + 2j + 3k, and w = i + (2 ¡ t)j + (t + 1)k, find the value of t such that u, v and w are coplanar M S P 13 Determine the measure of angle QDM given that M is the midpoint of [PS] of the rectangular prism Q A cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 B black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_14\419IB_HL-2_14.CDR Thursday, 11 March 2010 10:30:15 AM PETER 10 cm R cm D cm C IB_HL-2ed (420) 420 VECTORS IN AND DIMENSIONS (Chapter 14) REVIEW SET 14E Show that A(¡2, ¡1, 3), B(4, 0, ¡1) and C(¡2, 1, ¡4) are vertices of an isosceles triangle ! à ! à ! à s¡1 r+1 t Find scalars r, s and t if 4s 3r r = r ¡1 s + Find two points on the Z-axis which are units from P(¡4, 2, 5) à ! à ! ¡1 ¡2 If a = ¡1 and b = , find x if: a a ¡ x = 2b b b ¡ 2x = ¡a Find a and b if J(¡4, 1, 3), K(2, ¡2, 0) and L(a, b, 2) are collinear Given p = 2i ¡ j + 4k and q = ¡i ¡4j + 2k, find: jp ² qj a b the angle between p and q à ! à ! r a Find r and s if ¡5 10 s and are parallel b Find a vector of length units which is parallel to 3i ¡ 2j + k k a Find k given that @ p2 A is a unit vector à ¡k b Find the vector which is units long and has the opposite direction to à If u = ¡4 ! à and v = ¡1 ¡2 ! u²v a ¡1 ! , find: b the angle between u and v D(1,-4,¡3) 10 For the given tetrahedron, find the measure of angle DMC A(-2,¡1,-3) C(3,-3,¡2) M B(2,¡5,-1) 11 Consider the points A(0, 1, 1), B(¡2, 2, 3), C(1, ¡1, 2) and D(¡1, 3, k) a Find a vector of length 10 units which is perpendicular to the plane defined by points A, B and C b Find the area of triangle ABC c If point D lies on plane ABC, find k à 12 a Find t given that 2¡t t ! à and t t+1 ! are perpendicular cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b Show that K(4, 3, ¡1), L(¡3, 4, 2) and M(2, 1, ¡2) are vertices of a right angled triangle black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_14\420IB_HL-2_14.CDR Monday, 12 November 2007 4:39:18 PM PETERDELL IB_HL-2ed (421) 15 Chapter Complex numbers Contents: A B C D E Complex numbers as 2-D vectors Modulus, argument, polar form De Moivre’s Theorem Roots of complex numbers Further complex number problems cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 15A Review set 15B Review set 15C black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\421IB_HL-2_15.CDR Monday, November 2007 1:45:30 PM PETERDELL IB_HL-2ed (422) 422 COMPLEX NUMBERS (Chapter 15) A COMPLEX NUMBERS AS 2-D VECTORS Recall from Chapter that a complex number can be written in Cartesian form as z = a+bi where a = Re(z) and b = Im(z) are both real numbers Hence, there exists a one-to-one relationship between any complex number a + bi and any point (a, b) in the Cartesian Plane When we view points in a plane as complex numbers, we refer to the plane as the complex plane or the Argand plane The x-axis is called the real axis and the y-axis is called the imaginary axis All real numbers with b = lie on the real axis, and all purely imaginary numbers with a = lie on the imaginary axis The origin (0, 0) lies on both axes and it corresponds to z = 0, a real number Complex numbers that are neither real nor pure imaginary (a and b both 6= 0) lie in one of the four quadrants Now recall from Chapter 14 that any point P in the Cartesian plane corresponds uniquely to ¡ ! a vector The position vector of the point P is OP We know that vectors have magnitude and direction, so we can in turn attribute a magnitude and direction to complex numbers We can apply the vector operations of addition, subtraction, and scalar multiplication to give the correct answers for these operations with complex numbers For those studying the option “sets, relations and groups” this means there is an isomorphic relationship between complex numbers and 2-D vectors under the binary operation of + ³ ´ ³ ´ ³ ´ a c a+c + = for vectors b d b+d Note: and (a + bi) + (c + di) = (a + c) + (b + d)i ´ ³ a+c ´ (a + c) + (b + d)i so b+d for complex numbers The plane of complex numbers (also called the complex plane or Argand plane), has a horizontal real axis and a vertical imaginary axis When we illustrate complex numbers on the Argand plane, we call it an Argand diagram P(2, 3) R Q(4,-1) cyan magenta yellow represents ¡ i represents ¡3i represents ¡3 ¡ i 95 100 50 95 100 50 represents + 3i R is the real axis, I is the imaginary axis 75 25 95 100 50 75 25 95 100 50 75 25 R(0,-3) 75 S(-3,-1) 25 I ¡! ³ ´ OP = ¡! ³ ´ OQ = ¡1 ¡! ³ ´ OR = ¡3 ¡! ³ ¡3 ´ OS = ¡1 For example: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\422IB_HL-2_15.CDR Thursday, November 2007 3:45:07 PM PETERDELL IB_HL-2ed (423) 423 COMPLEX NUMBERS (Chapter 15) In general, I ¡! ³ x ´ OP = y represents x + yi P(x, y) y R x Example I zc Illustrate the positions of: z1 = 3, z2 = + 3i, z3 = 5i, z4 = ¡4 + 2i, z5 = ¡3 ¡ i and z6 = ¡2i in the complex plane zx zv -4 zz zb R zn -3 Example If z1 = + i and z2 = ¡ 4i find algebraically and vectorially: b z1 ¡ z2 a z1 + z2 a z1 + z2 = + i + ¡ 4i = ¡ 3i I zz-zx -zx zz b z1 ¡ z2 = + i ¡ (1 ¡ 4i) = + i ¡ + 4i = + 5i R zx zz+zx Reminder: Draw z1 first and at its arrow end draw z2 z1 + z2 goes from the start of z1 to the end of z2 Example If z = + 2i and w = ¡ i, find both algebraically and vectorially: a 2z + w b z ¡ 2w a 2z + w = 2(1 + 2i) + ¡ i = + 4i + ¡ i = + 3i w z z 2z+w cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 5 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\423IB_HL-2_15.CDR Monday, November 2007 3:04:15 PM PETERDELL IB_HL-2ed (424) 424 COMPLEX NUMBERS (Chapter 15) b z ¡ 2w = + 2i ¡ 2(3 ¡ i) = + 2i ¡ + 2i = ¡5 + 4i -w -w z-2w z -5 EXERCISE 15A.1 On an Argand diagram, illustrate the complex numbers: b z2 = ¡1 + 2i a z1 = e z5 = ¡ i d z4 = ¡6i c f z3 = ¡6 ¡ 2i z6 = 4i If z = + 2i and w = ¡ i, find both algebraically and vectorially: a z+w b z¡w c 2z ¡ w d w ¡ 3z If z1 = ¡ i and z2 = + 3i, find both algebraically and vectorially: z1 + a z1 + b z1 + 2i c z2 + 12 z1 d If z is any complex number, explain with illustration how to find geometrically: d 3i ¡ z a 3z b ¡2z c z¤ z + z¡4 e 2¡z f z¤ + i g h REPRESENTING CONJUGATES If z = x + iy, then z ¤ = x ¡ iy ¡ ! ³x´ ¡! ³ x ´ This means that if OP = y represents z, then OQ = ¡y represents z ¤ For example: ¡¡! OP1 ¡¡! OQ1 ¡¡! OP3 ¡¡! OQ3 I Pz (2, 4) Qc (0, 3) represents + 4i and Qv (-3, 2) represents ¡ 4i represents ¡3i and Px (4, 1) Pv (-3,-2) represents 3i etc Qx (4,-1) Pc (0,-3) R Qz (2,-4) It is clear that: ¡ ! ¡! ¡! ¡! If z is OP, its conjugate z ¤ is OQ where OQ is a reflection of OP in the real axis EXERCISE 15A.2 Show on an Argand diagram: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a z = + 2i and its conjugate z ¤ = ¡ 2i b z = ¡2 + 5i and its conjugate z ¤ = ¡2 ¡ 5i black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\424IB_HL-2_15.CDR Monday, November 2007 3:07:01 PM PETERDELL IB_HL-2ed (425) COMPLEX NUMBERS (Chapter 15) If z = ¡ i we can add z + z ¤ as shown in the diagram We notice that z + z ¤ is which is real Explain, by illustration, that z + z ¤ is always real I z+z* z* z Explain, by illustration, that z ¡ z ¤ is always purely imaginary or zero What distinguishes these two cases? 425 R z* If z is real, what is z ¤ ? Given that (a + i)(3 + bi) = 40 ¡ 74i find the values of a and b where a and b Z : B MODULUS, ARGUMENT, POLAR FORM MODULUS The modulus of the complex number z = a + bi is the length of the corresponding ³ ´ a vector b We denote the modulus of z by j z j The modulus of the complex number z = a + bi is the real number j z j = p a2 + b2 Notice that if z = a + bi then j z j gives the distance of the point (a, b) from the origin This is consistent with the definition in Chapter of j x j for x R We stated there that j x j is the distance of real number x from the origin O Consider the complex number z = + 2i I The distance from O to P is its modulus, j z j p So, j z j = 32 + 22 fPythagorasg P(3, 2) z R Example cyan magenta yellow 95 100 50 75 25 jzj p = (¡3)2 + (¡2)2 p = 9+4 p = 13 95 c 100 jzj p = 32 + (¡2)2 p = 9+4 p = 13 50 b 75 jzj p = 32 + 22 p = 13 25 a ¡3 ¡ 2i c 95 ¡ 2i 100 b 50 + 2i 75 a 25 95 100 50 75 25 Find j z j for z equal to: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\425IB_HL-2_15.CDR Monday, November 2007 3:07:27 PM PETERDELL IB_HL-2ed (426) 426 COMPLEX NUMBERS (Chapter 15) Example Prove that j z1 z2 j = j z1 j j z2 j for all complex numbers z1 and z2 Let z1 = a + bi and z2 = c + di where a, b, c and d are real ) z1 z2 = (a + bi)(c + di) = [ac ¡ bd] + i[ad + bc] p Thus j z1 z2 j = (ac ¡ bd)2 + (ad + bc)2 p = a2 c2 ¡ 2abcd + b2 d2 + a2 d2 + 2abcd + b2 c2 p = a2 (c2 + d2 ) + b2 (c2 + d2 ) p = (c2 + d2 )(a2 + b2 ) p p = a2 + b2 £ c2 + d2 = j z1 j j z2 j EXERCISE 15B.1 Find j z j for z equal to: a ¡ 4i b ¡8 + 2i c + 12i If z = + i and w = ¡1 + 3i find: a jzj b j z¤ j d j z ¤ j2 ¯ z ¯ ¯ ¯ ¯ ¯ w ¯ 3¯ ¯z ¯ c e j zw j f jzjjwj g i ¯ 2¯ ¯z ¯ j j z j2 k 3i e ¡4 d zz ¤ h jzj jwj l j z j3 From 2, suggest five possible rules for modulus If z = a + bi is a complex number, show that: a j z ¤ j = j z j b j z j = zz ¤ If z = cos µ + i sin µ, find j z j ¯z ¯ ¯ ¯ Simplify ¯ ¯ £ j w j using the result of Example and use it to show that w ¯z ¯ j z j ¯ ¯ provided w 6= ¯ ¯= w jwj a Use the result j z1 z2 j = j z1 j j z2 j to show that: ¯ ¯ i j z1 z2 z3 j = j z1 j j z2 j j z3 j and that ¯ z ¯ = j z j3 ¯ ¯ ii j z1 z2 z3 z4 j = j z1 j j z2 j j z3 j j z4 j and that ¯ z ¯ = j z j4 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b What is the generalisation of the results in a? c Use the Principle of mathematical induction to prove your conjecture in b d Hence prove that j z n j = j z jn for all n Z + ¯ ¯ p e Use the result of d to find ¯ z 20 ¯ for z = ¡ i black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\426IB_HL-2_15.CDR Monday, November 2007 3:10:31 PM PETERDELL IB_HL-2ed (427) 427 COMPLEX NUMBERS (Chapter 15) ¯ z ¯ jzj ¯ ¯ Given j z j = 3, use the rules j zw j = j z j j w j and ¯ ¯ = to find: w jwj a j 2z j b d j iz j e a If w = z+1 z¡1 b If w = z+1 z¡1 j¡3z j ¯ ¯ ¯1¯ ¯ ¯ ¯z¯ c f j (1 + 2i)z j ¯ ¯ ¯ 2i ¯ ¯ ¯ ¯ z2 ¯ where z = a + bi, find w in the form X + Y i when X and Y involve a and b and j z j = 1, find Re(w): SUMMARY OF MODULUS DISCOVERIES ² j z¤ j = j z j ² j z j2 = zz ¤ ² ¯ ¯ z1 j z1 z2 j = j z1 j j z2 j and ¯¯ z2 ² j z1 z2 z3 :::: zn j = j z1 j j z2 j j z3 j ::::: j zn j and j z n j = j z jn ¯ ¯ j z1 j ¯= ¯ j z2 j provided z2 6= for n Z + Example Find j z j given that j z ¡ j = j z ¡ 25 j where z is a complex number j z ¡ j = j z ¡ 25 j ) 25 j z ¡ j2 = j z ¡ 25 j2 25(z ¡ 1)(z ¡ 1)¤ = (z ¡ 25)(z ¡ 25)¤ fas zz ¤ = j z j g ) 25(z ¡ 1)(z ¤ ¡ 1) = (z ¡ 25)(z ¤ ¡ 25) fas (z § w)¤ = z ¤ § w¤ g 25zz ¤ ¡ 25z ¡ 25z ¤ + 25 = zz ¤ ¡ 25z ¡ 25z ¤ + 625 ) 24zz ¤ = 600 ) zz ¤ = 25 ) ) j z j = 25 ) jzj = fas j z j > 0g 10 Find j z j for the complex number z if j z + j = j z + j ¯ ¯ ¯ z+4 ¯ ¯ = 11 Find j z j for the complex number z if ¯¯ z+1 ¯ cyan magenta yellow 95 100 50 w z =¡ ¤ ¤ z w 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 12 If j z + w j = j z ¡ w j deduce that black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\427IB_HL-2_15.CDR Thursday, November 2007 3:45:57 PM PETERDELL IB_HL-2ed (428) 428 COMPLEX NUMBERS (Chapter 15) DISTANCE IN THE NUMBER PLANE P2(x2, ¡y2) P1(x1, ¡y1) Suppose P1 and P2 are two points in the complex plane which correspond to the complex numbers z1 and z2 zx zz Now j z1 ¡ z2 j = j (x1 + y1 i) ¡ (x2 + y2 i) j = j (x1 ¡ x2 ) + (y1 ¡ y2 )i j p = (x1 ¡ x2 )2 + (y1 ¡ y2 )2 Alternatively: Px O which we recognise as the distance between P1 and P2 ¡¡! ¡¡! ¡¡! P2 P1 = P2 O + OP1 = ¡z2 + z1 Pz = z1 ¡ z2 ¯¡¡!¯ ¯ ¯ ) jz1 ¡ z2 j = ¯P2 P1 ¯ = distance between P1 and P2 z1-z2 zx zz O Thus, ¡¡! ¡¡! jz1 ¡ z2 j is the distance between points P1 and P2 , where z1 ´ OP1 and z2 ´ OP2 : Note: The point corresponding to z1 ¡ z2 can be found by drawing a vector equal to ¡¡! P2 P1 emanating (starting) from the origin Can you explain why? CONNECTION TO COORDINATE GEOMETRY I R z There is a clear connection between complex numbers, Q OPRQ is a vector geometry and coordinate geometry w parallelogram M w For example: w+z ¡! ¡! P Notice that OR ´ w + z and OM ´ R z as the diagonals of the parallelogram bisect each other ¡! ¡ ! OR and PQ give the diagonals of the parallelogram formed by w and z Example P(2, 3) and Q(6, 1) are two points on the Cartesian plane Use complex numbers to find: a distance PQ b the midpoint of PQ cyan magenta yellow I P(2, 3) z Q(6, 1) w R 95 100 50 75 25 ) the midpoint of PQ is (4, 2) 95 100 50 75 25 95 100 50 75 25 z+w + 3i + + i = = + 2i 2 95 100 50 75 25 b If z = + 3i and w = + i then z ¡ w = + 3i ¡ ¡ i = ¡4 + 2i p p ) j z ¡ w j = (¡4)2 + 22 = 20 p ) PQ = 20 units a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\428IB_HL-2_15.CDR Monday, November 2007 3:26:36 PM PETERDELL IB_HL-2ed (429) 429 COMPLEX NUMBERS (Chapter 15) Example What transformation moves z to iz? If z = x + iy I then iz = i(x + iy) = xi + i2 y = ¡y + xi p We notice that j z j = x2 + y p and j iz j = (¡y)2 + x2 p = x2 + y P'(-y, x) P(x, y) R ) OP0 = OP So, (x, y) ! (¡y, x) under an anti-clockwise rotation of ¼ about O The transformation found in Example can be found more easily later (See Example 12.) EXERCISE 15B.2 Use complex numbers to find: i a A(3, 6) and B(¡1, 2) distance AB ii the midpoint of AB for b A(¡4, 7) and B(1, ¡3) ¡ ! OPQR is a parallelogram as shown OP ¡! represents z and OR represents w where z and w are complex numbers a In terms of z and w, what are: ¡! ¡ ! i OQ ii PR? I Q R w z P R b Explain from triangle OPQ, why j z + w j j z j + j w j : It is important to discuss when the equality case occurs c Explain from triangle OPR, why j z ¡ w j > j w j ¡ j z j Once again discuss when the equality case occurs What transformation moves: b z to ¡z a z to z ¤ c z to ¡z ¤ cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find the complex number z that satisfies the equation p j z j = 10 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\429IB_HL-2_15.CDR Thursday, November 2007 3:46:53 PM PETERDELL d z to ¡iz? 50 10 ¡ = + 9i given z¤ z IB_HL-2ed (430) 430 COMPLEX NUMBERS (Chapter 15) ARGUMENT ³ ´ a b The direction of the vector can be described by its angle from the positive real axis I Suppose the complex number z = a + bi is ¡! represented by vector OP as shown alongside a + bi is the Cartesian form of z: ¡ ! Suppose that OP makes an angle of µ with the positive real axis P( a, b) b q R a The angle µ is called the argument of z, or simply arg z arg z = µ has infinitely many possibilities i.e., z 7! arg z = µ not a function Can you explain why? is one-to-many and is To avoid the infinite number of possibilities for µ, we may choose to use µ ] ¡ ¼, ¼ ] which covers one full revolution and guarantees that z 7! arg z = µ is a function ² ² Note: Real numbers have argument of or ¼ Pure imaginary numbers have argument of ¼ or ¡ ¼2 POLAR FORM The polar form representation of a complex number is an alternative to Cartesian form, and has many useful applications r Any point P which lies on a circle with centre O(0, 0) and radius r, has Cartesian coordinates (r cos µ, r sin µ) I P(rcos¡q, rsin¡q) So, z = r cos µ + ir sin µ = r(cos µ + i sin µ) q -r r R But r = j z j and if we define cis µ = cos µ + i sin µ -r then z = j z j cis µ Consequently: Any complex number z has Cartesian form z = x + yi or polar form z = j z j cis µ where j z j is the modulus of z, µ is the argument of z, and cis µ = cos µ + i sin µ We will soon see that polar form is extremely powerful for dealing with multiplication and division of complex numbers, as well as quickly finding powers and roots of numbers (see De Moivre’s theorem) A useful identity is: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 If z = j z j cis µ, then z ¤ = j z j cis (¡µ) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\430IB_HL-2_15.CDR Monday, November 2007 4:04:21 PM PETERDELL IB_HL-2ed (431) 431 COMPLEX NUMBERS (Chapter 15) EULER FORM where r = j z j and µ = arg z z can also be written in Euler form as z = reiµ cis µ = eiµ In other words, We will derive this identity in Chapter 27 For example, consider the complex number z = + i: p j z j = and µ = ¼4 , p ¡ ¢ p ¼ ) + i = cis ¼4 = 2ei p ¡ ¢ is the polar form of + i So, cis ¼4 p i¼ and 2e is the Euler form of + i (1,¡1) I p R Example b ¡3 a 2i Write in polar form: a b c I 1¡i c I I p p -3 R - p4 R -i j 2i j = j¡3 j = ¼ µ=¼ µ= ) 2i = cis j1 ¡ ij = R 1-i p p 1+1= µ = ¡ ¼4 p ¡ ¢ ) ¡ i = cis ¡ ¼4 ) ¡3 = cis ¼ ¼ Example 10 p ¡ ¢ Convert cis 5¼ to Cartesian form ( - I , 12 ) 5p R p ¡ ¢ cis 5¼ p £ ¡ 5¼ ¢ â đô = cos + i sin 5¼ i p h p3 = ¡ +i£ = ¡ 32 + p i EXERCISE 15B.3 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find the modulus and argument of the following complex numbers and hence write them in polar form: a b 2i c ¡6 d ¡3i p p h + 2i e 1+i f ¡ 2i g ¡ 3+i black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\431IB_HL-2_15.CDR Wednesday, 14 November 2007 12:34:41 PM PETERDELL IB_HL-2ed (432) 432 COMPLEX NUMBERS (Chapter 15) What complex number cannot be written in polar form? Why? Convert k + ki to polar form (Careful! You must consider k > 0, k = 0, k < 0.) Convert to Cartesian form without using a calculator: ¡ ¢ ¡ ¢ b cis ¼4 a cis ¼2 p p ¡ ¢ ¡ ¢ e cis ¡ ¼4 cis 2¼ d ¡¼¢ c cis f cis ¼ a Find the value of cis b Find the modulus of cis µ, i.e., j cis µ j : c Show that cis ® cis ¯ = cis (® + ¯) MULTIPLYING AND DIVIDING IN POLAR FORM cis µ has three useful properties These are: ² cis µ £ cis Á = cis (µ + Á) cis µ = cis (µ ¡ Á) cis Á cis (µ + k2¼) = cis µ for all k Z ² ² The first two of these are similar to index laws: aµ aÁ = aµ+Á Proof: ² and aµ = aµ¡Á aÁ cis µ £ cis Á = (cos µ + i sin µ)(cos Á + i sin Á) = [cos µ cos Á ¡ sin µ sin Á] + i[sin µ cos Á + cos µ sin Á] = cos(µ + Á) + i sin(µ + Á) fcompound angle identitiesg = cis (µ + Á) cis µ cis (¡Á) cis µ = £ cis Á cis Á cis (¡Á) ² = ² cis (µ + k2¼) = cis µ £ cis (k2¼) = cis µ £ = cis µ cis (µ ¡ Á) cis = cis (µ ¡ Á) fas cis = 1g (1,¡0) The above results are even easier to prove if we use Euler’s form Using cis µ = eiµ and cis Á = eiÁ , we find: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 = ei(µ+Á) = cis (µ + Á) cis µ eiµ = iÁ cis Á e = eiµ¡iÁ = ei(µ¡Á) = cis (µ ¡ Á) ² cis µ cis Á = eiµ eiÁ ² black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\432IB_HL-2_15.CDR Thursday, November 2007 3:51:42 PM PETERDELL IB_HL-2ed (433) 433 COMPLEX NUMBERS (Chapter 15) Example 11 a cis Use the properties of cis to simplify: a cis = cis = cis = cis = cos ¡¼¢ ¡¼ cis + ¡ 5¼ ¢ ¡ 3¼ ¢ ¢ 10 ¼ ¼ cis + i sin ¼2 = + i(¡1) = ¡i = + i(1) =i Example 12 ¡ 3¼ ¢ 10 ¡ ¢ cis ¼5 ¡ ¢ cis 7¼ 10 b ¡ ¢ cis ¼5 ¡ 7¼ ¢ cis 10 ¢ ¡ = cis ¼5 ¡ 7¼ 10 ¡ ¢ = cis ¡ ¼2 ¡ ¢ ¡ ¢ = cos ¡ ¼2 + i sin ¡ ¼2 b 10 3¼ 10 ¡¼¢ I (0,¡1) p R -p (0,-1) (see Example earlier) What transformation moves z to iz? Let z = r cis µ and use i = cis ) iz = r cis µ £ cis ¼2 ¢ ¡ = r cis µ + ¼2 ¼ So, z has been rotated anti-clockwise by Example 13 Simplify cis ¡ 107¼ ¢ = 17 56 ¼ = 18¼ ¡ ¼6 ¢ ¡ ¢ ¡ = cis 18¼ ¡ ¼6 ) cis 107¼ ¡ ¢ fcis (µ + k2¼) = cis µg = cis ¡ ¼6 = about O : 107¼ p ¼ I -p ¡ 12 i R ( , -21) Example 14 p ¡ ¢ Write z = + 3i in polar form and then multiply it by cis ¼6 Illustrate what has happened on an Argand diagram ¡ ¢ What transformations have taken place when multiplying by cis ¼6 ? magenta yellow 95 100 50 75 95 50 25 95 100 50 75 25 95 100 50 75 25 cyan 75 R 25 ~`3 q p p If z = + 3i, then j z j = 12 + ( 3)2 ³ p ´ =2 ) z = 12 + 23 i ¡ ¢¢ ¡ ¡ ¢ ) z = cos ¼3 + i sin ¼3 ¡ ¢ ) z = cis ¼3 I a 100 a b c black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\433IB_HL-2_15.CDR Tuesday, November 2007 12:22:17 PM PETERDELL IB_HL-2ed (434) 434 COMPLEX NUMBERS (Chapter 15) (1 + p 3i) £ cis ¼ ¡¼¢ £ cis ¡ ¼3 ¼ ¢ + ¡ ¼3 ¢ = cis = cis = cis ¡¼¢ = 4(0 + 1i) = 4i ¡ ¢ c When z was multiplied by cis ¼6 its modulus (length) was doubled and it was rotated through ¼6 I b (0,¡4) p 4i 1+~`3\\i p R If a complex number z is multiplied by r cis µ then its modulus is multiplied by r and its argument is increased by µ Example 15 Use complex numbers to write cos cos = cis = cis ¡ 7¼ ¢ 12 ¡ 7¼ ¢ + i sin 12 and sin 12 ¡ 7¼ ¢ in simplest surd form 12 ¡ 7¼ ¢ I 12 + 4¼ 12 (, ) ( 12 ¡ 3¼ ¡ 7¼ ¢ ¢ 2 ¡ ¢ ¡ ¢ = cis ¼4 £ cis ¼3 fcis (µ + Á) = cis µ £ cis Ág ³ ´³ p ´ = p12 + p12 i 12 + 23 i ³p ´ ³ p ´ 1 p3 p3 + p ¡ + i = 2p 2 2 2 Equating real parts: cos 7¼ 12 = ³ p ´ 1¡p 2 Equating imaginary parts: sin 7¼ 12 = £ p p2 = p 3+1 p 2 = p p 6+ p p , ) R p p 2¡ EXERCISE 15B.4 cyan magenta yellow 95 f [cis µ]3 ¡ ¢ ¡ 8¼ ¢ cis 2¼ £ cis i £p â đô cis ¼8 c 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Use the properties of cis to simplify: cis 3µ a cis µ cis 2µ b cis µ ¡¼¢ ¡¼¢ ¡ ¢ ¡¼¢ e cis 12 cis ¼6 d cis 18 cis p ¡ ¢ ¡¼¢ 32 cis ¼8 cis 12 ¡ ¢ ¢ ¡ h p g cis 7¼ cis ¡ 7¼ 12 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\434IB_HL-2_15.CDR Wednesday, 14 November 2007 12:36:38 PM PETERDELL IB_HL-2ed (435) 435 COMPLEX NUMBERS (Chapter 15) Use the property cis (µ + k2¼) = cis µ a b cis 17¼ to evaluate: c cis (¡37¼) cis ¡ 91¼ ¢ 3 If z = cis µ: a What is j z j and arg z? b Write z ¤ in polar form c Write ¡z in polar form Note: ¡2 cis µ is not in polar form as the coefficient of cis µ is a length and so must be positive d Write ¡z ¤ in polar form a b c d Write i in polar form z = r cis µ is any complex number Write iz in polar form Explain why iz is the anti-clockwise rotation of z about O through ¼2 radians What transformation maps z onto ¡iz? Give reasoning in polar form Write in polar form: a cos µ ¡ i sin µ b Use a above to complete this sentence: If z = r cis µ then z ¤ = ::::: in polar form sin µ ¡ i cos µ Use complex number methods to find, in simplest surd form: ¡¼¢ ¡¼¢ ¡ ¢ ¡ ¢ a cos 12 and sin 12 b cos 11¼ and sin 11¼ 12 12 PROPERTIES OF ARGUMENT ² The basic properties of argument are: ² arg(zw) = arg z + arg w ² n arg ³z´ w = arg z ¡ arg w arg (z ) = n arg z Notice that they are identical to the laws of logarithms, with arg replaced by log or ln Properties of modulus and argument can be proved jointly using polar form Example 16 Use polar form to establish that j zw j = j z j j w j and arg(zw) = arg z + arg w Let z = j z j cis µ and w = j w j cis Á Now zw = j z j cis µ £ j w j cis Á = j z j j w j cis (µ + Á) fproperty of cisg | {z } non-negative ) j zw j = j z j j w j fthe non-negative number multiplied by cis ( ) g cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 and arg(zw) = µ + Á = arg z + arg w black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\435IB_HL-2_15.CDR Thursday, November 2007 3:52:21 PM PETERDELL IB_HL-2ed (436) 436 COMPLEX NUMBERS (Chapter 15) Example 17 If z = a p cis µ, find the modulus and argument of: a 2z p 2z = 2 cis µ b b iz p j 2z j = 2 and arg (2z) = µ ) ¼ i = cis p iz = cis ¼2 £ cis µ p p ¡ ¢ = cis ¼2 + µ So, j iz j = and arg (iz) = ) c c (1 ¡ i)z ¼ +µ p ¡ ¢ cis ¡ ¼4 p ¡ ¢ p (1 ¡ i)z = cis ¡ ¼4 £ cis µ ¡ ¢ = cis ¡ ¼4 + µ 1¡i = I -p ) R -i ) j (1 ¡ i)z j = and arg ((1 ¡ i) z) = µ ¡ ¼ Example 18 Suppose z = cis Á where Á is acute Find the modulus and argument of z + 1: jzj = ) z lies on the unit circle ¡! z + is OB ffound vectoriallyg I A z B z+1 R C OABC is a rhombus ) arg(z + 1) = A M z O fdiagonals bisect the angles of the rhombusg ¡ ¢ OM Also cos µ2 = ¡ ¢ ) OM = cos µ2 ¡ ¢ ) OB = cos µ2 ¡ ¢ ) j z + j = cos µ2 B q µ C EXERCISE 15B.5 Use polar form to establish: ¯z ¯ ³z´ jzj ¯ ¯ and arg = arg z ¡ arg w, provided w 6= 0: ¯ ¯= w jwj w cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Suppose z = cis µ Determine the modulus and argument of: c iz d a ¡z b z¤ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\436IB_HL-2_15.CDR Tuesday, November 2007 10:03:19 AM PETERDELL (1 + i)z IB_HL-2ed (437) 437 COMPLEX NUMBERS (Chapter 15) a If z = cis Á where Á is acute, determine the modulus and argument of z ¡ b Using a, write z ¡ in polar form c Hence write (z ¡ 1)¤ in polar form ¡! ¡! ABC is an equilateral triangle Suppose z1 represents OA, z2 represents OB and z3 ¡! represents OC a Explain what vectors represent z2 ¡ z1 and z3 ¡ z2 ¯ ¯ ¯ z2 ¡ z1 ¯ ¯ ¯: b Find ¯ z3 ¡ z2 ¯ µ ¶ z2 ¡ z1 c Determine arg : z3 ¡ z2 µ ¶3 z2 ¡ z1 : d Use b and c to find the value of z3 ¡ z2 B I A C R Given that z = a ¡ i where a is real, find the exact value of a if arg z = ¡ 5¼ i¼ to find the values of ei¼ and e cis µ b Prove that cis µ cis Á = cis (µ + Á) and = cis (µ ¡ Á) cis Á p z ii iz iii ¡iz iv c If z = cis µ, find the argument of i a Use eiµ = cos µ + i sin µ i z FURTHER CONVERSION BETWEEN CARTESIAN AND POLAR FORMS I r| cis {z µ} then: Polar form y tan µ = , x y x , sin µ = p cos µ = p 2 x +y x + y2 If z = x + iy = | {z } Cartesian form p r = x2 + y2 , P(x, y) r q y R x POLAR TO CARTESIAN ¡ ¢ z = cis ¼8 ¢ ¡ = cos ¼8 + i sin ¼8 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ¼ 1:85 + 0:765i black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\437IB_HL-2_15.CDR Thursday, November 2007 4:16:22 PM PETERDELL IB_HL-2ed (438) 438 COMPLEX NUMBERS (Chapter 15) CARTESIAN TO POLAR p p ¡3 z = ¡3+i has r = (¡3)2 + 12 = 10 and cos µ = p , sin µ = p 10 10 {z } | I ³ ´ ) µ = ¼ ¡ sin¡1 p110 fquadrant 2g q ³ ´ R ) (or cos¡1 p¡3 10 ) ¡3 + i ¼ p 10 cis (2:82) TI Alternatively (TI-83): C MATH CPX takes us to the complex number menu Pressing brings up abs( for calculating the modulus (absolute value) To find the modulus of ¡3 + i, press + 2nd i ) ENTER ~`1`0 MATH CPX then brings up angle( for calculating the argument To find the argument of ¡3 + i, press + ) ENTER 2nd i EXERCISE 15B.6 Use your calculator to convert to Cartesian form: p p ¢ ¡ a cis (2:5187) b 11 cis ¡ 3¼ c 2:836 49 cis (¡2:684 32) Use your calculator to convert to polar form: a ¡ 4i b ¡5 ¡ 12i c ¡11:6814 + 13:2697i Add the following using a + bi surd form and convert your answer to polar form: ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¡2¼ ¢ a cis ¼4 + cis ¡3¼ b cis 2¼ + cis Use the sum and product of roots to find the real quadratic equations with roots of: p ¡ 4¼ ¢ ¡ ¢ ¡ ¢ p ¡ ¢ b a cis 2¼ cis ¼4 , cis ¡¼ , cis C DE MOIVRE’S THEOREM Polar form enables us to easily calculate powers of complex numbers Notice that if z = j z j cis µ then z = j z j cis µ £ j z j cis µ z3 = z2 z and = j z j2 cis (µ + µ) = j z j2 cis 2µ £ j z j cis µ = j z j cis 2µ = j z j cis (2µ + µ) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 = j z j3 cis 3µ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\438IB_HL-2_15.CDR Wednesday, 14 November 2007 12:37:13 PM PETERDELL IB_HL-2ed (439) 439 COMPLEX NUMBERS (Chapter 15) The generalisation of this process is De Moivre’s Theorem: (j z j cis µ)n = j z jn cis nµ Proof for n Z + : (using mathematical induction) Pn is: (j z j cis µ)n = j z jn cis nµ (1) If n = 1, then (j z j cis µ)1 = j z j cis µ ) P1 is true k (2) If Pk is true, then (j z j cis µ)k = j z j cis kµ Thus (j z j cis µ)k+1 = (j z j cis µ)k £ j z j cis µ k = j z j cis kµ £ j z j cis µ k+1 cis (kµ + µ) k+1 cis (k + 1) µ = jzj = jzj findex lawg fusing Pk g findex law and cis propertyg Thus Pk+1 is true whenever Pk is true and P1 is true ) Pn is true fPrinciple of mathematical inductiong cis (¡nµ) = cis (0 ¡ nµ) = We observe also that cis nµ = [ cis µ ]n ) cis (¡nµ) = cis cis nµ fas cis µ = cis (µ ¡ Á)g cis Á fas cis = 1g ffor n a positive integerg = [ cis µ ]¡n Also [cis ¡µ¢ n ¡ ¡ µ ¢¢ = cis µ n ] = cis n n so the theorem is true for all n Z ¡ ¢ and so [cis µ] n = cis nµ So, De Moivre’s theorem seems to hold for any integer n and for n DE MOIVRE’S THEOREM (j z j cis µ)n = j z jn cis nµ for all rational n Example 19 p Find the exact value of ( + i)8 Check your answer by calculator using De Moivre’s theorem qp p p + i has modulus ( 3)2 + 12 = = ³p ´ p + i = 23 + 12 i ) = cis I R q ¼ cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ~`3 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\439IB_HL-2_15.CDR Thursday, November 2007 3:58:29 PM PETERDELL IB_HL-2ed (440) 440 COMPLEX NUMBERS (Chapter 15) p ¡ ) ( + i)8 = cis ¢ ¼ 8 ¡ 8¼ ¢ ¡ 4¼ ¢ = cis I = cis ³ = 28 ¡ 12 ¡ 4p R p ´ i ( -1 - 2 p = ¡128 ¡ 128 3i , ) Example 20 By considering cos 2µ + i sin 2µ, deduce the double angle formulae for cos 2µ and sin 2µ Now cos 2µ + i sin 2µ = cis 2µ fDe Moivre’s theoremg = [cis µ]2 = [cos µ + i sin µ]2 = [cos2 µ ¡ sin2 µ] + i[2 sin µ cos µ] Equating imaginary parts, Equating real parts, sin 2µ = sin µ cos µ cos 2µ = cos2 µ ¡ sin2 µ EXERCISE 15C Use De Moivre’s theorem to simplify: ¢10 ¡p ¡ a cis ¼5 b cis d p cis ¼ ¢ ¼ 36 12 p cis e ¼ 2 Use De Moivre’s theorem to find the exact value of: p a (1 + i)15 b (1 ¡ i 3)11 p d (¡1 + i)¡11 e ( ¡ i) c ¡p cis ¢ ¼ 12 f ¡ cis ¢ 53 c f ¼ p p ( ¡ i 2)¡19 p (2 + 2i 3)¡ Use your calculator to check the answers to a Suppose z = j z j cis µ where ¡¼ < µ ¼ p Use De Moivre’s theorem to find z in terms of j z j and µ p b What restrictions apply to Á = arg( z)? p c True or false? “ z has a non-negative real part.” Use De Moivre’s theorem to explain why j z n j = j z jn and arg (z n ) = n arg z: Show that cos µ ¡ i sin µ = cis (¡µ) Hence, simplify (cos µ ¡ i sin µ)¡3 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Write z = + i in polar form and hence write z n in polar form Find all values of n for which: a z n is real b z n is purely imaginary black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\440IB_HL-2_15.CDR Thursday, November 2007 4:01:34 PM PETERDELL IB_HL-2ed (441) 441 COMPLEX NUMBERS (Chapter 15) If j z j = and arg z = µ, determine the modulus and argument of: i a z3 b iz c d ¡ z z z2 ¡ If z = cis µ, prove that = i tan µ z2 + 10 a Use complex number methods to deduce that: ii sin 3µ = sin µ ¡ sin3 µ i cos 3µ = cos3 µ ¡ cos µ b Hence, find tan 3µ in terms of tan µ only c Using a and b above, solve the equations: 4x3 ¡ 3x = ¡ p12 i ii p p x3 ¡ 3x2 ¡ 3x + = 0: 11 Points A, B and C form an isosceles triangle with a right angle at B Let the points A, B and C be represented by the complex numbers z1 , z2 , and z3 respectively a Show that (z1 ¡ z2 )2 = ¡(z3 ¡ z2 )2 b If ABCD forms a square, what complex number represents the point D? Give your answer in terms of z1 , z2 and z3 12 Find a formula for a b B A cos 4µ in terms of cos µ sin 4µ in terms of cos µ and sin µ prove that z n + a If z = cis µ 13 C = cos nµ: zn = cos µ z c Use the binomial theorem to expand (z + )3 , and simplify your result z d By using a, b and c above show that cos3 µ = 14 cos 3µ + 34 cos µ p p ¡ ¢ is ¡5 2¡3 e Hence show the exact value of cos3 13¼ Hint: 13¼ 12 16 12 = b Hence, explain why z + then z n ¡ 14 Show that if z = cis µ that sin3 µ = sin µ ¡ = 2i sin nµ zn + ¼3 and hence sin 3µ 15 Use the results of 13 and 14 to prove that sin3 µ cos3 µ = D 3¼ 32 (3 sin 2µ ¡ sin 6µ) ROOTS OF COMPLEX NUMBERS SOLVING zn¡¡=¡¡c We will examine solutions of equations of the form z n = c where n is a positive integer and c is a complex number magenta yellow 95 100 50 75 25 95 100 50 75 25 95 50 75 25 95 100 50 75 25 cyan 100 The nth roots of complex number c are the n solutions of z n = c Definition: black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_15\441IB_HL-2_15.CDR Friday, 12 December 2008 12:16:19 PM TROY IB_HL-2ed (442) 442 COMPLEX NUMBERS (Chapter 15) For example, the 4th roots of 2i are the four solutions of z = 2i The roots may be found by factorisation, but this is sometimes difficult It is therefore desirable to have an alternative method such as the nth roots method presented in the following example Example 21 a factorisation Find the four 4th roots of by: b the ‘nth roots method’ We need to find the solutions of z = I R a By factorisation, z4 = b By the ‘nth roots method’, ) z4 ¡ = z4 = 2 ) (z + 1)(z ¡ 1) = ) z = cis (0 + k2¼) fpolar formg (z + i)(z ¡ i)(z + 1)(z ¡ 1) = ) z = [cis (k2¼)] ¢ ¡ ) z = §i or §1 fDe Moivreg ) z = cis k2¼ ¡ k¼ ¢ ) z = cis ) z = cis 0, cis ¼2 , cis ¼, cis fletting k = 0, 1, 2, 3g ) z = 1, i, ¡1, ¡i Note: ² 3¼ The factorisation method is fine provided the polynomial factorises easily This is not usually the case The substitution of k = 0, 1, 2, to find the roots could be done using any consecutive integers for k Why? ² EXERCISE 15D.1 Find the three cube roots of using: Solve for z: z = ¡8i a a b factorisation the ‘nth roots method’ z = ¡27i b Find the three cube roots of ¡1, and display them on an Argand diagram Solve for z: z = 16 a z = ¡16 b Find the four fourth roots of ¡i, and display them on an Argand diagram Solve the following and display the roots on an Argand diagram: a d z = + 2i p z4 = + i b z = ¡2 + 2i c e z = ¡4 ¡ 4i f p + 23 i p z = ¡2 ¡ 2i z2 = Example 22 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find the fourth roots of ¡4 in the form a + bi and then factorise z + into linear factors Hence, write z + as a product of real quadratic factors black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\442IB_HL-2_15.CDR Thursday, November 2007 4:03:27 PM PETERDELL IB_HL-2ed (443) 443 COMPLEX NUMBERS (Chapter 15) The fourth roots of ¡4 are solutions of z = ¡4 ) z = cis (¼ + k2¼) I p ) z = [4 cis (¼ + k2¼)] µ ¶ ¼ + k2¼ z = 4 cis ) R -4 1 ) 5¼ 7¼ 2 z = 2 cis ¼4 , 2 cis 3¼ fletting k = 0, 1, 2, 3g , cis , cis ´ p ³ ´ p ³ ´ p ³ ´ p ³ z = p2 + p12 i , ¡ p12 + p12 i , ¡ p12 ¡ p12 i , p12 ¡ p12 i ) z = + i, ¡1 + i, ¡1 ¡ i, ¡ i ) Roots § i have sum = and product = (1 + i)(1 ¡ i) = and ) come from the quadratic factor z ¡ 2z + Roots ¡1 § i have sum = ¡2 and product = (¡1 + i)(¡1 ¡ i) = and ) come from the quadratic factor z + 2z + In examples such as this we observe the connection between polynomial methods and complex number theory Thus z + = (z ¡ 2z + 2)(z + 2z + 2) Find the four solutions of z4¡+¡1¡=¡0 giving each in the form a¡+¡bi, and display them on an Argand diagram Hence write z4¡+¡1 as the product of two real quadratic factors ³p ´2 ¡ i 2 Consider z = ¡ ¢ ¡ ¢ ¼ ¼ ¼ ¼ 25 cos 10 ¡ i sin 10 cos 30 + i sin 30 a Using polar form and De Moivre’s theorem, find the modulus and argument of z b Hence show that z is a cube root of c By simplifying and without using a calculator, show that (1 ¡ 2z)(2z ¡ 1) is a real number a Write ¡16i in polar form b The fourth root of ¡16i which lies in the second quadrant is denoted by z Express z exactly in: i polar form ii Cartesian form SUMMARY OF SOLUTIONS OF zn = c (nth roots of c) ² ² ² There are exactly n nth roots of c If c R , the complex roots must occur in conjugate pairs If c = R , the complex roots not all occur in conjugate pairs ² The roots of z n will all have the same modulus which is j c j n Thus on an Argand diagram, the roots will all be the same distance from the origin 1 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 and hence lie on a circle radius j c j n black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\443IB_HL-2_15.CDR Wednesday, 14 November 2007 12:37:53 PM PETERDELL IB_HL-2ed (444) 444 COMPLEX NUMBERS (Chapter 15) ² The roots on the circle r = j c j n will be equally spaced around the circle If you join all the points you will get a geometric shape that is a regular polygon For example, n = (equilateral ¢), n = (square) n = (regular pentagon) n = (regular hexagon) etc THE nth ROOTS OF UNITY The nth roots of unity are the solutions of z n = Example 23 Find the three cube roots of unity and display them on an Argand diagram If w is the root with smallest positive argument, show that the roots are 1, w and w2 and that + w + w2 = The cube roots of unity are the solutions of z = But = cis = cis (0 + k2¼) for all k Z ) z = cis (k2¼) ) ) z = [cis (k2¼)] ¢ ¡ z = cis k2¼ fDe Moivre’s theoremg ¡ ¢ ¡ 2¼ ¢ fletting k = 0, 1, 2g z = cis 0, cis , cis 4¼ ) z = 1, ¡ 12 + ) ¡ 12 ¡ p i ( -1 ¡ 2¼ ¢ ¡ 4¼ ¢ ] = cis ¡ ¢ ) the roots are 1, w and w where w = cis 2¼ ³ ³ p ´ p ´ and + w + w2 = + ¡ 12 + 23 i + ¡ 12 ¡ 23 i = w = cis ¡ 2¼ ¢ p i, , ) I and w2 = [cis R ( -1 - 2 , ) EXERCISE 15D.2 In Example 23 we showed that the cube roots of are 1, w, w2 where w = cis ¡ 2¼ ¢ a Use this fact to solve the following equations, giving your answers in terms of w: i (z + 3)3 = ii (z ¡ 1)3 = iii (2z ¡ 1)3 = ¡1 b Show by vector addition that + w + w2 = if 1, w and w2 are the cube roots of unity In Example 21 we showed that the four fourth roots of unity were 1, i, ¡1, ¡i a Is it true that the four fourth roots of unity can be written in the form 1, w, w2 , w3 where w = cis ¼2 ? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b Show that + w + w2 + w3 = c Show by vector addition that b is true black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_15\444IB_HL-2_15.CDR Thursday, 11 March 2010 10:30:39 AM PETER IB_HL-2ed (445) COMPLEX NUMBERS (Chapter 15) 445 a Find the fifth roots of unity and display them on an Argand diagram b If w is the root with smallest positive argument, show that the roots are 1, w, w2 , w3 and w4 c Simplify (1+w+w2 +w3 +w4 )(1¡w) and hence show that 1+w+w2 +w3 +w4 must be zero d Show by vector addition that + w + w2 + w3 + w4 = ¡ ¢ is the nth root of unity with the smallest positive argument, show If w = cis 2¼ n that: a the n roots of z n = are 1, w, w2 , w3 , ., wn¡1 b + w + w2 + w3 + :::::: + wn¡1 = Note: The roots of unity lie on the unit circle, equally spaced around the unit circle Show that for any complex number ®, the sum of the n zeros of z n = ® is E FURTHER COMPLEX NUMBER PROBLEMS The following questions combine complex number theory with topics covered in previous chapters EXERCISE 15E Solve for z: z ¡ (2 + i)z + (3 + i) = Find the Cartesian equation for the locus of P(x, y) if z = x + iy and b arg(z ¡ i) = ¼6 c jz + 3j + jz ¡ 3j = a z ¤ = ¡iz Use the binomial expansion of (1 + i)2n to prove that: ¡2n¢ ¡2n¢ ¡2n¢ ¡2n¢ ¡ ¢ ¡ n¼ ¢ n 2n n + ¡ + ¡ + +(¡1) 2n = cos , n Z : By considering + cis µ + cis 2µ + cis 3µ + + cis nµ n X cos rµ find r=0 Prove that + cis µ = cos n X ¡n¢ r cos(rµ) ¡µ¢ cis ¡µ¢ as a geometric series, and hence determine the sum of the series r=0 REVIEW SET 15A Find the real and imaginary parts of (i ¡ p 3) If z = x+yi and P(x, y) moves in the complex plane, find the Cartesian equation for: a jz ¡ ij = jz + + ij b z ¤ ¡ iz = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find j z j if z is a complex number and j z + 16 j = j z + j black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\445IB_HL-2_15.CDR Tuesday, November 2007 11:39:07 AM PETERDELL IB_HL-2ed (446) 446 COMPLEX NUMBERS (Chapter 15) Points A and B arepthe representations in the complex plane of the numbers z = 2¡2i and w = ¡1 ¡ 3i respectively a Given that the origin is O, find the angle AOB in terms of ¼ b Calculate the argument of zw in terms of ¼ p c k ¡ ki where k < Write in polar form: a ¡5i b ¡ 2i Given that z = (1 + bi)2 arg z = ¼3 where b is real and positive, find the exact value of b if a Prove that cis µ £ cis Á = cis (µ + Á) p b If z = 2 cis ®, write (1 ¡ i)z in polar form Hence find arg[(1 ¡ i)z] ¡! ¡! z1 ´ OA and z2 ´ OB represent two sides of a right angled isosceles triangle OAB z2 a Determine the modulus and argument of 12 A z2 z1 b Hence, deduce that z12 + z22 = p ¡ ¢ ¢ p ¡ and w = b cos ¼4 ¡ i sin ¼4 Let z = a cos ¼6 + i sin ¼6 B z2 O ³ z ´4 Find in terms of a and b the exact values of the real and imaginary parts of w ¡ 2¼ ¢ 10 a List the five fifth roots of in terms of w where w = cis b Display the roots in a on an Argand diagram c By considering the factorisation of z ¡ in two different ways, show that: z + z + z + z + = (z ¡ w)(z ¡ w2 )(z ¡ w3 )(z ¡ w4 ) d Hence, find the value of (2 ¡ w)(2 ¡ w2 )(2 ¡ w3 )(2 ¡ w4 ) 11 Find the cube roots of ¡8i, giving your answers in the form a + bi where a and b not involve trigonometric ratios REVIEW SET 15B Let z1 = cos ¼6 + i sin ¼6 and z2 = cos ¼4 + i sin ¼4 µ ¶3 z1 in the form z = a + bi Express z2 If z = + i and w = ¡ 3i, find: b j w ¡ z¤ j a 2w¤ ¡ iz ¯ 10 ¯ ¯z ¯ c d arg(w ¡ z) ¡ 3i = + 2i 2a + bi Find rationals a and b such that cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 If z = x + yi and P(x, y) moves in the complex plane, find the Cartesian equation for the curve: ¯ ¯ ¯ z+2 ¯ ¯=2 b ¯¯ a arg(z ¡ i) = ¼2 z¡2 ¯ p p Write ¡ 3i in polar form Hence find all values of n for which (2 ¡ 3i)n is real black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\446IB_HL-2_15.CDR Tuesday, November 2007 11:47:15 AM PETERDELL IB_HL-2ed (447) COMPLEX NUMBERS (Chapter 15) 447 Determine the cube roots of ¡27 If z = cis µ, find the modulus and argument of: b a z3 c iz ¤ z Prove the following: a arg(z n ) = n arg z for all complex numbers z and rational n ³ z ´¤ z¤ = ¤ for all z and for all w 6= 0: b w w Find n such that each of the following can be written in the form [cis µ]n : a cos 3µ + i sin 3µ b c cos µ ¡ i sin µ cos 2µ + i sin 2µ 10 Determine the fifth roots of + 2i is real, prove that either j z j = or z is real z 12 If z = cis µ, prove that: a jzj = b z¤ = c sin4 µ = 18 (cos 4µ ¡ cos 2µ + 3) z 11 If z + 13 If w is the root of z = with smallest positive argument, find real quadratic equations with roots of: b w + w4 and w2 + w3 a w and w4 14 If j z + w j = j z ¡ w j prove that arg z and arg w differ by ¼ 15 The complex number z is a root of the equation j z j = j z + j a Show that the real part of z is ¡2 b Let v and w be two possible values of z such that j z j = i On an Argand diagram, sketch the points that represent v and w, given that v is in the 2nd quadrant ii Show that arg v = 2¼ iii Find arg w where ¡¼ < arg w ¼ µ m ¶ v w in terms of m and ¼ i Find arg c i vmw ii Hence find a value of m for which is a real number i REVIEW SET 15C What single transformation maps z onto: a z¤ b ¡z c iz? z and w are non-real complex numbers with the property that both z + w and zw are real Prove that z ¤ = w cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 If (x + iy)n = X + Y i where n is a positive integer, show that X + Y = (x2 + y2 )n black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\447IB_HL-2_15.CDR Wednesday, 14 November 2007 12:40:13 PM PETERDELL IB_HL-2ed (448) 448 COMPLEX NUMBERS (Chapter 15) ³ ´ Prove that j z ¡ w j2 + j z + w j2 = j z j2 + j w j2 z = has roots 1, ®, ®2 , ®3 and ®4 where ® = cis ¡ 2¼ ¢ : a Prove that + ® + ®2 + ®3 + ®4 = µ ¶5 z+2 = in terms of ® b Solve the equation z¡1 ¯ ¯ ¯ z+1 ¯ ¯ ¯ = 1, prove that z is purely imaginary If z 6= and ¯ z¡1 ¯ If z = cis Á and w = 1+z , show that w = cis Á also + z¤ p Write ¡1 + i in polar form and hence find the values of m for which p (¡1 + i 3)m is real ³ ´ ³ ´ µ+Á Prove that cis µ + cis Á = cos µ¡Á cis and hence show that 2 µ ¶5 ¡ ¢ z+1 for n = 1, 2, and = has solutions of the form z = i cot n¼ z¡1 10 Find the cube roots of ¡64i, giving your answers in the form a + bi where a and b are real a (2z)¡1 11 If z = cis µ, find the modulus and argument of: b 1¡z 12 Illustrate the region defined by fz: j z j and ¡ ¼4 < arg z ¼2 g Show clearly all included boundary points ¯ ¯ µ ¶ ¯1¯ 1 ¯ ¯ for z 6= and arg = ¡ arg z: 13 Use polar form to deduce that ¯ ¯ = z jzj z 14 If z = cis ®, write + z in polar form and hence determine the modulus and argument of + z ¡ ¢ ¡ ¢ Hint: sin µ = sin µ2 cos µ2 15 ¢P1 P2 P3 is an equilateral triangle, as illustrated ¡¡! ¡¡! O is an origin such that OP1 ´ z1 , OP2 ´ z2 ¡¡! and OP3 ´ z3 Suppose arg(z2 ¡ z1 ) = ® a Show that arg(z3 ¡ z2 ) = ® ¡ P2 2¼ b Find the modulus and argument of P1 z2 ¡ z1 z3 ¡ z2 P3 16 Two of the zeros of P (z) = z + az + bz + c where a, b, c R , are z = ¡3 and z = + i Find the values of a, b and c cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 17 State the five fifth roots of unity and hence solve: a (2z ¡ 1)5 = 32 b z + 5z + 10z + 10z + 5z = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_15\448IB_HL-2_15.CDR Tuesday, November 2007 12:16:20 PM PETERDELL c (z + 1)5 = (z ¡ 1)5 IB_HL-2ed (449) Chapter 16 Lines and planes in space Contents: Lines in 2-D and 3-D Applications of a line in a plane Relationship between lines Planes and distances Angles in space The intersection of two or more planes A B C D E F cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 16A Review set 16B Review set 16C Review set 16D black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\449IB_HL-2_16.CDR Friday, November 2007 9:48:14 AM PETERDELL IB_HL-2ed (450) 450 LINES AND PLANES IN SPACE (Chapter 16) INTRODUCTION ³ ´ Suppose the vector represents a displacement of km due East and ³ ´ represents a displacement of km due North The diagram shows the path of a yacht relative to a yacht club which is situated at (0, 0) At 12:00 noon the yacht is at the point A(2, 20) y 20 12:00 noon A 1:00 pm (-3 ) 2:00 pm The yacht is travelling in the ³ ´ direction ¡3 with a constant 10 speed of km h¡1 ¯³ ´¯ ¯ ¯ Since ¯ ¡3 ¯ = we can see that ³ ´ gives both the direction and ¡3 sea x speed of travel ³ ´ So, ¡3 is called the velocity 10 15 land vector of the yacht In order to define the position of the yacht at any time t hours after 12 noon, we can use the parametric equations x = + 4t and y = 20 ¡ 3t where t is called the parameter Note: If t = 0, x = and y = 20, so the yacht is at (2, 20) If t = 1, x = and y = 17, so the yacht is at (6, 17) If t = 2, x = 10 and y = 14, so the yacht is at (10, 14) We can find a vector equation for the yacht’s path as follows: Suppose the yacht is at R(x, y) at time t hours after 12:00 noon ¡! ¡! ¡! ) OR = OA + AR ³ ´ ³ ´ for t > ) r = 20 + t ¡3 ³ ´ ³ ´ ³ ´ x = 20 + t ¡3 ) y y A(2, 20) R(x,¡y) r which is the vector equation of the yacht’s path Notice how the parametric equations are easily found from the vector equation: cyan magenta ´ x yellow 95 100 50 75 25 95 100 50 75 then x = + 4t and y = 20 ¡ 3t 25 95 ¡3 ³ +t 100 ´ 50 20 75 ³ = 95 50 75 25 100 x y 25 ³ ´ If (-43 ) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\450IB_HL-2_16.CDR Wednesday, 14 November 2007 1:34:44 PM PETERDELL IB_HL-2ed (451) 451 LINES AND PLANES IN SPACE (Chapter 16) We can also find the Cartesian equation of the yacht’s path: x¡2 : µ ¶ x¡2 Substituting into y = 20 ¡ 3t gives y = 20 ¡ As x = + 4t, 4t = x ¡ and so t = ) 4y = 80 ¡ 3(x ¡ 2) ) 4y = 80 ¡ 3x + ) 3x + 4y = 86 where x > because t > A LINES IN 2-D AND 3-D In both 2-D and 3-D geometry we can determine the equation of a line by its direction and a fixed point through which it passes Suppose a line passes through a fixed point A such ¡! that OA = a, and its direction is given by vector b (i.e., the line is parallel to b) ¡! Let any point R be on the line so that OR = r ¡! ¡! ¡! By vector addition, OR = OA + AR ¡! ¡! Since AR k b, AR = tb for some t R ) r = a + tb r = a + tb, t R So, A (fixed point) b a R (any point) O (origin) line is the vector equation of the line LINES IN 2-D ² In 2-D we are dealing with a line in a plane ³ ´ ³ ´ ³ ´ x a1 b1 = + t is the vector equation of the line y a2 b2 where R(x, y) is any point on the line A(a1 , a2 ) is the known (fixed) point on the line ³ ´ b b = b1 is the direction vector of the line µ ¶ ³ ´ b2 rise b1 also enables us to calculate the slope of the line, m = Note: b = b b1 run ¾ x = a1 + b1 t , t R , are the parametric equations of the line y = a2 + b2 t where t is called the parameter With these equations each point on the line corresponds to exactly one value of t, so every point has a one-to-one correspondence with a real number ² ² cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 We can convert these equations into Cartesian form by equating t values y ¡ a2 x ¡ a1 = we obtain b2 x ¡ b1 y = b2 a1 ¡ b1 a2 Using t = b1 b2 which is the Cartesian equation of the line 95 100 50 75 25 ² black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\451IB_HL-2_16.CDR Tuesday, November 2007 1:46:44 PM PETERDELL IB_HL-2ed (452) 452 LINES AND PLANES IN SPACE (Chapter 16) Example Find a the vector b the parametric c the Cartesian equation ³ ´ of the line passing through the point (1, 5) with direction a b ³ ´ ¡! ³ ´ a = OA = and b = ³ ´ ³ ´ ³ ´ x = + t , t2R But r = a + tb ) y From a, x = + 3t and y = + 2t, t R R x¡1 y¡5 Now t = = ) 2x ¡ = 3y ¡ 15 ) 2x ¡ 3y = ¡13 fgeneral formg c ( 32) A a r O Example A particle at P(x(t), y(t)) moves such that x(t) = ¡ 3t and y(t) = 2t + 4, t > The distance units are metres and t is in seconds a Find the initial position of P b Illustrate the motion showing points where t = 0, 1, and c Find the speed of P ) the initial position of P is (2, 4) a x(0) = 2, y(0) = b x(1) = ¡1, y(1) = x(2) = ¡4, y(2) = x(3) = ¡7, y(3) = 10 y t=3 10 t=2 t=1 c ~`2`X`+`\3`X Every second P moves with x-step ¡3 and y-step 2, p which is a distance of 13 m ) the speed is constant and p is 13 m s¡1 t=0 -10 x -5 EXERCISE 16A.1 Find i ii the vector equation the parametric equations of the line: ³ ´ a passing through (3, ¡4) with direction ³ b passing through (5, 2) with direction ¡8 ´ cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 c cutting the x-axis at ¡6 with direction 3i + 7j d with direction ¡2i + j that passes through (¡1, 11) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\452IB_HL-2_16.CDR Tuesday, 13 November 2007 9:39:04 AM PETERDELL IB_HL-2ed (453) 453 LINES AND PLANES IN SPACE (Chapter 16) Find the parametric equations of the line passing through (¡1, 4) with direction vector ³ ´ and parameter ¸ Find the points on the line when ¸ = 0, 1, 3, ¡1, ¡4 ¡1 a Does (3, ¡2) lie on the line with parametric equations x = t + 2, y = ¡ 3t? Does (0, 6) lie on this line? b (k, 4) lies on the line with parametric equations x = ¡ 2t, y = + t Find k A particle at P(x(t), y(t)) moves such that x(t) = + 2t and y(t) = ¡ 5t, t > The distances are in centimetres and t is in seconds a Find the initial position of P b Illustrate the initial part of the motion of P where t = 0, 1, 2, c Find the speed of P LINES IN 3-D ² In 3-D we are dealing with a line in space à ! à ! à ! x y z ² = a1 a2 a3 b1 b2 b3 +¸ where is the vector equation of the line R(x, y, z) is any point on the line A(a1 , a2 ; a3 ) is the known (fixed) point on the line à ! b1 b2 b3 b= is the direction vector of the line We not talk about the slope of a line in 3-D We describe its direction only by its direction vector x = a1 + ¸b1 = y = a2 + ¸b2 are the parametric equations of the line ; z = a3 + ¸b3 where ¸ R is called the parameter Note: ² Every point on the line corresponds to exactly one value of ¸ x ¡ a1 y ¡ a2 z ¡ a3 = = (= ¸) are the Cartesian equations of the line b1 b2 b3 ² Example Find the vector equation and the parametric equations of the line through (1, ¡2, 3) in the direction 4i + 5j ¡ 6k à ! x y z The vector equation is à = ¡2 ! à +¸ ¡6 ! , ¸2R cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The parametric equations are: x = + 4¸, y = ¡2 + 5¸, z = ¡ 6¸, ¸ R black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\453IB_HL-2_16.CDR Tuesday, November 2007 2:10:24 PM PETERDELL IB_HL-2ed (454) 454 LINES AND PLANES IN SPACE (Chapter 16) Example Find the parametric equations of the line through A(2, ¡1, 4) and B(¡1, 0, 2) ¡! ¡! We require a direction vector for the line, either AB or BA ! à ! à ¡1 ¡ ¡3 ¡! AB = ¡ ¡1 = 2¡4 ¡2 Using the point A, the equations are: x = ¡ 3¸, y = ¡1 + ¸, z = ¡ 2¸, ¸ R Note: Using the point B, the equations are: x = ¡1 ¡ 3¹, y = ¹, z = ¡ 2¹, ¹ R These sets of equations are actually equivalent, and generate the same set of points They are related by ¹ = ¸ ¡ EXERCISE 16A.2 Find the vector equation of the line: à ! a parallel to and through the point (1, 3, ¡7) b through (0, 1, 2) and with direction vector i + j ¡ 2k c parallel to the X-axis and through the point (¡2, 2, 1) Find the parametric equations of the line: à ! ¡1 a parallel to and through the point (5, 2, ¡1) b parallel to 2i ¡ j + 3k and through the point (0, 2, ¡1) c perpendicular to the XOY plane and through (3, 2, ¡1) Find the parametric equations of the line through: a A(1, 2, 1) and B(¡1, 3, 2) b C(0, 1, 3) and D(3, 1, ¡1) c E(1, 2, 5) and F(1, ¡1, 5) d G(0, 1, ¡1) and H(5, ¡1, 3) Find the coordinates of the point where the line with parametric equations x = ¡ ¸, y = + ¸ and z = ¡ 2¸ meets: b the Y OZ plane c the XOZ plane a the XOY plane Find points on the line pwith parametric equations x = ¡ ¸, y = + 2¸ and z = + ¸ which are units from the point (1, 0, ¡2) The perpendicular from a point to a line minimises the distance from the point to that line Use quadratic theory to find the coordinates of the foot of the perpendicular: a from (1, 1, 2) to the line with equations x = + ¸, y = ¡ ¸, z = + ¸ à ! à ! à ! x y z cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b from (2, 1, 3) to the line with vector equation black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\454IB_HL-2_16.CDR Friday, November 2007 9:58:44 AM PETERDELL = +¹ ¡1 IB_HL-2ed (455) 455 LINES AND PLANES IN SPACE (Chapter 16) THE ANGLE BETWEEN TWO LINES (2-D and 3-D) In Chapter 14 we saw that the angle between two vectors is measured in the range 0o µ 180o We used the b1 ² b2 formula cos µ = j b1 j j b2 j In the case of lines which continue infinitely in both directions, we agree to talk about the acute angle between them We therefore use the formula q bx bx bz bz lz j b1 ² b2 j cos µ = j b1 j j b2 j q lx where b1 and b2 are the direction vectors of the given lines l1 and l2 respectively Example Find the angle between the lines l1 : x = ¡ 3t, y = ¡1 + t and l2 : x = + 2s, y = ¡4 + 3s ³ ´ ³ ´ ¡3 b2 = b1 = j¡6 + j cos µ = p p 10 13 ) ) cos µ ¼ 0:2631 and so µ ¼ 74:7o (1:30 radians) Example Find the angle between the lines l1 : x = ¡ 3¸, y = ¡1 + ¸, z = ¡ 2¸ 1¡x 2¡z and l2 : =y= à b1 = ¡3 ¡2 ! à and b2 = ¡3 ¡2 ! j9 + + 4j cos µ = p p 14 14 ) ) cos µ = and so µ = 0o In fact these lines are coincident, i.e., l1 and l2 are the same line EXERCISE 16A.3 ³ cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find the angle between the lines: l1 passing through (¡6, 3) parallel to l2 cutting the y-axis at (0, 8) with direction 5i + 4j black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\455IB_HL-2_16.CDR Tuesday, 29 January 2008 11:03:39 AM PETERDELL ¡3 ´ and IB_HL-2ed (456) 456 LINES AND PLANES IN SPACE (Chapter 16) l1 : x = ¡4 + 12t, y = + 5t and l2 : x = 3s, y = ¡6 ¡ 4s Find the angle between the lines: x = + 5p, y = 19 ¡ 2p and x = + 4r, y = + 10r are perpendicular Show that the lines: x¡8 9¡y z ¡ 10 = = 16 Find the angle between the lines: and x = 15 + 3¹, y = 29 + 8¹, z = ¡ 5¹ B APPLICATIONS OF A LINE IN A PLANE THE VELOCITY VECTOR OF A MOVING OBJECT ³ In Example we considered a particle which moves ³ ´ ¡3 is called the velocity vector of the particle ¡3 ´ every second ¯³ ´¯ p p ¯ ¡3 ¯ Since ¯ ¯ = (¡3)2 + 22 = 13, ³ ´ p ¡3 , and the velocity of the particle is 13 metres per second in the direction p the speed of the particle is 13 metres per second In general, ³ ´ a b if is the velocity vector of a moving object, then it is travelling ¯³ ´¯ p ³ ´ ¯ a ¯ a at a speed of ¯ b ¯ = a2 + b2 in the direction b Example ³ ´ x y ³ ´ = ³ +t ¡8 ´ is the vector equation of the path of an object t is the time in seconds, t > The distance units are metres Find the: a object’s initial position b velocity vector of the object c object’s speed ³ ´ x y ³ ´ At t = 0, b The velocity vector is c ¯³ ´¯ p p ¯ ¯ The speed is ¯ ¡8 ¯ = 36 + 64 = 100 = 10 m s¡1 magenta ´ ³ yellow 95 100 50 75 25 95 because the object moves 100 50 25 95 100 50 75 25 95 100 50 75 25 cyan ¡8 75 = ³ ) the object is at (7, 5) a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\456IB_HL-2_16.CDR Tuesday, November 2007 3:14:34 PM PETERDELL ¡8 ´ every second IB_HL-2ed (457) 457 LINES AND PLANES IN SPACE (Chapter 16) CONSTANT VELOCITY PROBLEMS Suppose an object moves with constant velocity b If the object is initially at A (when time t = 0) and at time t it is at R, then ¡! AR = tb fdistance = time £ speedg ¡! ¡! Now r = OA + AR ) r = a + tb b A R a r O Thus if a body has initial position vector a, and moves with constant velocity b, its position at time t is given by r = a + tb for t > Example An object is initially at (5, 10) and moves with velocity vector 3i ¡ j Find: a the position of the object at any time t where t is in minutes b the position at t = c the time when the object is due east of (0, 0) a y (5, 10) b= r = a + tb ³ ´ ³ ´ ³ ´ x = + t , t2R ) y 10 ¡1 ¶ ³ ´ µ + 3t x ) = 10 ¡ t y ³ ´ ¡1 P a r ) P is at (5 + 3t, 10 ¡ t) x b At t = 3, + 3t = 14 and 10 ¡ t = ) it is at (14, 7) c When the object is due east of (0, 0), y must be zero ) 10 ¡ t = ) t = 10 The object is due east of (0, 0) after 10 minutes THE CLOSEST DISTANCE FROM A POINT TO A LINE A ship sails through point A in the direction b and continues past a port P At what time will the ship R be closest to the port? b The ship is closest when PR is perpendicular to AR, ¡ ! ) PR ² b = fthe scalar product is zerog yellow 95 100 50 75 25 95 50 75 25 95 100 50 75 25 95 100 50 75 25 100 P A magenta P R b cyan A R black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\457IB_HL-2_16.CDR Tuesday, November 2007 4:00:11 PM PETERDELL IB_HL-2ed (458) 458 LINES AND PLANES IN SPACE (Chapter 16) Example sea If distances are measured in kilometres and a ³ ´ ship R moves in the direction at a speed ³ ´ R(x,¡y) ¡1 of 10 km h , find: a an expression for the position of the ship in terms of t where t is the number of hours after leaving port A b the time when it is closest to port P(10, 2) A (-8,¡3) O (0,¡0) P (10,¡2) land a ¯³ ´¯ p ¯ ¯ ¯ ¯ = 32 + 42 = ) since the speed is 10 km h¡1 , the ship’s ³ ´ ³ ´ velocity vector must be = ¡! ¡! ¡! Now OR = OA + AR ³ ´ ³ ´ ³ ´ x ¡8 ) = + t y ) R is at (¡8 + 6t, + 8t) b ¡ ! ³3´ The ship is closest to P when PR ? ¡ ! ³3´ PR ² = ¶ ³ ´ µ ¡8 + 6t ¡ 10 ² =0 + 8t ¡ ) ) ) 3(6t ¡ 18) + 4(1 + 8t) = ) 18t ¡ 54 + + 32t = ) 50t ¡ 50 = ) t=1 So, the ship is closest to port P hour after leaving A EXERCISE 16B.1 Each of the following vector equations represents the path of a moving object t is measured in seconds and t > Distances are measured in metres In each case find the: i initial position ii velocity vector iii speed of the object ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ x ¡4 12 x x ¡2 ¡6 = + t b = + t c = ¡7 + t ¡4 a y y ¡6 ¡4 y Find the velocity vector of a speed boat moving parallel to: ³ ´ ³ ´ 24 ¡1 a b with a speed of 150 km h with a speed of 12:5 km h¡1 ¡3 c 2i + j with a speed of 50 km h¡1 d ¡3i + 4j with a speed of 100 km h¡1 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Yacht A moves according to x(t) = + t, y(t) = ¡ 2t where the distance units are kilometres and the time units are hours Yacht B moves according to x(t) = + 2t, y(t) = ¡8 + t, t > 0: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\458IB_HL-2_16.CDR Tuesday, November 2007 4:22:11 PM PETERDELL IB_HL-2ed (459) 459 LINES AND PLANES IN SPACE (Chapter 16) a c d e Find the initial position of each yacht b Find the velocity vector of each yacht Show that the speed of each yacht is constant and state the speeds If they start at 6:00 am, find the time when the yachts are closest to each other Prove that the paths of the yachts are at right angles to each other ³ ´ Submarine P is at (¡5, 4) and fires a torpedo with velocity vector ¡1 at 1:34 pm ³ ´ ¡4 Submarine Q is at (15, 7) and a minutes later fires a torpedo in the direction ¡3 Distances are measured in kilometres and time is in minutes a Show that the position of P’s torpedo can be written as P(x1 (t), y1 (t)) where x1 (t) = ¡5 + 3t and y1 (t) = ¡ t b What is the speed of P’s torpedo? c Show that the position of Q’s torpedo can be written in the form x2 (t) = 15 ¡ 4(t ¡ a), y2 (t) = ¡ 3(t ¡ a): d Q’s torpedo is successful in knocking out P’s torpedo At what time did Q fire its torpedo and at what time did the explosion occur? ³ ´ Let represent a km displacement due east y ³ ´ 100 and represent a km displacement due north The control tower of an airport is at (0, 0) Aircraft within 100 km of (0, 0) will become visible on the radar screen at the control tower At 12:00 noon an aircraft is 200¡km east and 100¡km north of the control tower It is flying parallel to the ³ ´ p ¡3 vector b = ¡1 with a speed of 40 10 km h¡1 x -100 100 -100 a Write down the velocity vector of the aircraft b Write a vector equation for the path of the aircraft using t to represent the time in hours that have elapsed since 12:00 noon c Find the position of the aircraft at 1:00 pm d Show that the aircraft first becomes visible on the radar screen at 1:00 pm e Find the time when the aircraft is closest to the control tower and find the distance between the aircraft and the control tower at this time f At what time will the aircraft disappear from the radar screen? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Boat A’s position is given by x(t) = ¡ t, y(t) = 2t ¡ where the distance units are kilometres and the time units are hours Boat B’s position is given by x(t) = ¡ 3t, y(t) = ¡ 2t a Find the initial position of each boat b Find the velocity vector of each boat c What is the angle between the paths of the boats? d At what time are the boats closest to each other? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\459IB_HL-2_16.CDR Wednesday, 14 November 2007 12:48:11 PM PETERDELL IB_HL-2ed (460) 460 LINES AND PLANES IN SPACE (Chapter 16) GEOMETRIC APPLICATIONS OF r = a + tb Vector equations of two intersecting lines can be solved simultaneously to find the point where the lines meet Example 10 ³ ´ ³ ´ ³ ´ ¡2 = +s and ³ ´ ³ ´ ³ ´ x 15 ¡4 = + t , where s and t are scalars line has vector equation y x y Line has vector equation Use vector methods to find where the two lines meet ³ The lines meet where ¡2 ´ ³ ´ +s ) ¡2 + 3s = 15 ¡ 4t ) 3s + 4t = 17 (1) 3s + 4t = 17 8s ¡ 4t = 16 ) 11s ³ = 15 ´ ³ +t ¡4 ´ and + 2s = + t and 2s ¡ t = (2) f(2) is multiplied by 4g = 33 So, s = and in (2): 2(3) ¡ t = so t = ³ ´ ³ ´ ³ ´ ³ ´ x ¡2 Using line 1, = + = y ³ ´ ³ ´ ³ ´ ³ ´ x 15 ¡4 = + = Checking in line 2, y ) they meet at (7, 7) EXERCISE 16B.2 The triangle formed by the three lines is ABC ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ x ¡1 x Line (AB) is = + r , line (AC) is = + s y ¡2 y ³ ´ ³ ´ ³ ´ x 10 ¡2 = ¡3 + t where r, s and t are scalars and line (BC) is y a b c d Draw the three lines accurately on a grid Hence, find the coordinates of A, B and C Prove that ¢ABC is isosceles Use vector methods to check your answers to b A parallelogram is defined by four lines as follows: ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ x ¡4 x ¡4 Line (AB) is = + r , line (AD) is = + s , y y ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ x 22 ¡7 x 22 ¡1 = + t , line (CB) is = + u , line (CD) is y 25 ¡3 y 25 ¡2 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 where r, s, t and u are scalars black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\460IB_HL-2_16.CDR Tuesday, November 2007 4:27:01 PM PETERDELL IB_HL-2ed (461) 461 LINES AND PLANES IN SPACE (Chapter 16) a Draw an accurate sketch of the four lines and the parallelogram formed by them Label the vertices b From your diagram find the coordinates of A, B, C and D c Use vector methods to confirm your answers to b An isosceles triangle ABC is formed by these lines: ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ x x ¡1 = + r , line (BC) is = + s and Line (AB) is y y ¡2 ³ ´ ³ ´ ³ ´ x = + t where r, s and t are scalars line (AC) is y ¡1 a Use vector methods to find the coordinates of A, B and C b Which two sides of the triangle are equal in length? Find their lengths ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ 14 x 17 + r 10 , line QR is = + s and y ¡1 ¡9 ³ ´ ³ ´ ³ ´ x = + t where r, s and t are scalars line PR is y 18 ¡7 x y Line QP is ³ = ¡1 ´ Triangle PQR is formed by these lines a Use vector methods to find the coordinates of P, Q and R ¡! ¡ ¡! ¡ ! ! b Find vectors PQ and PR and evaluate PQ ² PR: c Hence, find the size of ]QPR d Find the area of ¢PQR Quadrilateral ABCD is formed by these lines: ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ x x 18 ¡8 = + r , line (BC) is = + s , Line (AB) is y y 32 ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ x 14 ¡8 x ¡3 = + t and line (AD) is = + u line (CD) is y 25 ¡2 y 12 where r, s, t and u are scalars a Use vector methods to find the coordinates of A, B, C and D ¡! ¡! b Write down vectors AC and DB and hence find: ¡! ¡! ¡! ¡! i j AC j ii j DB j iii AC ² DB c What the answers to b tell you about quadrilateral ABCD? C RELATIONSHIP BETWEEN LINES cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 the same line infinitely many solutions 95 lines not meet no solutions 100 one point of intersection unique solution 50 coincident 75 parallel 25 intersecting 95 100 50 75 25 LINE CLASSIFICATION IN DIMENSIONS black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\461IB_HL-2_16.CDR Wednesday, 14 November 2007 12:49:01 PM PETERDELL IB_HL-2ed (462) 462 LINES AND PLANES IN SPACE (Chapter 16) LINE CLASSIFICATION IN DIMENSIONS I I the lines are coplanar, or lie in the same plane: ² intersecting ² parallel ² coincident the lines are not coplanar and are hence skew Skew lines are any lines which are neither parallel nor intersecting ² If the lines are parallel, the angle between them is 0o q ² If the lines are intersecting, the angle between them is µ, as shown point of intersection ² If the lines are skew, there is still an angle that one line makes with the other If we translate one line to intersect the other, the angle between the original lines is defined as the angle between the intersecting lines, i.e., angle µ line line translated q line meet lines and are skew Example 11 Line has equations x = ¡1 + 2s, y = ¡ 2s and z = + 4s Line has equations x = ¡ t, y = t and z = ¡ 2t Show that the lines are parallel à ! à ! à ! à ! x y z Line is ¡1 1 = ¡2 +s with direction vector à ! ¡1 ¡2 Likewise, line has direction vector à ! à ! ¡2 Since ¡1 ¡2 = ¡2 ¡2 , the lines are parallel fIf a = kb for some scalar k, then a k b.g Example 12 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 x = ¡1 + 2s, y = ¡ 2s and z = + 4s x = ¡ t, y = t and z = ¡ 2t x = + 2u, y = ¡1 ¡ u, z = + 3u and line intersect and find the angle between them and line are skew 100 50 75 25 95 100 50 75 25 Line has equations Line has equations Line has equations a Show that line b Show that line black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\462IB_HL-2_16.CDR Wednesday, 14 November 2007 12:49:32 PM PETERDELL IB_HL-2ed (463) 463 LINES AND PLANES IN SPACE (Chapter 16) a Equating x, y and z values in lines and gives and ¡ 2t = + 3u ¡ t = + 2u t = ¡1 ¡ u ) t = ¡2u, ) t = ¡1 ¡ u, and 3u + 2t = ¡1 (1) {z } | Solving these we get ¡2u = ¡1 ¡ u ) ¡u = ¡1 ) u = and so t = ¡2 Checking in (1): 3u + 2t = 3(1) + 2(¡2) = ¡ = ¡1 X ) u = 1, t = ¡2 satisfies all three equations, a common solution Using u = 1, lines and meet at (1 + 2(1), ¡1 ¡ (1), + 3(1)) i.e., (3, ¡2, 7) à ! à ! ¡1 ¡2 Direction vectors for lines and are a = respectively Now cos µ = and b = ¡1 ja ² bj j¡2 ¡ ¡ 6j p , where µ is the acute = p jajjbj 1+1+4 4+1+9 angle between a and b ) cos µ = p 84 and so µ ¼ 10:89o ) the angle between lines and is about 10:9o b Equating x, y and z values in lines and gives and + 4s = + 3u ¡1 + 2s = + 2u ¡ 2s = ¡1 ¡ u and 4s ¡ 3u = (1) ) 2s ¡ 2u = 2, ) ¡2s + u = ¡2, {z } | Solving these we get 2s ¡ 2u = ) ¡2s + u = ¡2 ) ¡u = fadding themg ) u = and so 2s = i.e., s = Checking in (1), 4s ¡ 3u = 4(1) ¡ 3(0) = 6= So, there is no simultaneous solution to all equations ) the lines cannot meet, and as they are not parallel they must be skew Example 13 ½ Discuss the solutions to 3x ¡ y = 6x ¡ 2y = k for k R Give a geometric interpretation of the solutions · In augmented matrix form If k ¡ = 0, ¡1 ¡2 k ¸ · » ¡1 k¡4 ¸ R2 ¡ 2R1 the system is consistent, and the lines 3x ¡ y = and 6x ¡ 2y = are coincident cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Let x = t, so y = ¡2 + 3t There are infinitely many solutions of the form x = t, y = ¡2 + 3t, t R black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\463IB_HL-2_16.CDR Tuesday, November 2007 4:35:11 PM PETERDELL IB_HL-2ed (464) 464 LINES AND PLANES IN SPACE (Chapter 16) ³ ´ This line has direction vector If k ¡ 6= or slope and passes through the point (0, ¡2) i.e., k 6= the system is inconsistent In this case the lines 3x ¡ y = and 6x ¡ 2y = k are parallel and have no points of intersection PERPENDICULAR AND PARALLEL TESTS (for 2-D AND 3-D) I I Non-zero vectors v and w are perpendicular if v ² w = parallel if v = kw for some scalar k If v ² w = then j v j j w j cos µ = ) cos µ = fv, w non-zerog ) µ = 90o Proof of perpendicular case: EXERCISE 16C Classify the following line pairs as either parallel, intersecting or skew, and in each case find the measure of the acute angle between them: a x = + 2t, y = ¡ t, z = + t and x = ¡2 + 3s, y = ¡ s, z = + 2s b x = ¡1 + 2¸, y = ¡ 12¸, z = + 12¸ and x = 4¹ ¡ 3, y = 3¹ + 2, z = ¡¹ ¡ c x = 6t, y = + 8t, z = ¡1 + 2t and x = + 3s, y = 4s, z = + s d x = ¡ y = z + and x = + 3s, y = ¡2 ¡ 2s, z = 2s + 12 e x = + ¸, y = ¡ ¸, z = + 2¸ and x = + 3¹, y = ¡ 2¹, z = ¹ ¡ f x = ¡ 2t, y = + t, z = and x = + 4s, y = ¡1 ¡ 2s, z = Consider the two lines whose equations are 3x ¡ y = and 6x ¡ 2y = k where k is some real number Discuss the nature of the intersection of these lines for different values of k Discuss for the different values of a, the geometric solutions of the equations 4x + 8y = and 2x ¡ ay = 11 SHORTEST DISTANCE FROM A POINT TO A LINE (2-D AND 3-D) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The coordinates of any point P on a line can be expressed in terms of the parameter t ¡ ! We can hence find the vector AP in terms of t, where A is a point which is not on the line ¡ ! The shortest distance d occurs when AP is perpendicular to b, ¡ ! so we find t for which AP ² b = ¡! The shortest distance d = j AP j for this value of t black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\464IB_HL-2_16.CDR Tuesday, November 2007 4:39:32 PM PETERDELL b P d A IB_HL-2ed (465) 465 LINES AND PLANES IN SPACE (Chapter 16) Example 14 Find the shortest distance from P(¡1, 2, 3) to the line x¡1 y+4 = = z ¡ 3 The equation of the line in parametric form is x = + 2¸, y = ¡4 + 3¸, z = + ¸ ) any point A on this line has coordinates (1 + 2¸, ¡4 + 3¸, + ¸) ! à ! à + 2¸ ¡ ! and the direction vector of the line is b = ) PA = ¡6 + 3¸ ¸ ¡ ! Now for the shortest distance, PA ² b = fperpendicular distanceg ) + 4¸ ¡ 18 + 9¸ + ¸ = ) 14¸ = 14 and so ¸ = à ! p ¡ ! ¡ ! and d = j PA j = 26 units Thus PA = ¡3 A P(-1, 2, 3) Find the shortest distance from the point (2, ¡3) to the line 3x ¡ y = 4: Find the shortest distance from the point (3, 0, ¡1) to the line with equation r = 2i ¡ j + 4k + ¸(3i + 2j + k): à ! à ! Find the shortest distance from (1, 1, 3) to the line r = ¡1 +¸ THE SHORTEST DISTANCE BETWEEN SKEW LINES (EXTENSION) Example 15 Find the shortest distance between the skew lines x = t, y = ¡ t, z = + t and x = ¡ s, y = ¡1 + 2s, z = ¡ s à The lines have direction vectors v = ¯ ¯ i ¯ so v £ w = ¯ ¯ ¡1 ¯ k ¯ ¯ ¯ ¡1 ¯ j ¡1 ¯ ¯ ¡1 =¯ ¡1 ! à and w = ¯ ¯ ¯ 1 ¯ i ¡ ¯ ¡1 ¯ ¡1 ¡1 ¡1 ! ¯ ¯ ¯ 1 ¯ j + ¯ ¡1 ¯ ¡1 ¯ ¡1 ¯ k= ¯ à ¡1 ! Let A and B be points on the skew lines ) A is (t, ¡ t, + t) and B is (3 ¡ s, ¡1 + 2s, ¡ s) where s and t are scalars ¡! To find the shortest distance between the two skew lines, we need to find j AB j ¡! where AB is parallel to v £ w, i.e., perpendicular to both skew lines ! à ! à 3¡s¡t ¡1 ¡! ¡1 + 2s ¡ + t =k for some scalar k As AB k v £ w, cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 4¡s¡2¡t black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\465IB_HL-2_16.CDR Wednesday, November 2007 9:19:28 AM PETERDELL IB_HL-2ed (466) 466 LINES AND PLANES IN SPACE (Chapter 16) ) ¡ s ¡ t = ¡k, Thus 2s + t = and ) 2s + t = and ¡2 + 2s + t = and ¡ s ¡ t = k ¡ s ¡ t = ¡2 + s + t 2s + 2t = Solving simultaneously, t = and s = ¡ 12 1 q ¡! ¡! @ A and j AB j = 14 + + ) AB = ¡ 12 = p1 units p1 ) the shortest distance is units Find the shortest distance between the skew lines: a x = + 2t, y = ¡t, z = + 3t and x = y = z b x = ¡ t, y = + t, z = ¡ t and x = + s, y = ¡ 2s, z = s Find the shortest distance between the lines given in Exercise 16C question ² Note: ² To find the distance between parallel lines, find the distance from a point on one line to the other line What is the shortest distance between intersecting/coincident lines? D PLANES AND DISTANCES To find the equation of a plane, we need to know a point on the plane and also its orientation in space The orientation of a plane cannot be given by a single parallel vector because infinitely many planes of different orientation are parallel to a single direction vector We require two non-parallel vectors to define the orientation uniquely Any point R(x, y, z) on the plane with a known point A(a1 , a2 , a3 ) and two non-parallel à ! à ! b1 b2 b3 vectors b = and c = c1 c2 c3 must satisfy the vector equation ¡! AR = ¸b + ¹c for some scalars ¸ and ¹ ¡! ¡! ) OR ¡ OA = ¸b + ¹c ¡! ¡! ) OR = OA + ¸b + ¹c r = a + ¸b + ¹c is the vector equation of the plane cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 where r is the position vector of any point on the plane, a is the position vector of the known point A(a1 , a2 , a3 ) on the plane b and c are any two non-parallel vectors that are parallel to the plane ¸, ¹ R are two independent parameters black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\466IB_HL-2_16.CDR Wednesday, November 2007 9:21:06 AM PETERDELL IB_HL-2ed (467) LINES AND PLANES IN SPACE (Chapter 16) 467 Another way of defining the direction of a plane is to consider the vector cross product of the two vectors b and c which are parallel to the plane The vector n = b £ c is called a normal vector to the plane n is perpendicular to b and c and is hence perpendicular to any vector or line in or parallel to the plane This is because any vector parallel to the plane can be written in the form ¸b + ¹c Challenge: Show n ? ¸b + ¹c à ! a b c Suppose a plane in space has normal vector n = and that it includes the fixed point A(x1 , y1 , z1 ) R(x, y, z) is any other point in the plane ¡! Now AR is perpendicular to n ¡! ) n ² AR = ! à ! à a b c ) x ¡ x1 y ¡ y1 z ¡ z1 ² à ! n= a b c R(x, y, z) A(xz, yz, zz) =0 ) a(x ¡ x1 ) + b(y ¡ y1 ) + c(z ¡ z1 ) = ) ax + by + cz = ax1 + by1 + cz1 where the RHS is a constant ¡! n ² AR = is another form of the vector equation of the plane It could also be written as n ² (r ¡ a) = 0, which implies r ² n = a ² n à ! a b c If a plane has normal vector n = and passes through (x1 , y1 , z1 ) then it has equation ax + by + cz = ax1 + by1 + cz1 = d, where d is a constant This is the Cartesian equation of the plane Example 16 à ! Find the equation of the plane with normal vector and containing (¡1, 2, 4) à ! Since n = and (¡1, 2, 4) lies on the plane, the equation is cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 x + 2y + 3z = (¡1) + 2(2) + 3(4) i.e., x + 2y + 3z = 15 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\467IB_HL-2_16.CDR Friday, November 2007 10:12:59 AM PETERDELL IB_HL-2ed (468) 468 LINES AND PLANES IN SPACE (Chapter 16) Example 17 Find the equation of the plane through A(¡1, 2, 0), B(3, 1, 1) and C(1, 0, 3): a in vector form b in Cartesian form à ! à ! ¡! ¡! and CB = a AB = ¡1 ¡2 ¡! ¡! Now AB and CB are two non-parallel vectors both parallel to the plane à ! à ! à ! à ! ) r= x y z = +¸ ¡1 ¡2 +¹ , ¸, ¹ R using C as the known (fixed) point on the plane à ! à ! ¡! ¡! If n is the normal vector, then n = AB £ AC = ¡1 £ ¡2 b ¯ ¯ ¡1 ) n = ¯ ¡2 ¯ 1¯ 3¯ ¯ ¯4 i ¡ ¯2 ¯ 1¯ 3¯ ¯ ¯4 j + ¯2 à ¯ ¡1 ¯ ¡2 ¯ k= ¡1 ¡10 ¡6 ! à =¡ 10 ! Thus the plane has equation x + 10y + 6z = (¡1) + 10(2) + 6(0) fusing Ag ) x + 10y + 6z = 19 Note: Check that all points satisfy this equation Example 18 Find the parametric equations of the line through A(¡1, 2, 3) and B(2, 0, ¡3) and hence find where this line meets the plane with equation x ¡ 2y + 3z = 26: à ! ¡! so line AB has parametric equations AB = ¡2 ¡6 x = ¡1 + 3t, y = ¡ 2t, z = ¡ 6t (¤) This line meets the plane x ¡ 2y + 3z = 26 where ¡1 + 3t ¡ 2(2 ¡ 2t) + 3(3 ¡ 6t) = 26 ) ¡ 11t = 26 ) ¡11t = 22 and so t = ¡2 ) it meets the plane at (¡7, 6, 15) fsubstituting t = ¡2 into ¤g Example 19 Find the coordinates of the foot of the normal from A(2, ¡1, 3) to the plane x ¡ y + 2z = 27 Hence find the shortest distance from A to the plane à ! A x ¡ y + 2z = 27 has normal vector n yellow 95 100 50 75 25 95 50 75 25 95 50 75 25 95 100 50 75 25 100 magenta 100 ) the parametric equations of AN are x = + t, y = ¡1 ¡ t, z = + 2t N cyan ¡1 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\468IB_HL-2_16.CDR Tuesday, 13 November 2007 9:40:28 AM PETERDELL IB_HL-2ed (469) LINES AND PLANES IN SPACE (Chapter 16) 469 and this line meets the plane x ¡ y + 2z = 27 where + t ¡ (¡1 ¡ t) + 2(3 + 2t) = 27 ) + t + + t + + 4t = 27 ) 6t + = 27 ) 6t = 18 ) t=3 ) N is (5, ¡4, 9) p The shortest distance, AN = (5 ¡ 2)2 + (¡4 ¡ ¡1)2 + (9 ¡ 3)2 p = 54 units Example 20 Find the coordinates of the foot of the normal N from A(2, ¡1, 3) to the plane with equation r = i + 3k + ¸(4i ¡ j + k) + ¹(2i + j ¡ 2k), ¸, ¹ R : The normal to the plane has direction vector given by ¯ à ! ¯ ¯ ¯i j ¯ ¡1 k1 ¯ (4i ¡ j + k) £ (2i + j ¡ 2k) = ¯ ¯ = 10 ¯ ¡2 ¯ à ! à à ! x y z The equation of the normal through A is = ¡1 +t 10 ! so N must have coordinates of the form (2 + t, ¡ + 10t, + 6t) ! à ! à ! à à 2+t ¡1 + 10t + 6t But N lies in the plane, so < + t = + 4¸ + 2¹ ¡1 + 10t = ¡¸ + ¹ ) : + 6t = + ¸ ¡ 2¹ = < +¸ ¡1 +¹ ¡2 ! 4¸ + 2¹ ¡ t = ¡¸ + ¹ ¡ 10t = ¡1 : ¸ ¡ 2¹ ¡ 6t = and so Solving simultaneously with technology gives ¸ = ¢ ¡ 47 54 ) N is the point 137 , ¡ 137 , 137 40 137 , ¹= ¡7 137 , t= 137 Check by substituting for ¸ and ¹ in the equation of the plane EXERCISE 16D Find the equation of the plane: à ! a with normal vector ¡1 and through (¡1, 2, 4) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b perpendicular to the line connecting A(2, 3, 1) and B(5, 7, 2) and through A c perpendicular to the line connecting A(1, 4, 2) and B(4, 1, ¡4) and containing P such that AP : PB = : d containing A(3, 2, 1) and the line x = + t, y = ¡ t, z = + 2t black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\469IB_HL-2_16.CDR Wednesday, November 2007 10:33:31 AM PETERDELL IB_HL-2ed (470) 470 LINES AND PLANES IN SPACE (Chapter 16) State the normal vector to the plane with equation: a 2x + 3y ¡ z = b 3x ¡ y = 11 c z=2 d x=0 Find the equation of the: a XOZ-plane b plane perpendicular to the Z-axis and through (2, ¡1, 4) Find the equation of the plane in i vector form ii Cartesian form, through: a A(0, 2, 6), B(1, 3, 2) and C(¡1, 2, 4) b A(3, 1, 2), B(0, 4, 0) and C(0, 0, 1) c A(2, 0, 3), B(0, ¡1, 2) and C(4, ¡3, 0) Find the equations of the following lines: a through (1, ¡2, 0) and normal to the plane x ¡ 3y + 4z = b through (3, 4, ¡1) and normal to the plane x ¡ y ¡ 2z = 11 Find the parametric equations of the line through A(2, ¡1, 3) and B(1, 2, 0) and hence find where this line meets the plane with equation x + 2y ¡ z = Find the parametric equations of the line through P(1, ¡2, 4) and Q(2, 0, ¡1) and hence find where this line meets: b a the Y OZ-plane c the line with equations the plane with equation y + z = x¡3 y+2 z ¡ 30 = = ¡1 In the following, find the foot of the normal from A to the given plane and hence find the shortest distance from A to the plane: b A(2, ¡1, 3); x ¡ y + 3z = ¡10 a A(1, 0, 2); 2x + y ¡ 2z + 11 = c A(1, ¡4, ¡3); 4x ¡ y ¡ 2z = Find the coordinates of the mirror image of A(3, 1, 2) when reflected in the plane x + 2y + z = 10 Does the line through (3, 4, ¡1) and normal to x + 4y ¡ z = ¡2 intersect any of the coordinate axes? 11 Find the equations of the plane through A(1, 2, 3) and B(0, ¡1, 2) which is parallel to: a the X-axis b the Y -axis c the Z-axis y¡2 = z + and x + = y ¡ = 2z + are coplanar and find the equation of the plane which contains them 12 Show that the lines x ¡ = 13 A(1, 2, k) lies on the plane x + 2y ¡ 2z = Find: a the value of k b the coordinates of B such that AB is normal to the plane and units from it 14 In the following, find the foot of the normal from A to the given plane and hence find the shortest distance from A to the plane: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a A(3, 2, 1); r = 3i + j + 2k + ¸(2i + j + k) + ¹(4i + 2j ¡ 2k) b A(1, 0, ¡2); r = i ¡ j + k + ¸(3i ¡ j + 2k) + ¹(¡i + j ¡ k) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\470IB_HL-2_16.CDR Friday, November 2007 10:15:41 AM PETERDELL IB_HL-2ed (471) 471 LINES AND PLANES IN SPACE (Chapter 16) 15 Q is any point in the plane Ax + By + Cz + D = d is the distance from P(x1 , y1 , z1 ) to the given plane ¡! j QP ² n j a Explain why d = jnj P(xz, yz, zz) Ax + By + Cz + D =¡0 d Q N j Ax1 + By1 + Cz1 + D j p A2 + B + C b Hence, show that d = c Check your answers to question using the formula of b 16 Find the distance from: a (0, 0, 0) to x + 2y ¡ z = 10 Note: ² ² (1, ¡3, 2) to x + y ¡ z = b To find the distance between two parallel planes, find a point on one of the planes and use the method in Example 20 To find the distance between a line and a plane, both of which are parallel, find a point on the line and use the method in Example 20 17 Find the distance between the parallel planes: a x + y + 2z = and 2x + 2y + 4z + 11 = b ax + by + cz + d1 = and ax + by + cz + d2 = 18 Show that the line x = + t, y = ¡1 + 2t, z = ¡3t is parallel to the plane 11x ¡ 4y + z = 0, and find its distance from the plane 19 Find the equations of the two planes which are parallel to 2x ¡ y + 2z = and units from it E ANGLES IN SPACE THE ANGLE BETWEEN A LINE AND A PLANE Suppose a line has direction vector d and a plane has normal vector n We allow n to intersect the line making an angle of µ with it The required acute angle is Á and line n sin Á = cos µ = j n²d j j n jj d j d q q ) ¡1 Á = sin µ j n²d j j n jj d j ¶ f cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 plane black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\471IB_HL-2_16.CDR Wednesday, November 2007 10:40:47 AM PETERDELL IB_HL-2ed (472) 472 LINES AND PLANES IN SPACE (Chapter 16) Example 21 Find the angle between the plane x + 2y ¡ z = and the line with equations x = t, y = ¡ t, z = + 2t à n= n q d ¡1 ! à and d = ¡1 ! µ j1 ¡ ¡ 2j p p Á = sin 1+4+1 1+1+4 ¶ µ ¡1 p p = sin 6 ¡ ¢ ¡1 o = sin = 30 ¡1 q f ¶ THE ANGLE BETWEEN TWO PLANES nz nx q nx nz q plane 180-q plane q view here plane plane cos µ = j n1 ² n2 j is the cosine of the acute angle between two planes j n1 j j n2 j if two planes have normal vectors n1 and n2 and µ is the µ ¶ j n1 ² n2 j ¡1 acute angle between them then µ = cos j n1 j j n2 j So, Example 22 Find the angle between the planes with equations x + y ¡ z = and 2x ¡ y + 3z = ¡1 à nx x + y ¡ z = has normal vector n1 = nz 1 ¡1 à Px 2x ¡ y + 3z = ¡1 has normal vector n2 = Pz ! ¡1 and ! cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 If µ is the acute angle between the planes then µ ¶ j n1 ² n2 j ¡1 µ = cos j n1 j j n2 j black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\472IB_HL-2_16.CDR Friday, November 2007 10:17:40 AM PETERDELL IB_HL-2ed (473) 473 LINES AND PLANES IN SPACE (Chapter 16) µ j2 + ¡1 + ¡3j p p = cos 1+1+1 4+1+9 ¶ µ ¡1 p = cos 42 ¼ 72:0o ¶ ¡1 EXERCISE 16E Find the angle between: y+1 x¡1 = = z+2 b the plane 2x ¡ y + z = and the line x = t + 1, y = ¡1 + 3t, z = t c the plane 3x + 4y ¡ z = ¡4 and the line x ¡ = ¡ y = 2(z + 1) d the plane rp = 2i ¡ j + k + ¸(3i ¡ 4j ¡ k) + ¹(i + j ¡ 2k) and the line rl = 3i + 2j ¡ k + t(i ¡ j + k) a the plane x ¡ y + z = and the line Find the acute angle between the planes with equations: a 2x ¡ y + z = b x ¡ y + 3z = x + 3y + 2z = 3x + y ¡ z = ¡5 c 3x ¡ y + z = ¡11 2x + 4y ¡ z = d r1 = 3i + 2j ¡ k ¡¸(i ¡ j + k) + ¹(2i ¡ 4j + 3k) and r2 = i + j ¡ k ¡ ¸(2i + j + k) + ¹(i + j + k) à ! à ! à ! e 3x ¡ 4y + z = ¡2 and r = F ¡1 +¸ ¡1 +¹ 1 THE INTERSECTION OF TWO OR MORE PLANES ² Two planes in space could be: (1) intersecting (2) parallel (3) coincident Pz=Px ² Three planes in space could be: (1) all coincident (2) two coincident and one intersecting Pc Pz=Px=Pc Pz=Px (3) two coincident and one parallel Pz=Px cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Pc black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\473IB_HL-2_16.CDR Wednesday, November 2007 10:58:09 AM PETERDELL IB_HL-2ed (474) 474 LINES AND PLANES IN SPACE (Chapter 16) (4) two parallel and one intersecting (5) all parallel (7) all meet in a common line (8) the line of intersection of any is parallel to the third plane all meet at the one point (6) The program Winplot by Peanuts software displays these cases This work should be linked to Chapter 13 on matrices Solutions can be found by using inverse matrices (where there is a unique solution) and/or row operations on the augmented matrix The use of row operations on the augmented matrix is essential when the solution is not unique This is when the matrix of coefficients is singular (determinant = 0) and thus not invertible Example 23 a b a x + 3y ¡ z = 3x + 5y ¡ z = x ¡ 5y + (2 ¡ m)z = ¡ m2 for any real number m Give geometric interpretations of your results Hence solve x + 3y ¡ z = 0, giving a geometric interpretation 3x + 5y ¡ z = x ¡ 5y + z = Use elementary row operations to solve the system: Augmented matrix ·1 ¡1 » » ¡5 ¡1 2¡m 0 ¡ m2 0 ¡4 ¡8 ¡1 3¡m 0 ¡ m2 ¡1 ¡1 ¡m ¡ 0 ¡ m2 ·1 ·1 0 ¸ ¸ R2 ! R2 ¡ 3R1 R3 ! R3 ¡ R1 ¸ R2 ! R2 £ ¡ 12 R3 ! R3 ¡ 2R2 · Case (1) If m = ¡1, the augmented matrix becomes 0 ¡1 ¡1 0 ¸ The system is inconsistent, so there are no solutions ) the three planes not have a common point of intersection à ! à ! à ! magenta yellow 95 , n3 = 100 50 25 75 ¡1 n2 = 95 , 100 50 25 95 100 50 75 25 95 100 50 75 25 cyan 75 ¡1 The normals are n1 = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\474IB_HL-2_16.CDR Friday, November 2007 10:23:07 AM PETERDELL ¡5 IB_HL-2ed (475) 475 LINES AND PLANES IN SPACE (Chapter 16) None of the planes are parallel ) the line of intersection of any two is parallel to the third plane (diag (8)) (¡m ¡ 1)z = ¡ m2 ) (m + 1)z = m2 ¡ m2 ¡ ) z= m+1 Case (2) If m 6= ¡1 there is a unique solution From row 2, 2y ¡ z = ) y = 12 z = From row 1, x = ¡3y + z = m2 ¡ 2(m + 1) ¡ m2 ¡3(m2 ¡ 9) 2(m2 ¡ 9) + = 2(m + 1) 2(m + 1) 2(m + 1) The three planes meet at a unique point (diag (6)) ¶ µ ¡ m2 m2 ¡ m2 ¡ , , provided m 6= ¡1 2(m + 1) 2(m + 1) m + b Comparing with a, ¡ m = and ¡ m2 = ) m = ¡ ¡8 ¡8 ¢ ) the three planes meet at the point , , , i.e., (2, ¡2, ¡4) Example 24 a Find the intersection of the planes: b Hence solve: x + y + 2z = 2x + y ¡ z = x¡y¡z =5 · a x + y + 2z = 2x + y ¡ z = The augmented matrix 1 Give a geometric interpretation of your result ¡1 ¸ · » 1 ¸ The system is consistent with infinite solutions Let z = t so y = ¡5t, x = + 3t, t R are the solutions, ) the two planes meet in the line x = + 3t, y = ¡5t, z = t, t R : à ! Note: This line passes through (2, 0, 0) has direction vector cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 The solutions of the first equations are x = + 3t, y = ¡5t, z = t ffrom ag Substitution into the third equation gives + 3t + 5t ¡ t = ) 7t = ) t = 37 23 15 ) x= 7, y=¡7, z= ¢ ¡ 15 ) the three planes meet at the unique point 23 , ¡ , 75 25 95 100 50 75 25 b ¡5 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\475IB_HL-2_16.CDR Wednesday, November 2007 11:47:51 AM PETERDELL IB_HL-2ed (476) 476 LINES AND PLANES IN SPACE (Chapter 16) EXERCISE 16F a How many solutions are possible when solving simultaneously: a1 x + b1 y + c1 z = d1 a2 x + b2 y + c2 z = d2 ? b Under what conditions will the planes in a be: i parallel ii coincident? c Solve the following using elementary row operations and interpret each system of equations geometrically: i x ¡ 3y + 2z = ii 2x + y + z = iii x + 2y ¡ 3z = 3x ¡ 9y + 2z = x¡y+z = 3x + 6y ¡ 9z = 18 Discuss the possible solutions of the following systems where k is a real number Interpret the solutions geometrically: b x ¡ y + 3z = a x + 2y ¡ z = 2x + 4y + kz = 12 2x ¡ 2y + 6z = k For the eight possible geometric solutions of three planes in space, comment on the possible solutions in each case For example, has infinitely many solutions where x, y and z are expressed in terms of two parameters s and t Pz=Px=Pc Solve the following systems using elementary row operations and in each case state the geometric meaning of your solution: b x ¡ y + 2z = c x + 2y ¡ z = a x + y ¡ z = ¡5 2x + y ¡ z = 2x ¡ y ¡ z = x ¡ y + 2z = 11 4x + y ¡ 5z = ¡18 5x ¡ 2y + 5z = 11 3x ¡ 4y ¡ z = x¡y+z = 2x ¡ 2y + 2z = 11 x + 3y ¡ z = ¡2 d x + y ¡ 2z = x¡y+z = 3x + 3y ¡ 6z = e f x¡y¡z = x+y+z = 5x ¡ y + 2z = 17 x ¡ y + 3z = 2x ¡ 3y ¡ z = 3x ¡ 5y ¡ 5z = k where k takes all real values State the geometrical meaning of the different solutions Solve the system of equations Find all values of m for which x + 2y + mz = ¡1 2x + y ¡ z = mx ¡ 2y + z = has a unique solution In the cases where there is no unique solution, solve the system Give geometrical meanings to all possible solutions Illustrate each case Find if and where the following planes meet: à P1 : r1 = 2i ¡ j + ¸(3i + k) + ¹(i + j ¡ k) P3 : r3 = ¡1 ! à +t ¡1 ! à ¡u ¡1 ! cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 P2 : r2 = 3i ¡ j + 3k + r(2i ¡ k) + s(i + j) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\476IB_HL-2_16.CDR Tuesday, 13 November 2007 9:48:38 AM PETERDELL IB_HL-2ed (477) LINES AND PLANES IN SPACE (Chapter 16) 477 REVIEW SET 16A (2-D) Find a the vector equation b the parametric equations ³ ´ of the line that passes through (¡6, 3) with direction ¡3 Find the vector equation of the line which cuts the y-axis at (0, 8) and has direction 5i + 4j ³ ´ ³ ´ ³ ´ x 18 ¡7 = + t Find m (¡3, m) lies on the line with vector equation y ¡2 4 Find the velocity vector of an object that is moving in the direction 3i ¡ j with a speed of 20 km h¡1 A particle at P(x(t), y(t)) moves such that x(t) = ¡4 + 8t and y(t) = + 6t, where t > is the time in seconds The distance units are metres Find the: a initial position of P b position of P after seconds c speed of P d velocity vector of P p A yacht is sailing at a constant speed of 10 km h¡1 in the direction ¡i ¡ 3j Initially it is at point (¡6, 10) A beacon is at (0, 0) at the centre of a tiny atoll a Find in terms of i and j : i the initial position vector of the yacht ii the direction vector of the yacht iii the position vector of the yacht at any time t hours, t > b Find the time when the yacht is closest to the beacon c If there is a reef of radius km around the atoll, will the yacht hit the reef? ³ ´ Submarine X23 is at (2, 4) It fires a torpedo with velocity vector ¡3 at exactly ³ ´ ¡1 2:17 pm Submarine Y18 is at (11, 3) It fires a torpedo with velocity vector a at 2:19 pm to intercept the torpedo from X23 a b c d Find x1 (t) and y1 (t) for the torpedo fired from submarine X23 Find x2 (t) and y2 (t) for the torpedo fired from submarine Y18 At what time does the interception occur? What was the direction and speed of the interception torpedo? Trapezoid (trapezium) KLMN is formed by the lines: ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ x x 33 ¡11 line (KL) is = +p ; line (ML) is = +q ; y 19 ¡2 y ¡5 16 ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ x x 43 ¡5 = + r 10 ; line (MN) is = ¡9 + s line (NK) is y y where p, q, r and s are scalars cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 Which two lines are parallel? Why? Which lines are perpendicular? Why? Use vector methods to find the coordinates of K, L, M and N Calculate the area of trapezium KLMN 75 25 95 100 50 75 25 a b c d black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\477IB_HL-2_16.CDR Wednesday, November 2007 12:01:05 PM PETERDELL IB_HL-2ed (478) 478 LINES AND PLANES IN SPACE (Chapter 16) REVIEW SET 16B (3-D) Show that A(1, 0, 4), B(3, 1, 12), C(¡1, 2, 2) and D(¡2, 0, ¡5) are coplanar Find: a the equation of the plane b the coordinates of the nearest point on the plane to E(3, 3, 2) A is (3, 2, ¡1) and B is (¡1, 2, 4) a Write down the vector equation of the line through A and B b Find the equation of the plane through B with normal AB p c Find two points on the line AB which are 41 units from A P1 is the plane 2x ¡ y ¡ 2z = and P2 is the plane x + y + 2z = l is the line with parametric equations x = t, y = 2t ¡ 1, z = ¡ t b between P1 and P2 Find the acute angle: a that l makes with P1 For A(3, ¡1, 1) and B(0, 2, ¡1), find the: a vector equation of the line passing through A and B b the coordinates of P which divides BA in the ratio : 5 For C(¡3, 2, ¡1) and D(0, 1, ¡4) find the coordinates of the point(s) where the line passing through C and D meets the plane with equation 2x ¡ y + z = y+9 z ¡ 10 x¡8 = = and x = 15 + 3t, y = 29 + 8t, z = ¡ 5t: ¡16 a show that they are skew b find the acute angle between them c find the shortest distance between them Given the lines a How far is X(¡1, 1, 3) from the plane x ¡ 2y ¡ 2z = 8? b Find the coordinates of the foot of the perpendicular from Q(¡1, 2, 3) to the line ¡ x = y ¡ = ¡ 12 z P(2, 0, 1), Q(3, 4, ¡2) and R(¡1, 3, 2) are three points in space Find: ¡ ! ¡! ¡! à ! a PQ, j PQ j and QR b the parametric equations of line PQ c Use a to find the vector equation of the plane PQR Given the point A(¡1, 3, 2), the plane 2x ¡ y + 2z = 8, and the line defined by x = ¡ 2t, y = ¡6 + t, z = + 5t, find: a the distance from A to the plane b the coordinates of the point on the plane nearest to A c the shortest distance from A to the line cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 a Find the equation of the plane through A(¡1, 0, 2), B(0, ¡1, 1) and C(1, 2, ¡1) b Find the equation of the line, in parametric form, which passes through the origin and is normal to the plane in a c Find the point where the line in b intersects the plane in a 100 50 75 25 10 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\478IB_HL-2_16.CDR Friday, November 2007 10:24:11 AM PETERDELL IB_HL-2ed (479) 479 LINES AND PLANES IN SPACE (Chapter 16) x¡y+z = 2x + y ¡ z = ¡1 7x + 2y + kz = ¡k 11 Solve the system for any real number k using elementary row operations Give geometric interpretations of your results à ! à ! ¡1 12 p = ¡1 and q = a Find p £ q b Find m if p £ q is perpendicular to the line l with equation à ! à ! ¡2 r= m +¸ c Find the equation of the plane P containing l which is perpendicular to p £ q d Find t if the point A(4, t, 2) lies on the plane P e If B is the point (6, ¡3, 5), find the exact value of the sine of the angle between the line AB and the plane P REVIEW SET 16C For a b c A(¡1, 2, 3), B(2, 0, ¡1) and C(¡3, 2, ¡4) find: the equation of the plane defined by A, B and C the measure of angle CAB r given that D(r, 1, ¡r) is a point such that angle BDC is a right angle a Find where the line through L(1, 0, 1) and M(¡1, 2, ¡1) meets the plane with equation x ¡ 2y ¡ 3z = 14 b Find the shortest distance from L to the plane Given A(¡1, 2, 3), B(1, 0, ¡1) and C(1, 3, 0), find: a the normal vector to the plane containing A, B and C b D, the fourth vertex of parallelogram ACBD c the coordinates of the foot of the perpendicular from C to the line AB z¡3 y+2 = is parallel to the plane 6x +7y ¡5z = and find the distance between them Show that the line x¡ = x¡3 z+1 Consider the lines with equations =y¡4= and x = ¡1 + 3t, ¡2 y = + 2t, z = ¡ t a Are the lines parallel, intersecting or skew? Justify your answer b Determine the cosine of the acute angle between the lines cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 For A(2, ¡1, 3) and B(0, 1, ¡1), find: a the vector equation of the line through A and B, and hence b the coordinates of C on AB which is units from A black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\479IB_HL-2_16.CDR Wednesday, November 2007 12:22:22 PM PETERDELL IB_HL-2ed (480) 480 LINES AND PLANES IN SPACE (Chapter 16) Find the equation of the plane through A(¡1, 2, 3), B(1, 0, ¡1) and C(0, ¡1, 5) If X is (3, 2, 4), find the angle that AX makes with this plane a Find all vectors of length units which are normal to the plane x ¡ y + z = b Find a unit vector parallel to i + rj + 3k and perpendicular to 2i ¡ j + 2k c The distance from A(¡1, 2, 3) to the plane with equation 2x ¡ y + 2z = k is units Find k M P S Q Use vector methods to determine the measure of angle QDM given that M is the midpoint of PS R cm A D cm 10 cm B C 10 P(¡1, 2, 3) and Q(4, 0, ¡1) are two points in space Find: ¡ ! ¡! a PQ b the angle that PQ makes with the X-axis 11 ABC is a triangle in space M is the midpoint of side [BC] and O is the origin ¡! ¡! ¡! ¡! P is a point such that OP = 13 (OA + OB + OC) ¡! ¡! ¡! a Write OM in terms of OB and OC ¡! ¡! ¡ ! b Hence, show that OP = 13 (OA + 2OM) c Show that P lies on [AM] d Find the ratio in which P divides [AM] 12 Lines l1 and l2 are given by à à ! ¡2 ¡2 l1 : r = ¡1 +s ! à and l2 : r = ¡1 ! à +t ¡1 ¡1 ! a b c d e f cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find the coordinates of A, the point of intersection of the lines Show that the point B(0, ¡3, 2) lies on the line l2 Find the equation of the line BC given that C(3, ¡2, ¡2) lies on l1 Find the equation of the plane containing A, B and C Find the area of triangle ABC Show that the point D(9, ¡4, 2) lies on the normal to the plane passing through C g Find the volume of the pyramid ABCD black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\480IB_HL-2_16.CDR Friday, November 2007 10:25:05 AM PETERDELL IB_HL-2ed (481) LINES AND PLANES IN SPACE (Chapter 16) 481 REVIEW SET 16D Given the points A(4, 2, ¡1), B(2, 1, 5), and C(9, 4, 1): ¡! ¡! a Show that AB is perpendicular to AC b Find the equation of the plane containing A, B and C and hence determine the distance from this plane to the point (8, 1, 0) c Find the equation of the line through A and B d Determine the distance from D(8, 11, ¡5) to the line through A and B x = 3t ¡ 4, y = t + 2, z = 2t ¡ y¡5 ¡z ¡ l2 : x = = : 2 a Find the point of intersection of l1 and the plane 2x + y ¡ z = b Find the point of intersection of l1 and l2 c Find the equation of the plane that contains l1 and l2 The equations of two lines are: l1 : a Show that the plane 2x + y + z = contains the line l1 : x = ¡2t + 2, y = t, z = 3t + 1, t R b For what values of k does the plane x + ky + z = contain l1 ? c Without using row operations, find the values of p and q for which the following system of equations has an infinite number of solutions Clearly explain your reasoning 2x + y + z = x¡y+z = ¡2x + py + 2z = q: a Consider two unit vectors a and b Prove that the vector a + b bisects the angle between vector a and vector b b Consider the points H(9, 5, ¡5), J(7, 3, ¡4) and K(1, 0, 2) Find the equation of the line l that passes through J and bisects angle HJK c Find the coordinates of the point where l meets HK p x2 + y + z = 26 is the equation of a sphere, centre (0, 0, 0) and radius 26 units Find the point(s) where the line through (3, ¡1, ¡2) and (5, 3, ¡4) meets the sphere Find the angle between the plane 2x + 2y ¡ z = and the line x = t ¡ 1, y = ¡2t + 4, z = ¡t + Let r = 2i ¡ 2j ¡ k, s = 2i + j + 2k, t = i + 2j ¡ k, be the position vectors of the points R, S and T, respectively Find the area of the triangle ¢RST cyan magenta yellow B(4,¡4,-2) C(10,¡2,¡0) X Y 95 A(1,¡3,-4) 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 In the figure ABCD is a parallelogram X is the midpoint of BC, and Y is on AX such that AY : YX = : a Find the coordinates of X and D b Find the coordinates of Y c Show that B, Y and D are collinear black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\481IB_HL-2_16.CDR Wednesday, November 2007 12:41:16 PM PETERDELL D IB_HL-2ed (482) 482 LINES AND PLANES IN SPACE (Chapter 16) Let the vectors a, b and c be a = 3i + 2j ¡ k, b = i + j ¡ k, c = 2i ¡ j + k a Show that b £ c = ¡3j ¡ 3k b Verify for the given vectors that a £ (b £ c) = b (a ² c) ¡ c (a ² b) à 10 Given the vectors a = ¡2 ! à and b = a a and b are perpendicular ¡t 1+t 2t ! , find t if: b a and b are parallel x¡8 y+9 z ¡ 10 = = and ¡16 à ! à ! à ! 11 Line has equation x y z line has vector equation = 15 29 +¸ ¡5 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a Show that lines and are skew b Line is a translation of line which intersects line Find the equation of the plane containing lines and c Use b to find the shortest distance between lines and d Find the coordinates of the two points where the common perpendicular meets the lines and black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_16\482IB_HL-2_16.CDR Wednesday, 14 November 2007 12:50:19 PM PETERDELL IB_HL-2ed (483) Chapter 17 Descriptive statistics Contents: Continuous numerical data and histograms Measuring the centre of data Cumulative data Measuring the spread of data Statistics using technology Variance and standard deviation The significance of standard deviation A B C D E F G cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 17A Review set 17B black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\483IB_HL-2_17.CDR Monday, 12 November 2007 9:33:22 AM PETERDELL IB_HL-2ed (484) 484 DESCRIPTIVE STATISTICS (Chapter 17) BACKGROUND KNOWLEDGE IN STATISTICS Before starting this chapter you should make sure that you have a good understanding of the necessary background knowledge BACKGROUND KNOWLEDGE Click on the icon alongside to obtain a printable set of exercises and answers on this background knowledge THE PEA PROBLEM A farmer wishes to investigate the effect of a new organic fertiliser on his crops of peas He is hoping to improve the crop yield by using the fertiliser He divided a small garden into two equal plots and planted many peas in each Both plots were treated the same except the fertiliser was used on one but not the other A random sample of 150 pods was harvested from each plot at the same time, and the number of peas in each pod was counted The results were: Without fertiliser 6 6 5 6 6 8 6 6 6 8 5 6 7 6 7 6 7 6 7 7 4 6 6 5 6 5 6 65554446756 65675868676 34663767686 5 5 8 7 6 8 7 7 7 6 7 10 7 10 10 7 7 8 8 8 84877766863858767496668478 6 10 13 11 With fertiliser 9 7 7 4 9 For you to consider: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Can you state clearly the problem that the farmer wants to solve? How has the farmer tried to make a fair comparison? How could the farmer make sure that his selection was at random? What is the best way of organising this data? What are suitable methods of display? Are there any abnormally high or low results and how should they be treated? How can we best describe the most typical pod size? How can we best describe the spread of possible pod sizes? Can a satisfactory conclusion be made? 95 100 50 75 25 ² ² ² ² ² ² ² ² ² black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_17\484IB_HL-2_17.CDR Friday, 12 March 2010 1:47:07 PM PETER IB_HL-2ed (485) DESCRIPTIVE STATISTICS (Chapter 17) A 485 CONTINUOUS NUMERICAL DATA AND HISTOGRAMS A continuous numerical variable can take any value on part of the number line Continuous variables often have to be measured so that data can be recorded Examples of continuous numerical variables are: The height of Year 11 students: the variable can take any value from about 140 cm to 200 cm The speed of cars on a stretch of highway: the variable can take any value from km h¡1 to the fastest speed that a car can travel, but is most likely to be in the range 60 km h¡1 to 160 km h¡1 ORGANISATION AND DISPLAY OF CONTINUOUS DATA When data is recorded for a continuous variable there are likely to be many different values We organise the data by grouping it into class intervals A special type of graph called a histogram is used to display the data frequency A histogram is similar to a column graph but, to account for the continuous nature of the variable, a number line is used for the horizontal axis and the ‘columns’ are joined together Histogram no gaps An example is given alongside Notice that: ² the modal class (the class of values that appears most often) is easy to identify from a histogram ² the class intervals are the same size ² the frequency is represented by the height of the ‘columns’ data values SUMMARY: COLUMN GRAPHS AND HISTOGRAMS Column graphs and histograms both have the following features: ² the frequency of occurrence is on the vertical axis ² the range of scores is on the horizontal axis ² column widths are equal and the height varies according to frequency cyan magenta yellow 95 100 continuous data 50 discrete data 75 Histogram 25 Column Graph 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Histograms are used for continuous data They have no gaps between the columns black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\485IB_HL-2_17.CDR Monday, 12 November 2007 9:47:28 AM PETERDELL IB_HL-2ed (486) 486 DESCRIPTIVE STATISTICS (Chapter 17) CASE STUDY DRIVING A GOLF BALL While Norm Gregory was here for the golf championship, I measured how far he hit 30 drives on the practice fairway The results are given below in metres: 244:6 251:1 255:9 263:1 265:5 270:5 245:1 251:2 257:0 263:2 265:6 270:7 248:0 253:9 260:6 264:3 266:5 272:9 248:8 254:5 262:8 264:4 267:4 275:6 250:0 254:6 262:9 265:0 269:7 277:5 This type of data must be grouped before a histogram can be drawn In forming groups, find the lowest and highest values, and then choose a group width so that there are about to 12 groups In this case the lowest value is 244:6 m and the largest is 277:5 m This gives a range of approximately 35 m, so a group width of m will give eight groups of equal width We will use the following method of grouping The first group ‘240 - < 245’ includes any data value which is at least 240 m but less than 245 m Similarly the group ‘260 - < 265’ includes data which is at least 260 m but < 265 m This technique creates a group for every distance > 240 m but < 280 m A tally is used to count the data that falls in each group Do not try to determine the number of data values in the ‘240 - < 245’ group first off Simply place a vertical stroke in the tally column to register an entry as you work through the data from start to finish Every fifth entry in a group is marked with a diagonal line through the previous four so groups of five can be counted quickly A frequency column summarises the number of data values in each group The relative frequency column measures the percentage of the total number of data values that are in each group Norm Gregory’s 30 drives Frequency % Relative Distance (m) Tally (f) Frequency 240 - < 245 j 3:3 245 - < 250 jjj 10:0 © j 250 - < 255 © jjjj 20:0 255 - < 260 jj 6:7 © jj 260 - < 265 © jjjj 23:3 © j 265 - < 270 © jjjj 20:0 270 - < 275 jjj 10:0 275 - < 280 jj 6:7 Totals 30 100:0 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 From this table two histograms can be drawn: a frequency histogram and a relative frequency histogram They look as follows: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\486IB_HL-2_17.CDR Monday, 19 November 2007 11:07:41 AM PETERDELL IB_HL-2ed (487) 487 DESCRIPTIVE STATISTICS (Chapter 17) A frequency histogram displaying the distribution of 30 of Norm Gregory’s drives A relative frequency histogram displaying the distribution of 30 of Norm Gregory’s drives frequency 30 25 20 15 10 240 245 250 255 260 265 270 275 280 distance (m) relative frequency (%) 240 245 250 255 260 265 270 275 280 distance (m) The advantage of the relative frequency histogram is that we can easily compare it with other distributions with different numbers of data values Using percentages allows for a fair comparison Notice how the axes are both labelled and the graphs have titles The left edge of each bar is the first possible entry for that group Example The weight of parcels sent on a given day from a post office were, in kilograms: 2:1, 3:0, 0:6, 1:5, 1:9, 2:4, 3:2, 4:2, 2:6, 3:1, 1:8, 1:7, 3:9, 2:4, 0:3, 1:5, 1:2 Organise the data using a frequency table and graph the data The data is continuous since the weight could be any value from 0:1 kg up to kg The lowest weight was 0:3 kg and the heaviest was 4:2 kg, so we will use class intervals of kg The class interval ‘1 - < 2’ includes all weights from kg up to, but not including, kg A histogram is used to graph this continuous data Frequency 4 frequency Weights of parcels yellow weight (kg) Leaf 36 255789 1446 0129 Scale: j means 1:2 kg 95 100 75 25 95 100 50 75 25 95 magenta Stem The modal class is (1 - < 2) kg as this occurred most frequently 100 50 75 25 95 100 50 75 25 cyan A stemplot could also be used to organise the data: Note: 50 Weight (kg) 0-<1 1-<2 2-<3 3-<4 4-<5 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\487IB_HL-2_17.CDR Monday, 12 November 2007 9:56:16 AM PETERDELL IB_HL-2ed (488) 488 DESCRIPTIVE STATISTICS (Chapter 17) EXERCISE 17A A frequency table for the heights of a basketball squad is given below Height (cm) 170 175 180 185 190 195 200 - Frequency < 175 < 180 < 185 < 190 < 195 < 200 < 205 11 3 a b Explain why ‘height’ is a continuous variable Construct a histogram for the data The axes should be carefully marked and labelled, and you should include a heading for the graph c What is the modal class? Explain what this means Describe the distribution of the data d A school has conducted a survey of 60 students to investigate the time it takes for them to travel to school The following data gives the travel times to the nearest minute: 12 45 35 28 a b c d 15 40 31 16 14 21 20 18 25 10 12 15 17 10 27 32 25 10 19 46 34 27 32 14 42 16 15 18 37 12 20 24 45 14 18 18 15 20 45 16 10 10 33 26 16 25 38 32 14 22 Is travel time a discrete or continuous variable? Construct an ordered stemplot for the data using stems 0, 1, 2, Describe the distribution of the data What is the modal travelling time? For the following data, state whether a histogram or a column graph should be used and draw the appropriate graph a The number of matches in 30 match boxes: Number of matches per box Frequency 47 49 50 51 12 52 53 55 b The heights of 25 hockey players (to the nearest cm): Height (cm) Frequency 130 - 139 140 - 149 150 - 159 14 160 - 169 A plant inspector takes a random sample of six month old seedlings from a nursery and measures their height to the nearest mm The results are shown in the table alongside a Represent the data on a histogram b How many of the seedlings are 400 mm or more? c What percentage of the seedlings are between 349 and 400 mm? d The total number of seedlings in the nursery is 1462 Estimate the number of seedlings which measure i less than 400 mm ii between 374 and 425 mm cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 frequency 12 18 42 28 14 100 50 (mm) 324 349 374 399 424 449 75 25 95 Height 300 325 350 375 400 425 - 100 50 75 25 120 - 129 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\488IB_HL-2_17.CDR Monday, 12 November 2007 10:00:28 AM PETERDELL IB_HL-2ed (489) DESCRIPTIVE STATISTICS (Chapter 17) B 489 MEASURING THE CENTRE OF DATA We can get a better understanding of a data set if we can locate the middle or centre of the data and get an indication of its spread Knowing one of these without the other is often of little use There are three statistics that are used to measure the centre of a data set These are: the mode, the mean and the median THE MODE For discrete numerical data, the mode is the most frequently occurring value in the data set For continuous numerical data, we cannot talk about a mode in this way because no two data values will be exactly equal Instead we talk about a modal class, which is the class that occurs most frequently THE MEAN The mean of a data set is the statistical name for the arithmetic average mean = sum of all data values the number of data values The mean gives us a single number which indicates a centre of the data set It is usually not a member of the data set For example, a mean test mark of 73% tells us that there are several marks below 73% and several above it 73% is at the centre, but it does not necessarily mean that one of the students scored 73% If we let x n x ¹ be a data value be the number of data values in the sample or population represent the mean of a sample and ‘m’ reads ‘mu’ represent the mean of a population P P x x or x = then the mean is either: ¹ = n n THE MEDIAN The median is the middle value of an ordered data set An ordered data set is obtained by listing the data, usually from smallest to largest The median splits the data in halves Half of the data are less than or equal to the median and half are greater than or equal to it For example, if the median mark for a test is 73% then you know that half the class scored less than or equal to 73% and half scored greater than or equal to 73% cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 For an odd number of data, the median is one of the data For an even number of data, the median is the average of the two middle values and may not be one of the original data 75 25 95 100 50 75 25 Note: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\489IB_HL-2_17.CDR Monday, 12 November 2007 10:06:12 AM PETERDELL IB_HL-2ed (490) 490 DESCRIPTIVE STATISTICS (Chapter 17) Here is a rule for finding the median: If there are n data values, find n+1 The median is the µ ¶ n+1 th data value For example: DEMO If n = 13, 13 + = 7, so the median = 7th ordered data value If n = 14, 14 + = 7:5, so the median = average of 7th and 8th ordered data values THE MERITS OF THE MEAN AND MEDIAN AS MEASURES OF CENTRE The median is the only measure of centre that will locate the true centre regardless of the data set’s features It is unaffected by the presence of extreme values It is called a resistant measure of centre The mean is an accurate measure of centre if the distribution is symmetrical or approximately symmetrical If it is not, then unbalanced high or low values will drag the mean toward them and cause it to be an inaccurate measure of the centre It is called a non-resistant measure of centre because it is influenced by all data values in the set If it is considered inaccurate, it should not be used in discussion THE RELATIONSHIP BETWEEN THE MEAN AND THE MEDIAN FOR DIFFERENT DISTRIBUTIONS For distributions that are symmetric, the mean or median will be approximately equal mean and median mean and median If the data set has symmetry, both the mean and the median should accurately measure the centre of the distribution If the data set is not symmetric, it may be positively or negatively skewed: positively skewed distribution negatively skewed distribution mode mode median mean mean median cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Notice that the mean and median are clearly different for these skewed distributions black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\490IB_HL-2_17.CDR Monday, 12 November 2007 10:12:27 AM PETERDELL IB_HL-2ed (491) 491 DESCRIPTIVE STATISTICS (Chapter 17) INVESTIGATION MERITS OF THE MEAN AND MEDIAN Recall the data gained from Norm Gregory while he was here for the golf championship The data was as follows: 244:6 253:9 262:9 265:5 245:1 254:5 272:9 265:6 248:0 254:6 263:1 266:5 248:8 270:7 263:2 267:4 250:0 255:9 264:3 269:7 270:5 251:1 251:2 257:0 260:6 262:8 264:4 265:0 275:6 277:5 TI C STATISTICS PACKAGE What to do: Enter the data as a List into a graphics calculator or use the statistics package supplied a Produce a histogram of the data Set the X values from 240 to 280 with an increment of Set the Y values from to 30 b Comment on the shape of the distribution c Find i the median ii the mean d Compare the mean and the median Is the mean an accurate measure of the centre? Since we have continuous numerical data, we have a modal class rather than an individual mode a What is the modal class? b What would the modal class be if our intervals were m starting at 240 m? Now suppose Norm had hit a few very bad drives Let us say that his three shortest drives were very short! a Change the three shortest drives to 82:1 m, 103:2 m and 111:1 m b Repeat a, b, c and d but set the X values from 75 to 300 with an increment of 25 for the histogram c Describe the distribution as symmetric, positively skewed, or negatively skewed What effect have the changed values had on the mean and median as measures of the centre of the data? What would have happened if Norm had hit a few really long balls in addition to the very bad ones? Let us imagine that the longest balls he hit were very long a Change the three longest drives to 403:9 m, 415:5 m and 420:0 m b Repeat a, b, c and d but set the X values from 50 to 450 with an increment of 50 for the histogram c Describe the distribution as symmetric, positively skewed, or negatively skewed What effect have the changed values had on the mean and median as measures of the centre of the data? While collecting the data from Norm, I decided to have a hit as well I hit 30 golf balls with my driver The relative frequency histogram reveals the results below The relative frequency 0.4 distribution is clearly positively skewed median cyan magenta yellow mean 0.2 95 100 50 140 150 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The mean would not be a good measure of the centre of this distribution due to the few higher scores Indeed the mean is 163:66 m compared to the median of 157:50 m black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\491IB_HL-2_17.CDR Monday, 12 November 2007 10:23:48 AM PETERDELL 160 170 180 190 200 210 distance (m) IB_HL-2ed (492) 492 DESCRIPTIVE STATISTICS (Chapter 17) UNGROUPED DATA Example The number of trucks using a road over a 13-day period is : 5 6 For this data set, find: a the mean b the median c the mode sum of the data 13 data values 4+6+3+2+7+8+3+5+5+7+6+6+4 13 + 5:08 trucks a mean = b The ordered data set is: 3 4 5 6 7 ) median = trucks c is the score which occurs the most often ) mode = trucks fas n = 13, n+1 = 7g For the truck data of Example 2, how are the measures of the middle affected if on the 14th day the number of trucks was 7? We expect the mean to rise as the new data value is greater than the old mean In fact, the new mean = 73 66 + = ¼ 5:21 trucks 14 14 3 4 |{z} 56 667778 two middle scores The new ordered data set would be: 5+6 = 5:5 trucks ) median = fas n = 14, n+1 = 7:5g This new data set has two modes The modes are and trucks and we say that the data set is bimodal Note: ² If a data set has three or more modes, we not use the mode as a measure of the middle ² Consider the data: 5 The dot plot of this data is: For this data the mean, median and mode are all Equal values (or approximately equal values) of the mean, mode and median can indicate a symmetrical distribution of data Example sum of scores = 12:2 cyan magenta yellow 95 100 50 75 25 95 ) sum of scores = 12:2 £ = 61 The sum of the scores is 61 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The mean of five scores is 12:2 What is the sum of the scores? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\492IB_HL-2_17.CDR Monday, 12 November 2007 10:26:26 AM PETERDELL IB_HL-2ed (493) 493 DESCRIPTIVE STATISTICS (Chapter 17) EXERCISE 17B.1 Find the i mean ii median iii mode for each of the following data sets: a 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 8, 8, 8, 9, b 10, 12, 12, 15, 15, 16, 16, 17, 18, 18, 18, 18, 19, 20, 21 c 22:4, 24:6, 21:8, 26:4, 24:9, 25:0, 23:5, 26:1, 25:3, 29:5, 23:5 Consider the following two data sets: a b c d Data set A: Data set B: 3, 4, 4, 5, 6, 6, 7, 7, 7, 8, 8, 9, 10 3, 4, 4, 5, 6, 6, 7, 7, 7, 8, 8, 9, 15 Find the mean for both Data set A and Data set B Find the median of both Data set A and Data set B Explain why the mean of Data set A is less than the mean of Data set B Explain why the median of Data set A is the same as the median of Data set B The annual salaries of ten office workers are: $23 000, $46 000, $23 000, $38 000, $24 000, $23 000, $23 000, $38 000, $23 000, $32 000 a Find the mean, median and modal salaries of this group b Explain why the mode is an unsatisfactory measure of the middle in this case c Is the median a satisfactory measure of the middle of this data set? The following raw data is the daily rainfall (to the nearest millimetre) for the month of July 2007 in the desert: 3, 1, 0, 0, 0, 0, 0, 2, 0, 0, 3, 0, 0, 0, 7, 1, 1, 0, 3, 8, 0, 0, 0, 42, 21, 3, 0, 3, 1, 0, a Find the mean, median and mode for the data b Give a reason why the median is not the most suitable measure of centre for this set of data c Give a reason why the mode is not the most suitable measure of centre for this set of data d Are there any outliers in this data set? e On some occasions outliers are removed because they must be due to errors in observation and/or calculation If the outliers in the data set were accurately found, should they be removed before finding the measures of the middle? A basketball team scored 43, 55, 41 and 37 points in their first four matches a What is the mean number of points scored for the first four matches? b What score will the team need to shoot in the next match so that they maintain the same mean score? c The team scores only 25 points in the fifth match What is the mean number of points scored for the five matches? d The team scores 41 points in their sixth and final match Will this increase or decrease their previous mean score? What is the mean score for all six matches? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The mean of 10 scores is 11:6 What is the sum of the scores? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\493IB_HL-2_17.CDR Monday, 19 November 2007 11:31:13 AM PETERDELL IB_HL-2ed (494) 494 DESCRIPTIVE STATISTICS (Chapter 17) While on an outback safari, Bill drove an average of 262 km per day for a period of 12 days How far did Bill drive in total while on safari? The mean monthly sales for a clothing store are $15 467 Calculate the total sales for the store for the year Find x if 5, 9, 11, 12, 13, 14, 17 and x have a mean of 12 10 Find a given that 3, 0, a, a, 4, a, 6, a and have a mean of 11 Over the complete assessment period, Aruna averaged 35 out of a possible 40 marks for her maths tests However, when checking her files, she could only find of the tests For these she scored 29, 36, 32, 38, 35, 34 and 39 Determine how many marks out of 40 she scored for the eighth test 12 A sample of 10 measurements has a mean of 15:7 and a sample of 20 measurements has a mean of 14:3 Find the mean of all 30 measurements 13 Jana had seven spelling tests, each with twelve words, but she could only find the results of five of them These were: 9, 5, 7, and 10 She asked her teacher for the other two results and the teacher said that the mode of her scores was and the mean was What are the two missing results, given that Jana knows that her worst result was a 5? DISCUSSION Which of the measures of the middle is more affected by the presence of an outlier? Develop at least two examples to show how the measures of the middle can be altered by outliers MEASURES OF THE CENTRE FROM OTHER SOURCES When the same data appears several times we often summarise the data in table form Consider the data of the given table: Data value (x) Frequency (f) Product (f x) 1 15 1£3=3 1£4=4 £ = 15 £ = 42 15 £ = 105 £ = 64 £ = 45 P f x = 278 We can find the measures of the centre directly from the table The mode There are 15 of data value which is more than any other data value The mode is therefore P Total f = 40 The mean Adding a ‘Product’ column to the table helps to add all scores For example, there are 15 data of value and these add to £ 15 = 105 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 278 = 6:95 40 75 25 95 100 50 75 25 So, the mean = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\494IB_HL-2_17.CDR Monday, 12 November 2007 10:34:32 AM PETERDELL IB_HL-2ed (495) DESCRIPTIVE STATISTICS (Chapter 17) 495 The median There are 40 data values, an even number, so there are two middle data values As the sample size n = 40, Data Value Total n+1 41 = = 20:5 2 So, the median is the average of the 20th and 21st data values In the table, the blue numbers show us accumulated values We can see that the 20th and 21st data values (in order) are both 7’s, 7+7 =7 ) median = Frequency 1 12 15 27 40 one number is two numbers are or less five numbers are or less 12 numbers are or less 27 numbers are or less Notice that we have a skewed distribution even though the mean, median and mode are nearly equal So, we must be careful if we use measures of the middle to call a distribution symmetric Example The table below shows the number of aces served by tennis players in their first sets of a tournament Number of aces Frequency 11 18 13 Determine the mean number of aces for these sets Freq (f) 11 18 13 P f = 55 cyan magenta yellow 179 55 ¼ 3:25 aces 95 100 50 75 25 fi 95 50 75 25 95 100 50 75 25 95 100 50 75 25 i=1 = P fx has been abbreviated to x ¹= P f Note: x ¹= fi xi i=1 P P fx x= P f Product (fx) 22 54 52 35 12 P f x = 179 P 100 No of aces (x) Total black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\495IB_HL-2_17.CDR Monday, 12 November 2007 10:37:08 AM PETERDELL IB_HL-2ed (496) 496 DESCRIPTIVE STATISTICS (Chapter 17) Example Score 10 Total In a class of 20 students the results of a spelling test out of 10 are shown in the table Calculate the: a mean b median c mode Number of students 20 P a f = 20 P and fx = £ + £ + £ + £ + £ + £ 10 = 157 P fx 157 ) x= P = = 7:85 f 20 b There are 20 scores, so the median is the average of the 10th and 11th Score 10 c Number of Students 1st student 2nd and 3rd student 4th, 5th, 6th and 7th student 8th, 9th, 10th, 11th, 12th, 13th, 14th student The 10th and 11th students both scored ) median = POTTS © Jim Russell, General Features Pty Ltd Looking down the ‘number of students’ column, the highest frequency is This corresponds to a score of 8, so the mode = The publishers acknowledge the late Mr Jim Russell, General Features for the reproduction of this cartoon EXERCISE 17B.2 cyan magenta yellow 95 Number of times occurred 12 11 30 100 50 75 25 95 Number of heads Total 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The table alongside shows the results when coins were tossed simultaneously 30 times Calculate the: a mode b median c mean black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\496IB_HL-2_17.CDR Monday, 12 November 2007 10:40:58 AM PETERDELL IB_HL-2ed (497) DESCRIPTIVE STATISTICS (Chapter 17) 497 The following frequency table records the number of phone calls made in a day by 50 fifteen-year-olds a For this data, find the: No of phone calls Frequency i mean ii median iii mode b Construct a column graph for the data and show the position of the measures of centre 13 (mean, median and mode) on the horizontal axis c Describe the distribution of the data d Why is the mean larger than the median for this data? e Which measure of centre would be the most 11 suitable for this data set? A company claims that their match boxes contain, on average, 50 matches per box On doing a survey, the Consumer Protection Society recorded the following results: Number in a box 47 48 49 50 51 52 Total a Frequency 11 30 For the data calculate the: i mode ii median iii mean Do the results of this survey support the company’s claim? In a court for ‘false advertising’, the company won their case against the Consumer Protection Society Suggest why and how they did this b c Families at a school in Australia were surveyed Number of children Frequency The number of children in each family was recorded The results of the survey are shown 28 alongside 15 a Calculate the: i mean ii mode iii median b The average Australian family has 2:2 Total 59 children How does this school compare to the national average? c The data set is skewed Is the skewness positive or negative? d How has the skewness of the data affected the measures of the centre of the data set? For the data displayed in the following stem-and-leaf plots find the: i mean ii median iii mode cyan magenta 95 50 75 25 95 50 75 25 100 yellow 100 b Leaf 356 0124679 3368 47 where j means 53 95 100 50 Stem 75 25 95 100 50 75 25 a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\497IB_HL-2_17.CDR Monday, 12 November 2007 10:42:31 AM PETERDELL Stem Leaf 0488 00136789 036777 069 where j means 3:7 IB_HL-2ed (498) 498 DESCRIPTIVE STATISTICS (Chapter 17) Revisit The Pea Problem on page 484 a Use a frequency table for the Without fertiliser data to find the: i mean ii mode iii median number of peas per pod b Use a frequency table for the With fertiliser data to find the: i mean ii mode iii median number of peas per pod c Which of the measures of the centre is appropriate to use in a report on this data? d Has the application of fertiliser significantly improved the number of peas per pod? The selling $146 400, $256 400, $131 400, prices of the last 10 houses sold in a certain district were as follows: $127 600, $211 000, $192 500, $132 400, $148 000, $129 500, $162 500 a Calculate the mean and median selling prices and comment on the results b Which measure would you use if you were: i a vendor wanting to sell your house ii looking to buy a house in the district? The table alongside compares the mass at birth of some guinea pigs with their mass when they were two weeks old Guinea Pig A B C D E F G H a What was the mean birth mass? b What was the mean mass after two weeks? c What was the mean increase over the two weeks? Mass (g) at birth 75 70 80 70 74 60 55 83 Mass (g) at weeks 210 200 200 220 215 200 206 230 15 of 31 measurements are below 10 cm and 12 measurements are above 11 cm Find the median if the other measurements are 10:1 cm, 10:4 cm, 10:7 cm and 10:9 cm 10 Two brands of toothpicks claim that their boxes contain an average of 50 toothpicks per box In a survey the Consumer Protection Society (C.P.S.) recorded the following results: Brand A number in a box 46 47 48 49 50 51 52 53 55 Brand B frequency 1 10 20 15 number in a box frequency 48 49 17 50 30 51 52 53 54 a Find the average contents of Brands A and B b Would it be fair for the C.P.S to prosecute the manufacturers of either brand, based on these statistics? 11 Towards the end of season, a netballer had played 14 matches and had an average of 16:5 goals per game In the final two matches of the season she threw 21 goals and 24 goals Find the netballer’s new average cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 12 The mean and median of a set of measurements are both 12 If of the measurements are 7, 9, 11, 13, 14, 17 and 19, find the other two measurements black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\498IB_HL-2_17.CDR Monday, 12 November 2007 10:45:30 AM PETERDELL IB_HL-2ed (499) DESCRIPTIVE STATISTICS (Chapter 17) 499 13 In an office of 20 people there are only salary levels paid: $50 000 (1 person), $42 000 (3 people), $35 000 (6 people), $28 000 (10 people) a Calculate: i the median salary ii the modal salary iii the mean salary b Which measure of central tendency might be used by the boss who is against a pay rise for the other employees? DATA IN CLASSES When information has been gathered in classes we use the midpoint of the class to represent all scores within that interval We are assuming that the scores within each class are evenly distributed throughout that interval The mean calculated will therefore be an approximation to the true value Example Find the approximate mean of the following ages of bus drivers data, to the nearest year: age (yrs) frequency 21-25 11 26-30 14 age (yrs) 21-25 26-30 31-35 36-40 41-45 46-50 51-55 Total frequency (f ) 11 14 32 27 29 17 P f = 137 31-35 32 36-40 27 midpoint (x) 23 28 33 38 43 48 53 41-45 29 46-50 17 51-55 P fx x ¹= P f fx 253 392 1056 1026 1247 816 371 P f x = 5161 5161 137 ¼ 37:7 = EXERCISE 17B.3 50 students sit a mathematics test and the results are as follows: Score Frequency 0-9 10-19 20-29 30-39 27 40-49 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The table shows the petrol sales in one day by a number of city service stations a How many service stations were involved in the survey? b Estimate the total amount of petrol sold for the day by the service stations c Find the approximate mean sales of petrol for the day black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\499IB_HL-2_17.CDR Monday, 12 November 2007 10:56:38 AM PETERDELL Find an estimate of the mean score Litres (L) 2000 to 2999 3000 to 3999 4000 to 4999 5000 to 5999 6000 to 6999 7000 to 7999 frequency 4 14 23 16 IB_HL-2ed (500) 500 DESCRIPTIVE STATISTICS (Chapter 17) This histogram illustrates the results of an 50 frequency 40 aptitude test given to a group of people 30 seeking positions in a company 20 a How many people sat for the test? 10 b Find an estimate of the mean score 80 90 100 110 120 130 140 150 160 for the test score c What fraction of the people scored less than 100 for the test? d If the top 20% of the people are offered positions in the company, estimate the minimum mark required C CUMULATIVE DATA Sometimes it is useful to know the number of scores that lie above or below a particular value In such situations we can construct a cumulative frequency distribution table and use a graph called an ogive or cumulative frequency polygon to represent the data Example The data shown gives the weights of 120 male footballers a Construct a cumulative frequency distribution table b Represent the data on an ogive c Use your graph to estimate the: i median weight ii number of men weighing less than 73 kg iii number of men weighing more than 92 kg cyan magenta yellow frequency 12 14 19 37 22 This is + This is + + 12 This 50 means that there are 50 players who weigh less than 80 kg 95 100 50 75 Note: 25 95 100 50 75 25 95 17 31 50 87 109 117 119 120 100 12 14 19 37 22 50 55 w < 60 60 w < 65 65 w < 70 70 w < 75 75 w < 80 80 w < 85 85 w < 90 90 w < 95 95 w < 100 100 w < 105 75 cumulative frequency 25 frequency Weight (w kg) 95 100 50 75 25 a Weight (w kg) 55 w < 60 60 w < 65 65 w < 70 70 w < 75 75 w < 80 80 w < 85 85 w < 90 90 w < 95 95 w < 100 100 w < 105 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\500IB_HL-2_17.CDR Monday, 19 November 2007 11:32:14 AM PETERDELL The cumulative frequency gives a running total of the number of players up to certain weights IB_HL-2ed (501) DESCRIPTIVE STATISTICS (Chapter 17) b 501 Ogive of footballers’ weights cumulative frequency 120 112 100 80 c i The median is estimated using the 50th percentile As 50% of 120 = 60, we start with the cumulative frequency 60 and find the corresponding weight So, the median ¼ 81:5 There are 25 men who weigh less than 73 kg There are 120 ¡ 112 = men who weigh more than 91:3 kg 60 40 25 20 73 60 70 81.5 80 ii 91.3 90 100 110 iii weight median ¼ 81.5 EXERCISE 17C The following frequency distribution was obtained by asking 50 randomly selected people the size of their shoes Shoe size 5 12 6 12 7 12 8 12 9 12 10 frequency 1 13 17 Draw an ogive for the data and use it to find: a the median shoe size b how many people had a shoe size of: i The following data shows the lengths of 30 competition The measurements were taken to 31 38 34 40 24 33 30 36 40 34 37 44 38 36 34 33 12 or more ii or less trout caught in a lake during a fishing the nearest centimetre 38 32 35 32 36 27 35 31 38 35 36 33 33 28 a Construct a cumulative frequency table for trout lengths, x cm, using the intervals: 24 x < 27, 27 x < 30, and so on cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b Draw an ogive for the data c Use b to find the median length d Use the original data to find its median and compare your answer with c Comment! black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_17\501IB_HL-2_17.CDR Friday, 12 December 2008 12:17:08 PM TROY IB_HL-2ed (502) 502 DESCRIPTIVE STATISTICS (Chapter 17) In an examination the following scores were achieved by Draw an ogive for the data and use it to find: a the median examination mark b how many students scored less than 65 marks c how many students scored between 50 and 70 marks d how many students failed, given that the pass mark was 45 e the credit mark, given that the top 16% of students were awarded credits a group of students: Score 10 x < 20 20 x < 30 30 x < 40 40 x < 50 50 x < 60 60 x < 70 70 x < 80 80 x < 90 90 x < 100 frequency 21 36 40 27 The following table gives the age groups of car drivers involved in an accident in a city for a given year Draw a cumulative frequency Age (in years) No of accidents polygon for the data and use it to find: 16 x < 20 59 a the median age of the drivers involved in 20 x < 25 82 the accidents 25 x < 30 43 b the percentage of drivers involved in 30 x < 35 21 accidents who had an age of 23 or less 35 x < 40 19 c Estimate the probability that a driver 40 x < 50 11 involved in an accident is: 50 x < 60 24 i aged less than or equal to 27 years 60 x < 80 41 ii aged 27 years The table below gives the distribution of the life of electric light globes Draw an ogive for the data and use it to Life (hours) Number of globes estimate: l < 500 a the median life of a globe 500 l < 1000 17 b the percentage of globes which had a 1000 l < 2000 46 life of 2700 hours or less 2000 l < 3000 79 c the number of globes which had a life 3000 l < 4000 27 between 1500 and 2500 hours 4000 l < 5000 D MEASURING THE SPREAD OF DATA To accurately describe a distribution we need to measure both its centre and its spread or variability A B The given distributions have the same mean, but clearly they have different spreads The A distribution has most scores close to the mean whereas the C distribution has the greatest spread C mean cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 We will examine four different measures of spread: the range, the interquartile range (IQR), the variance and the standard deviation black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\502IB_HL-2_17.CDR Monday, 12 November 2007 11:48:04 AM PETERDELL IB_HL-2ed (503) 503 DESCRIPTIVE STATISTICS (Chapter 17) THE RANGE The range is the difference between the maximum (largest) and the minimum (smallest) data value Example A greengrocer chain considers purchasing apples from two different wholesalers They take six random samples of 50 apples to examine them for skin blemishes The counts for the number of blemished apples are: Wholesaler “Orchard Road”: Wholesaler “Red Tree”: 17 15 11 10 13 12 11 12 11 What is the range from each wholesaler? Range = 17 ¡ = 14 Orchard Road: Red Tree: Range = 13 ¡ 10 = The range is not considered to be a particularly reliable measure of spread as it uses only two data values It may be influenced by extreme values or outliers THE QUARTILES AND THE INTERQUARTILE RANGE The median divides the ordered data set into two halves and these halves are divided in half again by the quartiles The middle value of the lower half is called the lower quartile or 25th percentile One quarter or 25% of the data have values less than or equal to the lower quartile 75% of the data have values greater than or equal to the lower quartile The middle value of the upper half is called the upper quartile or 75th percentile One quarter or 25% of the data have values greater than or equal to the upper quartile 75% of the data have values less than or equal to the upper quartile The interquartile range is the range of the middle half (50%) of the data interquartile range = upper quartile ¡ lower quartile The data set is thus divided into quarters by the lower quartile (Q1 ), the median (Q2 ), and the upper quartile (Q3 ) IQR = Q3 ¡ Q1 So, the interquartile range, Example For the data set: 6, 4, 7, 5, 3, 4, 2, 6, 5, 7, 5, 3, 8, 9, 3, 6, find the: a median b lower quartile c upper quartile d interquartile range The ordered data set is: 3 4 5 5 6 7 (17 of them) cyan magenta yellow 95 100 50 75 25 ) the median = 9th score = 5 95 100 50 75 n+1 =9 25 95 100 50 As n = 17, 75 25 95 100 50 75 25 a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\503IB_HL-2_17.CDR Monday, 12 November 2007 12:01:40 PM PETERDELL IB_HL-2ed (504) 504 DESCRIPTIVE STATISTICS (Chapter 17) b/c d As the median is a data value we now ignore it and split the remaining data into two 3+4 lower upper Q1 = median of lower half = = 3:5 z }| { z }| { 23334455 56667789 6+7 Q3 = median of upper half = = 6:5 IQR = Q3 ¡ Q1 = 6:5 ¡ 3:5 = Example 10 For the data set: 11, 6, 7, 8, 13, 10, 8, 7, 5, 2, 9, 4, 4, 5, 8, 2, 3, find: a the median b Q1 c Q3 d the interquartile range The ordered data set is: 2 4 5 6 7 8 10 11 13 (18 of them) n+1 = 9:5 9th value + 10th value 6+7 ) median = = = 6:5 2 As the median is not a data value we split the data into two lower upper z }| { z }| { Note: 2 4 5 6 7 8 10 11 13 a As n = 18, b/c Some computer packages calculate quartiles in a different way to this example ) Q1 = 4, Q3 = IQR = Q3 ¡ Q1 =8¡4 =4 d EXERCISE 17D.1 For each of the following data sets, make sure the data is ordered and then find: i the median ii the upper and lower quartiles iii the range iv the interquartile range a 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 8, 8, 8, 9, b 10, 12, 15, 12, 24, 18, 19, 18, 18, 15, 16, 20, 21, 17, 18, 16, 22, 14 c 21:8, 22:4, 23:5, 23:5, 24:6, 24:9, 25, 25:3, 26:1, 26:4, 29:5 The times spent (in minutes) by 20 people waiting in a queue at a bank for a teller were: 3:4 2:1 3:8 2:2 4:5 1:4 0 1:6 4:8 1:5 1:9 3:6 5:2 2:7 3:0 0:8 3:8 5:2 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a Find the median waiting time and the upper and lower quartiles black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\504IB_HL-2_17.CDR Monday, 12 November 2007 12:12:41 PM PETERDELL IB_HL-2ed (505) DESCRIPTIVE STATISTICS (Chapter 17) 505 b Find the range and interquartile range of the waiting times c Copy and complete the following statements: i “50% of the waiting times were greater than minutes.” ii “75% of the waiting times were less than minutes.” iii “The minimum waiting time was minutes and the maximum waiting time was minutes The waiting times were spread over minutes.” Stem Leaf 347 034 003 137 For a c e g 678 56999 j means 37 the data set given, find: the minimum value b the median d the upper quartile f the interquartile range The heights of 20 ten year olds are recorded in the following stem-and-leaf plot: a Find: i the median height ii the upper and lower quartiles of the data Stem 10 11 12 13 10 j b Copy and complete the following statements: i “Half of the children are no more than cm tall.” ii “75% of the children are no more than cm tall.” the maximum value the lower quartile the range Leaf 134489 22446899 12588 reads 109 cm c Find the: i range ii interquartile range for the height of the ten year olds d Copy and complete: “The middle 50% of the children have heights spread over cm.” Revisit The Pea Problem on page 484 a For the Without fertiliser data, find: i the range iii the lower quartile v the interquartile range ii iv the median the upper quartile b Repeat a for the With fertiliser data c Reconsider the questions posed in The Pea Problem Amend your solutions where appropriate BOX-AND-WHISKER PLOTS cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A box-and-whisker plot (or simply a boxplot) is a visual display of some of the descriptive statistics of a data set It shows: ² the minimum value > > > ² the lower quartile (Q1 ) > = These five numbers form the ² the median (Q2 ) > > five-number summary of the data set ² the upper quartile (Q3 ) > > > ; ² the maximum value black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\505IB_HL-2_17.CDR Monday, 12 November 2007 12:14:32 PM PETERDELL IB_HL-2ed (506) 506 DESCRIPTIVE STATISTICS (Chapter 17) For the data set in Example 10 on page 504, the five-number summary and boxplot are: minimum Q1 median Q3 maximum =2 =4 = 6:5 =8 = 13 lower whisker upper whisker Q1 median 10 11 12 Q3 13 max The rectangular box represents the ‘middle’ half of the data set The lower whisker represents the 25% of the data with smallest values The upper whisker represents the 25% of the data with greatest values Example 11 For a c d the data set: construct the five-number summary b draw a boxplot find the i range ii interquartile range find the percentage of data values greater than 6: a The ordered data set is 3 4 5 5 7 (16 of them) Q1 = median = Q3 = 6:5 Q1 = < minimum = median = Q3 = 6:5 So the 5-number summary is: : maximum = b range = maximum ¡ minimum =9¡1 =8 c i ii d 25% of the data values are greater than 6: 10 IQR = Q3 ¡ Q1 = 6:5 ¡ = 3:5 EXERCISE 17D.2 10 20 30 40 50 60 70 80 points scored by a basketball team a The boxplot given summarises the goals scored by a basketball team Locate: i the median ii the maximum value iii the minimum value iv the upper quartile v the lower quartile cyan magenta 95 100 50 75 25 95 50 75 25 100 yellow the interquartile range ii the range 95 i 100 50 75 25 95 100 50 75 25 b Calculate: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\506IB_HL-2_17.CDR Monday, 12 November 2007 12:21:17 PM PETERDELL IB_HL-2ed (507) DESCRIPTIVE STATISTICS (Chapter 17) 507 test scores 10 20 30 40 50 60 70 80 90 100 The boxplot above summarises the class results for a test out of 100 marks Copy and complete the following statements about the test results: a The highest mark scored for the test was , and the lowest mark was b Half of the class scored a mark greater than or equal to c The top 25% of the class scored at least marks for the test d The middle half of the class had scores between and for this test e Find the range of the data set f Find the interquartile range of the data set g Estimate the mean mark for these test scores For the following data sets: i construct a 5-number summary iii find the range ii iv draw a boxplot find the interquartile range a 3, 5, 5, 7, 10, 9, 4, 7, 8, 6, 6, 5, 8, b 3, 7, 0, 1, 4, 6, 8, 8, 8, 9, 7, 5, 6, 8, 7, 8, 8, 2, c Stem 11 12 13 14 15 Leaf 03668 01113557 47799 11 j represents 117 The following side-by-side boxplots compare the time students in years and 12 spend on homework Year Year 12 10 a Copy and complete: Statistic minimum Q1 median Q3 maximum 15 Year 20 time Year 12 b Determine the: i range ii interquartile range for each group c True or false: i On average, Year 12 students spend about twice as much time on homework than Year students cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ii Over 25% of Year students spend less time on homework than all Year 12 students black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\507IB_HL-2_17.CDR Monday, 19 November 2007 11:33:17 AM PETERDELL IB_HL-2ed (508) 508 DESCRIPTIVE STATISTICS (Chapter 17) Enid examines a new variety of bean and does a count on the number of beans in 33 pods Her results were: 5, 8, 10, 4, 2, 12, 6, 5, 7, 7, 5, 5, 5, 13, 9, 3, 4, 4, 7, 8, 9, 5, 5, 4, 3, 6, 6, 6, 6, 9, 8, 7, a Find the median, lower quartile and upper quartile of the data set b Find the interquartile range of the data set c Draw a boxplot of the data set Ranji counts the number of bolts in several boxes and tabulates the data as follows: Number of bolts Frequency 33 34 35 36 13 37 12 38 39 40 a Find the five-number summary for this data set b Find the i range ii IQR for this data set c Construct a boxplot for the data set PERCENTILES A percentile is the score below which a certain percentage of the data lies ² the 85th percentile is the score below which 85% of the data lies ² If your score in a test is the 95th percentile, then 95% of the class have scored less than you For example: Notice that: ² ² ² the lower quartile (Q1 ) is the 25th percentile the median (Q2 ) is the 50th percentile the upper quartile (Q3 ) is the 75th percentile 100% 180 75% 120 50% 60 25% cyan magenta yellow 95 Qc 100 50 Qx 75 95 100 50 Qz 75 25 95 100 50 75 25 95 100 50 75 25 25 We can use the percentage scale to help find the quartiles black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\508IB_HL-2_17.CDR Monday, 12 November 2007 12:32:20 PM PETERDELL score percentage 240 cumulative frequency Ogive One way to display percentiles is to add a separate scale to an ogive On the graph alongside, the cumulative frequency is read from the axis on the left side, and each value corresponds to a percentage on the right side IB_HL-2ed (509) DESCRIPTIVE STATISTICS (Chapter 17) 509 A botanist has measured the heights of 60 seedlings and has presented her findings on the ogive below Heights of seedlings cumulative frequency a How many seedlings have heights of cm or less? b What percentage of seedlings are taller than cm? c What is the median height? 60 55 50 45 40 35 d What is the interquartile range for the heights? e Find the 90th percentile for the data and explain what your answer means 30 25 20 15 10 height (cm) 10 11 12 13 14 The following ogive displays the performance of 80 competitors in a cross-country race Cross-country race times Find: a the lower quartile time b the median c the upper quartile d the interquartile range e an estimate of the 40th percentile cumulative frequency 80 70 60 50 40 30 20 10 cyan time (min) magenta yellow 95 100 50 75 35 25 95 100 50 30 75 25 95 100 50 25 75 20 25 95 100 50 75 25 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\509IB_HL-2_17.CDR Monday, 12 November 2007 12:41:43 PM PETERDELL IB_HL-2ed (510) 510 DESCRIPTIVE STATISTICS (Chapter 17) E STATISTICS USING TECHNOLOGY GRAPHICS CALCULATOR A graphics calculator can be used to find descriptive statistics and to draw some types of graphs Consider the data set: 3 7 No matter what brand of calculator you use you should be able to: ² ² Enter the data as a list Enter the statistics calculation part of the menu and obtain the descriptive statistics like these shown x is the mean ² Obtain a box-and-whisker plot ² Obtain a vertical barchart if required ² Enter a second data set into another list and obtain a side-by-side boxplot for comparison with the first one Use the data: 5 7 4 You will need to change the viewing window as appropriate STATISTICS FROM A COMPUTER PACKAGE Click on the icon to load our statistics package Enter data set 1: 3 7 Enter data set 2: 5 7 4 Examine the side-by-side column graphs Click on the Box-and-Whisker tab to view the side-by-side boxplots Click on the Statistics tab to obtain the descriptive statistics STATISTICS PACKAGE EXERCISE 17E Use technology to answer the following questions: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 a Enter the data set: 3 7 and obtain the mean and the 5-number summary This is the data used in the screendumps above, so you can use them to check your results 100 50 75 25 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\510IB_HL-2_17.CDR Monday, 12 November 2007 12:49:09 PM PETERDELL IB_HL-2ed (511) DESCRIPTIVE STATISTICS (Chapter 17) 511 b Obtain the boxplot for the data in a c Obtain the vertical bar chart for the data in a d Enter the data set: 5 7 4 into a second list Find the mean and 5-number summary Create a side-by-side boxplot for both sets of data Shane and Brett play in the same cricket team and are fierce but comes to bowling During a season the number of wickets taken bowlers were: Shane: 4 3 4 3 5 Brett: 5 4 3 1 2 a b c d friendly rivals when it in each innings by the 4 4 3 Is the variable discrete or continuous? Enter the data into a graphics calculator or statistics package Produce a vertical column graph for each data set Are there any outliers? Should they be deleted before we start to analyse the data? Describe the shape of each distribution Compare the measures of the centre of each distribution Compare the spreads of each distribution Obtain side-by-side boxplots What conclusions, if any, can be drawn from the data? e f g h i A manufacturer of light globes claims that their new design has a 20% longer life than those they are presently selling Forty of each globe are randomly selected and tested Here are the results to the nearest hour: 103 96 113 111 126 100 122 110 84 117 103 113 104 104 Old type: 111 87 90 121 99 114 105 121 93 109 87 118 75 111 87 127 117 131 115 116 82 130 113 95 108 112 146 131 132 160 128 119 133 117 139 123 109 129 109 131 New type: 191 117 132 107 141 136 146 142 123 144 145 125 164 125 133 124 153 129 118 130 134 151 145 131 133 135 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 Is the variable discrete or continuous? Enter the data into a graphics calculator or statistics package Are there any outliers? Should they be deleted before we start to analyse the data? Compare the measures of centre and spread Obtain side-by-side boxplots Describe the shape of each distribution What conclusions, if any, can be drawn from the data? 75 25 95 100 50 75 25 a b c d e f g black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\511IB_HL-2_17.CDR Monday, 12 November 2007 12:55:12 PM PETERDELL IB_HL-2ed (512) 512 DESCRIPTIVE STATISTICS (Chapter 17) F VARIANCE AND STANDARD DEVIATION The problem with using the range and the IQR as measures of spread or variation is that both of them only use two values in their calculation Some data sets can therefore have their spread characteristics hidden when the range or IQR are quoted, and so we need a better way of describing variation Consider a data set of n values: x1 , x2 , x3 , x4 , , xn , with mean x xi ¡ x measures how far xi deviates from the mean, so one might suspect that the mean n P of the deviations n1 (xi ¡ x) would give a good measure of variation However, this i=1 value turns out to always be zero n P Instead we adopt the formula sn2 = values i=1 (xi ¡ x)2 and call it the variance of the n data n Notice in this formula that: ² (xi ¡ x)2 is also a measure of how far xi deviates from x However, the square ensures that each of these is positive, which is why the sum turns out not to be zero n P (xi ¡ x)2 is small, it will indicate that most of the data values are close to x ² If i=1 ² Dividing by n gives an indication of how far, on average, the data is from the mean v n uP u (xi ¡ x)2 t i=1 For a data set of n values, sn = is called the standard deviation n The square root in the standard deviation is used to correct the units For example, if xi is the weight of a student in kg, s2 would be in kg2 For this reason the standard deviation is more frequently quoted than the variance The standard deviation is a non-resistant measure of spread This is due to its dependence on the mean of the sample and because extreme data values will give large values for (x ¡ x)2 It is only a useful measure if the distribution is approximately symmetrical However, the standard deviation is particularly useful when the data from which it came is normally distributed This will be discussed later cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The IQR and percentiles are more appropriate tools for measuring spread if the distribution is considerably skewed black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\512IB_HL-2_17.CDR Monday, 12 November 2007 12:56:04 PM PETERDELL IB_HL-2ed (513) 513 DESCRIPTIVE STATISTICS (Chapter 17) Example 12 Find the means and standard deviations for the apple samples of Example What these statistics tell us? Orchard Road (x ¡ x)2 25 49 25 49 1 150 rP (x ¡ x)2 s = n r 150 = 60 x = = 10 ) x¡x ¡5 ¡7 ¡1 Total x 17 15 11 60 Red Tree ) x 10 13 12 11 12 11 69 x¡x ¡1:5 1:5 0:5 ¡0:5 0:5 ¡0:5 Total (x ¡ x)2 2:25 2:25 0:25 0:25 0:25 0:25 5:5 rP (x ¡ x)2 s = n r 5:5 = 69 x = = 11:5 =5 = 0:957 The wholesaler Red Tree supplied apples with more blemishes but with less variability (smaller standard deviation) than those supplied by Orchard Road EXERCISE 17F.1 The column graphs show two distributions: Sample A 12 Sample B 12 frequency frequency 10 10 8 6 4 2 10 11 12 10 11 12 13 14 a By looking at the graphs, which distribution appears to have wider spread? b Find the mean of each sample c Find the standard deviation for each sample Comment on your answers The number of points scored by Andrew and Brad in the last basketball matches are tabulated below Points by Andrew 23 17 31 25 25 19 28 32 Points by Brad 29 41 26 14 44 38 43 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a Find the mean and standard deviation for the number of points scored by each player b Which of the two players is more consistent? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\513IB_HL-2_17.CDR Monday, 19 November 2007 11:39:19 AM PETERDELL IB_HL-2ed (514) 514 DESCRIPTIVE STATISTICS (Chapter 17) Two baseball coaches compare the number of runs scored by their teams in their last ten matches: Rockets 10 Bullets 4 11 11 7 12 a Show that each team has the same mean and range of runs scored b Which team’s performance you suspect is more variable over the period? c Check your answer to b by finding the standard deviation for each distribution d Does the range or the standard deviation give a better indication of variability? A manufacturer of soft drinks employs a statistician for quality control He needs to check that 375 mL of drink goes into each can The machine which fills the cans may malfunction or slightly change its delivery due to constant vibration and other factors a Would you expect the standard deviation for the whole production run to be the same for one day as it is for one week? Explain your answer b If samples of 125 cans are taken each day, what measure would be used to: i check that an average of 375 mL of drink goes into each can ii check the variability of the volume of drink going into each can? c What is the significance of a low standard deviation in this case? The weights in kg of seven footballers are: 79, 64, 59, 71, 68, 68 and 74 a Find the mean and standard deviation for this group b Surprisingly, each footballer’s weight had increased by exactly 10 kg when measured five years later Find the new mean and standard deviation c Comment on your findings from a and b in general terms The weights of ten young turkeys to the nearest 0:1 kg are: 0:8, 1:1, 1:2, 0:9, 1:2, 1:2, 0:9, 0:7, 1:0, 1:1 a Find the mean and standard deviation for the turkeys b After being fed a special diet for one month, the weights of the turkeys doubled Find the new mean and standard deviation c Comment, in general terms, on your findings from a and b A sample of integers has a mean of and a variance of 5:25 The integers are: 1, 3, 5, 7, 4, 5, p, q Find p and q given that p < q A sample of 10 integers has a mean of and a variance of 3:2 The integers are: 3, 9, 5, 5, 6, 4, a, 6, b, Find a and b given that a > b a Prove that n P i=1 (xi ¡ x)2 = n P (xi2 ) ¡ n(x)2 i=1 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b Find the mean of the data set x1 , x2 , , x25 given that and the standard deviation is 5:2 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\514IB_HL-2_17.CDR Tuesday, 20 November 2007 9:14:31 AM PETERDELL 25 P i=1 xi2 = 2568:25 IB_HL-2ed (515) DESCRIPTIVE STATISTICS (Chapter 17) 515 10 The following table shows the change in cholesterol levels in volunteers after a two week trial of special diet and exercise Volunteer Change in cholesterol A 0:8 B 0:6 C 0:7 D 0:8 E 0:4 F 2:8 a Find the standard deviation of the data b Recalculate the standard deviation with the outlier removed c Discuss the effect of an extreme value on the standard deviation SAMPLING FROM A POPULATION Populations are often huge, and gathering data from every individual is impossible due to time constraints and cost Consequently, a random sample is taken from the population with the hope that it will truly reflect the characteristics of the population To ensure this, the sample must be sufficiently large, and be taken in such a way that the results are unbiased From the sample we hope to make inferences about the population’s mean and possibly other key features To help distinguish between a sample and the whole population, we use different notation for the mean, variance, and standard deviation This is shown in the table opposite sample population mean x ¹ variance sn2 ¾2 standard deviation sn ¾ Given statistics from a sample, we can make inferences about the population using the following results which are assumed without proof: When a sample of size n is used to draw inference about a population: ² the mean of the sample x is an unbiased estimate of ¹ ² sn¡1 = µ ¶ n s2 n¡1 n is an unbiased estimate of the variance ¾ 2 Note: Even if sn¡1 is an unbiased estimate of ¾ , this does not imply that sn¡1 is an unbiased estimate of ¾ NOTE ON PARAMETERS AND STATISTICS A parameter is a numerical characteristic of a population A statistic is a numerical characteristic of a sample Parameter opulation Sample tatistic cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 For example, if we are examining the mean age of people in retirement villages throughout Canada, the mean age found would be a parameter If we take a random sample of 300 people from the population of all retirement village persons, then the mean age would be a statistic black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\515IB_HL-2_17.CDR Friday, 30 November 2007 9:25:24 AM PETERDELL IB_HL-2ed (516) 516 DESCRIPTIVE STATISTICS (Chapter 17) Example 13 A random sample of 48 sheep was taken from a flock of over 2000 sheep The sample mean of their weights was 48:6 kg with variance 17:5 kg a Find the standard deviation of the sample b Find an unbiased estimation of the mean weight of sheep in the flock c Find an unbiased estimation of the variance of the population from which the sample was taken p p variance = 17:5 ¼ 4:18 kg a sn = b ¹ is estimated by x = 48:6 kg c ¾ is estimated by sn¡1 µ = ¶ 48 n s2 ¼ £ 17.5 ¼ 17:9 kg n¡1 n 47 EXERCISE 17F.2 A random sample of 87 deer from a huge herd had a mean weight of 93:8 kg with a variance of 45:9 kg a Find the standard deviation of the sample b Find an unbiased estimation of the mean and variance of the entire herd from which the sample was taken The weights (in grams) of a random sample of sparrows are as follows: 87 75 68 69 81 89 73 66 91 77 84 83 77 74 80 76 67 a Find the mean and standard deviation of the sample b Find unbiased estimates of the mean and variance of the population from which the sample was taken Jacko drives down to the beach every morning to go surfing On 16 randomly chosen trips he recorded his travel time xi in minutes 16 16 P P He finds that xi = 519 and (xi2 ) = 16 983 i=1 i=1 a the mean and Calculate unbiased estimates of times to the beach b the variance of the driving STANDARD DEVIATION FOR GROUPED DATA cyan magenta yellow 95 100 50 s is the standard deviation x is any score, x is the mean f is the frequency of each score 75 25 where 95 100 50 f (x ¡ x)2 P f 75 95 100 50 75 25 95 100 50 75 25 For grouped data s = 25 sP black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_17\516IB_HL-2_17.CDR Friday, 12 December 2008 12:18:01 PM TROY IB_HL-2ed (517) DESCRIPTIVE STATISTICS (Chapter 17) 517 Example 14 score frequency Find the standard deviation of the distribution: x Total f 10 (x ¡ x ¹)2 1 x¡x ¹ ¡2 ¡1 fx 12 30 1 2 4 P fx 30 x ¹= P = =3 f 10 sP f (x ¡ x ¹)2 P s= f r 12 = 10 f (x ¡ x ¹)2 2 12 ¼ 1:10 EXERCISE 17F.3 Below is a sample of family sizes taken at random from people in a city Number of children, x Frequency, f 14 18 13 5 a Find the sample mean and standard deviation b Find unbiased estimates of the mean and variance of the population from which the sample was taken Below is a random sample of the ages of squash players at the Junior National Squash Championship Age Frequency 11 12 13 14 15 16 17 18 a Find the mean and standard deviation of the ages b Find unbiased estimates of the mean and variance of the population from which the sample was taken The number of toothpicks in a random sample of 48 boxes was counted and the results tabulated Number of toothpicks Frequency 33 35 36 37 13 38 12 39 40 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a Find the mean and standard deviation of the number of toothpicks in the boxes b Find unbiased estimates of the mean and variance of the population from which the sample was taken black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\517IB_HL-2_17.CDR Thursday, 29 November 2007 9:37:08 AM PETERDELL IB_HL-2ed (518) 518 DESCRIPTIVE STATISTICS (Chapter 17) The lengths of 30 randomly selected 12-day old babies were measured to the nearest cm and the following data obtained: Length (cm) 40 - 41 42 - 43 44 - 45 46 - 47 48 - 49 50 - 51 52 - 53 a Find estimates of the mean length and the standard deviation of the lengths b Find unbiased estimates of the mean and variance of the population from which the sample was taken The weekly wages (in dollars) of 200 randomly selected steel workers are given alongside: Wage ($) 360 - 369:99 370 - 379:99 380 - 389:99 390 - 399:99 400 - 409:99 410 - 419:99 420 - 429:99 430 - 439:99 a Find estimates of the mean and the standard deviation of the wages b Find unbiased estimates of the mean and variance of the population from which the sample was taken G Frequency 1 11 Number of workers 17 38 47 57 18 10 10 THE SIGNIFICANCE OF STANDARD DEVIATION Consider the volumes of liquid in different cans of a particular brand of soft drink The distribution of volumes is symmetrical and bell-shaped This is due to natural variation produced by the machine which has been set to produce a particular volume Random or chance factors cause roughly the same number of cans to be overfilled as underfilled The resulting bell-shaped distribution is called the normal distribution It will be discussed in more detail in Chapter 29, but it is worth noting here how it relates to standard deviation If a large sample from a typical bell-shaped data distribution is taken, what percentage of the data values would lie between x ¡ s and x + s? DEMO Click on the icon and try to answer this question Repeat the sampling many times bell-shaped distribution Now try to determine the percentage of data values which would lie between x ¡ 2s and x + 2s, and then between x ¡ 3s and x + 3s It can be shown that for any measured variable from any population that is normally distributed, no matter the values of the mean and standard deviation: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ² approximately 68% of the population will measure between standard deviation either side of the mean ² approximately 95% of the population will measure between standard deviations either side of the mean ² approximately 99.7% of the population will measure between standard deviations either side of the mean black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\518IB_HL-2_17.CDR Thursday, 29 November 2007 9:37:56 AM PETERDELL IB_HL-2ed (519) DESCRIPTIVE STATISTICS (Chapter 17) 519 Example 15 A sample of 200 cans of peaches was taken from a warehouse and the contents of each can measured for net weight The sample mean was 486¡g with standard deviation 6:2¡g What proportion of the cans might lie within: a standard deviation from the mean b standard deviations from the mean? a About 68% of the cans would be expected to have contents between 486 § 6:2 g i.e., 479:8 g and 492:2 g b Nearly all of the cans would be expected to have contents between 486 § £ 6:2 g i.e., 467:4 and 504:6 g THE NORMAL CURVE The smooth curve that models normally distributed data is asymptotic to the horizontal axis, so in theory there are no limits within which all the members of the population will fall concave convex convex ¹¡-¡¾ ¹¡+¡¾ ¹ In practice, however, it is rare to find data outside of standard deviations from the mean, and exceptionally rare to find data beyond standard deviations from the mean Note that the position of standard deviation either side of the mean corresponds to the point where the normal curve changes from a concave to a convex curve EXERCISE 17G The mean height of players in a basketball competition is 184 cm If the standard deviation is cm, what percentage of them are likely to be: a taller than 189 cm b taller than 179 cm d over 199 cm tall? c between 174 cm and 199 cm The mean average rainfall of Claudona for August is 48 mm with a standard deviation of mm Over a 20 year period, how many times would you expect there to be less than 42 mm of rainfall during August in Claudona? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Two hundred lifesavers competed in a swimming race The mean time was 10 minutes 30 seconds The standard deviation was 15 seconds Find the number of competitors who probably: a took longer than 11 minutes b took less than 10 minutes 15 seconds c completed the race in a time between 10 15 sec and 10 45 sec black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\519IB_HL-2_17.CDR Monday, 12 November 2007 2:52:04 PM PETERDELL IB_HL-2ed (520) 520 DESCRIPTIVE STATISTICS (Chapter 17) The weights of babies born at Prince Louis Maternity Hospital last year averaged 3:0 kg with a standard deviation of 200 grams If there were 545 babies born at this hospital last year, estimate the number that weighed: a less than 3:2 kg b between 2:8 kg and 3:4 kg REVIEW SET 17A The data supplied below is the diameter (in cm) of a number of bacteria colonies as measured by a microbiologist 12 hours after seeding 0:4 2:1 3:4 3:9 4:7 3:7 0:8 3:6 4:1 4:9 2:5 3:1 1:5 2:6 4:0 1:3 3:5 0:9 1:5 4:2 3:5 2:1 3:0 1:7 3:6 2:8 3:7 2:8 3:2 3:3 a Produce a stemplot for this data b Find the i median ii range of the data c Comment on the skewness of the data The data below shows the 71:2 65:1 68:0 84:3 77:0 82:8 90:5 85:5 90:7 a b c d e distance in metres that Thabiso threw a baseball 71:1 74:6 68:8 83:2 85:0 74:5 87:4 84:4 80:6 75:9 89:7 83:2 97:5 82:9 92:9 95:6 85:5 64:6 73:9 80:0 86:5 Determine the highest and lowest value for the data set Produce between and 12 groups in which to place all the data values Prepare a frequency distribution table Draw a frequency histogram for the data Determine: i the mean ii the median 5, 6, 8, a, 3, b, have a mean of and a variance of Find the values of a and b For the following distribution of continuous grouped data: Scores Frequency a c to 9:9 10 to 19:9 13 20 to 29:9 27 b d Construct an ogive Find the interquartile range 30 to 39:9 17 Find the median of the data Find the mean and standard deviation The back-to-back stemplot alongside represents the times for the 100 metre freestyle recorded by members of a swimming squad Girls 763 87430 8833 7666 a Copy and complete the following table: Distribution shape centre (median) spread (range) Girls Boys magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b Discuss the distributions of times for the boys and girls What conclusion can you make? cyan 40 to 49:9 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\520IB_HL-2_17.CDR Monday, 12 November 2007 2:55:05 PM PETERDELL 32 33 34 35 36 37 38 39 40 41 Boys 0227 13448 024799 788 leaf unit: 0:1 sec IB_HL-2ed (521) 521 DESCRIPTIVE STATISTICS (Chapter 17) The given parallel boxplots represent A the 100-metre sprint times for the B members of two athletics squads 11 12 13 14 time in seconds a Determine the number summaries for both A and B b Determine the i range ii interquartile range for each group c Copy and complete: i We know the members of squad generally ran faster because ii We know the times in squad are more varied because Katja’s golf scores for her last 20 rounds were: 90 106 84 103 112 100 105 81 104 98 107 95 104 108 99 101 106 102 98 101 a Find the i median ii lower quartile iii upper quartile b Find the interquartile range of the data set c Find the mean and standard deviation of her scores The number of litres of petrol purchased by a random sample of motor vehicle drivers is shown alongside: Litres 15 - < 20 20 - < 25 25 - < 30 30 - < 35 35 - < 40 40 - < 45 45 - < 50 a Find the mean and standard deviation of the number of litres purchased b Find unbiased estimates of the mean and variance for the population this sample comes from Number of vehicles 13 17 29 27 18 The average height of 17-year old boys was found to be normally distributed with a mean of 179 cm and a standard deviation of cm Calculate the percentage of 17-year old boys whose heights are: a more than 195 cm b between 163 cm and 195 cm c between 171 cm and 187 cm 10 80 senior students needed to run 400 metres in a Physical Education program Their times were recorded and the results were used to produce the following cumulative frequency graph Estimate: 80 cumulative frequency 60 40 a the median b the interquartile range 20 time (seconds) cyan magenta yellow 95 100 50 75 55 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 50 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\521IB_HL-2_17.CDR Thursday, 29 November 2007 9:38:26 AM PETERDELL 60 65 70 IB_HL-2ed (522) 522 DESCRIPTIVE STATISTICS (Chapter 17) 11 This cumulative frequency curve shows the times taken for 200 students to travel to school by bus a Estimate how many of the students spent between 10 and 20 minutes travelling to school b If 30% of the students spent more than m minutes travelling to school, estimate the value of m cumulative frequency 200 180 160 140 120 100 80 60 40 20 time (min) 10 15 20 25 30 35 40 REVIEW SET 17B The winning margin in 100 basketball games was recorded The results are given alongside: Margin (points) - 10 11 - 20 21 - 30 31 - 40 41 - 50 Draw a histogram to represent this information Frequency 13 35 27 18 The table alongside shows the Number 47 48 49 50 51 52 number of matches in a sample of Frequency 21 29 35 42 18 31 boxes Find the mean and standard deviation for this data Does this result justify a claim that the average number of matches per box is 50? No of customers 250 - 299 300 - 349 350 - 399 400 - 449 450 - 499 500 - 549 550 - 599 The table alongside shows the number of customers visiting a supermarket on various days Find the mean number of customers per day Frequency 14 34 68 72 54 23 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find the range, lower quartile, upper quartile and standard deviation for the following data: 120, 118, 132, 127, 135, 116, 122, 128 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\522IB_HL-2_17.CDR Wednesday, 14 November 2007 1:09:17 PM DAVID3 IB_HL-2ed (523) DESCRIPTIVE STATISTICS (Chapter 17) 523 Draw a box and whisker plot for the following data: 11, 12, 12, 13, 14, 14, 15, 15, 15, 16, 17, 17, 18 A random sample of the weekly supermarket bills for a number of families was observed and recorded in the table given a Find the mean bill and the standard deviation of the bills b Find unbiased estimates of the mean and variance of the population from which the data was taken Bill ($) 70 - 79:99 80 - 89:99 90 - 99:99 100 - 109:99 110 - 119:99 120 - 129:99 130 - 139:99 140 - 149:99 No of families 27 32 48 25 37 21 18 7 The mean and standard deviation of a normal distribution are 150 and 12 respectively What percentage of values lie between: a 138 and 162 b 126 and 174 c 126 and 162 d 162 and 174? The middle 68% of a normal distribution lies between 16:2 and 21:4 a What is the mean and standard deviation of the distribution? b Over what range of values would you expect the middle 95% of the data to spread? A bottle shop sells on average 2500 bottles per day with a standard deviation of 300 bottles Assuming that the number of bottles is normally distributed, calculate the percentage of days when: a less than 1900 bottles are sold b more than 2200 bottles are sold c between 2200 and 3100 bottles are sold 10 Masoumeh measured the width (xi cm) of 30 randomly selected cockle shells and observed that 30 X xi = 116:3 cm 30 X and i=1 xi2 = 452:57 i=1 a Calculate: i the mean width ii the variance of the widths b Calculate unbiased estimates of the mean and variance of the population from which this sample comes 11 7, 5, a, 8, 1, a, 4, 6, b have a mean of and a variance of 79 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a Find the values of a and b given a, b Z + b What is the median of the data set? c Find the interquartile range of the data set black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\523IB_HL-2_17.cdr Thursday, 29 November 2007 9:39:14 AM PETERDELL IB_HL-2ed (524) 524 DESCRIPTIVE STATISTICS (Chapter 17) 12 An examination worth 100 marks was given to 800 biology students The cumulative frequency graph for the students’ results follows: number of candidates 800 700 600 500 400 300 200 100 marks 10 cyan 30 40 50 70 60 80 90 100 magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find the number of students who scored 45 marks or less for the test Find the median score Between what values the middle 50% of test results lie? Find the interquartile range of the data What percentage of students obtained a mark of 55 or more? If a ‘distinction’ is awarded to the top 10% of students, what score is required to receive this honour? 95 100 50 75 25 a b c d e f 20 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_17\524IB_HL-2_17.CDR Monday, 12 November 2007 3:36:22 PM PETERDELL IB_HL-2ed (525) Chapter Probability Contents: 18 A B C D E F G H I J K L Experimental probability Sample space Theoretical probability Compound events Using tree diagrams Sampling with and without replacement Binomial probabilities Sets and Venn diagrams Laws of probability Independent events Probabilities using permutations and combinations Bayes’ theorem cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 18A Review set 18B Review set 18C Review set 18D black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\525IB_HL-2_18.CDR Friday, 16 November 2007 3:35:49 PM PETERDELL IB_HL-2ed (526) 526 PROBABILITY (Chapter 18) In the field of mathematics called probability theory we use a mathematical method to describe the chance or likelihood of an event happening This theory has vitally important applications in physical and biological sciences, economics, politics, sport, life insurance, quality control, production planning, and a host of other areas We assign to every event a number which lies between and inclusive We call this number a probability An impossible event which has 0% chance of happening is assigned a probability of A certain event which has 100% chance of happening is assigned a probability of All other events can be assigned a probability between and The number line below shows how we could interpret different probabilities: not likely to happen likely to happen 0.5 impossible certain very unlikely to happen very likely to happen equal chance of happening as not happening The assigning of probabilities is usually based on either: ² observing the results of an experiment (experimental probability), ² using arguments of symmetry (theoretical probability) or HISTORICAL NOTE The development of modern probability theory began in 1653 when gambler Chevalier de Mere contacted mathematician Blaise Pascal with a problem on how to divide the stakes when a gambling game is interrupted during play Pascal involved Pierre de Fermat, a lawyer and amateur mathematician, and together they solved the problem In the process they laid the foundations upon which the laws of probability were formed Blaise Pascal Pierre de Fermat cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 In the late 17th century, English mathematicians compiled and analysed mortality tables These tables showed the number of people who died at different ages From these tables they could estimate the probability that a person would be alive at a future date This led to the establishment of the first life-insurance company in 1699 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\526IB_HL-2_18.CDR Tuesday, 13 November 2007 10:25:54 AM PETERDELL IB_HL-2ed (527) PROBABILITY (Chapter 18) 527 OPENING PROBLEM LIFE TABLE Male Number Expected Number Expected Age remaining surviving surviving remaining life life Age Life Insurance Companies use statistics on life expectancy and death rates to work out the premiums to charge people who insure with them 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 99 The life table shown is from Australia It shows the number of people out of 100 000 births who survive to different ages, and the expected years of remaining life at each age Female 100 000 98 809 98 698 98 555 98 052 97 325 96 688 96 080 95 366 94 323 92 709 89 891 85 198 78 123 67 798 53 942 37 532 20 998 8416 2098 482 73:03 68:90 63:97 59:06 54:35 49:74 45:05 40:32 35:60 30:95 26:45 22:20 18:27 14:69 11:52 8:82 6:56 4:79 3:49 2:68 2:23 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 99 100 000 99 307 99 125 98 956 98 758 98 516 98 278 98 002 97 615 96 997 95 945 94 285 91 774 87 923 81 924 72 656 58 966 40 842 21 404 7004 1953 79:46 75:15 70:22 65:27 60:40 55:54 50:67 45:80 40:97 36:22 31:59 27:10 22:76 18:64 14:81 11:36 8:38 5:97 4:12 3:00 2:36 Notice that out of 100 000 births, 98 052 males are expected to survive to the age of 20 and at that age the survivors are expected to live a further 54:35 years Things to think about: ² Can you use the life table to estimate how many years you can expect to live? ² What is the estimated probability of a new-born boy or girl reaching the age of 15? ² Can the table be used to estimate the probability that: I a 15 year old boy will reach the age of 75 I a 15 year old girl will not reach the age of 75? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ² An insurance company sells policies to people to insure them against death over a 30-year period If the person dies during this period, the beneficiaries receive the agreed payout figure Why are such policies cheaper to take out for a 20 year old than for a 50 year old? ² How many of your classmates would you expect to be alive and able to attend a 30 year class reunion? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\527IB_HL-2_18.CDR Tuesday, 13 November 2007 10:27:55 AM PETERDELL IB_HL-2ed (528) 528 PROBABILITY (Chapter 18) A EXPERIMENTAL PROBABILITY In experiments involving chance we use the following terms to describe what we are doing and the results we are obtaining ² The number of trials is the total number of times the experiment is repeated ² The outcomes are the different results possible for one trial of the experiment ² The frequency of a particular outcome is the number of times that this outcome is observed ² The relative frequency of an outcome is the frequency of that outcome expressed as a fraction or percentage of the total number of trials When a small plastic cone was tossed into the air 279 times it fell on its side 183 times and on its base 96 times The relative frequencies of side and base are 183 96 279 ¼ 0:656 and 279 ¼ 0:344 respectively side base In the absence of any further data, the relative frequency of each event is our best estimate of the probability of each event occurring Experimental probability = relative frequency We write: Experimental P(side) = 0:656, Experimental P(base) = 0:344 INVESTIGATION TOSSING DRAWING PINS If a drawing pin tossed in the air finishes on its back If it finishes If two drawing pins are tossed simultaneously the possible results are: we say it has finished we say it has finished on its side two backs back and side two sides What to do: Obtain two drawing pins of the same shape and size Toss the pair 80 times and record the outcomes in a table Obtain relative frequencies (experimental probabilities) for each of the three events Pool your results with four other people and so obtain experimental probabilities from 400 tosses Note: The others must have pins with the same shape Which gives the more reliable estimates, your results or the group’s? Why? Keep your results as they may be useful later in this chapter cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 In some cases, such as in the investigation above, experimentation is the only way of obtaining probabilities black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\528IB_HL-2_18.CDR Tuesday, 13 November 2007 10:30:55 AM PETERDELL IB_HL-2ed (529) 529 PROBABILITY (Chapter 18) EXERCISE 18A When a batch of 145 paper clips was dropped onto cm by cm squared paper it was observed that 113 fell completely inside squares and 32 finished up on the grid lines Find, to decimal places, the estimated probability of a clip falling: a inside a square b on a line Length - 19 20 - 39 40 - 59 60+ on cm inside cm Jose surveyed the length of TV commercials (in seconds) Find to decimal places the estimated probability that a randomly chosen TV commercial will last: a 20 to 39 seconds b more than a minute c between 20 and 59 seconds (inclusive) Frequency 17 38 19 number of days Betul records the number of phone calls she receives over a period of consecutive days a For how many days did the survey last? b Estimate Betul’s chance of receiving: i no phone calls on one day ii or more phone calls on a day iii less than phone calls on a day 10 4 Pat does a lot of travelling in her car and she keeps records on how often she fills her car with petrol The table alongside shows the frequencies of the number of days between refills Estimate the likelihood that: a there is a four day gap between refills b there is at least a four day gap between refills INVESTIGATION 2 number of calls per day Days between refills Frequency 37 81 48 17 COIN TOSSING EXPERIMENTS The coins of most currencies have two distinct faces, usually referred to as “heads” and “tails” When we toss a coin in the air, we expect it to finish on a head or tail with equal likelihood In this investigation the coins not have to be all the same type What to do: Toss one coin 40 times Record the number of heads resulting in a table: cyan magenta yellow 95 100 50 75 25 Frequency 95 100 50 75 Tally 25 95 100 50 75 25 95 100 50 75 25 Result head head black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\529IB_HL-2_18.CDR Monday, 19 November 2007 11:08:14 AM PETERDELL Relative frequency IB_HL-2ed (530) 530 PROBABILITY (Chapter 18) Toss two coins 60 times Record the number of heads resulting in a table Result heads head head Tally Frequency Relative frequency Toss three coins 80 times Record the number of heads resulting in a table Result heads heads head head Tally Frequency Relative frequency Share your results to 1, and with several others Comment on any similarities and differences Pool your results and find new relative frequencies for tossing one coin, two coins, tossing three coins COIN TOSSING Click on the icon to examine a coin tossing simulation Set it to toss one coin 10 000 times Run the simulation ten times, each time recording the % frequency for each possible result Comment on these results Do your results agree with what you expected? Repeat but this time with two coins and then with three coins From the previous investigation you should have observed that, when tossing two coins, there are roughly twice as many ‘one head’ results as there are ‘no heads’ or ‘two heads’ The explanation for this is best seen using two different coins where you could get: two heads one head one head no heads This shows that we should expect the ratio two heads : one head : no heads to be : : However, due to chance, there will be variations from this when we look at experimental results INVESTIGATION DICE ROLLING EXPERIMENTS You will need: At least one normal six-sided die with numbers to on its faces.¡ Several dice would be useful to speed up the experimentation What to do: WORKSHEET List the possible outcomes for the uppermost face when the die is rolled cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Consider the possible outcomes when the die is rolled 60 times Copy and complete the Outcomes Expected frequency Expected rel frequency following table of your expected results: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\530IB_HL-2_18.CDR Friday, 16 November 2007 3:39:23 PM PETERDELL IB_HL-2ed (531) 531 PROBABILITY (Chapter 18) Roll the die 60 times Record the results in a table like the one shown: Outcome Tally Frequency Relative frequency Total 60 Pool as much data as you can with other students ² Look at similarities and differences from one set to another ² Summarise the overall pooled data in one table SIMULATION Compare your results with your expectation in Use the die rolling simulation on the CD to roll the die 10 000 times Repeat this 10 times On each occasion, record your results in a table like that in Do your results further confirm your expected results? The different possible results when a pair of dice is rolled are shown alongside There are 36 possible outcomes Notice that three of the outcomes, f1, 3g, f2, 2g and f3, 1g, give a sum of Using the illustration above, copy and complete the table of expected (theoretical) results: Sum ¢ ¢ ¢ 12 Fraction of total 36 Fraction as decimal 0:083 If a pair of dice is rolled 360 times, how many of each result (2, 3, 4, , 12) would you expect to get? Extend the table in by adding another row and writing your expected frequencies within it Toss two dice 360 times Record the sum of the two numbers for each toss in a table Sum WORKSHEET Tally Frequency Rel Frequency Total 360 12 10 Pool as much data as you can with other students and find the overall relative frequency of each sum cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 11 Use the two dice simulation on the CD to roll the pair of dice 10 000 times Repeat this 10 times and on each occasion record your results in a table like that in Are your results consistent with your expectations? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\531IB_HL-2_18.CDR Tuesday, 13 November 2007 10:54:01 AM PETERDELL SIMULATION IB_HL-2ed (532) 532 PROBABILITY (Chapter 18) B SAMPLE SPACE A sample space U is the set of all possible outcomes of an experiment There are a variety of ways of representing or illustrating sample spaces LISTING OUTCOMES Example List the sample space of possible outcomes for: a tossing a coin b rolling a die a b When a coin is tossed, there are two possible outcomes ) sample space = fH, Tg When a die is rolled, there are possible outcomes ) sample space = f1, 2, 3, 4, 5, 6g 2-DIMENSIONAL GRIDS When an experiment involves more than one operation we can still use listing to illustrate the sample space However, a grid can often be more efficient Example coin T Illustrate the possible outcomes when coins are tossed by using a 2-dimensional grid Each of the points on the grid represents one of the possible outcomes: coin fHH, HT, TH, TTg H H T TREE DIAGRAMS The sample space in Example could also be represented by a tree diagram The advantage of tree diagrams is that they can be used when more than two operations are involved Example Illustrate, using a tree diagram, the possible outcomes when: a tossing two coins b drawing two marbles from a bag containing many red, green, and yellow marbles b coin marble H T H T T R G cyan magenta yellow 50 25 Y 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Each “branch” gives a different outcome and the sample space is seen to be fHH, HT, TH, TTg 95 H marble R G Y R G Y R G Y 100 coin 75 a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\532IB_HL-2_18.CDR Friday, January 2008 9:47:18 AM DAVID3 What is the sample space here? IB_HL-2ed (533) 533 PROBABILITY (Chapter 18) EXERCISE 18B List a b c d the sample space for the following: twirling a square spinner labelled A, B, C, D the sexes of a 2-child family the order in which blocks A, B, C and D can be lined up the different 3-child families Illustrate on a 2-dimensional grid the sample space for: a rolling a die and tossing a coin simultaneously b rolling two dice c rolling a die and spinning a spinner with sides A, B, C, D d twirling two square spinners: one labelled A, B, C, D and the other 1, 2, 3, Illustrate on a tree diagram the sample space for: a tossing a 5-cent and a 10-cent coin simultaneously b tossing a coin and twirling an equilateral triangular spinner labelled A, B and C c twirling two equilateral triangular spinners labelled 1, and and X, Y and Z d drawing two tickets from a hat containing a number of pink, blue and white tickets C THEORETICAL PROBABILITY Consider the octagonal spinner alongside Since the spinner is symmetrical, when it is spun the arrowed marker could finish with equal likelihood on each of the sections marked to The likelihood of obtaining a particular number, for example 4, would be: 8, chance in 8, 12 12 % or 0:125 This is a mathematical or theoretical probability and is based on what we theoretically expect to occur It is a measure of the chance of that event occurring in any trial of the experiment If we are interested in the event of getting a result of or more from one spin of the octagonal spinner, there are three favourable results (6, or 8) out of the eight possible results Since each of these is equally likely to occur, P(6 or more) = 38 We read Ei_ as ‘3 chances in 8’ In general, for an event E containing equally likely possible results, the probability of E occurring is cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 the number of members of the event E n (E) = the total number of possible outcomes n (U ) 75 25 95 100 50 75 25 P(E) = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\533IB_HL-2_18.CDR Friday, January 2008 9:54:45 AM DAVID3 IB_HL-2ed (534) 534 PROBABILITY (Chapter 18) Example A ticket is randomly selected from a basket containing green, yellow and blue tickets Determine the probability of getting: a a green ticket b a green or yellow ticket c an orange ticket d a green, yellow or blue ticket The sample space is fG, G, G, Y, Y, Y, Y, B, B, B, B, Bg which has + + = 12 outcomes a b P(G) = = 12 c P(a G or a Y) = = P(O) 3+4 12 12 = d 12 P(G, Y or B) = =0 3+4+5 12 =1 In Example notice that in c an orange result cannot occur The calculated probability is 0, because the event has no chance of occurring Also notice in d that a green, yellow or blue result is certain to occur It is 100% likely so the theoretical probability is The two events of no chance of occurring with probability and certain to occur with probability are two extremes P(E) Consequently, for any event E, COMPLEMENTARY EVENTS Example An ordinary 6-sided die is rolled once Determine the chance of: a getting a b not getting a c getting a or d not getting a or The sample space of possible outcomes is f1, 2, 3, 4, 5, 6g a b P(6) = c P(not a 6) = P(1, 2, 3, or 5) = 56 In Example notice that P(1 or 2) = d P(not a or 2) = P(3, 4, 5, or 6) = 46 P(6) + P(not getting a 6) = and that P(1 or 2) + P(not getting a or 2) = This is no surprise as getting a and not getting a are complementary events where one of them must occur Two events are complementary if their probabilities add up to If E is an event, then E is the complementary event of E: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 P(E) + P(E ) = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\534IB_HL-2_18.CDR Friday, 16 November 2007 3:39:49 PM PETERDELL IB_HL-2ed (535) 535 PROBABILITY (Chapter 18) EXERCISE 18C.1 A marble is randomly selected from a box containing green, red and blue marbles Determine the probability that the marble is: a red b green c blue d not red e neither green nor blue f green or red A carton of a dozen eggs contains eight brown eggs The rest are white a How many white eggs are there in the carton? b What is the probability that an egg selected at random is: i brown ii white? A dart board has 36 sectors labelled to 36 Determine the probability that a dart thrown at the centre of the board hits: a a multiple of b a number between and inclusive c a number greater than 20 d e a multiple of 13 f an odd number that is a multiple of g a multiple of and h a multiple of or 32 31 30 29 28 27 26 25 24 23 33 22 36 34 35 21 20 19 18 17 16 15 14 10 11 12 13 What is the probability that a randomly chosen person has his or her next birthday: a on a Tuesday b on a weekend c in July d in January or February? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 a List, in systematic order, the 24 different orders in which four people A, B, C and D may sit in a row b Determine the probability that when the four people sit at random in a row: i A sits on one end ii B sits on one of the two middle seats iii A and B are seated together iv A, B and C are seated together, not necessarily in that order 75 25 a List the possible 3-child families according to the gender of the children For example, GGB means “the first is a girl, the second is a girl, the third is a boy” b Assuming that each of these is equally likely to occur, determine the probability that a randomly selected 3-child family consists of: i all boys ii all girls iii boy then girl then girl iv two girls and a boy v a girl for the eldest vi at least one boy 95 100 50 75 25 5 List the six different orders in which Antti, Kai and Neda may sit in a row If the three of them sit randomly in a row, determine the probability that: a Antti sits in the middle b Antti sits at the left end c Antti sits at the right end d Kai and Neda are seated together black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\535IB_HL-2_18.CDR Tuesday, 13 November 2007 11:45:38 AM PETERDELL IB_HL-2ed (536) 536 PROBABILITY (Chapter 18) USING GRIDS TO FIND PROBABILITIES Two-dimensional grids can give us excellent visual displays of sample spaces We can use them to count favourable outcomes and so calculate probabilities coin B T This point represents ‘a tail from coin A’ and ‘a tail from coin B’ This point represents ‘a tail from coin A’ and ‘a head from coin B’ There are four members of the sample space H H T coin A Example Use a two-dimensional grid to illustrate the sample space for tossing a coin and rolling a die simultaneously From this grid determine the probability of: a tossing a head b getting a tail and a c getting a tail or a coin There are 12 members in the sample space T die 6 12 a P(head) = c P(tail or a ‘5’) = H = b P(tail and a ‘5’) = 12 12 fthe enclosed pointsg EXERCISE 18C.2 Draw the grid of the sample space when a 5-cent and a 10-cent coin are tossed simultaneously Hence determine the probability of getting: a two heads b two tails c exactly one head d at least one head A coin and a pentagonal spinner with sectors 1, 2, 3, and are tossed and spun respectively a Draw a grid to illustrate the sample space of possible outcomes b How many outcomes are possible? c Use your grid to determine the chance of getting: i iii ii iv a tail and a an odd number a head and an even number a head or a cyan magenta yellow 95 die 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A pair of dice is rolled The 36 different possible results are illustrated in the 2-dimensional grid Use the grid to determine the probability of getting: a two 3’s b a and a c a or a d at least one e exactly one f no sixes g a sum of h a sum greater than i a sum of or 11 j a sum of no more than black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\536IB_HL-2_18.CDR Tuesday, 13 November 2007 11:52:22 AM PETERDELL die IB_HL-2ed (537) 537 PROBABILITY (Chapter 18) DISCUSSION Read and discuss: Three children have been experimenting with a coin, tossing it in the air and recording the outcomes They have done this 10 times and have recorded 10 tails Before the next toss they make the following statements: “It’s got to be a head next time!” “No, it always has an equal chance of being a head or a tail The coin cannot remember what the outcomes have been.” “Actually, I think it will probably be a tail again, because I think the coin must be biased - it might be weighted somehow so that it is more likely to give a tail.” Jack: Sally: Amy: D COMPOUND EVENTS P(blue from X and red from Y) = B 16 R R R W Y W R R The question arises, “Is there a quicker, easier way to find this probability?” INVESTIGATION B G G X box Y Consider the following problem: Box X contains blue and green balls Box Y contains red and white ball A ball is randomly selected from each of the boxes Determine the probability of getting “a blue ball from X and a red ball from Y” By illustrating the sample space on the two-dimensional grid shown, we can see that of the 16 possibilities are blue from X and red from Y Each of the outcomes is equally likely, so R B B G G box X PROBABILITIES OF COMPOUND EVENTS The purpose of this investigation is to find a rule for calculating P(A and B) for two events A and B Suppose a coin is tossed and a die is rolled at the same time The result of the coin toss will be called outcome A, and the result of the die roll will be outcome B What to do: Copy and complete, using a 2-dimensional grid if necessary: P(A and B) P(a P(a P(a P(a P(A) P(B) head and a 4) head and an odd number) tail and a number larger than 1) tail and a number less than 3) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 What is the connection between P(A and B), P(A), and P(B)? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\537IB_HL-2_18.CDR Friday, 14 December 2007 11:01:23 AM PETERDELL IB_HL-2ed (538) 538 PROBABILITY (Chapter 18) INVESTIGATION REVISITING DRAWING PINS We cannot find by theoretical argument the probability that a drawing pin will land on its back We can only find this probability by experimentation So, when tossing two drawing pins can we use the rule for compound events: P(back and back) = P(back) £ P(back)? What to do: From Investigation on page 528, what is your estimate of P(back and back)? a Count the number of drawing pins in a full packet They must be identical to each other and the same ones that you used in Investigation b Drop the whole packet onto a solid surface and count the number of backs and sides Repeat this several times Pool results with others and finally estimate P(back) Find P(back) £ P(back) using 2b Is P(back and back) ¼ P(back) £ P(back)? From Investigations and 5, it seems that: If A and B are two events for which the occurrence of each one does not affect the occurrence of the other, then P(A and B) = P(A) £ P(B) Before we can formalise this as a rule, however, we need to distinguish between independent and dependent events INDEPENDENT EVENTS Events are independent if the occurrence of each of them does not affect the probability that the others occur Consider again the example on the previous page Suppose we happen to choose a blue ball from box X This in no way affects the outcome when we choose a ball from box Y So, the two events “a blue ball from X” and “a red ball from Y” are independent If A and B are independent events then P(A and B) = P(A) £ P(B) This rule can be extended for any number of independent events For example: If A, B and C are all independent events, then P(A and B and C) = P(A) £ P(B) £ P(C) Example A coin and a die are tossed simultaneously Determine the probability of getting a head and a without using a grid P(a head and a 3) = P(H) £ P(3) cyan magenta fevents are clearly physically independentg yellow 95 100 50 75 25 95 100 50 12 75 = 25 £ 95 100 50 75 25 95 100 50 75 25 = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\538IB_HL-2_18.CDR Friday, 16 November 2007 3:44:44 PM PETERDELL IB_HL-2ed (539) PROBABILITY (Chapter 18) 539 EXERCISE 18D.1 At a mountain village in Papua New Guinea it rains on average days a week Determine the probability that it rains on: a any one day b two successive days c three successive days A coin is tossed times Determine the probability of getting the following sequences of results: a head then head then head b tail then head then tail A school has two photocopiers On any one day, machine A has an 8% chance of malfunctioning and machine B has a 12% chance of malfunctioning Determine the probability that on any one day both machines will: a malfunction b work effectively A couple decide that they want children, none of whom will be adopted They will be disappointed if the children are not born in the order boy, girl, boy, girl Determine the probability that they will be: a happy with the order of arrival b unhappy with the order of arrival Two marksmen fire at a target simultaneously Jiri hits the target 70% of the time and Benita hits it 80% of the time Determine the probability that: a they both hit the target b they both miss the target c Jiri hits it but Benita misses d Benita hits it but Jiri misses An archer always hits a circular target with each arrow shot, and hits the bullseye on average out of every shots If arrows are shot at the target, determine the probability that the bullseye is hit: a every time b the first two times, but not on the third shot c on no occasion DEPENDENT EVENTS Suppose a hat contains red and blue tickets One ticket is randomly chosen, its colour is noted, and it is then put aside A second ticket is then randomly selected What is the chance that it is red? reds remaining If the first ticket was red, P(second is red) = 47 to choose from reds remaining If the first ticket was blue, P(second is red) = 57 to choose from So, the probability of the second ticket being red depends on what colour the first ticket was We therefore have dependent events cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Two or more events are dependent if they are not independent Dependent events are events for which the occurrence of one of the events does affect the occurrence of the other event black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\539IB_HL-2_18.CDR Tuesday, 13 November 2007 12:21:28 PM PETERDELL IB_HL-2ed (540) 540 PROBABILITY (Chapter 18) For compound events which are dependent, a similar product rule applies as to that for independent events: If A and B are dependent events then P(A then B) = P(A) £ P(B given that A has occurred) Example A box contains red and yellow tickets Two tickets are randomly selected from the box one by one without replacement Find the probability that: a both are red b the first is red and the second is yellow a P(both red) = P(first selected is red and second is red) = P(first selected is red) £ P(second is red given that the first is red) = 46 £ 35 reds remain out of a total of after a red is drawn first = b reds out of a total of tickets P(first is red and second is yellow) = P(first is red) £ P(second is yellow given that the first is red) = 46 £ 25 yellows remain out of a total of after a red is drawn first = 15 reds out of a total of tickets In each fraction the numerator is the number of outcomes in the event The denominator is the total number of possible outcomes Example A hat contains tickets with numbers 1, 2, 3, , 19, 20 printed on them If tickets are drawn from the hat, without replacement, determine the probability that all are prime numbers f2, 3, 5, 7, 11, 13, 17, 19g are primes ) there are 20 numbers of which are primes ) P(3 primes) = P(1st drawn is prime and 2nd is prime and 3rd is prime) = 20 £ 19 £ 18 primes out of 20 numbers primes out of 19 numbers after a successful first draw primes out of 18 numbers after two successful draws ¼ 0:0491 EXERCISE 18D.2 cyan magenta yellow 95 100 50 75 Drawing three chocolates simultaneously implies there is no replacement 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A bin contains 12 identically shaped chocolates of which are strawberry creams If chocolates are selected simultaneously from the bin, determine the probability that: a they are all strawberry creams b none of them are strawberry creams black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\540IB_HL-2_18.CDR Tuesday, 27 November 2007 10:35:27 AM PETERDELL IB_HL-2ed (541) 541 PROBABILITY (Chapter 18) A box contains red and green balls Two balls are drawn one after another from the box Determine the probability that: a both are red b the first is green and the second is red c a green and a red are obtained A lottery has 100 tickets which are placed in a barrel Three tickets are drawn at random from the barrel to decide prizes If John has tickets in the lottery, determine his probability of winning: a first prize b first and second prize c all prizes d none of the prizes A hat contains names of players in a tennis squad including the captain and the vice captain If a team of is chosen at random by drawing the names from the hat, determine the probability that it does not: a contain the captain b contain the captain or the vice captain E USING TREE DIAGRAMS Tree diagrams can be used to illustrate sample spaces if the alternatives are not too numerous Once the sample space is illustrated, the tree diagram can be used for determining probabilities Consider two archers firing simultaneously at a target Yuka’s results Li has probability 34 of hitting a target Li’s results t _ R H and Yuka has probability 45 H Er_ M The tree diagram for Q_t this information is: R_t H rQ_ H = hit M = miss M Notice ² ² ² ² tQ_ M outcome probability H and H Er_ £ tR_ = Qw_Wp_ H and M Er_ £ tQ_ = Dw_p_ M and H rQ_ £ tR_ = Fw_p_ M and M rQ_ £ tQ_ = Aw_p_ that: total The probabilities for hitting and missing are marked on the branches There are four alternative branches, each showing a particular outcome All outcomes are represented The probability of each outcome is obtained by multiplying the probabilities along its branch Example 10 Carl is not having much luck lately His car will only start 80% of the time and his motorbike will only start 60% of the time a Draw a tree diagram to illustrate this situation b Use the tree diagram to determine the chance that: i both will start ii Carl has no choice but to use his car C = car starts M = motorbike starts car C 0.8 cyan magenta yellow 0.6 motorbike M M' C and M' M C' and M 0.4 M' C' and M' 95 100 50 outcome C and M 0.4 0.6 C' 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 0.2 75 a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\541IB_HL-2_18.CDR Monday, 26 November 2007 2:49:14 PM PETERDELL probability 0.8£0.6 = 0.48 0.8£0.4 = 0.32 0.2£0.6 = 0.12 0.2£0.4 = 0.08 total 1.00 IB_HL-2ed (542) 542 PROBABILITY (Chapter 18) b i ii P(both start) = P(C and M) = 0:8 £ 0:6 = 0:48 P(car starts but motorbike does not) = P(C and M0 ) = 0:8 £ 0:4 = 0:32 If there is more than one outcome in an event then we need to add the probabilities of these outcomes Example 11 Bag A contains red and yellow tickets Bag B contains red and yellow tickets A bag is randomly selected by tossing a coin, and one ticket is removed from it Determine the probability that it is yellow Bag A Bag B –1 3R 1R 2Y 4Y 3– bag A –1 ticket outcome R A and R 2– 1– Y A and Y X R B and R 4– Y B and Y X B P(yellow) = P(A and Y) + P(B and Y) fbranches marked with a Xg = 12 £ 25 + 12 £ 45 = EXERCISE 18E Suppose this spinner is spun twice a Copy and complete the branches on the tree diagram shown B b c d e What What What What is is is is the the the the probability that probability that probability that probability that black appears on both spins? yellow appears on both spins? different colours appear on the two spins? black appears on either spin? The probability of rain tomorrow is estimated to be 15 If it does rain, Mudlark will start favourite with probability 12 of winning If it is fine he only has a in 20 chance of winning Display the sample space of possible results of the horse race on a tree diagram Hence determine the probability that Mudlark will win tomorrow Machine A makes 40% of the bottles produced at a factory Machine B makes the rest Machine A spoils 5% of its product, while Machine B spoils only 2% Determine the probability that the next bottle inspected at this factory is spoiled cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Jar A contains white and red discs and Jar B contains white and red disc A jar is chosen at random by the flip of a coin, and one disc is taken at random from it Determine the probability that the disc is red black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\542IB_HL-2_18.CDR Tuesday, 13 November 2007 12:42:16 PM PETERDELL IB_HL-2ed (543) 543 PROBABILITY (Chapter 18) Three bags contain different numbers of blue and red marbles A bag is selected using a die which has three A faces, two B faces, and one C face Red Blue A Red Blue B Red Blue C One marble is then selected randomly from the bag Determine the probability that it is: a blue b red F SAMPLING WITH AND WITHOUT REPLACEMENT Suppose we have a large group of objects If we select one of the objects at random and inspect it for particular features, then this process is known as sampling If the object is put back in the group, we call it sampling with replacement If the object is put to one side, we call it sampling without replacement Sampling is commonly used in the quality control of industrial processes Sometimes the inspection process makes it impossible to return the object to the large group Such processes include: ² Is a chocolate hard or soft-centred? Bite it or squeeze it to see ² Does an egg contain one or two yolks? Break it open and see ² Is the object correctly made? Pull it apart to see Consider a box containing red, blue and yellow marble Suppose we wish to sample two marbles: ² with replacement of the first before the second is drawn ² without replacement of the first before the second is drawn Examine how the tree diagrams differ: With replacement 1st Without replacement 2nd R Ey_ R B ( ) Wy_ Qy_ Ey_ Y ( ) R ( ) Ey_ B Wy_ Wt_ R Wt_ B ( ) Qt_ Y ( ) Et_ R ( ) Qt_ B Ey_ B Wy_ Wy_ Ey_ Y ( ) R ( ) Wy_ B ( ) Qy_ Y Qy_ Qy_ Y 2nd R 1st B Et_ Y ( ) R ( ) Wt_ B ( ) Qt_ Qy_ Y can’t have YY This branch represents a blue marble with the first draw and a red marble with the second draw We write this as BR magenta yellow 95 100 = 50 ² 75 25 95 100 50 75 with replacement P(two reds) = 36 £ 25 95 100 50 75 25 95 100 50 75 25 cyan ² Notice that: black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_18\543IB_HL-2_18.CDR Thursday, 11 March 2010 10:31:04 AM PETER without replacement P(two reds) = 36 £ 25 = IB_HL-2ed (544) 544 PROBABILITY (Chapter 18) Example 12 For the example of the box containing red, blue and yellow marble find the probability of getting two different colours: a if replacement occurs b if replacement does not occur a P(two different colours) = P(RB or RY or BR or BY or YR or YB) = = b £ 11 18 + £ 6 + £ + £ + fticked onesg £ P(two different colours) = P(RB or RY or BR or BY or YR or YB) = = £ 11 15 + £ + £ + £ + Notice that in b P(2 different colours) = ¡ P(2 the same) = ¡ P(RR or BB) = ¡ ( 36 £ 25 + 26 £ 15 ) = 11 15 + £ fcrossed onesg £ + £ Example 13 A bag contains red and blue marbles Two marbles are drawn simultaneously from the bag Determine the probability that at least one is red draw R draw u_R R Ti_ Eu_ B Ei_ P(at least one red) = P(RR or RB or BR) = 58 £ 47 + 58 £ 37 + B Tu_ R Wu_ B Alternatively, = = £ Drawing simultaneously is the same as sampling without replacement 20+15+15 56 25 28 P(at least one red) = ¡ P(no reds) fcomplementary eventsg = ¡ P(BB) etc EXERCISE 18F Two marbles are drawn in succession from a box containing purple and green marbles Determine the probability that the two marbles are different colours if: a the first is replaced b the first is not replaced cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 5 tickets numbered 1, 2, 3, and are placed in a bag Two are taken from the bag without replacement Determine the probability that: a both are odd b both are even c one is odd and the other even A Jar A contains red and green tickets Jar B contains red and green tickets A die has faces with A’s and faces with B’s, and when rolled it is used to select either jar A or jar B B When a jar has been selected, two tickets are randomly selected without replacement from it Determine the probability that: a both are green b they are different in colour A black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\544IB_HL-2_18.CDR Friday, 16 November 2007 3:57:19 PM PETERDELL IB_HL-2ed (545) 545 PROBABILITY (Chapter 18) Marie has a bag of sweets which are all identical in shape The bag contains orange drops and lemon drops She selects one sweet at random, eats it, and then takes another at random Determine the probability that: a both sweets were orange drops b both sweets were lemon drops c the first was an orange drop and the second was a lemon drop d the first was a lemon drop and the second was an orange drop Add your answers to a, b, c and d Explain why the answer must be A bag contains four red and two blue marbles Three marbles are selected simultaneously Determine the probablity that: a all are red b only two are red c at least two are red Bag A contains red and white marbles Bag B contains red and white marbles One marble is randomly selected from A and its colour noted If it is red, reds are added to B If it is white, whites are added to B A marble is then selected from B What are the chances that the marble selected from B is white? A man holds two tickets in a 100-ticket lottery in which there are two winning tickets If no replacement occurs, determine the probability that he will win: a both prizes b neither prize c at least one prize A container holds red balls, white balls, and black balls A ball is chosen at random from the container and is not replaced A second ball is then chosen Find the probability of choosing one white and one black ball in any order A bag contains yellow and n blue markers The probability of choosing yellow markers, without replacement after the first choice, How many blue markers are there in the bag? is 13 INVESTIGATION SAMPLING SIMULATION When balls enter the ‘sorting’ chamber shown they hit a metal rod and may go left or right This movement continues as the balls fall from one level of rods to the next The balls finally come to rest in collection chambers at the bottom of the sorter This sorter looks very much like a tree diagram rotated through 90o in A B C D E Click on the icon to open the simulation Notice that the sliding bar will alter the probabilities of balls going to the left or right at each rod cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 What to do: To simulate the results of tossing two coins, set the bar to 50% and the sorter to show Run the simulation 200 times and repeat this four more times Record each set of results black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\545IB_HL-2_18.CDR Tuesday, 13 November 2007 1:48:30 PM PETERDELL SIMULATION IB_HL-2ed (546) 546 PROBABILITY (Chapter 18) A bag contains blue and red marbles Two marbles are randomly selected from the bag, the first being replaced before the second is drawn = 70%, set the bar to 70% Since P(blue) = 10 The sorter should show: Run the simulation a large number of times Use the results to estimate the probability of getting: a two blues b one blue c no blues 2nd selection outcome probability Jq_p_ B BB &qJ_p_*X } B Jq_p_ Dq_p_ R BR &qJ_p_* &qD_p_* Jq_p_ B RB &qD_p_* &qJ_p_* Dq_p_ R RR &qD_p_*X {z 1st selection | The tree diagram representation of the marble selection in is: a The tree diagram gives us theoretical probabilities for the different outcomes Do they agree with the experimental results obtained in 2? R Dq_p_ b Write down the algebraic expansion of (a + b)2 : and b = 10 in the (a + b)2 expansion What you notice? c Substitute a = 10 From the bag of blue and red marbles, three marbles are randomly selected with replacement Set the sorter to levels and the bar to 70%: Run the simulation a large number of times to obtain experimental estimates of the probabilities of getting: a three blues b two blues c one blue d no blues a Use a tree diagram showing 1st selection, 2nd selection and 3rd selection to find theoretical probabilities of getting the results of b Show that (a + b)3 = a3 + 3a2 b + 3ab2 + b3 and use this expansion with and b = 10 to also check the results of and 5a a = 10 Consider the sampling simulator with the bar at 50% to explain why many distributions are symmetrical and bell-shaped G BINOMIAL PROBABILITIES Consider a die which has red faces and black faces We roll the die three times and record the results BRR RBB If R represents “the result is red” and B represents “the RBR BRB result is black”, the possible outcomes are as shown RRR RRB BBR BBB alongside: Notice that the ratio of possible outcomes is : : : One outcome is all red Three outcomes are two red and one black Three outcomes are one red and two black One outcome is all black Now for each die, P(R) = and P(B) = 23 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 So, for rolling the die times we have the following events and probabilities: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\546IB_HL-2_18.CDR Monday, 26 November 2007 2:51:49 PM PETERDELL IB_HL-2ed (547) 547 PROBABILITY (Chapter 18) R B R B B Notice that ¡ ¢3 +3 Outcome all red RRR BRR R B R B R B R B R Event RBR red and black RRB RBB ¡ ¢2 ¡ ¢ 3 +3 red and black BRB all black BBB ¡ ¢ ¡ ¢2 3 + BBR ¡ ¢3 Probabilities ¡1¢ ¡1¢ ¡1¢ 3 ¡2¢ ¡1¢ ¡1¢ ¡ 31 ¢ ¡ 32 ¢ ¡ 31 ¢ 3 ¡1¢ ¡1¢ ¡2¢ 3 Total Probability ¡ ¢3 = 27 3 ¡ ¢2 ¡ ¢ = 27 ¡ ¢ ¡ ¢2 = 12 27 3 ¡1¢ ¡2¢ ¡2¢ 3 ¡2¢ ¡1¢ ¡2¢ 3 ¡2¢ ¡2¢ ¡1¢ 3 ¡2¢ ¡2¢ ¡2¢ 3 3 ¡ ¢3 = is the binomial expansion for ¡1 27 + ¢ 3 If E is an event with probability p of occurring and its complement E has probability q = ¡ p of occurring, then the probability generator for the various outcomes over n independent trials is (p + q)n In general, For example: Suppose E is the event of a randomly chosen light globe being faulty, with P(E) = p = 0:03 and P(E ) = q = 0:97 If four independent samples are taken, the probability generator is (0:03 + 0:97)4 = (0:03)4 + 4(0:03)3 (0:97) + 6(0:03)2 (0:97)2 + 4(0:03)(0:97)3 + (0:97)4 Es Notice that Es and E Es and E s E and E s E0s ¡ ¢ P(E occurs x times and E occurs n ¡ x times) = nx px q n¡x : Example 14 An archer has a 90% chance of hitting the target with each arrow If arrows are fired, determine the probability generator and hence the chance of hitting the target: a twice only b at most times Let H be the event of ‘hitting the target’, so P(H) = 0:9 and P(H ) = 0:1 The probability generator is (0:9 + 0:1)5 = (0:9)5 + 5(0:9)4 (0:1) + 10(0:9)3 (0:1)2 + 10(0:9)2 (0:1)3 + 5(0:9)(0:1)4 + (0:1)5 hits and miss hits hits and misses hits and misses hit and misses misses cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Let X be the number of arrows that hit the target a P(hits twice only) b P(hits at most times) = P(X = 2) = P(X = 0, 1, or 3) = 10(0:9)2 (0:1)3 = (0:1)5 + 5(0:9)(0:1)4 + 10(0:9)2 (0:1)3 + 10(0:9)3 (0:1)2 = 0:0081 ¼ 0:0815 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\547IB_HL-2_18.CDR Tuesday, 13 November 2007 2:05:53 PM PETERDELL IB_HL-2ed (548) 548 PROBABILITY (Chapter 18) A graphics calculator can be used to find binomial probabilities For example, to find the probabilities in Example 14 use: b P(X 3) = binomcdf(5, 0:9, 3) a P(X = 2) = binompdf(5, 0:9, 2) n p x n p x TI Use your calculator to check the answers given above C EXERCISE 18G a Expand (p + q)4 : b If a coin is tossed four times, what is the probability of getting heads? a Expand (p + q)5 b If five coins are tossed simultaneously, what is the probability of getting: ii heads and tails i heads and tail in any order iii heads and tail in that order? a Expand ( 23 + 13 )4 b Four chocolates are randomly taken (with replacement) from a box which contains strawberry creams and almond centres in the ratio : What is the probability of getting: i all strawberry creams ii two of each type iii at least strawberry creams? a Expand ( 34 + 14 )5 b In New Zealand in 1946, coins of value two shillings were of two types: normal kiwis and ‘flat back’ kiwis, in the ratio : From a batch of 1946 two shilling coins, five were selected at random with replacement What is the probability that: i two were ‘flat backs’ ii at least were ‘flat backs’ iii at most were normal kiwis? When rifle shooter Huy fires a shot, he hits the target 80% of the time If Huy fires shots at the target, determine the probability that he has: b at least hits a hits and misses in any order 5% of electric light bulbs are defective at manufacture If bulbs are tested at random with each one being replaced before the next is chosen, determine the probability that: a two are defective b at least one is defective In a multiple choice test there are 10 questions Each question has choices, one of which is correct If 70% is the pass mark and Raj (who knows nothing) guesses at each answer, determine the probability that he will pass Martina beats Jelena in games out of at tennis What is the probability that Jelena wins a set of tennis games to 4? Hint: What is the score after games? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 How many ordinary dice are needed for there to be a better than an even chance of at least one six when they are thrown together? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\548IB_HL-2_18.CDR Friday, 16 November 2007 4:02:01 PM PETERDELL IB_HL-2ed (549) PROBABILITY (Chapter 18) H 549 SETS AND VENN DIAGRAMS Venn diagrams are a useful way of representing the events in a sample space These diagrams usually consist of a rectangle which represents the complete sample space or universal set, and circles within it which represent particular events Venn diagrams can be used to solve certain types of probability questions and also to establish a number of probability laws The Venn diagram alongside shows the sample space for rolling a die A We can write the universal set U = f1, 2, 3, 4, 5, 6g since the sample space consists of the numbers from U to The event A is “a number less than 3” There are two outcomes which satisfy event A, and we can write A = f1, 2g SET NOTATION ² The universal set or sample space U is represented by a rectangle An event A is usually represented by a circle A U A0 (shaded green) is the complement of A (shaded purple) It represents the non-occurrence of A Note: P(A) + P(A0 ) = ² A A' U If U = f1, 2, 3, 4, 5, 6, 7g and A = f2, 4, 6g then A0 = f1, 3, 5, 7g ² x A reads ‘x is in A’ and means that x is an element of the set A ² n(A) reads ‘the number of elements in set A’ ² A [ B denotes the union of sets A and B This set contains all elements belonging to A or B or both A and B A [ B is shaded in purple A A [ B = fx j x A or x Bg B U ² A \ B denotes the intersection of sets A and B This is the set of all elements common to both sets A \ B is shaded in purple cyan magenta yellow 95 100 50 75 25 95 50 100 A \ B = fx j x A and x Bg U 75 25 95 B 100 50 75 25 95 100 50 75 25 A black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\549IB_HL-2_18.CDR Friday, 16 November 2007 4:03:21 PM PETERDELL IB_HL-2ed (550) 550 PROBABILITY (Chapter 18) ² Disjoint sets are sets which not have elements in common A B These two sets are disjoint A \ B = ? where ? represents an empty set A and B are said to be mutually exclusive U Note: We cannot determine whether two or more sets are independent by just looking at a Venn diagram Example 15 If A is the set of all factors of 36 and B is the set of all factors of 54, find: a A[B b A\B A = f1, 2, 3, 4, 6, 9, 12, 18, 36g and B = f1, 2, 3, 6, 9, 18, 27, 54g a A [ B = the set of factors of 36 or 54 = f1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54g b A \ B = the set of factors of both 36 and 54 = f1, 2, 3, 6, 9, 18g Example 16 a On separate Venn diagrams containing two events A and B that intersect, shade the region representing: a in A but not in B b neither in A nor B: A b B 15 27 26 cyan magenta yellow 95 100 50 75 25 95 100 50 75 Number who play at least one sport = 15 + 27 + 26 = 68 25 e Number who play neither sport =7 d Number who play both sports = 27 95 c 100 Number who play hockey = 27 + 26 = 53 50 b 75 Number in the club = 15 + 27 + 26 + = 75 25 a b who play hockey d who play neither sport 95 H a in the club c who play both sports e who play at least one sport 100 50 75 25 U T If the Venn diagram alongside illustrates the number of people in a sporting club who play tennis (T ) and hockey (H ), determine the number of people: B U Example 17 A black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\550IB_HL-2_18.CDR Tuesday, 13 November 2007 2:18:24 PM PETERDELL IB_HL-2ed (551) 551 PROBABILITY (Chapter 18) Example 18 The Venn diagram alongside represents the set U of all children in a class Each dot represents a student The event E shows all those students with blue eyes Determine the probability that a randomly selected child: a has blue eyes b does not have blue eyes E' E U n(U ) = 23, n(E) = E P(blue eyes) = b P(not blue eyes) = 15 U n(E) = n(U ) a 23 n(E ) = n(U ) 15 23 or P(not blue) = ¡ P(blue eyes) = ¡ 23 = 15 23 Example 19 In a class of 30 students, 19 study Physics, 17 study Chemistry, and 15 study both of these subjects Display this information on a Venn diagram and hence determine the probability that a randomly selected class member studies: a both subjects b at least one of the subjects c Physics but not Chemistry d exactly one of the subjects e neither subject f Chemistry if it is known that the student studies Physics P Let P represent the event of ‘studying Physics’ and C represent the event of ‘studying Chemistry’ C a b c a + b = 19 fas b + c = 17 fas b = 15 fas a + b + c + d = 30 fas b = 15, a = 4, c = 2, Now d P C 15 ) 19 study Physicsg 17 study Chemistryg 15 study bothg there are 30 in the classg d = 9 magenta yellow 10 P(C given P ) = 15 15+4 95 f = 100 95 100 50 75 25 = = P(studies exactly one) = 15 = 4+2 30 10 30 95 50 75 25 95 100 50 75 = 4+15+2 30 d 15 P(studies neither) = 25 30 50 e = P(P but not C) = or P(studies at least one subject) 75 c 15 30 25 = cyan b P(studies both) 100 a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\551IB_HL-2_18.CDR Tuesday, 13 November 2007 2:25:13 PM PETERDELL 15 19 IB_HL-2ed (552) 552 PROBABILITY (Chapter 18) EXERCISE 18H.1 If A is the set of all factors of and B is the set of all positive even integers < 11: a describe A and B using set notation b find: i n(A) ii A [ B iii A \ B On separate Venn diagrams containing two events A and B that intersect, shade the region representing: a in A b in B c in both A and B f in exactly one of A or B d in A or B e in B but not in A The Venn diagram alongside illustrates the number of H C students in a particular class who study Chemistry (C) 17 and History (H) Determine the number of students: a in the class b who study both subjects c who study at least one of the subjects d who only study Chemistry In a survey at an alpine resort, people were asked whether they liked skiing (S ) or snowboarding (B ) Use the Venn diagram to determine the number of people: a b c d U S 37 in the survey who liked both activities who liked neither activity who liked exactly one activity 15 B U In a class of 40 students, 19 play tennis, 20 play netball, and play neither of these sports A student is randomly chosen from the class Determine the probability that the student: a plays tennis b does not play netball c plays at least one of the sports d plays one and only one of the sports e plays netball but not tennis f plays tennis given he or she plays netball 50 married men were asked whether they gave their wife flowers or chocolates for their last birthday The results were: 31 gave chocolates, 12 gave flowers, and gave both chocolates and flowers If one of the married men was chosen at random, determine the probability that he gave his wife: a chocolates or flowers b chocolates but not flowers c neither chocolates nor flowers d flowers if it is known that he did not give her chocolates The medical records for a class of 30 children showed that 24 had previously had measles, 12 had previously had measles and mumps, and 26 had previously had at least one of measles or mumps If one child from the class is selected at random, determine the probability that he or she has had: a mumps b mumps but not measles c neither mumps nor measles d measles if it is known that the child has had mumps cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 If A and B are two non-disjoint sets, shade the region of a Venn diagram representing: b A0 \ B c A [ B0 d A0 \ B a A0 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\552IB_HL-2_18.CDR Friday, 16 November 2007 4:04:06 PM PETERDELL IB_HL-2ed (553) PROBABILITY (Chapter 18) The diagram alongside is the most general case for three events in the same sample space U On separate Venn diagram sketches, shade: c B\C a A b B0 d A[C e A\B\C f (A [ B) \ C A 553 B C U USING VENN DIAGRAMS TO VERIFY SET IDENTITIES Example 20 Verify that (A [ B)0 = A0 \ B this shaded region is (A [ B) A this shaded region is (A [ B)0 B represents A0 represents B A represents A0 \ B B Thus (A [ B)0 and A0 \ B are represented by the same regions, verifying that (A [ B)0 = A0 \ B : EXERCISE 18H.2 Verify that: a (A \ B)0 = A0 [ B b A [ (B \ C) = (A [ B) \ (A [ C) c A \ (B [ C) = (A \ B) [ (A \ C) Suppose S = fx: x is a positive integer < 100g Let A = fmultiples of in Sg and B = fmultiples of in Sg a How many elements are there in: i A ii B iii A \ B b If n(E) represents the number of elements in set E, verify that n(A [ B) = n(A) + n(B) ¡ n(A \ B): iv A[B? A B a c b From the Venn diagram, P(A) = a+b a+b+c+d d a Use the Venn diagram to find: i P(B) ii P(A and B) iii P(A or B) iv P(A) + P(B) ¡ P(A and B) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b What is the connection between P(A or B) and P(A) + P(B) ¡ P(A and B)? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\553IB_HL-2_18.CDR Friday, 16 November 2007 4:04:29 PM PETERDELL IB_HL-2ed (554) 554 PROBABILITY (Chapter 18) I LAWS OF PROBABILITY THE ADDITION LAW In the previous exercise we showed that for two events A and B, P(A [ B) = P(A) + P(B) ¡ P(A \ B) This is known as the addition law of probability, and can be written as P(either A or B) = P(A) + P(B) ¡ P(both A and B) Example 21 If P(A) = 0:6, P(A [ B) = 0:7 and P(A \ B) = 0:3, find P(B): P(A [ B) = P(A) + P(B) ¡ P(A \ B) ) 0:7 = 0:6 + P(B) ¡ 0:3 ) P(B) = 0:4 or a Using a Venn diagram with the probabilities on it, a + 0:3 = 0:6 and a + b + 0:3 = 0:7 ) a = 0:3 ) a + b = 0:4 ) 0:3 + b = 0:4 ) b = 0:1 ) P(B) = 0:3 + b = 0:4 b 0.3 A B MUTUALLY EXCLUSIVE EVENTS (DISJOINT EVENTS) If A and B are mutually exclusive events then P(A \ B) = and so the addition law becomes P(A [ B) = P(A) + P(B): Example 22 A box of chocolates contains with hard centres (H) and 12 with soft centres (S) a Are the events H and S mutually exclusive? b Find i P(H) ii P(S) iii P(H \ S) iv P(H [ S): a Chocolates cannot have both a hard and a soft centre ) H and S are mutually exclusive b i P(H) = = 18 ii P(S) = = 12 18 iii P(H \ S) =0 P(H [ S) iv = 18 18 =1 CONDITIONAL PROBABILITY If we have two events A and B, then cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A j B is used to represent that ‘A occurs knowing that B has occurred’ A j B is often read as ‘A given B’ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\554IB_HL-2_18.CDR Tuesday, 29 January 2008 11:56:04 AM PETERDELL IB_HL-2ed (555) 555 PROBABILITY (Chapter 18) Example 23 In a class of 25 students, 14 like pizza and 16 like iced coffee One student likes neither and students like both One student is randomly selected from the class What is the probability that the student: a likes pizza b likes pizza given that he or she likes iced coffee? P The Venn diagram of the situation is shown C 14 25 P(pizza) = b P(pizza j iced coffee) = 16 fof the 16 who like iced coffee, like pizzag If A and B are events then Proof: P(A j B) A a P(A \ B) P(B) P(A j B) = fof the 25 students, 14 like pizzag a 10 = b b+c = b=(a + b + c + d) (b + c)=(a + b + c + d) = P(A \ B) P(B) b c B d U fVenn diagramg P(A \ B) = P(A j B) P(B) or P(A \ B) = P(B j A) P(A) It follows that Example 24 In a class of 40 students, 34 like bananas, 22 like pineapples, and dislike both fruits If a student is randomly selected, find the probability that the student: a likes both fruits b likes at least one fruit c likes bananas given that he or she likes pineapples d dislikes pineapples given that he or she likes bananas B represents students who like bananas P represents students who like pineapples We are given that a + b = 34 b + c = 22 a + b + c = 38 P c B P 16 18 ) c = 38 ¡ 34 =4 P(likes at least one) magenta yellow 50 25 = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\555IB_HL-2_18.CDR Tuesday, 13 November 2007 2:55:16 PM PETERDELL and so b = 18 and a = 16 P(B j P ) c = 95 50 25 95 = 38 40 19 20 100 = 100 50 25 95 100 50 75 25 = 18 40 20 75 = cyan b P(likes both) 75 a 95 b 100 a 75 B 18 22 11 P(P j B) d = = 16 34 17 IB_HL-2ed (556) 556 PROBABILITY (Chapter 18) Example 25 Bin A contains red and white tickets Bin B contains red and white ticket A die with faces marked A and two faces marked B is rolled and used to select bin A or B A ticket is then selected from this bin Determine the probability that: a the ticket is red b the ticket was chosen from B given it is red bin a ticket Et_ R Wt_ W Rt_ R = = A Ry_ Wy_ P(R) £ b P(B j R) = = B W Qt_ + = £ fpath + path g P(B \ R) P(R) £ path 2 EXERCISE 18I In a group of 50 students, 40 study Mathematics, 32 study Physics, and each student studies at least one of these subjects a Use a Venn diagram to find how many students study both subjects b If a student from this group is randomly selected, find the probability that he or she: i studies Mathematics but not Physics ii studies Physics given that he or she studies Mathematics In a group of 40 boys, 23 have dark hair, 18 have brown eyes, and 26 have dark hair, brown eyes or both One of the boys is selected at random Determine the probability that he has: a dark hair and brown eyes b neither dark hair nor brown eyes c dark hair but not brown eyes d brown eyes given that he has dark hair 50 students go bushwalking 23 get sunburnt, 22 get bitten by ants, and are both sunburnt and bitten by ants Determine the probability that a randomly selected student: a escaped being bitten b was either bitten or sunburnt c was neither bitten nor sunburnt d was bitten, given that he or she was sunburnt e was sunburnt, given that he or she was not bitten cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 400 families were surveyed It was found that 90% had a TV set and 60% had a computer Every family had at least one of these items If one of these families is randomly selected, find the probability it has a TV set given that it has a computer black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\556IB_HL-2_18.CDR Friday, 16 November 2007 4:08:19 PM PETERDELL IB_HL-2ed (557) PROBABILITY (Chapter 18) 557 In a certain town three newspapers are published 20% of the population read A, 16% read B, 14% read C, 8% read A and B, 5% read A and C, 4% read B and C, and 2% read all newspapers A person is selected at random Use a Venn diagram to help determine the probability that the person reads: a none of the papers b at least one of the papers c exactly one of the papers d either A or B e A, given that the person reads at least one paper f C, given that the person reads either A or B or both Urn A contains red and blue marbles, and urn B contains red and blue marble Peter selects an urn by tossing a coin, and takes a marble from that urn a Determine the probability that it is red b Given that the marble is red, what is the probability that it came from B? The probability that Greta’s mother takes her shopping is 25 When Greta goes shopping with her mother she gets an icecream 70% of the time When Greta does not go shopping with her mother she gets an icecream 30% of the time Determine the probability that: a Greta’s mother buys her an icecream when shopping b Greta went shopping with her mother, given that her mother buys her an icecream On a given day, photocopier A has a 10% chance of malfunctioning and machine B has a 7% chance of the same Given that at least one of the machines malfunctioned today, what is the chance that machine B malfunctioned? On any day, the probability that a boy eats his prepared lunch is 0:5 The probability that his sister eats her lunch is 0:6 The probability that the girl eats her lunch given that the boy eats his is 0:9 Determine the probability that: a both eat their lunch b the boy eats his lunch given that the girl eats hers c at least one of them eats lunch 10 The probability that a randomly selected person has cancer is 0:02 The probability that he or she reacts positively to a test which detects cancer is 0:95 if he or she has cancer, and 0:03 if he or she does not Determine the probability that a randomly tested person: a reacts positively b has cancer given that he or she reacts positively 11 A double-headed, a double-tailed, and an ordinary coin are placed in a tin can One of the coins is randomly chosen without identifying it The coin is tossed and falls “heads” Determine the probability that the coin is the “double-header” 12 The English Premier League consists of 20 teams Tottenham is currently in 8th place on the table It has 20% chance of winning and 60% chance of losing against any team placed above it If a team is placed below it, Tottenham has a 50% chance of winning and a 30% chance of losing Find the probability that Tottenham will draw its next game cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 13 If events A and B are not mutually exclusive, P(A [ B) = P(A) + P(B) ¡ P(A \ B) Use a Venn diagram to find a corresponding rule for P(A [ B [ C) where A, B and C are not mutually exclusive black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\557IB_HL-2_18.CDR Tuesday, 13 November 2007 3:16:42 PM PETERDELL IB_HL-2ed (558) 558 PROBABILITY (Chapter 18) J INDEPENDENT EVENTS A and B are independent events if the occurrence of each one of them does not affect the probability that the other occurs, i.e., P(A j B) = P(A) and P(B j A) = P(B) So, as P(A \ B) = P(A j B) P(B), A and B are independent events , P(A \ B) = P(A) P(B) Example 26 When two coins are tossed, A is the event of getting heads When a die is rolled, B is the event of getting a or Show that A and B are independent events P(A) = die and P(B) = 26 Therefore, P(A) P(B) = £ = 12 P(A \ B) = P(2 heads and a or a 6) HH HT TH coins TT = 24 = 12 So, as P(A \ B) = P(A) P(B), the events A and B are independent Example 27 P(A) = 12 , P(B) = and P(A [ B) = p Find p if: a A and B are mutually exclusive b A and B are independent a If A and B are mutually exclusive, A \ B = ? and so P(A \ B) = But P(A [ B) = P(A) + P(B) ¡ P(A \ B) ) p= b + ¡0 = If A and B are independent, P(A \ B) = P(A) P(B) = ¡ Given P(A) = 25 , P(B j A) = ) P(A [ B) = + and hence p = and P(B j A0 ) = £ = Example 28 b P(A \ B ) so P(B \ A) = P(B j A) P(A) = magenta yellow £ = 95 £ = 15 20 100 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Similarly, P(B \ A0 ) = P(B j A0 ) P(A0 ) = 50 P(B \ A) P(A) 75 P(B j A) = 25 a P(B) cyan find: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\558IB_HL-2_18.CDR Friday, 16 November 2007 4:09:35 PM PETERDELL IB_HL-2ed (559) 559 PROBABILITY (Chapter 18) ) the Venn diagram is: 15 a P(B) = A B + 20 = 17 60 b P(A \ B ) = P(A) ¡ P(A \ B) qS_t_ wD_p_ = = 2 ¡ 15 15 EXERCISE 18J If P(R) = 0:4, P(S) = 0:5 and P(R [ S) = 0:7, are R and S independent events? If P(A) = 25 , P(B) = P(A \ B) a and P(A [ B) = 12 , find: P(B j A) b c P(A j B) Are A and B independent events? If P(X) = 0:5, P(Y ) = 0:7 and X and Y are independent events, determine the probability of the occurrence of: a both X and Y b X or Y c neither X nor Y e X given that Y occurs d X but not Y The probabilities that A, B and C can solve a particular problem are 35 , 23 and 12 respectively If they all try, determine the probability that at least one of the group solves the problem a Find the probability of getting at least one six when a die is rolled times b If a die is rolled n times, find the smallest n such that P(at least one in n throws) > 99%: A and B are independent events Prove that A0 and B are also independent events Two students, Karl and Hanna, play a game in which they take it in turns to select a card, with replacement, from a well-shuffled pack of 52 playing cards The first person to select an ace wins the game Karl has the first turn i Find the probability that Karl wins on his third turn ii Show that the probability that Karl wins prior to his (n + 1)th turn is ¡ ¢ 13 12 2n 25 ¡ ( 13 ) a iii Hence, find the probability that Karl wins the game b If Karl and Hanna play this game seven times, find the probability that Karl will win more games than Hanna Given that P(A \ B) = 0:1 and P(A \ B ) = 0:4, find P(A [ B ) if A and B are independent Given P(C) = 20 , P(C j D0 ) = and P(C j D) = 13 , cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a find i P(D) ii P(C [ D0 ) b Are C and D independent events? Give a reason for your answer black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\559IB_HL-2_18.CDR Tuesday, 13 November 2007 3:25:59 PM PETERDELL IB_HL-2ed (560) 560 PROBABILITY (Chapter 18) 10 A delirious man stands on the edge of a cliff and takes random steps either towards or away from the cliff’s edge The probability of him stepping away from the edge is 35 , and towards the edge is 25 Find the probability he does not step over the cliff in his first four steps K PROBABILITIES USING PERMUTATIONS AND COMBINATIONS Permutations and combinations can sometimes be used to find probabilities of various events They are particularly useful if the sample size is large It is useful to remember that: P(an event) = number of possibilities with the required properties of the event total number of unrestricted possibilities For example: Suppose we select at random a team of players from a squad of boys ¡ ¢ and girls The total number of unrestricted possibilities is 15 since we are choosing of the 15 available players The number of possibilities with the property of ‘4 boys and girls’ ¡ ¢¡ ¢ is 84 73 since we want any of the boys and any of the girls ¡8¢ ¡7¢ ) P(4 boys and girls) = ¡ 15 ¢3 The biggest difficulty in probability problems involving permutations or combinations seems to be in sorting out which to use ² ² Remember: permutations involve the ordering of objects or things, whereas combinations involve selections such as committees or teams Example 29 From a squad of 13 which includes brothers, a team of is randomly selected by drawing names from a hat Determine the probability that the team contains: a all the brothers b at least of the brothers ¡ 13 ¢ different teams of that can be chosen from 13 people There are ¡4¢ ¡9¢ a Of these teams, contain all brothers and any others ¡4¢ ¡9¢ ¡ 13 ¢3 ¼ 0:0490 ) P(team contains all the brothers) = b P(at least brothers) = P(2 brothers or brothers or brothers) ¡4¢ ¡9¢ ¡4¢ ¡9¢ ¡4¢ ¡9¢ ¡ 13 ¢5 = ¡ 13 ¢4 + ¡ 13 ¢3 + 7 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ¼ 0:783 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\560IB_HL-2_18.CDR Monday, 26 November 2007 3:01:51 PM PETERDELL IB_HL-2ed (561) PROBABILITY (Chapter 18) 561 Example 30 letters U, S, T, I, N are placed at random in a row What is the probability that the word UNITS is spelled out? There are 5! different permutations of the letters, one of which spells UNITS ) P(UNITS is spelled) = 5! = 120 Notice that counting permutations is not essential here We could have used: P(UNITS) = £ to choose from and we want only U £ £ £ 1 = 120 now are left and we want N EXERCISE 18K A committee of is chosen from 11 people by random selection What is the chance that sisters X and Y are on the committee? alphabet blocks D, A, I and S are placed at random in a row What is the likelihood that they spell out either AIDS or SAID? A team of is randomly chosen from a squad of 12 Determine the probability that both the captain and vice-captain are chosen Of the 22 people on board a plane, are professional golfers If the plane crashes and people are killed, determine the chance that all three golfers survive 5 boys sit at random on seats in a row Determine the probability that the two friends Keong and James sit: a at the ends of the row b together A committee of is randomly selected from men and women Determine the likelihood that it consists of: a all men b at least men c at least one of each sex people including friends A, B and C are randomly seated on a row of chairs Determine the likelihood that A, B and C are seated together cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A school committee of is to be chosen at random from 11 senior students and junior students Find the probability that: a only senior students are chosen b all three junior students are chosen black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\561IB_HL-2_18.CDR Tuesday, 13 November 2007 3:36:55 PM PETERDELL IB_HL-2ed (562) 562 PROBABILITY (Chapter 18) L BAYES’ THEOREM Suppose a sample space U is partitioned into two mutually exclusive regions by an event A and its complement A0 We can show this on a Venn diagram as A A' or A U A' U Now consider another event B in the sample space U We can show this on a Venn diagram as A A B A' or B U U P(A j B) = Bayes’ theorem states that P(B j A) P(A) P(B) where P(B) = P(B j A) P(A) + P(B j A0 ) P(A0 ) Proof: P(A j B) = P(A \ B) P(B j A) P(A) = P(B) P(B) where P (B) = P(B \ A) + P(B \ A0 ) = P(B j A) P(A) + P(B j A0 ) P(A0 ) Example 31 A can contains blue and green marbles One marble is randomly drawn from the can without replacement and its colour is noted A second marble is then drawn What is the probability that: a the second marble is blue b the first was green given that the second is blue? Let A be the event that the first marble is green Wy_ yellow 95 100 50 A' 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Ry_ magenta B Qt_ B' Et_ B Wt_ B' A Let B be the event that the second marble is blue cyan Rt_ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\562IB_HL-2_18.CDR Friday, 16 November 2007 4:13:13 PM PETERDELL IB_HL-2ed (563) 563 PROBABILITY (Chapter 18) a b P(second marble is blue) = P(B) = P(B j A) P(A) + P(B j A0 ) P(A0 ) = = £ + £ P(first was green j second is blue) = P(A j B) = = = P(B j A) P(A) P(B) £ fBayes’ theoremg fusing ag EXERCISE 18L Coffee making machines Alpha and Beta produce coffee in identically shaped plastic cups Alpha produces 65% of the coffee sold each day, and Beta produces the remainder Alpha underfills a cup 4% of the time while Beta underfills a cup 5% of the time a If a cup of coffee is chosen at random, what is the probability it is underfilled? b A cup of coffee is randomly chosen and is found to be underfilled What is the probability it came from Alpha? 54% of the students at a university are females 8% of the male students are colour-blind and 2% of the female students are colour-blind If a randomly chosen student: a is colour-blind, find the probability that the student is male b is not colour-blind, find the probability that the student is female A marble is randomly chosen from a can containing red and blue marbles It is replaced by two marbles of the other colour Another marble is then randomly chosen from the can If the marbles chosen are the same colour, what is the probability that they are both blue? 35% of the animals in a deer herd carry the TPC gene 58% of these deer also carry the SD gene, while 23% of the deer without the TPC gene carry the SD gene If a deer is randomly chosen and is found to carry the SD gene, what is the probability it does not carry the TPC gene? A new blood test has been shown to be effective in the early detection of a form of cancer The probability that the test correctly identifies someone with the cancer is 0:97, and the probability that the test correctly identifies someone without the cancer is 0:93 Approximately 0:1% of the general population are known to contract this cancer A patient had a blood test and the test results were positive for the cancer Find the probability that the patient actually had the cancer A man drives his car to work 80% of the time; otherwise he rides his bicycle When he rides his bicycle to work he is late 25% of the time, whereas when he drives his car to work he is late 15% of the time On a particular day, the man arrives on time Find the probability that he rode his bicycle to work on that day cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The probabilities that Hiran’s mother and father will be alive in one year’s time are 0:99 and 0:98 respectively What is the probability that if only one of them is alive in 12 months, it is his mother? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\563IB_HL-2_18.CDR Tuesday, 13 November 2007 3:55:55 PM PETERDELL IB_HL-2ed (564) 564 PROBABILITY (Chapter 18) A manufacturer produces drink bottles He uses machines which produce 60% and 40% of the bottles respectively 3% of the bottles made by the first machine are defective and 5% of the bottles made by the second machine are defective What is the probability that a defective bottle came from: a the first machine b the second machine? Az Suppose a sample space U is partitioned into three by the mutually exclusive events A1 , A2 and A3 as shown in the B Venn diagram Ac U The sample space also contains another event B Ax a Show that P(B) = P(B j A1 ) P(A1 ) + P(B j A2 ) P(A2 ) + P(B j A3 ) P(A3 ) b Hence show that Bayes’ theorem for the case of three partitions is P(Ai j B) = P P(B j Ai ) P(Ai ) P(B j Aj ) P(Aj ) , i f1, 2, 3g where P(B) = P(B) j=1 10 A printer has three presses A, B and C which print 30%, 40% and 30% of daily production respectively Due to the age of the machines and other problems, the presses cannot be used 3%, 5% and 7% of the day, respectively A printed page is randomly chosen a What is the probability that the press which printed the page is in use? b If the press in a is in use, what is the probability it is press A? c If the press in a is not in use, what is the probability it is either A or C? 11 12% of the over-60 population of Agento have lung cancer Of those with lung cancer, 50% were heavy smokers, 40% were moderate smokers, and 10% were non-smokers Of those without lung cancer, 5% were heavy smokers, 15% were moderate smokers, and 80% were non-smokers If a member of the over-60 population of Agento is chosen at random, what is the probability that: a the person was a heavy smoker b the person has lung cancer given the person was a moderate smoker c the person has lung cancer given the person was a non-smoker? REVIEW SET 18A List the different orders in which people A, B, C and D could line up If they line up at random, determine the probability that: a A is next to C b there is exactly one person between A and C cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A coin is tossed and a square spinner labelled A, B, C, D, is twirled Determine the probability of obtaining: a a head and consonant b a tail and C c a tail or a vowel black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_18\564IB_HL-2_18.CDR Friday, 12 December 2008 12:29:43 PM TROY IB_HL-2ed (565) PROBABILITY (Chapter 18) 565 A class contains 25 students 13 play tennis, 14 play volleyball, and plays neither of these sports If a student is randomly selected from the class, determine the probability that the student: a plays both tennis and volleyball b plays at least one of these sports c plays volleyball given that he or she does not play tennis Niklas and Rolf play tennis with the winner being the first to win two sets Niklas has a 40% chance of beating Rolf in any set Draw a tree diagram showing the possible outcomes and hence determine the probability that Niklas will win the match The probability that a man will be alive in 25 years is 35 , and the probability that his wife will be alive is 23 Determine the probability that in 25 years: a both will be alive b at least one will be alive c only the wife will be alive Each time Mae and Ravi play chess, Mae has probability 45 of winning If they play games, determine the probability that: a Mae wins of the games b Mae wins either or of the games If I buy tickets in a 500 ticket lottery, determine the probability that I win: a the first prizes b at least one of the first prizes A school photocopier has a 95% chance of working on any particular day Find the probability that it will be working on at least one of the next two days A team of five is randomly chosen from six doctors and four dentists Determine the likelihood that it consists of: a all doctors b at least two doctors 10 girls and boys sit at random on seats in a row Determine the probability that: a they alternate with girls sitting between boys b the girls are seated together 11 The students in a school are all vaccinated against measles 48% of the students are males, of whom 16% have an allergic reaction to the vaccine 35% of the girls also have an allergic reaction If a student is randomly chosen from the school, what is the probability that the student: a has an allergic reaction b is female given that a reaction occurs? REVIEW SET 18B Systematically list the possible sexes of a 4-child family Hence determine the probability that a randomly selected 4-child family consists of two children of each sex cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 In a group of 40 students, 22 study Economics, 25 study Law, and study neither of these subjects Determine the probability that a randomly chosen student studies: a both Economics and Law b at least one of these subjects c Economics given that he or she studies Law black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\565IB_HL-2_18.CDR Tuesday, 13 November 2007 4:07:57 PM PETERDELL IB_HL-2ed (566) 566 PROBABILITY (Chapter 18) A bag contains red, yellow and blue marbles Two marbles are randomly selected from the bag without replacement What is the probability that: a both are blue b both are the same colour c at least one is red d exactly one is yellow? What is meant by: a b independent events disjoint events? On any one day it could rain with 25% chance and be windy with 36% chance Draw a tree diagram showing the possibilities with regard to wind and rain on a particular day Hence determine the probability that on a particular day there will be: a rain and wind b rain or wind A, B and C have 10%, 20% and 30% chance of independently solving a certain maths problem If they all try independently of one another, what is the probability that this group will solve the problem? Jon goes cycling on three random mornings of each week When he goes cycling he has eggs for breakfast 70% of the time When he does not go cycling he has eggs for breakfast 25% of the time Determine the probability that he: a has eggs for breakfast b goes cycling given that he has eggs for breakfast ¡3 a Expand + ¢ b A tin contains 20 pens of which 12 have blue ink Four pens are randomly selected (with replacement) from the tin What is the probability that: i two of them have blue ink ii at most two have blue ink? X plays Y at table tennis and from past experience wins sets in every played If they play sets, write down the probability generator and hence determine the probability that a Y wins of them b Y wins at least of them 10 With every attempt, Jack has an 80% chance of kicking a goal In one quarter of a match he has kicks for goal Determine the probability that he scores: a goals then misses twice b goals and misses twice REVIEW SET 18C cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 In a certain class, 91% of the students passed Mathematics and 88% of the students passed Chemistry 85% of students passed both Mathematics and Chemistry a Show that the events of passing Mathematics and passing Chemistry are not independent b A randomly selected student passed Chemistry Find the probability that this student did not pass Mathematics black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\566IB_HL-2_18.CDR Tuesday, 13 November 2007 4:15:33 PM PETERDELL IB_HL-2ed (567) PROBABILITY (Chapter 18) 567 A group of ten students included three from year 12 and four from year 11 The principal called a meeting with five of the group, and randomly selected students to attend Calculate the probability that exactly two year 12 and two year 11 students were called to the meeting Given P(Y ) = 0:35 and P(X [ Y ) = 0:8, and that X and Y are independent events, find: a P(X) b the probability that X occurs or Y occurs, but not both X and Y A person with a university degree has a 0:33 chance of getting an executive position A person without a university degree has a 0:17 chance of the same If 78% of all applicants for an executive position have a university degree, find the probability that the successful applicant does not have one Given P(X j Y ) = 23 , P(Y ) = and X \ Y = ?, find P(X): All of the 28 boys in an Australian school class play either Australian Rules football or soccer 13 of the boys are migrants and of these 13 play soccer Of the remaining 15 boys, play soccer A boy is selected at random and plays Australian rules Find the probability that he is a migrant The probability that a particular salesman will leave his sunglasses behind in any store is 15 Suppose the salesman visits two stores in succession and leaves his sunglasses behind in one of them What is the probability that the salesman left his sunglasses in the first store? How many tosses of a fair coin are necessary to have a better than even chance of getting at least four heads? An urn contains three red balls and six blue balls a A ball is drawn at random and found to be blue What is the probability that a second draw with no replacement will also produce a blue ball? b Two balls are drawn without replacement and the second is found to be red What is the probability that the first ball was also red? c Based on the toss of a coin, either a red ball or a blue ball is added to the urn Given that a random draw now produces a blue ball, what is the probability the added ball was i red ii blue? 10 Using a 52 card pack, a ‘perfect’ poker hand contains 10, J, Q, K, A of one suit What is the probability of dealing: a a ‘perfect’ poker hand in any order b a ‘perfect’ poker hand in the order 10, J, Q, K, A? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 11 With each pregnancy a particular woman will give birth to either a single baby or twins There is a 15% chance of having twins during each pregnancy If after pregnancies she has given birth to children, what is the probability she had twins first? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\567IB_HL-2_18.CDR Tuesday, 13 November 2007 4:17:48 PM PETERDELL IB_HL-2ed (568) 568 PROBABILITY (Chapter 18) REVIEW SET 18D a What is meant by saying that two events are independent? b If A and B are independent events, prove that A and B are also independent Two coins A and B are tossed and the probability of a ‘match’ (either two heads or two tails) is 12 Prove that at least one of the coins is unbiased ¡ ¢ a If P (x) = nx pn¡x q x where p + q = and x = 0, 1, 2, 3, ., n µ ¶µ ¶ n¡x q prove that P (x + 1) = P (x) where P (0) = pn x+1 p b If n = and p = 12 , use a to find P (0), P (1), P (2), ., P (5): A and B are independent events where P(A) = 0:8 and P(B) = 0:65 d P(B j A) Determine: a P(A [ B) b P(A j B) c P(A0 j B ) An unbiased coin is tossed n times Find the smallest value of n for which the probability of getting at least two heads is greater than 99% The independent probabilities that components of a TV set will need replacing within 1 one year are 20 , 50 and 100 respectively Calculate the probability that there will need to be a replacement of: a at least one component within a year b exactly one component within a year When Peter plays John at tennis, the probability that Peter wins his service game is p and the probability that John wins his service game is q where p > q, p + q > Which is more likely: A Peter will win at least two consecutive games out of when he serves first B Peter will win at least two consecutive games out of when John serves first? Four different numbers are randomly chosen from S = f1, 2, 3, 4, 5, 10g X is the second largest of the numbers selected Determine the probability that X is: a b c 9 Two different numbers were chosen at random from the digits to inclusive and it was observed that their sum was even Determine the probability that both numbers were odd 10 A dart thrower has a one in three chance of hitting the correct number with any throw He throws darts at the board and X is the number of successful hits Find the probability generator for X, and hence calculate the probability of him scoring an odd number of successful hits given that he has at least two successful hits 11 The diagram alongside shows an electrical circuit with switches The probability that any switch is open is 13 Determine the probability that the current flows from A to B A B cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 12 One letter is randomly selected from each of the names JONES, PETERS and EVANS a Determine the probability that the three letters are the same b What is the likelihood that only two of the letters are the same? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_18\568IB_HL-2_18.CDR Monday, 26 November 2007 3:08:19 PM PETERDELL IB_HL-2ed (569) 19 Chapter Introduction to calculus Contents: A B C D Limits Finding asymptotes using limits Trigonometric limits Calculation of areas under curves cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 19 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_19\569IB_HL-2_19.CDR Friday, 16 November 2007 4:28:44 PM PETERDELL IB_HL-2ed (570) 570 INTRODUCTION TO CALCULUS (Chapter 19) Calculus is a major branch of mathematics which builds on algebra, trigonometry, and analytic geometry It has widespread applications in science, engineering, and financial mathematics The study of calculus is divided into two fields, differential calculus and integral calculus, both of which we will study in this course These fields are linked by the Fundamental theorem of calculus which we will study in Chapter 24 HISTORICAL NOTE Calculus is a Latin word meaning ‘pebble’ Ancient Romans used stones for counting The history of calculus begins with the Egyptian Moscow papyrus from about 1850 BC Its study continued in Egypt before being taken up by the Greek mathematician Archimedes of Syracuse It was further developed through the centuries by mathematicians of many nations Two of the most important contributors were Gottfried Wilhelm Leibniz and Sir Isaac Newton who independently developed the fundamental theorem of calculus A LIMITS The idea of a limit is essential to differential calculus We will see that it is necessary for finding the slope of a tangent to a curve at any point on the curve Consider the following table of values for f (x) = x2 x f(x) 1 in the vicinity of x = 1:9 1:99 1:999 1:9999 2:0001 2:001 2:01 2:1 3:61 3:9601 3:996 00 3:999 60 4:000 40 4:004 00 4:0401 4:41 Notice that as x approaches from the left, then f (x) approaches from below Likewise, as x approaches from the right, then f (x) approaches from above We say that as x approaches from either direction, f (x) approaches a limit of 4, and write lim x2 = 4: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 x!2 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_19\570IB_HL-2_19.CDR Friday, 16 November 2007 4:32:18 PM PETERDELL IB_HL-2ed (571) 571 INTRODUCTION TO CALCULUS (Chapter 19) INFORMAL DEFINITION OF A LIMIT The following definition of a limit is informal but adequate for the purposes of this course: If f(x) can be made as close as we like to some real number A by making x sufficiently close to a, we say that f (x) approaches a limit of A as x approaches a, and we write lim f(x) = A x!a We also say that as x approaches a, f (x) converges to A Notice that we have not used the value of f (x) when x = a, i.e., f(a) This is very important to the concept of limits 5x + x2 and we wish to find the limit as x ! 0, it is tempting x for us to simply substitute x = into f (x) For example, if f (x) = Not only we get the meaningless value of Observe that if f (x) = 0, but also we destroy the basic limit method ( = + x if x 6= then f (x) is undefined if x = 5x + x2 x f(x) has the graph shown y It is the straight line y = x + with the point (0, 5) missing, called a point of discontinuity of the function However, even though this point is missing, the limit of f (x) as x approaches does exist In particular, as x ! from either direction, f (x) ! 5x + x2 f (x ) = x missing point -5 5x + x2 =5 ) lim x!0 x x Rather than graph functions each time to determine limits, most can be found algebraically Example a lim x2 a Evaluate: b x!2 lim x!0 5x + x2 x x2 can be made as close as we like to by making x sufficiently close to ) lim x2 = x!2 5x + x2 x b ( = + x if x 6= is undefined if x = ) 5x + x2 x!0 x = lim + x, x 6= lim x!0 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 =5 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_19\571IB_HL-2_19.CDR Wednesday, 14 November 2007 9:44:53 AM PETERDELL IB_HL-2ed (572) 572 INTRODUCTION TO CALCULUS (Chapter 19) RULES FOR LIMITS If f(x) and g(x) are functions and c is a constant: ² lim c = c x!a ² lim c f (x) = c lim f (x) x!a x!a lim [f(x) § g(x)] = lim f(x) § lim g(x) ² x!a x!a x!a lim [f(x)g(x)] = lim f (x) £ lim g(x) ² x!a ² x!a x!a f (x) = lim f(x) ¥ lim g(x) provided x!a g(x) x!a lim x!a lim g(x) 6= x!a Example Use these rules to evaluate: lim (x + 2)(x ¡ 1) a b x!3 lim x!1 x2 + x¡2 As x ! 3, x + ! and x ¡ ! a ) lim (x + 2)(x ¡ 1) = £ = 10 x!3 So, as x ! 3, (x + 2)(x ¡ 1) converges to 10 As x ! 1, x2 + ! and x ¡ ! ¡1 b ) lim x!1 x2 + = = ¡3 x¡2 ¡1 So, as x ! 1, x2 + x¡2 converges to ¡3 LIMITS AT INFINITY We can use the idea of limits to discuss the behaviour of functions for extreme values of x x!1 x ! ¡1 We write and to mean when x gets as large as we like and positive, to mean when x gets as large as we like and negative We read x ! as “x tends to plus infinity” and x ! ¡1 as “x tends to minus infinity” Notice that as x ! 1, < x < x2 < x3 < :::::: magenta yellow 95 100 50 75 25 95 100 50 75 25 95 x = 0, and so on x2 100 50 lim x!1 75 25 95 cyan = 0, x lim x!1 100 50 75 25 ) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_19\572IB_HL-2_19.CDR Wednesday, 14 November 2007 9:50:21 AM PETERDELL IB_HL-2ed (573) 573 INTRODUCTION TO CALCULUS (Chapter 19) Example Evaluate the following limits: 2x + a lim x!1 x ¡ a x!1 x!1 x2 ¡ 3x + ¡ x2 x = lim x!1 1¡ x = = lim x!1 fas x ! 1, !0 x and ! 0g x = fsince x2 ¡ 3x + ¡ x2 x!1 + x x ¡1 x2 x!1 ¡1 x!1 = lim fdividing each term by x2 g x!1 ! 0, x fas x ! 1, ! 0, and ! 0g x2 x = ¡1 lim or 1¡ = lim x!1 2x + x¡4 2(x ¡ 4) + + x¡4 µ ¶ 11 = lim + x!1 x¡4 fdividing each term in both numerator and denominator by xg =2 lim lim or 2+ = lim 2x + x¡4 lim b b lim 11 x!1 x¡4 = 0g x2 ¡ 3x + ¡ x2 ¡x2 + 3x ¡ x2 ¡ ¡(x2 ¡ 1) + 3x ¡ x!1 x2 ¡ µ ¶ 3x ¡ = lim ¡1 + x!1 x ¡1 = lim 3x¡3 x!1 x ¡1 = ¡1 fsince lim = 0g EXERCISE 19A Evaluate the limits: magenta yellow lim h2 (1 ¡ h) f x i h!0 lim x!¡2 l lim x2 ¡ x ¡ x2 ¡ 5x + o lim x3 ¡ 2x ¡ r h!0 x!3 x!2 95 h3 ¡ 8h h 100 lim 50 95 50 x!1 c 75 q 95 x3 ¡ x2 ¡ 100 lim x!2 50 n 75 x2 ¡ 2x x2 ¡ 25 lim k 2h2 + 6h h 100 lim lim (5 ¡ 2x) x!¡1 25 h 75 25 95 100 50 p 75 lim (x2 + 5) h!0 m 25 e x!0 j lim 5x2 ¡ 3x + x!2 g b d cyan lim (x + 4) x!3 a black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_19\573IB_HL-2_19.CDR Friday, 12 December 2008 12:32:32 PM TROY lim (3x ¡ 1) x!4 lim x!¡1 ¡ 2x x2 + lim x2 ¡ 3x x lim x2 ¡ x x2 ¡ x!0 x!1 lim 2x2 ¡ 50 + 13x ¡ 10 x!¡5 3x2 lim x!¡2 3x2 + 5x ¡ x2 ¡ 2x ¡ IB_HL-2ed (574) 574 INTRODUCTION TO CALCULUS (Chapter 19) Examine lim x!1 x Evaluate: a lim 3x ¡ x+1 b lim x2 + x2 ¡ e x!1 d x!1 B lim ¡ 2x 3x + c lim x2 ¡ 2x + x2 + x ¡ f x!1 x!1 lim x 1¡x lim x3 ¡ 3x3 + x2 ¡ 8x ¡ x!1 x!1 FINDING ASYMPTOTES USING LIMITS f (x) g(x) Rational functions are functions of the form where f (x), g(x) are polynomials Rational functions are characterised by the existence of asymptotes which may be vertical, horizontal, or oblique An oblique asymptote is neither horizontal nor vertical We can investigate the asymptotes of a function by using limits Consider the function f : x 7! 2x + x¡4 Clearly the domain of f is: fx R , x 6= 4g or x ] ¡ 1, [ [ ] 4, [ There is a vertical asymptote (VA) at x = To discuss the behaviour near the VA, we find what happens to f (x) as x ! from the left and right ² First we draw a sign diagram of f (x): + ² Hence as x ! (left), x ! (right), - + x -\Ew_ f (x) ! ¡1 f (x) ! +1 This describes the behaviour of the graph of f (x) near the VA x = Is there another type of asymptote? Now f(x) = 2(x ¡ 4) + 11 11 2x + = =2+ x¡4 x¡4 x¡4 ² As x ! +1, f (x) ! (above) ² As x ! ¡1, f (x) ! (below) 11 ! from aboveg x¡4 11 fas ! from belowg x¡4 fas Hence, there is a horizontal asymptote (HA) at y = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 This horizontal asymptote corresponds to the answer in Example part a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_19\574IB_HL-2_19.CDR Friday, 16 November 2007 4:33:09 PM PETERDELL IB_HL-2ed (575) INTRODUCTION TO CALCULUS (Chapter 19) 575 Example Find any asymptotes of the function f : x ! behaviour of f (x) near these asymptotes x2 ¡ 3x + ¡ x2 and discuss the ¡1 We notice that f (x) = (x ¡ 2)(x ¡ 1) ¡(x ¡ 2) = (1 ¡ x)(1 + x) 1+x provided x 6= So, there is a point discontinuity at x = Also, when x = ¡1, f (x) is undefined fdividing by zerog This indicates that x = ¡1 is a vertical asymptote The sign diagram for f(x) is: As x ! ¡1 (left), f(x) ! ¡1 As x ! ¡1 (right), f(x) ! +1 - + -1 x ) x = ¡1 is a VA ¡x + ¡(x + 1) + 3 ¡(x ¡ 2) = = = ¡1 + 1+x x+1 x+1 x+1 f (x) ! ¡1 (above) as ! and is > x+1 For x 6= 1, f (x) = As x ! 1, As x ! ¡1, f (x) ! ¡1 (below) ) HA is y = ¡1 as ! and is < x+1 (see Example part b) Example x3 + x2 + Determine all asymptotes and discuss the behaviour of f (x) = x +x+2 near its asymptotes First consider the quadratic denominator Its discriminant is negative so it is positive for all x ) no VAs exist For other asymptotes we need to carry out the division x x +x+2 x +x +4 ¡ 2x x3 + x2 + 2x Hence, f(x) = x + x +x+2 ¡2x + As x ! 1, As x ! ¡1, ¡ 2x ! and is < 0, +x+2 so f(x) ! x from below ¡ 2x ! and is > 0, +x+2 so f(x) ! x from above x2 x2 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 So, we have the oblique asymptote (OA) y = x Check this result by sketching the function on your GDC black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_19\575IB_HL-2_19.CDR Friday, 16 November 2007 4:46:09 PM PETERDELL IB_HL-2ed (576) 576 INTRODUCTION TO CALCULUS (Chapter 19) Example For each of the following, determine all asymptotes and discuss the behaviour of the graph near these asymptotes: b y = ln(x3 ¡ 9x) a y = ¡ 4e¡x a The domain is ] ¡ 1, [ or R , and so no VA exists fas ¡4e¡x < and As x ! 1, y ! (below) As x ! ¡1, y ! ¡1 lim 4e¡x = 0g x!1 So, we have a HA of y = b To find the domain we notice that ln(x3 ¡ 9x) is defined only when x3 ¡ 9x > ) x(x2 ¡ 9) > ) x(x + 3)(x ¡ 3) > + Now x(x + 3)(x ¡ 3) has sign diagram: -3 ) the domain is: + x ] ¡ 3, [ [ ] 3, [ x ! ¡3 (right), y ! ¡1 x ! (left), y ! ¡1 x ! (right), y ! ¡1 x ! 1, y ! +1 As As As As Check the asymptotes by sketching the graph EXERCISE 19B For each of the following, determine all asymptotes and discuss the behaviour of the graph near its asymptotes: f x3 ¡ x2 + 1 magenta yellow 95 1000 + 2e¡0:16t 100 50 75 25 95 100 50 25 75 y= 25 l y = ex ¡ x k f (x) = e2x ¡ 7ex + 12 95 j 100 y = x + ln x i 95 d 2x2 + 10 x x ¡1 y= x +1 x¡2 f (x) = x +x¡2 y= f (x) = ex¡ x y= cyan b h g 100 50 75 25 e 50 c 3x ¡ x+3 4900 W = 5000 ¡ , t>0 t +1 x f(x) = x +1 f(x) = 75 a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_19\576IB_HL-2_19.CDR Monday, 19 November 2007 10:08:50 AM PETERDELL IB_HL-2ed (577) 577 INTRODUCTION TO CALCULUS (Chapter 19) C TRIGONOMETRIC LIMITS INVESTIGATION EXAMINING sin µ NEAR µ = µ sin µ when µ is close to and µ is in radians µ We consider this ratio graphically, numerically, and geometrically This investigation examines What to do: sin µ is an even function What does this mean graphically? µ sin µ sin µ is even we need only examine for positive µ Since µ µ sin µ when µ = 0? a What is the value of µ sin µ GRAPHING for ¡ ¼2 µ ¼2 using a graphics b Graph y = PACKAGE µ calculator or graphing package sin µ c Explain why the graph indicates that lim = µ!0 µ Show that f(µ) = Copy and complete the given table, using your calculator: Extend your table to include negative values of µ which approach r Explain why the area of the shaded segment is A = 12 r2 (µ ¡ sin µ) q µ sin µ sin µ µ 0:5 0:1 0:01 0:001 Indicate how to use the given figure and the shaded area to show that, if µ is in radians and µ > 0, then sin µ lim = µ!0 µ sin µ using degrees rather than radians Repeat the above investigation to find lim µ!0 µ If µ is in radians, then lim Theorem: µ!0 Proof: y P(cos¡q, sin¡q) Suppose P(cos µ, sin µ) lies on the unit circle in the first quadrant PQ is drawn perpendicular to the x-axis, and arc QR with centre O is drawn Now, R -1 q Q T sin µ sin µ = 1, i.e., as µ ! 0, converges to µ µ x area of sector OQR area ¢OQP area sector OTP magenta yellow 95 £ µ 12 (OQ)(PQ) 12 (OT)2 £ µ 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 cyan 2 (OQ) ) -1 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_19\577IB_HL-2_19.CDR Tuesday, 22 January 2008 4:41:23 PM thomas @THOMAS IB_HL-2ed (578) 578 INTRODUCTION TO CALCULUS (Chapter 19) 2 µ cos ) µ6 cos µ sin µ 12 µ sin µ µ ) cos µ 6 cos µ fdividing throughout by 12 µ cos µ, which > 0g Now as µ ! 0, both cos µ ! and !1 cos µ ) as µ ! (right), sin µ ! µ sin µ is an even function, so as µ ! (left), µ sin µ = Thus lim µ!0 µ But sin µ ! also µ sin µ as µ ! from both the right and µ sin µ This is very important when we the left before concluding the value of lim µ!0 µ deduce limits Note: We have established the behaviour of Example Find lim µ!0 sin 3µ µ lim sin 3µ µ µ!0 = lim µ!0 sin 3µ £3 3µ = £ lim 3µ!0 sin 3µ 3µ fas µ ! 0, 3µ ! alsog =3£1 =3 EXERCISE 19C Find: a d lim sin 2µ µ b lim sin µ sin 4µ µ2 e µ!0 µ!0 µ sin µ ¡ ¢ sin h2 cos h lim h!0 h c lim µ!0 f A circle contains n congruent isosceles triangles, all with apex O and with base vertices on the circle as shown a Explain why the sum of the areas of the triangles ¡ ¢ is Sn = 12 nr2 sin 2¼ n b Find i ii lim Sn n!1 lim µ!0 tan µ µ lim n sin n!1 ¡ 2¼ ¢ n r geometrically algebraically cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 c What can be deduced from b? black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_19\578IB_HL-2_19.CDR Friday, 12 December 2008 12:33:50 PM TROY IB_HL-2ed (579) INTRODUCTION TO CALCULUS (Chapter 19) a Show that cos(A + B) ¡ cos(A ¡ B) = ¡2 sin A sin B b If A+B = S and A¡B = D, show that cos S¡cos D = ¡2 sin c Hence find lim h!0 ¡ S+D ¢ sin 579 ¡ S¡D ¢ cos(x + h) ¡ cos x h D CALCULATION OF AREAS UNDER CURVES Consider the function f (x) = x2 + y We wish to estimate the area A enclosed by y = f(x), the x-axis, and the vertical lines x = and x = f (x ) = x + 20 15 10 Suppose we divide the x-interval into three strips of width unit as shown A y The diagram alongside shows upper rectangles, which are rectangles with top edges at the maximum value of the curve on that interval x f (x ) = x + 20 15 The area of the upper rectangles, AU = £ f(2) + £ f (3) + £ f (4) = + 10 + 17 = 32 units2 10 17 10 The next diagram shows lower rectangles, which are rectangles with top edges at the minimum value of the curve on that interval x f (x ) = x + y 20 15 The area of the lower rectangles, AL = £ f(1) + £ f(2) + £ f (3) = + + 10 = 17 units2 10 17 10 2 x Now clearly AL < A < AU , so the required area lies between 17 units2 and 32 units2 If the interval x was divided into equal intervals, each of length 12 , then AU = 12 f (1 12 ) + f (2) = 12 ( 13 +5+ 29 + 1 f (2 ) 53 + 10 + f (3) + 1 f (3 ) 1 f(2 ) + f (3) + + f(4) + 17) = 27:875 units2 and AL = 12 f(1) + = 12 (2 + 13 1 f(1 ) +5+ 29 + f(2) + 10 + + + 1 f(3 ) 53 ) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 = 20:375 units2 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_19\579IB_HL-2_19.CDR Wednesday, 14 November 2007 10:17:36 AM PETERDELL IB_HL-2ed (580) 580 INTRODUCTION TO CALCULUS (Chapter 19) From this refinement we conclude that the required area lies between 20:375 and 27:875 units2 As we create more subdivisions, the estimates AL and AU will become more and more accurate In fact, as the subdivision width is reduced further and further, both AL and AU will converge to A Example 50 km h¡1 away from a city +1 The time t is the number of hours after the car leaves the city a Sketch a graph of the speed against time b Estimate the distance the car has travelled after hours c Explain how you could obtain a better estimate for the distance in b A car travels at the speed of v(t) = 100 ¡ a 100 The graph shows that the function 50 v(t) = 100 ¡ is increasing 5t + v 50 As t ! 1, t b 5t2 50 !0 +1 5t2 ) a HA is v = 100 Since distance travelled = speed £ time, the distance travelled is the area enclosed by v(t), the t-axis, and the vertical lines t = and t = We estimate the distance travelled by dividing the interval into four subdivisions and then using upper and lower rectangles 100 v 91.7 50 97.6 AU = £ v(1) + £ v(2) + £ v(3) + £ v(4) = £ 91:7 + £ 97:6 + £ 98:9 + £ 99:4 ¼ 387:6 98.9 99.4 t 100 v 91.7 50 97.6 AL = £ v(0) + £ v(1) + £ v(2) + £ v(3) = £ 50 + £ 91:7 + £ 97:6 + £ 98:9 ¼ 338:2 98.9 99.4 t cyan yellow 95 100 50 75 25 95 100 50 75 25 95 50 magenta 338:2 + 387:6 ¼ 363 km 100 A¼ 75 25 95 100 50 75 25 Now AL < distance travelled < AU , so the distance travelled is between 338:2 km and 387:6 km A good estimate might be the average of the distances, black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_19\580IB_HL-2_19.CDR Wednesday, 14 November 2007 10:24:31 AM PETERDELL IB_HL-2ed (581) INTRODUCTION TO CALCULUS (Chapter 19) c 581 We could obtain a better estimate by using more subdivisions For example, for the case of subdivisions we have: 100 v 100 50 v 50 t t AU ¼ 379:5 AL ¼ 354:8 ) a better estimate would be A = 354:8 + 379:5 ¼ 367 km EXERCISE 19D.1 At time t = hours a car starts from Port Wakefield, a distance of 95 km from Adelaide, and travels with speed given by v(t) = 50 + 50e¡t km h¡1 towards Adelaide Sketch a graph of the speed of the car for t What are the maximum and minimum speeds of the car for t 1? Show that the distance d of the car from Adelaide after one hour is less than 27 km By dividing the time of travel into half hour intervals, estimate the distance of the car from Adelaide after hour of travel e Improve the estimate you made in d by considering time intervals of a quarter of an hour each a b c d When items are sold, we say the marginal profit is the profit on the sale of each item The marginal profit usually increases as the number of articles sold increases Suppose that the marginal profit of selling the nth house is p(n) = 200n ¡ n2 dollars a Sketch a graph of the marginal profit for n 100 b Show that the total profit for selling 100 houses is less than $1 000 000 c By considering the maximum and minimum marginal profit of the first 50 and the second 50 houses sold, estimate the profit made for selling 100 houses d Improve the estimate you have made in c by considering four equal intervals e Suggest a way of finding the exact profit made by selling 100 houses USING TECHNOLOGY By subdividing the horizontal axis into small enough intervals, we can in theory find estimates for the area under a curve which are as close as we like to the actual value We illustrate this process by estimating the area A between the graph of y = x2 and the x-axis for x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 This example is of historical interest Archimedes (287 - 212 BC) found the exact area In an article that contains 24 propositions he developed the essential theory of what is now known as integral calculus black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_19\581IB_HL-2_19.CDR Monday, 19 November 2007 10:09:09 AM PETERDELL IB_HL-2ed (582) 582 INTRODUCTION TO CALCULUS (Chapter 19) Consider f (x) = x2 and divide the interval x into equal subdivisions ¦(x)¡=¡xX y ¦(x)¡=¡xX y (1, 1) (1, 1) x x AL = 14 (0)2 + 14 ( 14 )2 + 14 ( 12 )2 + 14 ( 34 )2 and AU = 14 ( 14 )2 + 14 ( 12 )2 + 14 ( 34 )2 + 14 (1)2 ¼ 0:219 ¼ 0:469 n and AU for Now suppose there are n subdivisions, each of width AREA FINDER We can use technology to help calculate AL large values of n Click on the appropriate icon to access our area finder software or instructions for the procedure on a graphics calculator The following table summarises the results you should obtain for n = 4, 10, 25 and 50 n 10 25 50 TI C AL 0:218 75 0:285 00 0:313 60 0:323 40 AU 0:468 75 0:385 00 0:353 60 0:343 40 Average 0:343 75 0:335 00 0:333 60 0:333 40 The exact value of A is in fact 13 Notice how both AL and AU are converging to this value as n increases EXERCISE 19D.2 Use rectangles to find lower and upper sums for the area between the graph of y = x2 and the x-axis for x Use n = 10, 25, 50, 100 and 500 Give your answers to decimal places As n gets larger, both AL and AU converge to the same number which is a simple fraction What is it? Use rectangles to find lower and upper sums for the areas between the graphs of each of the following functions and the x-axis for x Use values of n = 5, 10, 50, 100, 500, 1000 and 10 000 Give your answer to decimal places in each case 1 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a ii y = x iii y = x iv y = x i y = x3 b For each case in a, AL and AU converge to the same number which is a simple fraction What fractions are they? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_19\582IB_HL-2_19.CDR Monday, 19 November 2007 10:30:05 AM PETERDELL IB_HL-2ed (583) INTRODUCTION TO CALCULUS (Chapter 19) 583 c On the basis of your answer to b, conjecture what the area between the graph of y = xa and the x-axis for x might be for any number a > y Consider the quarter circle of centre (0, 0) and radius units illustrated 4 Its area is y = - x2 (full circle of radius 2) = £ ¼ £ 22 =¼ x a By calculating the areas of lower and upper rectangles for n = 10, 50, 100, 200, 1000, 10 000, find rational bounds for ¼ b Archimedes found the famous approximation 10 71 < ¼ < For what value of n is your estimate for ¼ better than that of Archimedes? THE DEFINITE INTEGRAL Consider the lower and upper rectangle sums for a function which is positive and increasing on the interval [ a, b ] b¡a We divide [ a, b ] into n subdivisions of width ±x = n y y y = ƒ(x) y = ƒ(x) a x1 x2 x3 x0 x xn-2 b xn-1 xn a x1 x2 x3 x0 xn-2 b xn-1 xn x Since the function is increasing, AL = ±x f (x0 ) + ±x f(x1 ) + ±x f (x2 ) + :::::: + ±x f (xn¡2 ) + ±x f (xn¡1 ) n¡1 P f(xi )±x = i=0 and AU = ±x f (x1 ) + ±x f(x2 ) + ±x f (x3 ) + :::::: + ±x f (xn¡1 ) + ±x f (xn ) n P = f (xi )±x i=1 AU ¡ AL = ±x (f (xn ) ¡ f (x0 )) = (b ¡ a) (f (b) ¡ f (a)) n Notice that lim (AU ¡ AL ) = ) n!1 lim AL = lim AU n!1 cyan magenta yellow 95 100 50 75 n!1 25 95 50 lim AL = A = lim AU n!1 75 25 95 100 50 75 25 95 100 50 75 25 ) since AL < A < AU , n!1 100 ) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_19\583IB_HL-2_19.CDR Monday, 19 November 2007 10:31:27 AM PETERDELL IB_HL-2ed (584) 584 INTRODUCTION TO CALCULUS (Chapter 19) This fact is true for all positive continuous functions on the interval [ a, b ] The value A is known as the “definite integral of f (x) from a to b”, written Z b A= f (x) dx y y = ƒ(x) a If f(x) > for all x [ a, b ], then Z b f (x) dx is the shaded area a b x a HISTORICAL NOTE The word integration means “to put together into a whole” An integral is the “whole” produced from integration, since the areas f (xi ) £ ±x of the thin rectangular strips are put together into one whole area Z The symbol is called an integral sign In the time of Newton and Leibniz it was the stretched out letter s, but it is no longer part of the alphabet Example Z a Sketch the graph of y = x for x Shade the area described by x4 dx b Use technology to calculate the lower and upper rectangle sums for n equal subdivisions where n = 5, 10, 50, 100 and 500 Z x4 dx to significant figures c Use the information in b to find a b n 10 50 100 500 y y¡¡=¡x4 0.8 0.6 A = ò x dx 0.4 AL 0:1133 0:1533 0:1901 0:1950 0:1990 AU 0:3133 0:2533 0:2101 0:2050 0:2010 0.2 x 0.2 0.4 0.6 0.8 c When n = 500, AL ¼ AU ¼ 0:20, to significant figures Z Z x4 dx < AU , x4 dx ¼ 0:20 (2 s.f.) ) since AL < cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 0 95 100 50 75 25 95 100 50 75 25 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_19\584IB_HL-2_19.CDR Monday, 19 November 2007 10:39:09 AM PETERDELL IB_HL-2ed (585) 585 INTRODUCTION TO CALCULUS (Chapter 19) Example 10 Z Use graphical evidence and known area facts to find: (2x + 1) dx y y = 2x + (2x + 1) dx =6 x p ¡ x2 dx = shaded area ¢ ¡ £2 = 1+5 (2,¡5) 1 b R a b Z a p ¡ x2 then y = ¡ x2 and so x2 + y2 = which is the p equation of the unit circle y = ¡ x2 is the upper half R1p y ¡ x2 dx If y = y = ~`1`-¡` !`X = shaded area = 14 (¼r2 ) where r = -1 x = ¼ EXERCISE 19D.3 p x for x Z p x dx Shade the area described by a Sketch the graph of y = b Find the lower and upper rectangle sums for n = 5, 10, 50, 100 and 500 Z p x dx to significant figures c Use the information in b to find a Sketch the graph of y = p + x3 and the x-axis for x b Find the lower and upper rectangle sums for n = 50, 100, 500 Z p c What is your best estimate for + x3 dx? Use graphical evidence and known area facts to find: Z Z (1 + 4x) dx b (2 ¡ x) dx a cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 ¡1 100 50 75 25 95 100 50 75 25 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_19\585IB_HL-2_19.CDR Monday, 19 November 2007 10:42:55 AM PETERDELL Z c ¡2 p ¡ x2 dx IB_HL-2ed (586) 586 INTRODUCTION TO CALCULUS (Chapter 19) Z INVESTIGATION ESTIMATING Z ¡x The integral e ¡ e x2 dx ¡3 dx is of considerable interest to statisticians ¡3 In this investigation we shall estimate the value of this integral using upper and lower rectangular sums for n = 4500 This value of n is too large for most calculators to handle in a single list, so we will perform it in sections TI What to do: ¡x Sketch the graph of y = e C for ¡3 x 3: Calculate the upper and lower rectangular sums for the three intervals x 1, x and x using n = 750 for each Combine the upper rectangular sums and the lower rectangular sums you found in to obtain an upper and lower rectangular sum for x for n = 2250 ¡x Use the fact that the function y = e is symmetric to find upper and lower rectangular sums for ¡3 x for n = 2250 Z AREA ¡x e dx Use your results of and to find an estimate for FINDER ¡3 How accurate is your estimate? p Compare your estimate in with 2¼ REVIEW SET 19 Evaluate the limits: a x2 ¡ lim x!2 ¡ x d lim x!4 3x2 ¡ 12 lim x!¡2 5x2 + 10x b x2 ¡ 16 x¡4 e lim x!¡1 c 2x + x2 ¡ f p x¡2 lim x!4 x ¡ lim x!1 ¡ 2x ¡ x2 2x2 ¡ Find any asymptotes of the following functions and discuss the behaviour of the graph near them: a y= x2 + x ¡ x¡2 b y = ex¡2 ¡ c y = ln(x2 + 3) d f (x) = e¡x ln x e y = x + ln(2x ¡ 3) f x 7! ln(¡x) + sin 4µ µ b cyan magenta 95 100 50 c 75 25 95 yellow 2µ sin 3µ lim µ!0 100 50 75 25 95 100 25 50 lim µ!0 95 100 50 75 25 a 75 Find: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_19\586IB_HL-2_19.CDR Monday, 19 November 2007 10:43:48 AM PETERDELL lim n sin( ¼n ) n!1 IB_HL-2ed (587) INTRODUCTION TO CALCULUS (Chapter 19) 587 and the x-axis for x 1: + x2 Divide the interval into equal parts and display the upper and lower rectangles a Sketch the region between the curve y = b Find the lower and upper rectangle sums for n = 5, 50, 100 and 500 Z c Give your best estimate for dx and compare this answer with ¼ 1+x The rate at which drugs are eliminated from the body is called the clearance rate The clearance rate depends on the individual person as well as the amount of drugs present Suppose that in a healthy adult the clearance rate r(t) of 110 mg of caffeine (about one cup of coffee) is given by r(t) = 25e¡0:22t mg h¡1 , where t is the number of hours after which the caffeine is taken a Sketch the graph of r(t) for t b Show that hours after the intake of 110 mg of caffeine, the amount of caffeine left in the body is between 10 and 70 mg c By considering the maximum and minimum clearance for each hour, estimate the amount of caffeine left in the body hours after an intake of 110 mg d Suggest a way of improving the accuracy of your estimate in c ¢ ¡ ¢ ¡ cos x + h2 sin h2 Find lim h!0 h State any assumptions made in finding your answer a Show that sin(A + B) ¡ sin(A ¡ B) = cos A sin B and A ¡ B = D, show that ¡ ¢ ¡ ¢ sin S ¡ sin D = cos S+D sin S¡D 2 b If A + B = S sin(x + h) ¡ sin x h!0 h c Hence, find lim a Sketch the graph of the function f (x) = x for x 6 for x 6 x Use these values to show that the area between the graph of f(x) and the x-axis for x 6 lies between 23 and b Find the maximum and minimum value of f(x) = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 c Divide the interval x 6 into smaller intervals of equal length By considering the smallest and the largest values of f on each of these subdivisions, find an estimate for the area between the graph of f and the x-axis for x 6 d Suggest a way of improving the accuracy of your estimate in c black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_19\587IB_HL-2_19.CDR Monday, 19 November 2007 10:44:02 AM PETERDELL IB_HL-2ed (588) 588 INTRODUCTION TO CALCULUS (Chapter 19) Consider the graph of f(x) = e¡x a Sketch the graph of y = f (x) for x Z e¡x dx using upper and lower rectangles b Find upper and lower bounds for with subdivisions c Use technology to find, correct to significant figures, upper and lower bounds Z for e¡x dx when n = 100 10 The graph of y = f(x) is illustrated: y Evaluate the following using area interpretation: Z Z a f (x) dx b f (x) dx semicircle 2 4 x -2 11 Prove that for µ in radians: sin µ a µ > ) lim =1 µ!0 µ sin µ b µ < ) lim = also µ!0 µ c the area of the shaded segment is given by 2 r (µ ¡ sin µ) q d Use c to give geometric evidence that a is true 12 a y ¦(x)¡=¡4¡-¡xX Use four upper and lower rectangles to find rationals A and B such that: Z (4 ¡ x2 ) dx < B A< b x Hence, find a good estimate of Z (4 ¡ x2 ) dx a Sketch the graph of x 7! sin2 x for x [0, ¼] Z ¼ sin2 x dx ¼ ¼2 by considering triangle and trapezium b Hence, show that magenta yellow 95 100 50 75 25 95 100 50 95 100 50 75 25 95 100 50 75 25 cyan 75 approximations 25 13 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_19\588IB_HL-2_19.CDR Thursday, 11 March 2010 10:44:47 AM PETER IB_HL-2ed (589) Chapter 20 Differential calculus Contents: A B C D E F G The derivative function Derivatives at a given x-value Simple rules of differentiation The chain rule Product and quotient rules Tangents and normals Higher derivatives cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 20A Review set 20B Review set 20C black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_20\589IB_HL-2_20.CDR Wednesday, 14 November 2007 1:41:55 PM PETERDELL IB_HL-2ed (590) 590 DIFFERENTIAL CALCULUS (Chapter 20) In the previous chapter we discussed how the topic of calculus is divided into two fields: differential calculus and integral calculus In this chapter we begin to examine differential calculus and how it relates to rate problems and the gradient of curves HISTORICAL NOTE The topic of differential calculus originated in the 17th century with the work of Sir Isaac Newton and Gottfried Wilhelm Leibniz These mathematicians developed the necessary theory while attempting to find algebraic methods for solving problems dealing with: ² the gradients of tangents to curves at any point on the curve, and ² finding the rate of change in one variable with respect to another Isaac Newton 1642 – 1727 Gottfried Leibniz 1646 – 1716 RATES OF CHANGE A rate is a comparison between two quantities with different units We often judge performances by using rates For example: ² ² ² Sir Donald Bradman’s batting rate at Test cricket level was 99:94 runs per innings Michael Jordan’s basketball scoring rate was 20:0 points per game Rangi’s typing speed is 63 words per minute with an error rate of 2:3 errors per page Speed is a commonly used rate It is the rate of change in distance per unit of time We are familiar with the formula average speed = distance travelled time taken However, if a car has an average speed of 60 km h¡1 for a journey, it does not mean that the car travels at exactly 60 km h¡1 the whole time distance travelled In fact, the speed will probably vary continuously throughout the journey So, how can we calculate the car’s speed at any particular time? 60 km cyan magenta yellow 1h 95 100 50 time 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Suppose we are given a graph of the car’s distance travelled against time taken If this graph is a straight line then we know the speed is constant and is given by the gradient of the line black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\590IB_HL-2_20.CDR Friday, 12 March 2010 1:51:27 PM PETER IB_HL-2ed (591) 591 DIFFERENTIAL CALCULUS (Chapter 20) If the graph is a curve, then the car’s instantaneous speed is given by the gradient of the tangent to the curve at that time distance travelled distance time time THE TANGENT TO A CURVE curve A chord or secant of a curve is a straight line segment which joins any two points on the curve The gradient of the chord [AB] measures the average rate of change of the function for the given change in x-values chord (secant) B A tangent is a straight line which touches a curve at a point The gradient of the tangent at point A measures the instantaneous rate of change of the function at point A A In the limit as B approaches A, the gradient of the chord [AB] will be the gradient of the tangent at A INVESTIGATION tangent THE GRADIENT OF A TANGENT y Given a curve f(x), how can we find the gradient of the tangent at the point (a, f(a))? ƒ(x)¡=¡xX For example, the point A(1, 1) lies on the curve f(x) = x2 What is the gradient of the DEMO tangent at A? A (1, 1) x What to do: Suppose B lies on f(x) = x2 and B has coordinates (x, x2 ) a Show that the chord [AB] has gradient x2 ¡ f(x) ¡ f(1) or x¡1 x¡1 y B¡(x,¡xX) A (1, 1) b x Copy and complete: x cyan magenta yellow 95 Point B (5, 25) 1:5 1:1 1:01 1:001 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Comment on the gradient of [AB] as x gets closer to Repeat the process as x gets closer to 1, but from the left of A Click on the icon to view a demonstration of the process What you suspect is the gradient of the tangent at A? black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\591IB_HL-2_20.CDR Friday, 12 March 2010 3:03:16 PM PETER gradient of [AB] IB_HL-2ed (592) 592 DIFFERENTIAL CALCULUS (Chapter 20) Fortunately we not have to use a graph and table of values each time we wish to find the gradient of a tangent Instead we can use an algebraic and geometric approach which involves limits LIMIT ARGUMENT From the investigation, the gradient of [AB] = ) gradient of [AB] = x2 ¡ x¡1 (x + 1)(x ¡ 1) =x+1 x¡1 provided that x 6= In the limit as B approaches A, x ! and the gradient of [AB] ! the gradient of the tangent at A y = x2 B1 So, the gradient of the tangent at the point A is B2 x2 ¡ x!1 x ¡ = lim x + 1, x 6= B3 mT = lim B4 As B approaches A, the gradient of [AB] approaches or converges to x!1 =2 tangent at A A Limit arguments like that above form the foundation of differential calculus A THE DERIVATIVE FUNCTION For a non-linear function with equation y_ =_ f(x), gradients of tangents at various points continually change y Our task is to determine a gradient function so that when we replace x by some value a then we will be able to find the gradient of the tangent at x = a y¡=¡ƒ(x) x Consider a general function y = f (x) where A is (x, f(x)) and B is (x + h, f(x + h)) y y¡=¡ƒ(x) B ƒ(x¡+¡h) The chord [AB] has gradient = A ƒ(x) = h x f (x + h) ¡ f (x) x+h¡x f (x + h) ¡ f (x) h If we now let B approach A, then the gradient of [AB] approaches the gradient of the tangent at A x¡+¡h x So, the gradient of the tangent at the variable point (x, f(x)) is the limiting value of cyan magenta yellow 95 100 50 75 f(x + h) ¡ f (x) h 25 h!0 95 lim 100 50 75 25 95 100 50 as h approaches 0, or 75 25 95 100 50 75 25 f(x + h) ¡ f (x) h black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\592IB_HL-2_20.CDR Friday, 12 March 2010 3:04:10 PM PETER IB_HL-2ed (593) 593 DIFFERENTIAL CALCULUS (Chapter 20) This formula gives the gradient for any value of the variable x Since there is only one value of the gradient for each value of x, the formula is actually a function The gradient function, also known as the derived function or derivative function or simply the derivative is defined as f (x + h) ¡ f (x) f (x) = lim h! h We read the derivative function as ‘eff dashed x’ INVESTIGATION FINDING GRADIENTS OF FUNCTIONS WITH TECHNOLOGY This investigation can be done by graphics calculator or by clicking on the icon to open the demonstration The idea is to find the gradients at various points on a simple curve and use these to predict the gradient function for the curve What to do: Use a graphical argument to explain why: a if f (x) = c where c is a constant, then f (x) = b if f (x) = mx + c where m and c are constants, then f (x) = m Consider f(x) = x2 Find f (x) for x = 1, 2, 3, 4, 5, using technology Predict f (x) from your results Use technology and modelling techniques to find f (x) for: b f (x) = x4 c f (x) = x5 a f (x) = x3 p 1 d f (x) = f f (x) = x = x e f (x) = x x Use the results of to complete the following: “if f (x) = xn , then f (x) = ::::::” CALCULUS DEMO TI C For more complicated functions the method presented in the investigation cannot be used to find the gradient function To find the gradient function f (x) for a general function f (x), we need to evaluate the limit lim h!0 f (x + h) ¡ f (x) We call this the method of first principles h Example f (x + h) ¡ f (x) h (x + h) ¡ x2 = lim h!0 h x2 + 2hx + h2 ¡ x2 = lim h!0 h h(2x + h) = lim fas h 6= 0g h!0 h = 2x f (x) = lim h!0 Use the definition of f (x) to find cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 the gradient function of f (x) = x2 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\593IB_HL-2_20.CDR Friday, 12 March 2010 3:41:29 PM PETER IB_HL-2ed (594) 594 DIFFERENTIAL CALCULUS (Chapter 20) Example Find, from first principles, f (x) if f (x) = If f (x) = x f (x + h) ¡ f(x) , f (x) = lim h!0 x h 13 ¡ (x + h)x = lim x + h x £ h!0 h (x + h)x = lim h!0 x ¡ (x + h) hx(x + h) ¡1 ¡h = lim h!0 hx(x + h) fas h 6= 0g =¡ x2 fas h ! 0, x + h ! xg Example Find, from first principles, the gradient function of f (x) = If f(x) = p x p f(x + h) ¡ f (x) x, f (x) = lim h!0 h p p x+h¡ x = lim h!0 h µp p ¶ µp p ¶ x+h¡ x x+h+ x p = lim p h!0 h x+h+ x = lim x+h¡x p p h( x + h + x) = lim h p p h( x + h + x) h!0 h!0 fas h 6= 0g p =p x+ x = p x EXERCISE 20A Find, from first principles, the gradient function of f (x) where f (x) is: d x4 a x b c x3 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Reminder: (a + b)3 = a3 + 3a2 b + 3ab2 + b3 (a + b)4 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\594IB_HL-2_20.CDR Friday, 12 March 2010 3:00:58 PM PETER IB_HL-2ed (595) 595 DIFFERENTIAL CALCULUS (Chapter 20) Find, from first principles, f (x) given that f (x) is: a x2 ¡ 3x b 2x + c x3 ¡ 2x2 + 3 Find, from first principles, the derivative of f(x) where f (x) is: 1 d a b c x+2 2x ¡ x2 x3 Find, from first principles, the derivative of f (x) equal to: p p x+2 b p 2x + a c x Using the results from the questions above, copy and complete: Derivative (in form kxn ) Function x x2 x3 x4 x¡1 x¡2 x¡3 2x = 2x1 Use your table to predict a formula for f (x) where f(x) = xn and n is rational 1 p x x2 ¡1 ¡1 = 12 x x Using first principles, prove that: “if f (x) = xn then f (x) = nxn¡1 for n Z + ” B DERIVATIVES AT A GIVEN x-VALUE Suppose we are given a function f (x) and asked to find its derivative at the point where x = a This is actually the gradient of the tangent to the curve at x = a, which we write as f (a) y¡=¡ƒ(x) y tangent at a There are two methods for finding f (a) using first principles: point of contact The first method is to start with the definition of the gradient function a x f (x + h) ¡ f(x) Since f (x) = lim , h!0 h f (a) = lim cyan magenta yellow 95 100 50 75 h! 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 we can simply substitute x = a to give black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\595IB_HL-2_20.CDR Friday, 12 March 2010 3:05:40 PM PETER f (a + h) ¡ f (a) h IB_HL-2ed (596) 596 DIFFERENTIAL CALCULUS (Chapter 20) The second method is to consider two points on the graph of y = f (x), a fixed point A(a, f(a)) and a variable point B(x, f(x)) f (x) ¡ f(a) The gradient of chord [AB] = y y¡=¡ƒ(x) x¡a B ƒ(x) (x,¡ƒ(x)) In the limit as B approaches A, x ! a and the gradient of chord [AB] ! gradient A ƒ(a) of the tangent at A ƒ(x) ) f (a) = lim x a tangent at A with gradient ƒ'(a) f (x) ¡ f (a) x¡a x!a f (x) ¡ f(a) x¡a Thus f (a) = lim Note: The gradient of the tangent at x = a is defined as the gradient of the curve at the point where x = a, and is the instantaneous rate of change in y with respect to x at that point x! a is an alternative definition for the gradient of the tangent at x = a Example Find, from first principles, the gradient of the tangent to: b y = ¡ x ¡ x2 a y = 2x2 + at x = a f (2) = lim f (x) ¡ f (2) x¡2 f (2) = lim 2x2 + ¡ 11 x¡2 x!2 ) x!2 = lim x!2 where f (2) = 2(2)2 + = 11 2x2 ¡ x¡2 2(x + 2)(x ¡ 2) x¡2 = 2£4 =8 fas x 6= 2g = lim x!2 b f (¡1) = lim x!¡1 f (x) ¡ f (¡1) x ¡ (¡1) = lim ¡ x ¡ x2 ¡ x+1 = lim ¡x ¡ x2 x+1 x!¡1 x!¡1 at x = ¡1 where f (¡1) = ¡ (¡1) ¡ (¡1)2 = 3+1¡1 =3 ¡x(1 + x) = lim x!¡1 x+1 fas x 6= ¡1g cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 =1 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\596IB_HL-2_20.CDR Friday, 12 March 2010 3:09:56 PM PETER IB_HL-2ed (597) 597 DIFFERENTIAL CALCULUS (Chapter 20) Example Find, from first principles, the derivative of: 2x ¡ a f(x) = at x = b f(x) = x x+3 a f(x) ¡ f(2) x¡2 µ9 9¶ ¡ = lim x x!2 x¡2 µ9 9¶ ¡ 2x = lim x x!2 x ¡ 2x at x = ¡1 f (2) = lim x!2 = lim 18 ¡ 9x 2x(x ¡ 2) = lim ¡9(x ¡ 2) 2x(x ¡ 2) x!2 f2x is the LCD of and 92 g fDo not ‘multiply out’ the denominator since we need to find and cancel the common factor.g x!2 x fas x 6= 2g = ¡ 94 b f (x) ¡ f (¡1) 2(¡1) ¡ where f (¡1) = = ¡ 32 x ¡ (¡1) (¡1) + à 2x¡1 ! x+3 + = lim x!¡1 x+1 à 2x¡1 ! 2(x + 3) x+3 + £ = lim x!¡1 x+1 2(x + 3) f (¡1) = lim x!¡1 = lim 2(2x ¡ 1) + 3(x + 3) 2(x + 1)(x + 3) = lim 4x ¡ + 3x + 2(x + 1)(x + 3) = lim 7x + 2(x + 1)(x + 3) = lim 7(x + 1) 2(x + 1)(x + 3) x!¡1 x!¡1 x!¡1 x!¡1 cyan yellow 95 100 50 75 25 95 100 50 75 magenta 25 95 = 100 50 75 25 95 100 50 75 25 = 2(2) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_20\597IB_HL-2_20.CDR Monday, 19 November 2007 11:53:18 AM PETERDELL IB_HL-2ed (598) 598 DIFFERENTIAL CALCULUS (Chapter 20) Example Find, using first principles, the instantaneous rate of change in y = p x at x = p p x and f(9) = = f(x) = f(x) ¡ f(9) x!9 x¡9 p x¡3 so f (9) = lim x!9 x ¡ p x¡3 p = lim p x!9 ( x + 3)( x ¡ 3) 1 = p 9+3 Now f (9) = lim ftreating x ¡ as the difference of two squares, x 6= 9g = ) the instantaneous rate of change in y = p x at x = is Example Use the first principles formula f (a) = lim h!0 a b a f (a + h) ¡ f(a) h to find: the gradient of the tangent to f (x) = x2 + 2x at x = the instantaneous rate of change of f(x) = at x = ¡3 x f(5 + h) ¡ f (5) h f (5) = lim h!0 where f(5) = 52 + 2(5) = 35 = lim (5 + h)2 + 2(5 + h) ¡ 35 h = lim 25 + 10h + h2 + 10 + 2h ¡ 35 h = lim h2 + 12h h h!0 h!0 h!0 h(h + 12) h!0 h1 = 12 fas h 6= 0g = lim ) the gradient of the tangent at x = is 12 cyan magenta yellow 95 100 50 75 where f (¡3) = 25 95 100 50 75 25 95 100 50 75 25 h!0 f (¡3 + h) ¡ f(¡3) h ! à 4 + = lim ¡3+h h!0 h f (¡3) = lim 95 100 50 75 25 b black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\598IB_HL-2_20.CDR Friday, 12 March 2010 3:11:19 PM PETER ¡3 = ¡ 43 IB_HL-2ed (599) DIFFERENTIAL CALCULUS (Chapter 20) à = lim h¡3 + h h!0 ! £ = lim 12 + 4(h ¡ 3) 3h(h ¡ 3) = lim 4h 3h(h ¡ 3) h!0 h!0 599 3(h ¡ 3) 3(h ¡ 3) fas h 6= 0g = ¡ 49 ) the instantaneous rate of change in f (x) at x = ¡3 is ¡ 49 EXERCISE 20B Questions 1, and may be done using either of the two methods given Find, from first principles, the gradient of the tangent to: b f (x) = 2x2 + 5x at x = ¡1 a f(x) = ¡ x2 at x = d f (x) = 3x + at x = ¡2 c f(x) = ¡ 2x2 at x = Find, from first principles, the derivative of: at x = b a f(x) = x d c f(x) = at x = x 4x + at x = f e f(x) = x¡2 at x = ¡2 x 4x f (x) = at x = x¡3 3x f (x) = at x = ¡4 x +1 f (x) = ¡ Find, from first principles, the instantaneous rate of change in: p p x at x = b x at x = 14 c p at x = d a x Use the first principles formula f (a) = lim h!0 f(a + h) ¡ f (a) h p x ¡ at x = 10 to find: a the gradient of the tangent to f (x) = x2 + 3x ¡ at x = b the gradient of the tangent to f (x) = ¡ 2x ¡ 3x2 at x = ¡2 c the instantaneous rate of change in f (x) = at x = ¡2 2x ¡ 1 d the gradient of the tangent to f (x) = at x = x p e the instantaneous rate of change in f(x) = x at x = f the instantaneous rate of change in f(x) = p at x = x cyan magenta yellow a b 95 100 50 find: 75 25 95 100 f (a + h) ¡ f (a) h 50 25 95 100 50 75 25 95 100 50 75 25 h!0 75 Using f (a) = lim black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\599IB_HL-2_20.CDR Friday, 12 March 2010 3:12:46 PM PETER f (2) for f(x) = x3 f (3) for f(x) = x4 IB_HL-2ed (600) 600 DIFFERENTIAL CALCULUS (Chapter 20) C SIMPLE RULES OF DIFFERENTIATION Differentiation is the process of finding a derivative or gradient function From questions and in Exercise 20A you should have discovered that if f (x) = xn then f (x) = nxn¡1 There are other rules which can be used to differentiate more complicated functions without having to resort to the tedious method of first principles We will discover some of these rules in the following investigation INVESTIGATION SIMPLE RULES OF DIFFERENTIATION In this investigation we attempt to differentiate functions of the form cxn where c is a constant, and functions which are a sum (or difference) of terms of the form cxn What to do: Find, from first principles, the derivatives of: b 2x3 c a 4x2 p x p Compare your results with the derivatives of x2 , x3 and x obtained earlier Copy and complete: “If f(x) = cxn , then f (x) = ::::::” Use first principles to find f (x) for: a f (x) = x2 +3x b f (x) = x3 ¡ 2x2 Use to copy and complete: “If f (x) = u(x) + v(x) then f (x) = ::::::” You should have discovered the following rules for differentiating functions: f (x) f (x) Name of rule c (a constant) differentiating a constant xn nxn¡1 differentiating xn c u(x) c u0 (x) constant times a function u(x) + v(x) u (x) + v (x) addition rule Each of these rules can be proved using the first limit definition of f (x) For example: ² If f(x) = cu(x) where c is a constant then f (x) = cu0 (x) f (x) = lim Proof: h!0 f (x + h) ¡ f (x) h cu(x + h) ¡ cu(x) h!0 h · ¸ u(x + h) ¡ u(x) = lim c h!0 h cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 = lim black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\600IB_HL-2_20.CDR Friday, 12 March 2010 3:13:59 PM PETER IB_HL-2ed (601) 601 DIFFERENTIAL CALCULUS (Chapter 20) u(x + h) ¡ u(x) h = c lim h!0 = c u0 (x) ² If f(x) = u(x) + v(x) then f (x) = u0 (x) + v (x) f (x + h) ¡ f(x) h!0 h µ ¶ u(x + h) + v(x + h) ¡ [u(x) + v(x)] = lim h!0 h µ ¶ u(x + h) ¡ u(x) + v(x + h) ¡ v(x) = lim h!0 h f (x) = lim Proof: u(x + h) ¡ u(x) v(x + h) ¡ v(x) + lim h!0 h!0 h h = u0 (x) + v (x) = lim Using the rules we have now developed we can differentiate sums of powers of x f (x) = 3x4 + 2x3 ¡ 5x2 + 7x + then For example, if f (x) = 3(4x3 ) + 2(3x2 ) ¡ 5(2x) + 7(1) + = 12x3 + 6x2 ¡ 10x + Example Find f (x) for f (x) equal to: a 5x3 + 6x2 ¡ 3x + f(x) = 5x3 + 6x2 ¡ 3x + a f(x) = 7x ¡ b ) f (x) = 5(3x2 ) + 6(2x) ¡ 3(1) = 15x2 + 12x ¡ b 7x ¡ + x x + x x = 7x ¡ 4x¡1 + 3x¡3 ) f (x) = 7(1) ¡ 4(¡1x¡2 ) + 3(¡3x¡4 ) = + 4x¡2 ¡ 9x¡4 = 7+ ¡ x x Example and hence find the gradient of the x tangent to the function at the point where x = Find the gradient function of f(x) = x2 ¡ x f (x) = 2x ¡ 4(¡1x¡2 ) ) = 2x + 4x¡2 cyan magenta yellow 50 x2 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 = 2x + 95 = x2 ¡ 4x¡1 100 f(x) = x2 ¡ black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\601IB_HL-2_20.CDR Friday, 12 March 2010 3:14:48 PM PETER Now f (2) = + = 5, So, the tangent has gradient of IB_HL-2ed (602) 602 DIFFERENTIAL CALCULUS (Chapter 20) Example 10 Find the gradient function of f(x) where f (x) is: p a f (x) = x + = 3x + 2x¡1 x p a x+ x b x2 ¡ p x ) f (x) = 3( 12 x¡ ) + 2(¡1x¡2 ) = 32 x¡ ¡ 2x¡2 = p ¡ 2 x x b f (x) = x2 ¡ p = x2 ¡ 4x¡ x ) f (x) = 2x ¡ 4(¡ 12 x¡ ) = 2x + 2x¡ 2 = 2x + p x x ALTERNATIVE NOTATION If we are given a function f(x) then f (x) represents the derivative function dy If we are given y in terms of x then y or are commonly used to represent the dx derivative Note: ² dy reads “dee y by dee x” or “the derivative of y with respect to x” dx ² dy is not a fraction dx ² d(:::::) reads “the derivative of ( ) with respect to x” dx The notation lim ±x!0 ±y ±x dy stems from the fact that the gradient function is found by considering dx where ±y and ±x are small changes in y and x respectively Example 11 If y = 3x2 ¡ 4x, find As y = 3x2 ¡ 4x, cyan dy = 6x ¡ dx magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 ² the gradient function or derivative of y = 3x2 ¡ 4x from which the gradient at any point can be found ² the instantaneous rate of change in y as x changes 25 95 100 50 75 25 dy is: dx dy and interpret its meaning dx black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\602IB_HL-2_20.CDR Friday, 12 March 2010 3:17:58 PM PETER IB_HL-2ed (603) 603 DIFFERENTIAL CALCULUS (Chapter 20) EXERCISE 20C Find f (x) given that f (x) is: a d x3 x2 + x b e 2x3 ¡ 2x2 c f g x3 + 3x2 + 4x ¡ h 5x4 ¡ 6x2 i k x3 + x l x3 + x ¡ x n (2x ¡ 1)2 o (x + 2)3 j m Find 2x ¡ x2 p x 7x2 x2 + 3x ¡ 3x ¡ x dy for: dx 5x2 a y = 2x3 ¡ 7x2 ¡ b y = ¼x2 c y= d y = 100x e y = 10(x + 1) f y = 4¼x3 p x x c (5 ¡ x)2 f x(x + 1)(2x ¡ 5) Differentiate with respect to x: a 6x + b 6x2 ¡ 9x4 3x d 4x ¡ e 4x Find the gradient of the tangent to: a y = x2 c y = 2x2 ¡ 3x + at x = ¡1 e y= at x = x2 ¡ x2 at x = b y= x2 d y= 2x2 ¡ x f y= x3 ¡ 4x ¡ x2 Find the gradient function of f(x) where f(x) is: p p a x+x b 3x c ¡p x p ¡5 x e magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 p x x2 at x = ¡1 p x d 2x ¡ h 2x ¡ p x x dy , find and interpret its meaning x dx b The position of a car moving along a straight road is given by S = 2t2 +4t metres dS where t is the time in seconds Find and interpret its meaning dt c The cost of producing x toasters each week is given by dC C = 1785 + 3x + 0:002x2 dollars Find and interpret its meaning dx 95 100 50 75 25 g at x = a If y = 4x ¡ cyan p 3x2 ¡ x x f at x = black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\603IB_HL-2_20.CDR Friday, 12 March 2010 3:18:51 PM PETER IB_HL-2ed (604) 604 DIFFERENTIAL CALCULUS (Chapter 20) D THE CHAIN RULE In Chapter we defined the composite of two functions f and g as f(g(x)) We can often write complicated functions as the composite of two or more simpler functions For example, consider y = (x2 + 3x)4 This could be rewritten as y = u4 where u = x2 + 3x, or as y = f (g(x)) where f(x) = x4 and g(x) = x2 + 3x Example 12 p x and g(x) = ¡ 3x f(x) and g(x) such that f(g(x)) = x ¡ x2 a f (g(x)) if f (x) = Find: b a b f (g(x)) = f(g(x)) = f(2 ¡ 3x) p = ¡ 3x ) f(x) = 1 = x¡x g(x) x and g(x) = x ¡ x2 EXERCISE 20D.1 Find f (g(x)) if: c f(x) = x2 and g(x) = 2x + p f(x) = x and g(x) = ¡ 4x e f(x) = a x d f (x) = 2x + and g(x) = x2 p f (x) = ¡ 4x and g(x) = x f f (x) = x2 + and g(x) = b and g(x) = x2 + Find f (x) and g(x) such that f(g(x)) is: a (3x + 10)3 b c 2x + p x2 ¡ 3x d x 10 (3x ¡ x2 )3 DERIVATIVES OF COMPOSITE FUNCTIONS The reason we are interested in writing complicated functions as composite functions is to make finding derivatives easier In the following investigation we look for a rule that will help us to differentiate composite functions INVESTIGATION DIFFERENTIATING COMPOSITES cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The purpose of this investigation is to gain insight into how we can differentiate composite functions dy = nxn¡1 ”, we might Based on our previous rule “if y = xn then dx dy suspect that if y = (2x + 1)2 then = 2(2x + 1)1 = 2(2x + 1) dx But is this so? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_20\604IB_HL-2_20.CDR Wednesday, 14 November 2007 3:55:32 PM PETERDELL IB_HL-2ed (605) DIFFERENTIAL CALCULUS (Chapter 20) 605 What to do: Consider y = (2x+1)2 Expand the brackets and then find dy dy Is = 2(2x+1)? dx dx Consider y = (3x+1)2 Expand the brackets and then find dy dy Is = 2(3x+1)1 ? dx dx Consider y = (ax + 1)2 Expand the brackets and find dy dy Is = 2(ax + 1)1 ? dx dx If y = u2 where u is a function of x, what you suspect dy will be equal to? dx dy dx Does your answer agree with the rule you suggested in 4? Consider y = (x2 + 3x)2 Expand it and find In the previous investigation you probably found that if y = u2 then dy du dy du = 2u £ = : dx dx du dx Now consider y = (2x + 1)3 which has the form y = u3 Expanding we have y = (2x + 1)3 = (2x)3 + 3(2x)2 + 3(2x)12 + 13 = 8x3 + 12x2 + 6x + dy = 24x2 + 24x + dx = 6(4x2 + 4x + 1) = 6(2x + 1)2 = 3(2x + 1)2 £ du = 3u2 £ which is again dx ) where u = 2x + fbinomial expansiong dy du : du dx From the investigation and from the above example we formulate the chain rule: If y = f (u) where u = u(x) then dy dy du = dx du dx This rule is extremely important and enables us to differentiate complicated functions much faster For example, we can readily see that for any function f(x): If y = [f (x)]n then dy = n[f (x)]n¡ £ f (x) dx cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A non-examinable proof of this rule is included for completeness black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_20\605IB_HL-2_20.CDR Wednesday, 14 November 2007 4:09:06 PM PETERDELL IB_HL-2ed (606) 606 DIFFERENTIAL CALCULUS (Chapter 20) Consider y = f (u) where u = u(x) Proof: u For a small change of ±x in x, there is a small change of u(x + ±x) ¡ u(x) = ±u in u and a small change of ±y in y y u¡=¡u(x) du y+dy u y dx ±y ±y ±u = £ ±x ±u ±x x x+d x x Now y=ƒ(u) u+du dy du u u+d u u ffraction multiplicationg As ±x ! 0, ±u ! also ) lim ±x!0 ) ±y ±y ±u = lim £ lim ±x ±u!0 ±u ±x!0 ±x flimit ruleg dy dy du = dx du dx f (x + h) ¡ f (x) , we replace h by ±x and f(x + h) ¡ f (x) h ±y dy by ±y, we have f (x) = = lim dx ±x!0 ±x Note: If in f (x) = lim h!0 Example 13 Find a dy if: dx b y=p ¡ 2x a y = (x2 ¡ 2x)4 y = (x2 ¡ 2x)4 ) y = u4 where u = x2 ¡ 2x dy dy du Now = fchain ruleg dx du dx Notice that the brackets around 2x¡¡¡2 are essential = 4u3 (2x ¡ 2) = 4(x2 ¡ 2x)3 (2x ¡ 2) y= p ¡ 2x b ) y = 4u¡ Now where u = ¡ 2x dy dy du = dx du dx fchain ruleg = £ (¡ 12 u¡ ) £ (¡2) = 4u¡ cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 = 4(1 ¡ 2x)¡ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_20\606IB_HL-2_20.CDR Monday, 19 November 2007 12:10:09 PM PETERDELL IB_HL-2ed (607) 607 DIFFERENTIAL CALCULUS (Chapter 20) EXERCISE 20D.2 Write in the form aun , clearly stating what u is: p b x2 ¡ 3x a (2x ¡ 1) p x3 ¡ x2 d (3 ¡ x)3 e dy dx Find the gradient function ¡ 2x y = (4x ¡ 5)2 b y= d y = (1 ¡ 3x)4 e y = 6(5 ¡ x)3 g y= (5x ¡ 4)2 h y= 3x ¡ x2 Find the gradient of the tangent to: p a y = ¡ x2 at x = 12 (2x ¡ 1)4 y= e p y= x+2 x 10 ¡3 f x2 c y= f y= b d at x = p 2x3 ¡ x2 µ ¶3 y = x2 ¡ x y = (3x + 2)6 at x = ¡1 p y = £ ¡ 2x at x = µ ¶3 y = x+ x f at x = i p 3x ¡ x2 at x = 1 If y = x3 a Find p ¡ x2 for: a c c then x = y dy dx dx dy and b Explain why and hence show that dy dx £ = dx dy dy dx £ = whenever these derivatives exist for any general dx dy function y = f (x) E PRODUCT AND QUOTIENT RULES If f(x) = u(x) + v(x) then f (x) = u0 (x) + v (x): So, the derivative of a sum of two functions is the sum of the derivatives But, what if f(x) = u(x)v(x)? Is f (x) = u0 (x)v (x)? In other words, is the derivative of a product of two functions equal to the product of the derivatives of the two functions? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The following example shows that this cannot be true: black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\607IB_HL-2_20.CDR Friday, 12 March 2010 3:19:40 PM PETER IB_HL-2ed (608) 608 DIFFERENTIAL CALCULUS (Chapter 20) p p If f(x) = x x we could say f (x) = u(x)v(x) where u(x) = x and v(x) = x Now f(x) = x 1 so f (x) = 32 x But u0 (x)v (x) = £ 12 x¡ = 12 x¡ 6= f (x) THE PRODUCT RULE If u(x) and v(x) are two functions of x and y = uv dy du dv = v+u dx dx dx then y = u0 (x)v(x) + u(x)v0 (x) or p Consider again the example f(x) = x x This is a product u(x)v(x) where v(x) = x u(x) = x and u0 (x) = and ) v (x) = 12 x¡ 1 f (x) = u0 v + uv0 = £ x + x £ 12 x¡ According to the product rule 1 = x + 12 x = 32 x which is correct X Example 14 Find a dy if: dx y= a y= p x(2x + 1)3 p x(2x + 1)3 b y = x2 (x2 ¡ 2x)4 is the product of u = x ) u0 = 12 x¡ and v = (2x + 1)3 and v = 3(2x + 1)2 £ = 6(2x + 1)2 dy fproduct ruleg = u0 v + uv0 dx 1 ¡ 12 = x (2x + 1)3 + x £ 6(2x + 1)2 Now 1 = 12 x¡ (2x + 1)3 + 6x (2x + 1)2 b y = x2 (x2 ¡ 2x)4 is the product of u = x2 and v = (x2 ¡ 2x)4 ) u0 = 2x and v = 4(x2 ¡ 2x)3 (2x ¡ 2) Now dy fproduct ruleg = u0 v + uv0 dx = 2x(x ¡ 2x) + x2 £ 4(x2 ¡ 2x)3 (2x ¡ 2) = 2x(x2 ¡ 2x)4 + 4x2 (x2 ¡ 2x)3 (2x ¡ 2) For completeness we now prove the product rule Proof: Let y = u(x)v(x) Suppose there is a small change of ±x in x which causes corresponding changes of ±u in u, ±v in v, and ±y in y cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 As y = uv, y + ±y = (u + ±u)(v + ±v) ) y + ±y = uv + (±u)v + u(±v) + ±u±v ) ±y = (±u)v + u(±v) + ±u±v black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_20\608IB_HL-2_20.CDR Thursday, 15 November 2007 9:29:32 AM PETERDELL IB_HL-2ed (609) 609 DIFFERENTIAL CALCULUS (Chapter 20) ±y = ±x ) ) ±y = ±x!0 ±x µ µ ±u ±x ¶ µ v+u ±u ±x!0 ±x lim ¶ lim ±v ±x ¶ µ + µ ±u ±x ¶ ±v ±x!0 ±x v+u fdividing each term by ±xg ±v ¶ lim +0 fas ±x ! 0, ±v ! alsog dy du dv = v+u dx dx dx ) EXERCISE 20E.1 dy using the product rule: dx b y = 4x(2x + 1)3 y = x2 (2x ¡ 1) p e y = 5x2 (3x2 ¡ 1)2 y = x(x ¡ 3)2 Find a d Find the gradient of the tangent to: y = x4 (1 ¡ 2x)2 at x = ¡1 p y = x ¡ 2x at x = ¡4 a c p x(3 ¡ x)2 b d c f p y = x2 ¡ x p y = x(x ¡ x2 )3 p x(x2 ¡ x + 1)2 at x = p y = x3 ¡ x2 at x = y= dy (3 ¡ x)(3 ¡ 5x) p = dx x p Find the x-coordinates of all points on y = x(3 ¡ x)2 where the tangent is horizontal If y = show that THE QUOTIENT RULE p x ¡ 3x x2 + , 2x ¡ Expressions like and Quotient functions have the form Q(x) = x3 (x ¡ x2 )4 are called quotients u(x) v(x) Notice that u(x) = Q(x) v(x) and by the product rule u0 (x) = Q0 (x) v(x) + Q(x) v (x) ) u0 (x) ¡ Q(x) v (x) = Q0 (x) v(x) u(x) ) Q0 (x) v(x) = u0 (x) ¡ v (x) v(x) ) Q0 (x) v(x) = u0 (x) v(x) ¡ u(x) v0 (x) v(x) ) Q0 (x) = u0 (x) v(x) ¡ u(x) v0 (x) [v(x)]2 cyan magenta u0 (x)v(x) ¡ u(x)v0 (x) [v(x)]2 yellow 95 100 50 75 25 95 100 where u and v are functions of x then 50 u v then Q0 (x) = 75 95 100 50 75 25 95 100 50 75 25 or if y = u(x) v(x) 25 So, if Q(x) = and this formula is called the quotient rule black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\609IB_HL-2_20.CDR Friday, 12 March 2010 3:20:28 PM PETER dy u0 v ¡ uv0 = dx v2 IB_HL-2ed (610) 610 DIFFERENTIAL CALCULUS (Chapter 20) Example 15 dy Use the quotient rule to find if: dx a + 3x x2 + y= Now is a quotient with u = + 3x and v = x2 + Now u0 = ) dy u0 v ¡ uv0 = dx v2 3(x2 + 1) ¡ (1 + 3x)2x (x2 + 1)2 = 3x2 + ¡ 2x ¡ 6x2 (x2 + 1)2 = ¡ 2x ¡ 3x2 (x2 + 1)2 is a quotient where u = x ¡ 12 ) u0 = 12 x dy u0 v ¡ uv0 = dx v2 = and v = 2x fquotient ruleg = p x y= (1 ¡ 2x)2 b p x b y= (1 ¡ 2x)2 + 3x a y= x +1 ¡ 12 (1 2x and v = (1 ¡ 2x)2 and v0 = 2(1 ¡ 2x)1 £ (¡2) = ¡4(1 ¡ 2x) fquotient ruleg ¡ 2x)2 ¡ x £ (¡4(1 ¡ 2x)) (1 ¡ 2x)4 ¡ 12 (1 2x ¡ 2x)2 + 4x (1 ¡ 2x) = (1 ¡ 2x)4 · µ p ¶¸ p ¡ 2x x p +4 x p (1 ¡ 2x) x x = (1 ¡ 2x)4 flook for common factorsg ¡ 2x + 8x = p x(1 ¡ 2x)3 6x + = p x(1 ¡ 2x)3 dy as in the above example is often unnecessary, especially if dx you want to find the gradient of a tangent at a given point In such cases you can substitute a value for x without simplifying cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Note: Simplification of black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\610IB_HL-2_20.CDR Friday, 12 March 2010 3:21:31 PM PETER IB_HL-2ed (611) 611 DIFFERENTIAL CALCULUS (Chapter 20) EXERCISE 20E.2 dy if: dx Use the quotient rule to find + 3x 2¡x p x y= ¡ 2x a y= d b y= x2 2x + c y= e y= x2 ¡ 3x ¡ x2 f x y=p ¡ 3x Find the gradient of the tangent to: x a y= at x = 1 ¡ 2x p x c y= at x = 2x + x3 x2 + b y= d x2 y=p x2 + x2 x ¡3 at x = ¡1 at x = ¡2 p dy x+1 x : , show that =p 1¡x dx x(1 ¡ x)2 dy b For what values of x is i zero ii undefined? dx a If y = x2 ¡ 3x + dy x2 + 4x ¡ : , show that = x+2 dx (x + 2)2 dy i zero ii undefined? b For what values of x is dx c What is the graphical significance of your answers in b? a If y = F TANGENTS AND NORMALS Consider a curve y = f (x) If A is the point with x-coordinate a, then the gradient of the tangent at this point is f (a) = mT y = ƒ(x) tangent point of contact The equation of the tangent is y ¡ f (a) = f (a) fequating gradientsg x¡a A (a, ƒ(a)) normal x=a or y ¡ f (a) = f (a)(x ¡ a) Alternatively, if the point A is at (a, b), then the equation of the tangent is y¡b = f (a) or y ¡ b = f (a)(x ¡ a) x¡a cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A normal to a curve is a line which is perpendicular to the tangent at the point of contact The gradients of perpendicular lines are negative reciprocals of each other black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\611IB_HL-2_20.CDR Friday, 12 March 2010 3:25:46 PM PETER IB_HL-2ed (612) 612 DIFFERENTIAL CALCULUS (Chapter 20) the gradient of a normal at x = a is mN = ¡ Thus, f (a) For example, if f (x) = x2 then f (x) = 2x = ¡ 14 At x = 2, f (2) = and ¡ f (2) So, at x = the tangent has gradient and the normal has gradient ¡ 14 Since f (2) = 4, the tangent has equation y ¡ = 4(x ¡ 2) or y = 4x ¡ and the normal has equation y ¡ = ¡ 14 (x ¡ 2) or y = ¡ 14 x + 92 Note: If a line has gradient say, and passes through (2, ¡3) say, another quick way to write down its equation is 4x ¡ 5y = 4(2) ¡ 5(¡3), i.e., 4x ¡ 5y = 23 If the gradient was ¡ 45 , we would have: 4x + 5y = 4(2) + 5(¡3), i.e., 4x + 5y = ¡7 Example 16 Find the equation of the tangent to f (x) = x2 + at the point where x = y Since f (1) = + = 2, the point of contact is (1, 2) ƒ(x)¡=¡xX¡+¡1 Now f (x) = 2x ) f (1) = y¡2 =2 x¡1 which is y ¡ = 2x ¡ or y = 2x ) the tangent has equation (1, 2) x Example 17 Find the equation of the normal to y = p x =4 (4,¡4) yellow 95 100 50 75 25 50 75 25 95 50 75 25 95 100 50 75 25 100 magenta ) the equation of the normal is 2x ¡ 1y = 2(4) ¡ 1(4) or 2x ¡ y = x cyan so the point of contact is (4, 4) dy Now as y = 8x¡ , = ¡4x¡ dx dy and when x = 4, = ¡4 £ 4¡ = ¡ 12 dx ) the normal at (4, 4) has gradient 21 = 95 y p8 100 When x = 4, y = at the point where x = black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\612IB_HL-2_20.CDR Friday, 12 March 2010 3:29:52 PM PETER IB_HL-2ed (613) 613 DIFFERENTIAL CALCULUS (Chapter 20) Example 18 Find the equations of any horizontal tangents to y = x3 ¡ 12x + dy = 3x2 ¡ 12 dx Horizontal tangents have gradient 0, so 3x2 ¡ 12 = ) 3(x2 ¡ 4) = ) 3(x + 2)(x ¡ 2) = ) x = ¡2 or Since y = x3 ¡ 12x + 2, When x = 2, y = ¡ 24 + = ¡14 When x = ¡2, y = ¡8 + 24 + = 18 ) the points of contact are (2, ¡14) and (¡2, 18) ) the tangents are y = ¡14 and y = 18 Example 19 Find the equation of the tangent to y = p 10 ¡ 3x at the point where x = Let f(x) = (10 ¡ 3x) When x = 3, y = ) f (x) = 12 (10 ¡ 3x)¡ £ (¡3) ) f (3) = 12 (1)¡ £ (¡3) = ¡ 32 p 10 ¡ = ) the point of contact is (3, 1) y¡1 =¡ x¡3 So, the tangent has equation i.e., 2y ¡ = ¡3x + or 3x + 2y = 11 Example 20 Find the coordinates of the point(s) where the tangent to y = x3 + x + at (1, 4) meets the curve again Let f (x) = x3 + x + ) f (x) = 3x2 + ) f (1) = + = ) the tangent at (1, 4) has gradient and its equation is 4x ¡ y = 4(1) ¡ or y = 4x Now y = 4x meets y = x3 + x + where x3 + x + = 4x ) x3 ¡ 3x + = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 This cubic must have a repeated zero of x = because of the tangent which touches the curve at x = black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\613IB_HL-2_20.CDR Friday, 12 March 2010 3:30:39 PM PETER (1,¡4) (-2,-8) IB_HL-2ed (614) 614 DIFFERENTIAL CALCULUS (Chapter 20) (x ¡ 1)2 (x + 2) = ) x2 £ x = x3 (¡1)2 £ = ) x = or ¡2 When x = ¡2, y = (¡2)3 + (¡2) + = ¡8 ) the tangent meets the curve again at (¡2, ¡8) Example 21 Find the equations of the tangents to y = x2 from the external point (2, 3) Let (a, a2 ) lie on f(x) = x2 y¡=¡xX y Now f (x) = 2x, (2,¡3) so f (a) = 2a ) at (a, a2 ) the gradient of the tangent is 2a ) its equation is 2ax ¡ y = 2a(a) ¡ (a2 ) i.e., 2ax ¡ y = a2 (a,¡aX) x But this tangent passes through (2, 3) ) 2a(2) ¡ = a2 ) a2 ¡ 4a + = ) (a ¡ 1)(a ¡ 3) = ) a = or If a = 1, the tangent has equation 2x ¡ y = with point of contact (1, 1) If a = 3, the tangent has equation 6x ¡ y = with point of contact (3, 9) EXERCISE 20F Find the equation of the tangent to: a y = x ¡ 2x2 + at x = c b y = x3 ¡ 5x at x = d p x + at x = 4 y=p at (1, 4) x y= Find the equation of the normal to: a c y = x2 at the point (3, 9) p y = p ¡ x at the point (1, 4) x b d y = x3 ¡ 5x + at x = ¡2 p y = x ¡ at x = x a Find the equations of the horizontal tangents to y = 2x3 + 3x2 ¡ 12x + p b Find all points of contact of horizontal tangents to the curve y = x + p x c Find k if the tangent to y_ =_ 2x3_ +_ kx2_ ¡_ at the point where x_ =_ has gradient cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 d Find the equation of the tangent to y = ¡ 3x + 12x2 ¡ 8x3 to the tangent at (1, 2) black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\614IB_HL-2_20.CDR Friday, 12 March 2010 3:34:01 PM PETER which is parallel IB_HL-2ed (615) DIFFERENTIAL CALCULUS (Chapter 20) 615 a The tangent to the curve y = x2 + ax + b where a and b are constants, is 2x + y = at the point where x = Find the values of a and b p b b The normal to the curve y = a x + p where a and b are constants, has x equation 4x + y = 22 at the point where x = Find the values of a and b Find the equation of the tangent to: p a y = 2x + at x = c f(x) = x ¡ 3x b at (¡1, ¡ 14 ) d Find the equation of the normal to: at (1, 14 ) a y= (x + 1)2 c f(x) = p x(1 ¡ x)2 at x = at x = ¡1 2¡x x2 f (x) = at (2, ¡4) 1¡x y= b y=p ¡ 2x d f (x) = at x = ¡3 x2 ¡ 2x + at x = ¡1 p y = a ¡ bx where a and b are constants, has a tangent with equation 3x + y = at the point where x = ¡1 Find a and b a Find where the tangent to the curve y = x3 at the point where x = 2, meets the curve again b Find where the tangent to the curve y = ¡x3 + 2x2 + at the point where x = ¡1, meets the curve again c Find where the tangent to the curve y = x3 + at the point where x = 1, x meets the curve again a Find the equation of the tangent to y = x2 ¡ x + at the point where x = a Hence, find the equations of the two tangents from (0, 0) to the curve State the coordinates of the points of contact b Find the equations of the tangents to y = x3 from the external point (¡2, 0) p c Find the equation(s) of the normal(s) to y = x from the external point (4, 0) x2 Sketch the graph of the function Find the equation of the tangent at the point where x = a If the tangent in b cuts the x-axis at A and the y-axis at B, find the coordinates of A and B Find the area of triangle OAB and discuss the area of the triangle as a ! 10 Consider f (x) = a b c d x 11 Consider f : x 7! p 2¡x b Show that f (x) = a State the domain of f 4¡x 2(2 ¡ x) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 c Find the equation of the normal to f at the point where f (x) = ¡1 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_20\615IB_HL-2_20.CDR Monday, 19 November 2007 12:33:36 PM PETERDELL IB_HL-2ed (616) 616 DIFFERENTIAL CALCULUS (Chapter 20) p p 12 The graphs of y = x + a and y = 2x ¡ x2 point of intersection Find a and the point of intersection have the same gradient at their 13 If P is at (¡2, 3) and Q is at (6, ¡3), the line segment [PQ] is a tangent to y = b (x + 1)2 Find b G HIGHER DERIVATIVES THE SECOND DERIVATIVE Given a function f (x), the derivative f (x) is known as the first derivative The second derivative of f(x) is the derivative of f (x), i.e., the derivative of the first derivative d2 y to represent the second derivative dx2 We use f 00 (x) or y00 or Note that: ² d2 y d = dx dx ² d2 y dx2 µ dy dx ¶ reads “dee two y by dee x squared ” THE SECOND DERIVATIVE IN CONTEXT Michael rides up a hill and down the other side to his friend’s house The dots on the graph show Michael’s position at various times t t=0 t=5 t = 15 t = 19 t = 17 t = 10 Michael’s place friend’s house The distance Michael has travelled at various times is given in the following table: Time (t min) Distance travelled (s m) 0 2:5 7:5 10 12:5 15 17 19 498 782 908 989 1096 1350 1792 2500 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The cubic model s ¼ 1:18t3 ¡ 30:47t2 + 284:52t ¡ 16:08 metres fits this data well However, notice that the model gives s(0) ¼ ¡16:08 m whereas the actual data gives s(0) = This sort of problem often occurs when modelling from data A graph of the data points and the cubic curve follows: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_20\616IB_HL-2_20.CDR Monday, 19 November 2007 12:34:20 PM PETERDELL IB_HL-2ed (617) 617 DIFFERENTIAL CALCULUS (Chapter 20) y 2500 2000 y=1.18x 3-30.47x 2+284.52x-16.08 1500 1000 500 10 15 x ds ¼ 3:54t2 ¡ 60:94t + 284:52 metres per minute is the instantaneous rate of change dt in displacement per unit of time, or instantaneous velocity The instantaneous rate of change in velocity at any point in time is Michael’s acceleration, µ ¶ d ds d2 s so = is the instantaneous acceleration, dt dt dt Now d2 s = 7:08t ¡ 60:94 metres per minute per minute dt2 i.e., Notice that when t = 12, s ¼ 1050 m ds ¼ 63 metres per minute and dt d2 s ¼ 24 metres per minute per minute dt2 We will examine displacement, velocity and acceleration in greater detail in the next chapter Example 22 Now f (x) = x3 ¡ 3x¡1 ) f (x) = 3x2 + 3x¡2 ) f 00 (x) = 6x ¡ 6x¡3 = 6x ¡ x Find f 00 (x) given that f (x) = x3 ¡ : x HIGHER DERIVATIVES d2 y is obtained by differentiating y = f(x) twice, it is clear that we dx2 can continue to differentiate to obtain the 3rd, 4th, 5th derivatives, and so on Given f 00 (x) = f (3) (x) = We call these: d3 y , dx3 f (4) (x) = d4 y , dx4 f (5) (x) = d5 y , dx5 respectively In general: the nth derivative of y with respect to x is obtained by differentiating y = f (x) n times cyan magenta yellow 95 100 50 for the nth derivative 75 25 dn y dxn 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 We use the notation f (n) (x) or black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_20\617IB_HL-2_20.CDR Thursday, 15 November 2007 11:07:42 AM PETERDELL IB_HL-2ed (618) 618 DIFFERENTIAL CALCULUS (Chapter 20) Example 23 f (x) = x7 + x3 + x ¡ 2x¡2 ) f (x) = 7x6 + 3x2 + + 4x¡3 ) f 00 (x) = 42x5 + 6x ¡ 12x¡4 Find f (3) (x) given f (x) = x7 + x3 + x ¡ x2 f (3) (x) = 210x4 + + 48x¡5 ) EXERCISE 20G Find f 00 (x) given that: a f(x) = 3x2 ¡ 6x + b f (x) = 2x3 ¡ 3x2 ¡ x + c f(x) = p ¡ x d f (x) = e f(x) = (1 ¡ 2x)3 f ¡ 3x x2 x+2 f (x) = 2x ¡ d3 y d2 y and given that: dx2 dx3 Find a y = x ¡ x3 d y= 4¡x x x2 b y = x2 ¡ e y = (x2 ¡ 3x)3 c y =2¡ p x f y = x2 ¡ x + 1¡x Find x when f 00 (x) = for: a f(x) = 2x3 ¡ 6x2 + 5x + a If y = 1¡x find b f (x) = x2 x +2 dy dx b Use the principle of mathematical induction to prove that if y = ¡ x n! dn y + = for all n Z dxn (1 ¡ x)n+1 then REVIEW SET 20A Find the equation of the tangent to y = ¡2x2 Find dy for: dx y = 3x2 ¡ x4 a b at the point where x = ¡1 y= x3 ¡ x x2 Find, from first principles, the derivative of f (x) = x2 + 2x Find the equation of the normal to y = ¡ 2x x2 at the point where x = Find where the tangent to y = 2x3 + 4x ¡ at (1, 5) cuts the curve again cyan magenta yellow 95 100 50 75 25 95 100 50 at x = is 2x ¡ y = Find a and b 75 25 ax + b p x 95 100 50 75 25 95 100 50 75 25 The tangent to y = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_20\618IB_HL-2_20.CDR Thursday, 15 November 2007 11:01:16 AM PETERDELL IB_HL-2ed (619) 619 DIFFERENTIAL CALCULUS (Chapter 20) Find a given that the tangent to y = at x = passes through (1, 0) (ax + 1)2 Find the equation of the normal to y = p at the point where x = x Determine the derivative with respect to t of: p t+5 b A= a M = (t + 3) t2 dy 10 Use the rules of differentiation to find for: dx b y = (x ¡ )4 a y = p ¡ 3x x x 11 If y = p ¡ 4x, find: a dy dx c d2 y dx2 b c y= p x2 ¡ 3x d3 y dx3 REVIEW SET 20B Differentiate with respect to x: b a 5x ¡ 3x¡1 (3x2 + x)4 c (x2 + 1)(1 ¡ x2 )3 Determine the equation of any horizontal tangents to the curve with equation y = x3 ¡ 3x2 ¡ 9x + x+1 Find the equation of the normal to y = at the point where x = x ¡2 Differentiate with respect to x: Find f 00 (x) for: a a f(x) = f(x) = 3x2 ¡ (x + 3)3 p x x b f (x) = b p f(x) = x4 x2 + p x p The tangent to y = x2 ¡ x at x = ¡3 cuts the axes at A and B Determine the area of triangle OAB y = 2x is a tangent to the curve y = x3 + ax + b at x = Find a and b The tangent to y = x3 + ax2 ¡ 4x + at x = is parallel to the line y = 3x Find the value of a and the equation of the tangent at x = Where does the tangent cut the curve again? The curve f (x) = 2x3 + Ax + B has a tangent with gradient 10 at the point (¡2, 33) Find the values of A and B 10 Use the product rule for differentiation to prove that: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a if y = uv where u and v are functions of x, then µ ¶ µ ¶ d u d v d2 y du dv = + u v + dx2 dx2 dx dx dx2 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\619IB_HL-2_20.CDR Friday, 12 March 2010 3:35:04 PM PETER IB_HL-2ed (620) 620 DIFFERENTIAL CALCULUS (Chapter 20) b if y = uvw where u, v and w are functions of x, then dy du dv dw = vw + u w + uv : dx dx dx dx 11 Prove using the principle of mathematical induction that if y = xn , n Z + , then dy You may assume the product rule of differentiation = nxn¡1 dx REVIEW SET 20C Differentiate with respect to x: p y = x3 ¡ x2 a b x2 ¡ 3x y= p x+1 Find x if f 00 (x) = and f (x) = 2x4 ¡ 4x3 ¡ 9x2 + 4x + 3x If the normal to f(x) = 1+x length of [BC] Find d2 y for: dx2 at (2, 2) cuts the axes at B and C, determine the y = 3x4 ¡ a x y = x3 ¡ x + p x b x y=p has a tangent with equation 5x + by = a at the point where x = ¡3 1¡x Find the values of a and b The curve f (x) = 3x3 +Ax2 +B has tangent with gradient at the point (¡2, 14) 00 Find A and B and hence f (¡2) a Find a (x + 2)2 p p Show that the curves whose equations are y = 3x + and y = 5x ¡ x2 have the same gradient at their point of intersection Find the equation of the common tangent at this point for x > a Sketch the graph of x 7! x b Find the equation of the tangent to the function at the point where x = k, k > c If the tangent in b cuts the x-axis at A and the y-axis at B, find the coordinates of A and B d What can be deduced about the area of triangle OAB? e Find k if the normal to the curve at x = k passes through the point (1, 1) The line joining A(2, 4) to B(0, 8) is a tangent to y = cyan yellow 95 for all n Z + 100 50 75 25 95 50 75 25 95 50 100 magenta 100 (¡2)n n! dn y = dxn (2x + 1)n+1 , prove that 2x + 75 25 95 100 50 75 25 10 For y = black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_20\620IB_HL-2_20.CDR Friday, 12 March 2010 3:43:04 PM PETER IB_HL-2ed (621) 21 Chapter Applications of differential calculus Contents: A B C D E F G H Time rate of change General rates of change Motion in a straight line Some curve properties Rational functions Inflections and shape Optimisation Implicit differentiation cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 21A Review set 21B Review set 21C black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\621IB_HL-2_21.CDR Thursday, 15 November 2007 12:34:12 PM PETERDELL IB_HL-2ed (622) 622 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) We saw in the previous chapter that one application of differential calculus is in finding the equations of tangents and normals to curves There are many other uses, however, including the following which we will consider in this chapter: ² ² ² ² functions of time curve properties ² ² rates of change optimisation A motion in a straight line applications in economics TIME RATE OF CHANGE There are countless quantities in the real world that vary with time ² ² ² For example: temperature varies continuously the height of a tree varies as it grows the prices of stocks and shares vary with each day’s trading Varying quantities can be modelled using functions of time For example, we could use: ² s(t) to model the distance travelled by a runner ² H(t) to model the height of a person riding in a Ferris wheel ² C(t) to model the capacity of a person’s lungs, which changes when the person breathes We saw in the previous chapter that if y = f(x) then dy f (x) or is the gradient of the tangent at any value dx of x, and also the rate of change in y with respect to x y y¡=¡ƒ(x) We can likewise find the derivative of a function of time to tell us the rate at which something is happening For the examples above: x ² ds or s0 (t) is the instantaneous speed of the runner It might have units metres per dt second or m s¡1 ² dH or H (t) is the instantaneous rate of ascent of the person in the Ferris wheel dt It might also have units metres per second or m s¡1 ² dC or C (t) is the person’s instantaneous rate of change in lung capacity It might dt have units litres per second or L s¡1 EXERCISE 21A cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The estimated future profits of a small business are given by P (t) = 2t2 ¡ 12t + 118 thousand dollars, where t is the time in years from now dP a What is the current annual profit? b Find and state its units dt black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_21\622IB_HL-2_21.CDR Friday, 12 March 2010 3:57:27 PM PETER IB_HL-2ed (623) 623 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) c What is the significance of dP ? dt d When will the profit i decrease ii increase? e What is the minimum profit and when does it occur? dP at t = 4, 10 and 25 What these figures represent? f Find dt Water is draining from a swimming pool such that the remaining volume of water after t minutes is V = 200(50 ¡ t)2 m3 Find: a the average rate at which the water leaves the pool in the first minutes b the instantaneous rate at which the water is leaving at t = minutes A ball is thrown vertically and its height above the ground is given by s(t) = 1:2 + 28:1t ¡ 4:9t2 metres where t is the time in seconds a b c d e From what distance above the ground was the ball released? Find s0 (t) and state what it represents Find t when s0 (t) = What is the significance of this result? What is the maximum height reached by the ball? Find the ball’s speed: i when released ii at t = s iii State the significance of the sign of the derivative f How long will it take for the ball to hit the ground? d2 s ? g What is the significance of dt2 at t = s A shell is accidentally fired vertically from a mortar at ground level and reaches the ground again after 14:2 seconds a Given that its height above the ground at any time t seconds is given by s(t) = bt ¡ 4:9t2 metres, show that the initial velocity of the shell is b m s¡1 b Find the initial velocity of the shell B GENERAL RATES OF CHANGE Earlier we discovered that: if s(t) is a distance function then s0 (t) or ds is the dt instantaneous rate of change in distance with respect to time, which is speed dy gives the rate of change in y with respect to x dx cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 dy will be negative dx 25 If y decreases as x increases, then dy will be positive dx 95 Note: If y increases as x increases, then 100 50 75 25 In general, black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\623IB_HL-2_21.CDR Thursday, 15 November 2007 12:52:31 PM PETERDELL IB_HL-2ed (624) 624 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) Example According to a psychologist, the ability of a person to understand spatial concepts p is given by A = 13 t where t is the age in years, t [5, 18] a b Find the rate of improvement in ability to understand spatial concepts when a person is: i years old ii 16 years old dA Explain why > 0, t [5, 18] Comment on the significance of this result dt c Explain why a A= ii As p t = 13 t ¡1 dA = 16 t = p dt t dA 1 When t = 9, ) the rate of improvement is 18 = 18 dt units per year for a year old dA 1 When t = 16, ) the rate of improvement is 24 = 24 dt units per year for a 16 year old i b d2 A < 0, t [5, 18] Comment on the significance of this result dt2 ) p t is never negative, p t is never negative dA > for all t [5, 18] dt This means that the ability to understand spatial concepts increases with age ) c d2 A dA 1 ¡ 32 = ¡ 12 t =¡ p = 16 t¡ so dt dt 12t t d A ) < for all t [5, 18] dt2 This means that while the ability to understand spatial concepts increases with time, the rate of increase slows down with age Example The cost of producing x items in a factory each day is given by C(x) = 0:000 13x3 + 0:002x2 + 5x + 2200 cost of labour cyan fixed or overhead costs such as heating, cooling, maintenance, rent magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find C (x), which is called the marginal cost function Find the marginal cost when 150 items are produced Interpret this result Find C(151) ¡ C(150) Compare this with the answer in b 95 100 50 75 25 a b c raw material costs black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\624IB_HL-2_21.CDR Monday, 26 November 2007 3:22:12 PM PETERDELL IB_HL-2ed (625) 625 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) a The marginal cost function is C (x) = 0:000 39x2 + 0:004x + b C (150) = $14:38 This is the rate at which the costs are increasing with respect to the production level x when 150 items are made per day It gives an estimate of the cost for making the 151st item chord (c answer) C(151) tangent (b answer) C(150) 150 151 C(151) ¡ C(150) ¼ $3448:19 ¡ $3433:75 ¼ $14:44 This is the actual cost of making the 151st item each week, so the answer in b gives a good estimate c EXERCISE 21B You are encouraged to use technology to graph the function for each question This is often useful in interpreting results The quantity of a chemical in human skin which is responsible for its ‘elasticity’ is given p by Q = 100 ¡ 10 t where t is the age of a person in years a Find Q at: i t=0 ii t = 25 iii t = 100 b At what rate is the quantity of the chemical changing at the ages of: i 25 years ii 50 years? c Show that the rate at which the skin loses the chemical is decreasing for all t > 97:5 t+5 metres, where t is the number of years after the tree was planted from an established seedling a How high is the tree at planting? b Find the height of the tree at t = 4, t = and t = 12 years c Find the rate at which the tree is growing at t = 0, and 10 years dH d Show that > for all t > What is the significance of this result? dt The height of pinus radiata, grown in ideal conditions, is given by H = 20 ¡ The total cost of running a train from Paris to Marseille is given by 200 000 euros where v is the average speed of the train in km h¡1 C(v) = 15 v2 + v a Find the total cost of the journey if the average speed is: ii 100 km h¡1 i 50 km h¡1 b Find the rate of change in the cost of running the train at speeds of: i 30 km h¡1 ii 90 km h¡1 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 c At what speed will the cost be a minimum? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\625IB_HL-2_21.CDR Thursday, 15 November 2007 1:55:15 PM PETERDELL IB_HL-2ed (626) 626 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) Alongside is a land and sea profile where the x-axis is sea level The function x(x ¡ 2)(x ¡ 3) km gives the y = 10 y sea hill lake x height of the land or sea bed relative to sea level a Find where the lake is located relative to the shore line of the sea dy b Find and interpret its value when x = 12 and when x = 12 km dx c Find the deepest point of the lake and the depth at this point A tank contains 50 000 litres of water The tap is left fully on and all the water drains from the tank in 80 minutes The volume of water remaining in the tank after t minutes µ ¶2 t where t 80 is given by V = 50 000 ¡ 80 dV dV and draw the graph of against t a Find dt dt b At what time was the outflow fastest? c Show that d2 V is always constant and positive Interpret this result dt2 A fish farm grows and harvests barramundi in a large dam The population of fish after t years is given by the function P (t) The rate of change in the population µ ¶ ³ dP dP P c ´ is modelled by = aP ¡ ¡ P dt dt b 100 where a, b and c are known constants a is the birth rate of the barramundi, b is the maximum carrying capacity of the dam and c is the percentage that is harvested each year dP = a Explain why the fish population is stable when dt b If the birth rate is 6%, the maximum carrying capacity is 24 000, and 5% is harvested each year, find the stable population c If the harvest rate changes to 4%, what will the stable population increase to? Seablue make denim jeans The cost model for making x pairs per day is C(x) = 0:0003x3 + 0:02x2 + 4x + 2250 dollars cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find the marginal cost function C (x): Find C (220) What does it estimate? Find C(221) ¡ C(220) What does this represent? Find C 00 (x) and the value of x when C 00 (x) = What is the significance of this point? 95 100 50 75 25 a b c d black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\626IB_HL-2_21.CDR Thursday, 15 November 2007 1:57:34 PM PETERDELL IB_HL-2ed (627) 627 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) C MOTION IN A STRAIGHT LINE DISPLACEMENT Suppose an object P moves along a straight line so that its position s from an origin O is given as some function of time t, i.e., s = s(t) where t > s(t) O origin P s(t) is a displacement function and for any value of t it gives the displacement from O s(t) is a vector quantity Its magnitude is the distance from O, and its sign indicates the direction from O if s(t) > 0, if s(t) = 0, if s(t) < 0, It is clear that P is located to the right of O P is located at O P is located to the left of O MOTION GRAPHS Consider s(t) = t2 + 2t ¡ cm s(0) = ¡3 cm, s(1) = cm, s(2) = cm, s(3) = 12 cm, s(4) = 21 cm To appreciate the motion of P we draw a motion graph DEMO t=1 t=0 t=2 10 15 20 t=4 t=3 25 Click on the demo icon to get a better idea of the motion Fully animated, we not only get a good idea of the position of P, but also of what is happening to its velocity and acceleration VELOCITY AND ACCELERATION AVERAGE VELOCITY The average velocity of an object moving in a straight line in the time interval from t = t1 to t = t2 is the ratio of the change in displacement to the time taken s(t2 ) ¡ s(t1 ) , where s(t) is the displacement function t2 ¡ t1 On a graph of s(t), the average velocity is the gradient of a chord average velocity = INSTANTANEOUS VELOCITY s(1 + h) ¡ s(1) approached a fixed value as h h approached and this value must be the instantaneous velocity at t = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 In Chapter 19 we established that black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_21\627IB_HL-2_21.CDR Friday, 12 March 2010 3:58:06 PM PETER IB_HL-2ed (628) 628 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) If s(t) is a displacement function of an object moving in a straight line, then s(t + h) ¡ s(t) h velocity function of the object at time t v(t) = s0 (t) = lim h! is the instantaneous velocity or On a graph of s(t), the instantaneous velocity is the gradient of a tangent AVERAGE ACCELERATION If an object moves in a straight line with velocity function v(t) then its average acceleration on the time interval from t = t1 to t = t2 is the ratio of the change in velocity to the time taken average acceleration = v(t2 ) ¡ v(t1 ) t2 ¡ t1 INSTANTANEOUS ACCELERATION If a particle moves in a straight line with velocity function v(t), then the instantaneous acceleration at time t is a(t) = v0 (t) = lim h! v(t + h) ¡ v(t) h Example A particle moves in a straight line with displacement from O given by s(t) = 3t ¡ t2 metres at time t seconds Find: a the average velocity in the time interval from t = to t = seconds b the average velocity in the time interval from t = to t = + h seconds c lim h!0 a s(2 + h) ¡ s(2) h and comment on its significance b average velocity s(5) ¡ s(2) = 5¡2 (15 ¡ 25) ¡ (6 ¡ 4) ¡10 ¡ = = ¡4 m s¡1 = c lim h!0 average velocity s(2 + h) ¡ s(2) = 2+h¡2 = 3(2 + h) ¡ (2 + h)2 ¡ h + 3h ¡ ¡ 4h ¡ h2 ¡ h ¡h ¡ h2 = h = ¡1 ¡ h m s¡1 provided h 6= = s(2 + h) ¡ s(2) = lim (¡1 ¡ h) h!0 h = ¡1 m s¡1 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 This is the instantaneous velocity of the particle at time t = seconds black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_21\628IB_HL-2_21.CDR Friday, 12 March 2010 3:59:30 PM PETER IB_HL-2ed (629) 629 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) EXERCISE 21C.1 A particle P moves in a straight line with a displacement function of s(t) = t2 + 3t ¡ metres, where t > 0, t in seconds a Find the average velocity from t = to t = seconds b Find the average velocity from t = to t = + h seconds s(1 + h) ¡ s(1) and comment on its significance h d Find the average velocity from time t to time t + h seconds and interpret c Find the value of lim h!0 lim h!0 s(t + h) ¡ s(t) h A particle P moves in a straight line with a displacement function of s(t) = ¡ 2t2 cm, where t > 0, t in seconds a Find the average velocity from t = to t = seconds b Find the average velocity from t = to t = + h seconds c Find the value of lim h!0 d Interpret lim h!0 s(2 + h) ¡ s(2) and state the meaning of this value h s(t + h) ¡ s(t) h p A particle moves in a straight line with velocity function v(t) = t + cm s¡1 , t > a Find the average acceleration from t = to t = seconds b Find the average acceleration from t = to t = + h seconds c Find the value of d Interpret lim h!0 lim h!0 v(1 + h) ¡ v(1) Interpret this value h v(t + h) ¡ v(t) h An object moves in a straight line with displacement function s(t) and velocity function v(t), t > State the meaning of: a lim h!0 s(4 + h) ¡ s(4) h b lim h!0 v(4 + h) ¡ v(4) h VELOCITY AND ACCELERATION FUNCTIONS If a particle P moves in a straight line and its position is given by the displacement function s(t), t > 0, then: ² the velocity of P at time t is given by v(t) = s0 (t) fthe derivative of the displacement functiong cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ² the acceleration of P at time t is given by a(t) = v (t) = s00 (t) fthe derivative of the velocity functiong black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\629IB_HL-2_21.CDR Thursday, 15 November 2007 2:21:23 PM PETERDELL IB_HL-2ed (630) 630 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) ² s(0), v(0) and a(0) give us the position, velocity and acceleration of the particle at time t = 0, and these are called the initial conditions SIGN INTERPRETATION Suppose a particle P moves in a straight line with displacement function s(t) relative to an origin O Its velocity function is v(t) and its acceleration function is a(t) We can use sign diagrams to interpret: ² ² ² where the particle is located relative to O the direction of motion and where a change of direction occurs when the particle’s velocity is increasing or decreasing SIGNS OF s(t): s(t) =0 >0 <0 Interpretation P is at O P is located to the right of O P is located to the left of O v(t) = lim h!0 If v(t) > then s(t + h) ¡ s(t) > ) s(t + h) > s(t) SIGNS OF v(t): v(t) =0 >0 <0 For h > the particle is moving from s(t) to s(t + h) Interpretation P is instantaneously at rest P is moving to the right P is moving to the left SIGNS OF a(t): a(t) >0 <0 =0 s(t + h) ¡ s(t) h s(t) s(t + h) ) P is moving to the right Interpretation velocity is increasing velocity is decreasing velocity may be a maximum or minimum A useful table: Phrase used in a question initial conditions at the origin stationary reverses maximum height constant velocity max or velocity t s v a 0 0 When a particle reverses direction, its velocity must change sign 0 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 We need a sign diagram of a to determine if the point is a maximum or minimum black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\630IB_HL-2_21.CDR Monday, 26 November 2007 3:34:43 PM PETERDELL IB_HL-2ed (631) 631 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) SPEED As we have seen, velocities have size (magnitude) and sign (direction) The speed of an object is a measure of how fast it is travelling regardless of the direction of travel the speed at any instant is the modulus of the object’s velocity, i.e., if S(t) represents speed then S = jvj Thus To determine when the speed of an object is increasing or decreasing, we need to employ a sign test This is: ² If the signs of v(t) and a(t) are the same (both positive or both negative), then the speed of P is increasing If the signs of v(t) and a(t) are opposite, then the speed of P is decreasing ² We prove the first of these as follows: ½ Proof: Let S = jvj be the speed of P at any instant ) S = Case 1: Case 2: v ¡v if v > if v < dS dv = = a(t) dt dt dS > which implies that S is increasing If a(t) > then dt dv dS =¡ = ¡a(t) If v < 0, S = ¡v and ) dt dt dS > which also implies that S is increasing If a(t) < then dt If v > 0, S = v and ) Thus if v(t) and a(t) have the same sign then the speed of P is increasing DISPLACEMENT, VELOCITY AND ACCELERATION GRAPHS INVESTIGATION In this investigation we examine the motion of a projectile which is fired in a vertical direction The projectile is affected by gravity, which is responsible for the projectile’s constant acceleration MOTION DEMO We then extend the investigation to consider other cases of motion in a straight line What to do: Click on the icon to examine vertical projectile motion in a straight line Observe first the displacement along the line, then look at the velocity or rate of change in displacement ² displacement v time ² acceleration v time Comment on the shape of these graphs Examine the three graphs ² velocity v time Pick from the menu or construct functions of your own choosing to investigate the relationship between displacement, velocity and acceleration cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 You are encouraged to use the motion demo above to answer questions in the following exercise black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\631IB_HL-2_21.CDR Thursday, 15 November 2007 2:35:19 PM PETERDELL IB_HL-2ed (632) 632 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) Example A particle moves in a straight line with position relative to some origin O given by s(t) = t3 ¡ 3t + cm, where t is the time in seconds (t > 0) a Find expressions for the particle’s velocity and acceleration, and draw sign diagrams for each of them b Find the initial conditions and hence describe the motion at this instant c Describe the motion of the particle at t = seconds d Find the position of the particle when changes in direction occur e Draw a motion diagram for the particle f For what time interval(s) is the particle’s speed increasing? Note: t > g What is the total distance travelled for t [0, 2]? ) critical value a t = ¡1 is not s(t) = t3 ¡ 3t + cm required fas v(t) = s0 (t)g ) v(t) = 3t2 ¡ = 3(t2 ¡ 1) - + v(t) = 3(t + 1)(t ¡ 1) cm s¡1 which has sign diagram t and a(t) = 6t cm s¡2 fas a(t) = v (t)g which has sign diagram + a(t) t b When t = 0, s(0) = cm ¡1 v(0) = ¡3 cm s a(0) = cm s¡2 ) the particle is cm to the right of O, moving to the left at a speed of cm s¡1 c s(2) = ¡ + = cm v(2) = 12 ¡ = cm s¡1 a(2) = 12 cm s¡2 ) the particle is cm to the right of O, moving to the right at a speed of cm s¡1 Since a and v have the same sign, the speed is increasing When t = 2, d Since v(t) changes sign when t = 1, a change of direction occurs at this instant s(1) = ¡ + = ¡1, so the particle changes direction when it is cm left of O e -1 Note: The motion is actually on the line, not above it as shown as t ! 1, s(t) ! and v(t) ! Speed is increasing when v(t) and a(t) have the same sign, i.e., t > f g Total distance travelled = + = cm cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Note: In later chapters on integral calculus another technique for finding the distances travelled and displacement over time will be explored black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\632IB_HL-2_21.CDR Thursday, 15 November 2007 2:43:31 PM PETERDELL IB_HL-2ed (633) 633 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) EXERCISE 21C.2 (Use a graphics calculator to check sign diagrams.) An object moves in a straight line with position given by s(t) = t2 ¡ 4t + cm from an origin O, where t is in seconds, t > a Find expressions for the object’s velocity and acceleration at any instant and draw sign diagrams for each function b Find the initial conditions and explain what is happening to the object at that instant c Describe the motion of the object at time t = seconds d At what time(s) does the object reverse direction? Find the position of the object at these instants e Draw a motion diagram for the object f For what time intervals is the speed of the object decreasing? A stone is projected vertically so that its position above ground level after t seconds is given by s(t) = 98t ¡ 4:9t2 metres, t > a Find the velocity and acceleration functions for the stone and draw sign diagrams for each function b Find the initial position and velocity of the stone c Describe the stone’s motion at times t = and t = 12 seconds d Find the maximum height reached by the stone e Find the time taken for the stone to hit the ground s(t) m A particle moves in a straight line with displacement function s(t) = 12t ¡ 2t3 ¡ centimetres where t is in seconds, t > a Find velocity and acceleration functions for the particle’s motion b Find the initial conditions and interpret their meaning c Find the times and positions when the particle reverses direction d At what times is the particle’s: i speed increasing ii velocity increasing? The position of a particle moving along the x-axis is given by x(t) = t3 ¡ 9t2 + 24t metres where t is in seconds, t > a Draw sign diagrams for the particle’s velocity and acceleration functions b Find the position of the particle at the times when it reverses direction, and hence draw a motion diagram for the particle c At what times is the particle’s: i speed decreasing ii velocity decreasing? d Find the total distance travelled by the particle in the first seconds of motion cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 5 An experiment to determine the position of an s(t) object fired vertically from the earth’s surface was performed From the results, a two dimensional graph of the position above the earth’s surface s(t) metres was plotted, where t was the time in t seconds It was noted that the graph was parabolic Assuming a constant gravitational acceleration g, When finding the total distance show that if the initial velocity is v(0) then: travelled, always look for direction reversals first b s(t) = v(0) £ t + 12 gt2 a v(t) = v(0) + gt black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\633IB_HL-2_21.CDR Thursday, 15 November 2007 3:00:20 PM PETERDELL IB_HL-2ed (634) 634 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) D SOME CURVE PROPERTIES In this section we consider some properties of curves which can be established using derivatives These include intervals in which curves are increasing and decreasing, and the stationary points of functions INCREASING AND DECREASING INTERVALS The concepts of increasing and decreasing are closely linked to intervals of a function’s domain Some examples of intervals and their graphical representations are: Algebraic form Means Alternative notation x>4 x [ 4, [ x>4 x ] 4, [ x62 x ] ¡1, ] x<2 x ] ¡1, [ 26x64 x [ 2, ] 26x<4 x [ 2, [ Suppose S is an interval in the domain of f (x), so f (x) is defined for all x in S ² f (x) is increasing on S , f(a) < f (b) for all a, b S such that a < b ² f (x) is decreasing on S , f(a) > f (b) for all a, b S such that a < b Reminder: , is read “if and only if” For example: y = x2 is decreasing for x and increasing for x > y @=!X x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Note: People often get confused about the point x = They wonder how the curve can be both increasing and decreasing at the same point when it is clear that the tangent is horizontal The answer is that increasing and decreasing are associated with intervals, not particular values for x We must clearly state that y = x2 is decreasing on the interval x and increasing on the interval x > black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\634IB_HL-2_21.CDR Thursday, 15 November 2007 3:09:19 PM PETERDELL IB_HL-2ed (635) 635 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) We can deduce when a curve is increasing or decreasing by considering f (x) on the interval in question For most functions that we deal with in this course: ² ² f (x) > for all x in S f (x) for all x in S , , f (x) is increasing on S f (x) is decreasing on S MONOTONICITY Many functions are either increasing or decreasing for all x R We say these functions are monotone increasing or monotone decreasing For example: y y y = -x y = 2x 1 x x y = 3¡x is decreasing for all x y = 2x is increasing for all x Notice that: ² ² for an increasing function, an increase in x produces an increase in y for a decreasing function, an increase in x produces a decrease in y decrease in y increase in y increase in x increase in x Example Find intervals where f(x) is: a increasing b decreasing (-1, 3) y y = ƒ(x) x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 f(x) is decreasing for ¡1 x 100 b 50 f(x) is increasing for x ¡1 and for x > since all tangents have gradients > on these intervals 75 a 25 95 100 50 75 25 (2,-4) black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_21\635IB_HL-2_21.CDR Friday, 12 March 2010 4:00:13 PM PETER IB_HL-2ed (636) 636 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) Sign diagrams for the derivative are extremely useful for determining intervals where a function is increasing or decreasing Consider the following examples: f(x) = x2 ² f (x) = 2x y which has sign diagram - DEMO decreasing x f(x) = ¡x2 ² y is and f (x) = ¡2x which has sign diagram DEMO + f(x) = x3 f (x) = 3x2 + increasing for all x (never negative) f(x) = x3 ¡ 3x + f (x) = 3x2 ¡ = 3(x2 ¡ 1) = 3(x + 1)(x ¡ 1) which has sign diagram DEMO + -1 decreasing + DEMO y which has sign diagram x ² decreasing for x increasing for x > increasing y increasing So f(x) = x2 x ² + - + -1 increasing decreasing increasing x Example Find the intervals where the following functions are increasing or decreasing: b f (x) = 3x4 ¡ 8x3 + a f(x) = ¡x3 + 3x2 + a f (x) = ¡x3 + 3x2 + ) f (x) = ¡3x2 + 6x ) f (x) = ¡3x(x ¡ 2) which has sign diagram - + - cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 So, f (x) is decreasing for x and for x > and is increasing for x black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\636IB_HL-2_21.CDR Monday, 26 November 2007 3:38:59 PM PETERDELL IB_HL-2ed (637) 637 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) f (x) = 3x4 ¡ 8x3 + ) f (x) = 12x3 ¡ 24x2 = 12x2 (x ¡ 2) which has sign diagram b - - + So, f (x) is decreasing for x and is increasing for x > Remember that f (x) must be defined for all x on an interval before we can classify the interval as increasing or decreasing We must exclude points where a function is undefined, and need to take care with vertical asymptotes Example Consider f (x) = 3x ¡ x2 ¡ x ¡ ¡3(x ¡ 5)(x ¡ 1) (x ¡ 2)2 (x + 1)2 a Show that f (x) = b Hence, find intervals where y = f (x) is increasing or decreasing a f(x) = f (x) = and draw its sign diagram 3x ¡ ¡x¡2 x2 3(x2 ¡ x ¡ 2) ¡ (3x ¡ 9)(2x ¡ 1) (x ¡ 2)2 (x + 1)2 = 3x2 ¡ 3x ¡ ¡ [6x2 ¡ 21x + 9] (x ¡ 2)2 (x + 1)2 = ¡3x2 + 18x ¡ 15 (x ¡ 2)2 (x + 1)2 = ¡3(x2 ¡ 6x + 5) (x ¡ 2)2 (x + 1)2 = ¡3(x ¡ 5)(x ¡ 1) (x ¡ 2)2 (x + 1)2 fquotient ruleg which has sign diagram cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 f(x) is increasing for x < and for < x f(x) is decreasing for x < ¡1 and for ¡1 < x and for x > 75 25 95 100 50 75 25 b ++ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\637IB_HL-2_21.CDR Thursday, 15 November 2007 3:34:24 PM PETERDELL IB_HL-2ed (638) 638 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) EXERCISE 21D.1 Find intervals where f(x) is a i b y ii increasing decreasing: c y y (2, 3) (-2, 2) x x x d (3,-1) e y f y y x=4 (5, 2) x x (2, 4) (1, -1) x Find intervals where f(x) is increasing or decreasing: b f(x) = ¡x3 a f (x) = x2 p c f (x) = 2x2 + 3x ¡ d f(x) = x e f (x) = p f f(x) = x3 ¡ 6x2 x f (x) = ¡2x3 + 4x f (x) = 3x4 ¡ 16x3 + 24x2 ¡ f (x) = x3 ¡ 6x2 + 3x ¡ f (x) = 3x4 ¡ 8x3 ¡ 6x2 + 24x + 11 g i k m h j l n f(x) = ¡4x3 + 15x2 + 18x + f(x) = 2x3 + 9x2 + 6x ¡ p f(x) = x ¡ x f(x) = x4 ¡ 4x3 + 2x2 + 4x + 4x : +1 ¡4(x + 1)(x ¡ 1) i Show that f (x) = (x2 + 1)2 a Consider f (x) = x2 and draw its sign diagram ii Hence, find intervals where y = f(x) is increasing or decreasing 4x (x ¡ 1)2 ¡4(x + 1) i Show that f (x) = (x ¡ 1)3 b Consider f (x) = and draw its sign diagram ii Hence, find intervals where y = f(x) is increasing or decreasing ¡x2 + 4x ¡ x¡1 ¡(x + 1)(x ¡ 3) i Show that f (x) = (x ¡ 1)2 c Consider f (x) = and draw its sign diagram cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ii Hence, find intervals where y = f(x) is increasing or decreasing black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\638IB_HL-2_21.CDR Thursday, 15 November 2007 3:35:31 PM PETERDELL IB_HL-2ed (639) 639 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) Find intervals where f(x) is increasing or decreasing if: x3 a f(x) = b f (x) = x2 + x ¡1 x¡1 STATIONARY POINTS A stationary point of a function is a point such that f (x) = MAXIMA AND MINIMA y Consider the following graph which has a restricted domain of ¡5 x 6 D(6,¡18) B(-2,¡4) -2 x C(2,-4) y¡=¡ƒ(x) A(-5,-16\Qw_\) A is a global minimum as it is the minimum value of y on the entire domain B is a local maximum as it is a turning point where the curve has shape at that point and f (x) = C is a local minimum as it is a turning point where the curve has shape at that point and f (x) = D is a global maximum as it is the maximum value of y on the entire domain Note: ² Local maxima and minima are stationary points where f (x) = The tangents at these points are horizontal ² It is possible for a local maximum or minimum to also be the global maximum or minimum of a function For example, for y = x2 the point (0, 0) is a local minimum and is also the global minimum HORIZONTAL INFLECTIONS OR STATIONARY POINTS OF INFLECTION (SPI) It is not always true that whenever we find a value of x where f (x) = we have a local maximum or minimum For example, f(x) = x3 has f (x) = 3x2 and f (x) = when x = 0: Notice that the x-axis is a tangent to the curve which actually crosses over the curve at O(0, 0) This tangent is horizontal but O(0, 0) is neither a local maximum nor a local minimum cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 It is called a horizontal inflection (or inflexion) as the curve changes its curvature or shape black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\639IB_HL-2_21.CDR Friday, 16 November 2007 2:08:27 PM PETERDELL y x IB_HL-2ed (640) 640 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) SUMMARY y A stationary point is a point where f (x) = It could be a local maximum, local minimum, or a horizontal inflection (SPI) horizontal inflection local maximum -2 Consider the following graph: local minimum The sign diagram of its gradient function is: + -2 local maximum Sign diagram of f (x) near x = a Stationary point + local maximum + a + + local horizontal minimum inflection Shape of curve near x = a x=a x=a + a + - - a - local minimum horizontal inflection or stationary inflection x - or a - or x=a x=a Example Find and classify all stationary points of f (x) = x3 ¡ 3x2 ¡ 9x + 5: f (x) = x3 ¡ 3x2 ¡ 9x + ) f (x) = 3x2 ¡ 6x ¡ = 3(x2 ¡ 2x ¡ 3) = 3(x ¡ 3)(x + 1) which has sign diagram: + + -1 So, we have a local maximum at x = ¡1 and a local minimum at x = f (¡1) = (¡1)3 ¡ 3(¡1)2 ¡ 9(¡1) + = 10 f (3) = 33 ¡ £ 32 ¡ £ + = ¡22 cyan magenta yellow 95 100 50 75 25 95 100 50 ) local minimum at (3, ¡22) 75 25 95 100 50 75 25 95 100 50 75 25 ) local maximum at (¡1, 10) black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_21\640IB_HL-2_21.CDR Friday, 12 March 2010 4:01:18 PM PETER IB_HL-2ed (641) 641 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) If we are asked to find the greatest or least value on an interval, then we should always check the endpoints We seek the global maximum or minimum on the given domain Example Find the greatest and least value of x3 ¡ 6x2 + on the interval ¡2 x First we graph y = x3 ¡ 6x2 + on [¡2, 5] In this case the greatest value is clearly at the dy local maximum when = dx dy = 3x2 ¡ 12x Now dx = 3x(x ¡ 4) = when x = or So, the greatest value is f(0) = when x = The least value is either f (¡2) or f (4), whichever is smaller Now f (¡2) = ¡27 and f(4) = ¡27 ) least value is ¡27 when x = ¡2 and x = EXERCISE 21D.2 B(-2,¡8) y y¡=¡ƒ(x) The tangents at points A, B and C are horizontal a Classify points A, B and C -4 b Draw a sign diagram for the gradient C function f (x) for all x c State intervals where y = f (x) is: i increasing ii decreasing A(3,¡-11) d Draw a sign diagram for f (x) for all x e Comment on the differences between the sign diagrams found above x For each of the following functions, find and classify the stationary points, and hence sketch the function showing all important features b f (x) = x3 + a f(x) = x2 ¡ d f (x) = x4 ¡ 2x2 c f(x) = x3 ¡ 3x + p e f(x) = x ¡ 6x + 12x + f f (x) = x + p h f (x) = x4 ¡ 6x2 + 8x ¡ g f(x) = x ¡ x p j f (x) = x4 ¡ 2x2 ¡ i f(x) = ¡ x x At what value of x does the quadratic function f (x) = ax2 + bx + c, a 6= 0, have a stationary point? Under what conditions is the stationary point a local maximum or a local minimum? f(x) = 2x3 + ax2 ¡ 24x + has a local maximum at x = ¡4 Find a: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 5 f(x) = x3 + ax + b has a stationary point at (¡2, 3) a Find the values of a and b b Find the position and nature of all stationary points black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_21\641IB_HL-2_21.CDR Friday, 12 March 2010 4:01:56 PM PETER IB_HL-2ed (642) 642 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) A cubic polynomial P (x) touches the line with equation y = 9x + at the point (0, 2), and has a stationary point at (¡1, ¡7) Find P (x) Find the greatest and least value of: a x3 ¡ 12x ¡ for ¡3 x ¡ 3x2 + x3 b for ¡2 x A manufacturing company makes door hinges The cost function for making x hinges per hour is C(x) = 0:0007x3 ¡0:1796x2 +14:663x+160 dollars where 50 x 150 The condition 50 x 150 applies as the company has a standing order filled by producing 50 each hour, but knows that production of more than 150 per hour is useless as they will not sell Find the minimum and maximum hourly costs and the production levels when each occurs E RATIONAL FUNCTIONS Rational functions have the form f(x) = For example, f(x) = 2x ¡ x2 + g(x) h(x) and f (x) = where g(x) and h(x) are polynomials x2 ¡ x2 ¡ 3x + are rational functions We have seen that one feature of a rational function is the presence of asymptotes These are lines (or curves) that the graph of the function approaches when x or y takes large values Vertical asymptotes can be found by solving h(x) = Horizontal asymptotes can be found by finding what value f(x) approaches as j x j ! Oblique asymptotes are neither horizontal nor vertical They can be found using the division process and then finding what function f(x) approaches as j x j ! When finding the position and nature of stationary points for rational functions, we usually begin by using the quotient rule to find the derivative Rational functions of the form y = linear linear FUNCTIONS OF THE FORM were covered earlier in this text y= linear quadratic For these functions the order of the polynomial in the denominator is higher than that in the numerator cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 As j x j ! 1, f (x) ! 0, and so they all have the horizontal asymptote y = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\642IB_HL-2_21.CDR Monday, 26 November 2007 3:39:54 PM PETERDELL IB_HL-2ed (643) 643 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) Example 10 Consider f (x) = 3x ¡ ¡x¡2 x2 a b c Determine the equations of any asymptotes Find f (x) and determine the position and nature of any stationary points Find the axes intercepts d Sketch the graph of the function a f(x) = 3x ¡ 3x ¡ = x2 ¡ x ¡ (x ¡ 2)(x + 1) Vertical asymptotes are x = and x = ¡1 fwhen the denominator is 0g Horizontal asymptote is y = fas jxj ! 1, f(x) ! 0g b f (x) = = 3(x2 ¡ x ¡ 2) ¡ (3x ¡ 9)(2x ¡ 1) (x ¡ 2)2 (x + 1)2 3x2 ¡ 3x ¡ ¡ [6x2 ¡ 21x + 9] (x ¡ 2)2 (x + 1)2 ¡3x2 + 18x ¡ 15 = (x ¡ 2)2 (x + 1)2 So, f (x) has sign diagram: - -+ -1 + ) a local maximum when x = and a local minimum when x = ¡3(x2 ¡ 6x + 5) = (x ¡ 2)2 (x + 1)2 The local max is (5, 13 ) ¡3(x ¡ 5)(x ¡ 1) = (x ¡ 2)2 (x + 1)2 c fQuotient ruleg The local is (1, 3) y d Cuts the x-axis when y = ) 3x ¡ = or x = 4\Qw_ min(1' 3) So, the x-intercept is max (5' Qe_) y¡=¡0 Cuts the y-axis when x = ¡9 ) y= = 12 ¡2 So, the y-intercept is 12 x x¡=¡-1 x¡=¡2 quadratic quadratic y= FUNCTIONS OF THE FORM Sign diagrams must show vertical asymptotes 2x2 ¡ x + have a horizontal asymptote x2 + x ¡ which can be found by dividing every term by x2 Functions such as y = + x x Notice that y = 1+ ¡ x x cyan magenta yellow 95 =2 100 50 75 25 95 so as j x j ! 1, y ! 100 50 75 25 95 100 50 75 25 95 100 50 75 25 2¡ black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_21\643IB_HL-2_21.CDR Thursday, 11 March 2010 10:44:55 AM PETER IB_HL-2ed (644) 644 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) Alternatively, by the division process ¡ 3x y = 2+ ! as j x j ! x +x¡2 2x ¡ x + 2x2 + 2x ¡ ¡3x + 2 x +x¡2 Example 11 For f (x) = a b c d a x2 ¡ 3x + x2 + 3x + Determine the equations of its asymptotes Find f (x) and determine the position and nature of any turning points Find the axes intercepts Sketch the graph of the function 1¡ x2 ¡ 3x + = f(x) = x + 3x + 1+ x ¡ 3x + ) f(x) = (x + 1)(x + 2) f (x) = b + so as j x j ! 1, y ! x x ) HA is y = + x x2 vertical asymptotes are x = ¡1 and x = ¡2 (2x ¡ 3)(x2 + 3x + 2) ¡ (x2 ¡ 3x + 2)(2x + 3) (x + 1)2 (x + 2)2 6x2 ¡ 12 fon simplifyingg (x + 1)2 (x + 2)2 -~`2 ~`2 p p 6(x + 2)(x ¡ 2) and has ++ - + = 2 sign diagram (x + 1) (x + 2) -2 -1 p p So, we have a local maximum at x = ¡ and a local minimum at x = p p The local max is (¡ 2, ¡33:971) The local is ( 2, ¡0:029) ) f (x) = c d Cuts the x-axis when y = ) x ¡ 3x + = ) (x ¡ 1)(x ¡ 2) = ) x = or So, the x-intercepts are and y y¡=¡1 local max (-~`2' -33"971) Cuts the y-axis when x = ) y = 22 = x local (~`2' -0"029) So, the y-intercept is cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 x¡=¡-2 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\644IB_HL-2_21.CDR Friday, 16 November 2007 2:10:20 PM PETERDELL x¡=¡-1 IB_HL-2ed (645) 645 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) quadratic linear y= FUNCTIONS OF THE FORM For these functions the order of the polynomial in the numerator is higher than that in the denominator This results in an oblique asymptote which is found by the division process For example, y = As j x j ! 1, x2 + 2x ¡ =x¡1+ x+3 x+3 on division ! and so y ! x ¡ Thus y = x ¡ is an oblique asymptote x+3 Example 12 ¡x2 + 4x ¡ : x¡1 For f (x) = a b c Determine the equation of its asymptotes Find f (x) and determine the position and nature of any turning points Find the axes intercepts d Sketch the graph of the function a f(x) = ¡x2 + 4x ¡ = ¡x + ¡ x¡1 x¡1 ¡1 ¡1 ¡7 ¡4 ¡1 ) a vertical asymptote is x = fas x ! 1, j f(x) j ! 1g and an oblique asymptote is y = ¡x + fas j x j ! 1, y ! ¡x + 3g b f (x) = = (¡2x + 4)(x ¡ 1) ¡ (¡x2 + 4x ¡ 7) £ (x ¡ 1)2 ¡2x2 + 6x ¡ + x2 ¡ 4x + (x ¡ 1)2 which has sign diagram: = ¡x + 2x + (x ¡ 1)2 - + -1 ¡(x2 ¡ 2x ¡ 3) = (x ¡ 1)2 =¡ c yellow d y x¡=¡1 local (-1,¡6) local max (3,-2) 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 (x + 1)(x ¡ 3) (x ¡ 1)2 Cuts the x-axis when y = 0, ) ¡x2 + 4x ¡ = ) x2 ¡ 4x + = ) ¢ = 16 ¡ £ £ < so there are no real roots ) does not cut the x-axis magenta ) a local maximum at (3, ¡2) and a local minimum at (¡1, 6) Cuts the y-axis when x = ¡7 = ) y-intercept is ) y= ¡1 cyan + black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\645IB_HL-2_21.CDR Monday, 26 November 2007 3:45:46 PM PETERDELL 3 x y¡=¡-x¡+¡3 IB_HL-2ed (646) 646 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) EXERCISE 21E Determine the equations of the asymptotes of: 2x 1¡x a y= b y= x ¡4 (x + 2)2 c y= 3x + x2 + d y= 2x2 ¡ x + x2 ¡ e y= ¡x2 + 2x ¡ x2 + x + f y= 3x2 ¡ x + (x + 2)2 g y= 3x3 + x2 ¡ x¡2 h y= 2x2 ¡ 5x ¡ x+1 i y= 3x2 + x 2x ¡ For each of the following functions: i determine the equation(s) of the asymptotes ii find dy and hence determine the position and nature of any stationary points dx iii find the axes intercepts iv sketch the function, showing all information obtained in i, ii and iii a y= c y= x2 ¡ 5x + x2 + 5x + d y= x2 ¡ 6x + (x + 1)2 of the following functions: determine the equation(s) of the asymptotes find f (x) and hence determine the position and nature of any stationary points find the axes intercepts sketch the graph of the function, showing all information in i, ii, and iii c f(x) = f (x) = 4x (x ¡ 1)2 d f (x) = magenta x2 4x ¡ 4x ¡ 3x ¡ (x + 2)2 yellow y = ¡2x + ¡ 95 100 50 75 25 c x¡2 x3 x2 + 95 x2 + 3x x+1 100 50 75 50 25 25 f (x) = e x3 x2 ¡ f(x) = 95 y= 100 b d b x2 + 4x + x+2 y= cyan 4x +1 x2 of the following functions: determine the equation(s) of the asymptotes find f (x) and hence determine the position and nature of any stationary points find the axes intercepts sketch the graph of the function a 95 x2 ¡ x2 + f(x) = 100 50 75 25 y= a For each i ii iii iv b 75 For each i ii iii iv x2 ¡ x ¡x¡6 x2 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_21\646IB_HL-2_21.CDR Thursday, 11 March 2010 10:53:11 AM PETER IB_HL-2ed (647) 647 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) F INFLECTIONS AND SHAPE When a curve, or part of a curve, has shape: we say that the shape is concave downwards we say that the shape is concave upwards TEST FOR SHAPE Consider the concave downwards curve: Wherever we are on the curve, as x is increased, the gradient of the tangent decreases m=0 m=1 m = -1 m=2 ) f (x) is decreasing, ) its derivative is negative, so f 00 (x) < m = -2 y¡=¡-xX Likewise, if the curve is concave upwards: y¡=¡xX m = -2 Wherever we are on the curve, as x is increased, the gradient of the tangent increases m=2 m = -1 ) f (x) is increasing, ) its derivative is positive, so f 00 (x) > m=1 m=0 POINTS OF INFLECTION (INFLEXION) A point of inflection is a point on a curve at which a change of curvature (shape) occurs, i.e., DEMO or point of inflection point of inflection cyan magenta yellow 95 stationary inflection tangent a gradient = y¡=¡ƒ(x) - 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Notes: ² If the tangent at a point of inflection is horizontal then this point is also a stationary point We say that we have a horizontal or stationary inflection (SPI) For example, SD ¦ '(x) black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_21\647IB_HL-2_21.CDR Friday, 12 March 2010 4:03:20 PM PETER a - x SD ¦ ''(x) + a x IB_HL-2ed (648) 648 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) ² If the tangent at a point of inflection is not horizontal we say that we have a non-horizontal or non-stationary inflection (NSPI) For example, non-stationary inflection tangent gradient ¹ y¡=¡ƒ(x) x¡=¡b x¡=¡a x¡=¡c + SD ¦ '(x) b - c + - + a SD ¦ ''(x) x x ² The tangent at the point of inflection, also called the inflecting tangent, crosses the curve at that point There is a point of inflection at x = a if f 00 (a) = and the sign of f 00 (x) changes on either side of x = a The point of inflection corresponds to a change in curvature + In the vicinity of a, f 00 (x) has sign diagram either Observe that if f (x) = x4 f (x) = 4x a - - or a + then + 00 and f (x) = 12x2 and f 00 (x) has sign diagram Although f 00 (0) = we not have a point of inflection at (0, 0) because the sign of f 00 (x) does not change on either side of x = In fact the graph of f (x) = x4 is: + y y¡=¡xV (-1, 1) (1, 1) x local minimum (0, 0) SUMMARY For a curve (or part curve) which is concave downwards on an interval S, f 00 (x) for all x in S For a curve (or part curve) which is concave upwards on an interval S, f 00 (x) > for all x in S If f 00 (x) changes sign at x = a, and f 00 (a) = 0, then we have a ² horizontal inflection if f (a) = ² non-horizontal inflection if f (a) 6= cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Click on the demo icon to examine some standard functions for turning points, points of inflection, and intervals where the function is increasing, decreasing, and concave up or down black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_21\648IB_HL-2_21.CDR Friday, 12 March 2010 4:15:04 PM PETER DEMO IB_HL-2ed (649) 649 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) Example 13 Find and classify all points of inflection of f (x) = x4 ¡ 4x3 + f (x) = x4 ¡ 4x3 + f (x) = 4x3 ¡ 12x2 f 00 (x) = 12x2 ¡ 24x = 12x(x ¡ 2) 00 f (x) = when x = or ) ) ) - + concave up ¦'(x) x + ¦''(x) x - + concave down concave up Since the signs of f 00 (x) change about x = and x = 2, these two points are points of inflection Also f (0) = 0, f (2) = 32 ¡ 48 6= and f(0) = 5, f (2) = 16 ¡ 32 + = ¡11 Thus (0, 5) is a horizontal inflection, and (2, ¡11) is a non-horizontal inflection Example 14 For a b c d e f (x) = 3x4 ¡ 16x3 + 24x2 ¡ : find and classify all points where f (x) = find and classify all points of inflection find intervals where the function is increasing or decreasing find intervals where the function is concave up or down Hence, sketch the graph showing all important features a f (x) = 3x4 ¡ 16x3 + 24x2 ¡ ) f (x) = 12x3 ¡ 48x2 + 48x = 12x(x2 ¡ 4x + 4) + + 2 = 12x(x ¡ 2) Now f (0) = ¡9 and f (2) = ) (0, ¡9) is a local minimum and (2, 7) is a horizontal inflection f 00 (x) = 36x2 ¡ 96x + 48 = 12(3x2 ¡ 8x + 4) = 12(x ¡ 2)(3x ¡ 2) + ) (2, 7) is a horizontal inflection ¡ ¢ and 23 , f( 23 ) or ( 23 , ¡2:48) is a non-horizontal inflection c f (x) is decreasing for x f (x) is increasing for x > d f (x) is concave up for x magenta yellow stationary y inflection (2' 7) e and x > 95 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 cyan + (0'-9) local x f (x) is concave down for We_ 100 b black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\649IB_HL-2_21.CDR Monday, 26 November 2007 3:54:17 PM PETERDELL x (We_\'-2"48) non-stationary inflection IB_HL-2ed (650) 650 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) f 00 (x) = corresponds to the stationary points of y = f (x), so f 00 (x) will change sign at a local maximum or minimum of y = f (x) Such points correspond to the points of inflection of y = f(x) If a local maximum or minimum of y = f (x) touches the x-axis, then it corresponds to a stationary point of inflection of y = f (x) Otherwise it corresponds to a non-stationary point of inflection Example 15 Sign diagram of f (x) is: The graph below shows a gradient function y_ =_ f'_(x) Sketch a graph which could be y = f(x), showing clearly the x-values corresponding to all stationary points and points of inflection - + -6 x f (x) is a maximum when x = ¡4 and a minimum when x ¼ 12 At these points f 00 (x) = but f (x) 6= 0, so they correspond to non-stationary points of inflection y¡=¡¦'(x) local max + -1 y 100 -6 - x ¦'(x) y 100 NSPI -50 -6 -1 -50 x ¦(x) local NSPI local Example 16 The local minimum corresponds to f (x) = and f 00 (x) 6= The NSPI corresponds to f (x) 6= and f 00 (x) = The SPI corresponds to f (x) = and f 00 (x) = Using the graph of y = f (x) below, sketch the graphs of y = f (x) and y = f 00 (x) y y y=¦''(x) x x y=¦(x) y=¦(x) y=¦'(x) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 NSPI black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_21\650IB_HL-2_21.CDR Friday, 12 March 2010 4:09:21 PM PETER SPI IB_HL-2ed (651) APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) 651 EXERCISE 21F Find and classify all points of inflection of: a c f(x) = x2 + f(x) = x3 ¡ 6x2 + 9x + b d e f(x) = ¡3x4 ¡ 8x3 + f For each i ii iii iv v of the following functions: find and classify all points where f (x) = find and classify all points of inflection find intervals where the function is increasing or decreasing find intervals where the function is concave up or down Sketch the graph showing all important features f (x) = x2 a d f (x) = x3 ¡ 3x2 ¡ 24x + f (x) = x4 ¡ 4x2 + g f(x) = ¡ x3 f(x) = x3 + 6x2 + 12x + f(x) = ¡ p x p x b f (x) = x3 c f (x) = e f (x) = 3x4 + 4x3 ¡ f (x) = ¡ p x f f (x) = (x ¡ 1)4 h For the graphs of y = f (x) below, sketch a graph which could be y = f (x) Show clearly the location of any stationary points and points of inflection a b y y y¡=¦'(x) y¡=¦'(x) x -3 -2 x Using the graphs of y = f (x) below, sketch the graphs of y = f (x) and y = f 00 (x) Show clearly any axes intercepts and turning points a b c y¡=¡¦(x) y y y y¡=¡¦(x) y¡=¡¦(x) cyan -2 magenta yellow 95 100 50 75 25 95 100 50 x 75 25 95 100 50 75 25 95 100 50 75 25 -2 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\651IB_HL-2_21.CDR Monday, 26 November 2007 3:57:38 PM PETERDELL x x IB_HL-2ed (652) 652 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) G OPTIMISATION There are many problems for which we need to find the maximum or minimum value of a function We can solve such problems using differential calculus techniques The solution is often referred to as the optimum solution and the process is called optimisation Consider the following problem: An industrial shed is to have a total floor space of 600 m2 and is to be divided into rectangular rooms of equal size The walls, internal and external, will cost $60 per metre to build What dimensions should the shed have to minimise the cost of the walls? We let each room be x m by y m as shown xm Clearly x > and y > ym The total length of wall material is L = 6x + 4y m We know that the total area is 600 m2 , 200 so 3x £ y = 600 and hence y = x Knowing this relationship enables us to write L in terms of one variable, in this case x ¶ µ ¶ µ 200 800 = 6x + m L = 6x + x x µ ¶ 800 The cost is $60 per metre, so the total cost is C(x) = 60 6x + dollars x Now C(x) = 360x + 48 000x¡1 ) C (x) = 360 ¡ 48 000x¡2 48 000 ) C (x) = when 360 = x2 48 000 i.e., x2 = ¼ 133:333 and so x ¼ 11:547 360 Now when x ¼ 11:547, y ¼ 200 ¼ 17:321 and C(11:547) ¼ 8313:84 dollars 11:547 So, the minimum cost is about $8310 when the shed is 34:6 m by 17:3 m WARNING The maximum or minimum value does not always occur when the first derivative is zero It is essential to also examine the values of the function at the endpoint(s) of the domain for global maxima and minima dy =0 dx y¡=¡ƒ(x) magenta yellow 95 50 25 95 x¡=¡b 100 50 25 95 100 50 75 25 95 100 50 75 25 cyan 75 x¡=¡p x¡=¡a 100 The maximum value of y occurs at the endpoint x = b The minimum value of y occurs at the local minimum x = p dy =0 dx 75 For example: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\652IB_HL-2_21.CDR Friday, 16 November 2007 2:13:27 PM PETERDELL IB_HL-2ed (653) APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) 653 TESTING OPTIMAL SOLUTIONS If one is trying to optimise a function f (x) and we find values of x such that f (x) = 0, how we know whether we have a maximum or a minimum solution? The following are acceptable tests: SIGN DIAGRAM TEST If near to x = a where f (a) = the sign diagram is: + ² a - ² we have a local maximum - + a we have a local minimum SECOND DERIVATIVE TEST If near x = a where f (a) = and: d2 y < we have dx2 ² ² shape, which is a local maximum d2 y > we have dx2 shape, which is a local minimum GRAPHICAL TEST If the graph of y = f (x) shows: ² ² we have a local maximum we have a local minimum OPTIMISATION PROBLEM SOLVING METHOD The following steps should be followed: Step 1: Draw a large, clear diagram(s) of the situation Step 2: Construct a formula with the variable to be optimised (maximised or minimised) as the subject It should be written in terms of one convenient variable, x say You should write down what restrictions there are on x Step 3: Find the first derivative and find the value(s) of x when it is zero Step 4: If there is a restricted domain such as a x b, the maximum or minimum may occur either when the derivative is zero or else at an endpoint Show by the sign diagram test, the second derivative test or the graphical test, that you have a maximum or a minimum situation Example 17 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A rectangular cake dish is made by cutting out squares from the corners of a 25 cm by 40 cm rectangle of tin-plate, and then folding the metal to form the container What size squares must be cut out to produce the cake dish of maximum volume? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\653IB_HL-2_21.CDR Friday, 16 November 2007 10:13:00 AM PETERDELL DEMO IB_HL-2ed (654) 654 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) Step 1: Let x cm be the side lengths of the squares that are cut out Step 2: Volume = length £ width £ depth = (40 ¡ 2x)(25 ¡ 2x)x cm3 ) V = (40 ¡ 2x)(25x ¡ 2x2 ) cm3 (25-2x) cm Notice that x > and 25 ¡ 2x > ) < x < 12:5 x cm (40-2x) cm dV fproduct ruleg = ¡2(25x ¡ 2x2 ) + (40 ¡ 2x)(25 ¡ 4x) dx = ¡50x + 4x2 + 1000 ¡ 50x ¡ 160x + 8x2 = 12x2 ¡ 260x + 1000 = 4(3x2 ¡ 65x + 250) = 4(3x ¡ 50)(x ¡ 5) which is when x = 50 = 16 or x = Step 3: Now Step 4: Sign diagram test dV + has sign diagram: dx 16 We_ 12.5 or Second derivative test d2 V d2 V = 24x ¡ 260 and at x = 5, = ¡140 which is < dx2 dx2 ) the shape is and we have a local maximum So, the maximum volume is obtained when x = 5, i.e., when cm squares are cut from the corners Example 18 open Find the most economical shape (minimum surface area) for a box with a square base, vertical sides and an open top, given that it must contain litres Step 1: Let the base lengths be x cm and the depth be y cm The volume V = length £ width £ depth ) V = x2 y ) 4000 = x2 y (1) fas litre ´ 1000 cm3 g y cm x cm x cm The total surface area A = area of base + (area of one side) = x2 + 4xy µ ¶ 4000 = x2 + 4x fusing (1)g x2 cyan magenta yellow 95 100 50 75 25 where x > 0 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ) A(x) = x2 + 16 000x¡1 Step 2: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\654IB_HL-2_21.CDR Friday, 16 November 2007 10:26:15 AM PETERDELL IB_HL-2ed (655) 655 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) A0 (x) = 2x ¡ 16 000x¡2 Step 3: 16 000 x2 ) 2x = 16 000 p ) x = 8000 = 20 ) A0 (x) = when 2x = Step 4: or Sign diagram test Second derivative test A00 (x) = + 32 000x¡3 32 000 =2+ x3 which is always positive as x3 > for all x > + - 20 if x = 10 A0 (10) = 20 ¡ if x = 30 A0 (30) = 60 ¡ 16 000 100 = 20 ¡ 160 = ¡140 16 000 900 ¼ 60 ¡ 17:8 ¼ 42:2 Both tests establish that the minimum material is used to make the container 4000 = 10 when x = 20 and y = 202 10 cm So, 20 cm 20 cm is the most economical shape Sometimes the variable to be optimised is in the form of a single square root function In these situations it is convenient to square the function and use the fact that if A > 0, the optimum value of A(x) occurs at the same value of x as the optimum value of [A(x)]2 Example 19 An animal enclosure is a right angled triangle with one A leg being a drain The farmer has 300 m of fencing available for the other two sides, AB and BC p a Show that AC = 90 000 ¡ 600x if AB = x m b Find the maximum area of the triangular enclosure B Hint: If the area is A m2 , find A2 in terms of x A is a maximum when A2 takes its maximum value (AC)2 + x2 = (300 ¡ x)2 fPythagorasg ) (AC)2 = 90 000 ¡ 600x + x2 ¡ x2 = 90 000 ¡ 600x p ) AC = 90 000 ¡ 600x a cyan magenta yellow xm (300-x) m < x < 300 95 100 50 75 25 95 50 25 95 100 50 75 25 95 100 50 75 25 = C A (AC £ x) p x 90 000 ¡ 600x 100 = C B The area of triangle ABC is A(x) = 12 (base £ altitude) 75 b drain black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\655IB_HL-2_21.CDR Monday, 26 November 2007 3:59:57 PM PETERDELL IB_HL-2ed (656) 656 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) ) [A(x)]2 = ) x2 (90 000 ¡ 600x) = 22 500x2 ¡ 150x3 d [A(x)]2 = 45 000x ¡ 450x2 dx = 450x(100 ¡ x) + with sign diagram: A(x) is maximised when x = 100 p so Amax = 12 (100) 90 000 ¡ 60 000 - 100 100 m 300 200 m ¼ 8660 m Example 20 A square sheet of metal has smaller squares cut from its corners as shown a cm What sized square should be cut out so that when the sheet is bent into an open box it will hold the maximum amount of liquid? a cm Let x cm by x cm squares be cut out Volume = length £ width £ depth = (a ¡ 2x) £ (a ¡ 2x) £ x ) V (x) = x(a ¡ 2x)2 x (a-2x)¡cm Now V (x) = 1(a ¡ 2x)2 + x £ 2(a ¡ 2x)1 £ (¡2) fproduct ruleg = (a ¡ 2x)[a ¡ 2x ¡ 4x] = (a ¡ 2x)(a ¡ 6x) a a or ) V (x) = when x = a However, a ¡ 2x must be > and so x < a a Thus x = is the only value in < x < with V (x) = Second derivative test: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Now V 00 (x) = ¡2(a ¡ 6x) + (a ¡ 2x)(¡6) fproduct ruleg = ¡2a + 12x ¡ 6a + 12x = 24x ¡ 8a ³a´ ) V 00 = 4a ¡ 8a = ¡4a which is < convex a ) the volume is maximised when x = downwards a Conclusion: When x = , the resulting container has maximum capacity black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\656IB_HL-2_21.CDR Friday, 16 November 2007 10:56:23 AM PETERDELL IB_HL-2ed (657) APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) 657 EXERCISE 21G Use calculus techniques in the following problems A manufacturer can produce x fittings per day where x 10 000 The costs are: ² E1000 per day for the workers ² E2 per day per fitting 5000 ² E per day for running costs and maintenance x How many fittings should be produced daily to minimise costs? For the cost function C(x) = 720 + 4x + 0:02x2 dollars and price function p(x) = 15 ¡ 0:002x dollars, find the production level that will maximise profits The total cost of producing x blankets per day is 14 x2 + 8x + 20 dollars, and for this production level each blanket may be sold for (23 ¡ 12 x) dollars How many blankets should be produced per day to maximise the total profit? v2 per hour where v is the speed of the boat 10 All other costs amount to $62:50 per hour Find the speed which will minimise the total cost per kilometre The cost of running a boat is $ A duck farmer wishes to build a rectangular enclosure of area 100 m2 The farmer must purchase wire netting for three of the sides as the fourth side is an existing fence Naturally, the farmer wishes to minimise the length (and therefore cost) of fencing required to complete the job a If the shorter sides have length x m, show that the required length of wire netting 100 to be purchased is L = 2x + : x 100 : b Use technology to help you sketch the graph of y = 2x + x c Find the minimum value of L and the corresponding value of x when this occurs d Sketch the optimum situation showing all dimensions Radioactive waste is to be disposed of in fully enclosed lead boxes of inner volume 200 cm3 The base of the box has dimensions in the ratio : h cm x cm a What is the inner length of the box? b Explain why x2 h = 100 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 600 c Explain why the inner surface area of the box is given by A(x) = 4x2 + cm2 x 600 d Use technology to help sketch the graph of y = 4x2 + : x e Find the minimum inner surface area of the box and the corresponding value of x f Sketch the optimum box shape showing all dimensions black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\657IB_HL-2_21.CDR Friday, 16 November 2007 11:03:14 AM PETERDELL IB_HL-2ed (658) 658 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) Consider the manufacture of cylindrical tin cans of L capacity where the cost of the metal used is to be minimised This means that the surface area must be as small as possible 1000 r cm a Explain why the height h is given by h = cm ¼r2 b Show that the total surface area A is given by h cm 2000 A = 2¼r2 + cm2 r c Use technology to help you sketch the graph of A against r d Find the value of r which makes A as small as possible e Sketch the can of smallest surface area Sam has sheets of metal which are 36 cm by 36 cm square He wants to cut out identical squares which are x cm by x cm from the corners of each sheet He will then bend the sheets along the dashed lines to form an open container a Show that the capacity of the container is given by V (x) = x(36 ¡ 2x)2 cm3 b What sized squares should be cut out to produce the container of greatest capacity? 36 cm 36 cm An athletics track has two ‘straights’ of length l m and two semicircular ends of radius x m The perimeter of the track is 400 m a Show that l = 200 ¡ ¼x and hence write down the possible values that x may have b Show that the area inside the track is A = 400x ¡ ¼x2 c What values of l and x produce the largest area inside the track? xm lm 10 A sector of radius 10 cm is bent to form a conical cup as shown sector q° 10 cm A join 10 cm becomes when edges AB and CB are joined with tape B C Suppose the resulting cone has base radius r cm and height h cm µ¼ : a Show that in the sector, arc AC = 18 µ : b If r is the radius of the cone, explain why r = 36 q ¡ µ ¢2 c If h is the height of the cone show that h = 100 ¡ 36 : cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 d Find the cone’s capacity V in terms of µ only e Use technology to sketch the graph of V (µ): f Find µ when V (µ) is a maximum black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\658IB_HL-2_21.CDR Monday, 26 November 2007 4:00:32 PM PETERDELL IB_HL-2ed (659) 659 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) 11 B is a row boat km out at sea from A AC C A x km X is a straight sandy beach, km long Peter km can row the boat at km h¡1 and run along the beach at 17 km h¡1 Suppose Peter rows directly from B to point X on [AC] such that km AX = x km a Explain why x 6 b If T (x) is the total time Peter takes to B row to X and then run along the beach p x2 + 25 ¡ x + hrs to C, show that T (x) = 17 dT = What is the significance of this value of x? Prove your c Find x such that dx statement A 12 A pumphouse is to be placed at some point X along a river B km Two pipelines will then connect the pumphouse to homesteads A and B km M How far from M should point X be so that the total length of pipeline is minimised? N X km river 13 Open cylindrical bins are to contain 100 litres Find the radius and height of the bin shape which requires the least amount of material (minimises the surface area) 14 Two lamps have intensities 40 and candle-power and are m apart If the intensity of illumination I at any point is directly proportional to the power of the source, and inversely proportional to the square of the 40 cp cp distance from the source, find the darkest point on the line joining the two lamps 6m 15 A right angled triangular pen is made from 24 m of fencing, all used for sides AB and BC Side AC is an existing brick wall a If AB = x m, find D(x) in terms of x d[D(x)]2 b Find and hence draw a sign diagram dx for it c Find the smallest and the greatest possible value of D(x) and the design of the pen in each case C D(x) metres B A wall 16 At 1:00 pm a ship A leaves port P, and sails in the direction 30o T at 12 km h¡1 At the same time, ship B is 100 km due east of P and is sailing at km h¡1 towards P a Show that the distance D(t) between the two ships is given by p D(t) = 304t2 ¡ 2800t + 10 000 km, where t is the number of hours after 1:00 pm cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b Find the minimum value of [D(t)]2 for all t > c At what time, to the nearest minute, are the ships closest? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\659IB_HL-2_21.CDR Friday, 16 November 2007 11:43:14 AM PETERDELL IB_HL-2ed (660) 660 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) 17 AB is a m high fence which is m from a vertical wall RQ An extension ladder PQ Q is placed on the fence so that it touches the wall ground at P and the wall at Q a If AP = x m, find QR in terms of x b If the ladder has length L m, show µ ¶ R that [L(x)]2 = (x + 2)2 + x p d[L(x)]2 = only when x = c Show that dx B 1m 2m A P d Find, correct to the nearest centimetre, the shortest length of the extension ladder You must prove that this length is the shortest Sometimes finding the zeros of the derivative is difficult, and in such cases we can use technology Use the graphing package or your graphics calculator to help solve the following problems GRAPHING PACKAGE TI C A 18 A, B and C are computers A printer P is networked to each computer Where should P be located so that the total cable length AP + BP + CP is a minimum? 8m P B y (3, 11) Caville 19 (7, 3) Bracken (1, 2) Aden N 5m C Three towns and their grid references are marked on the diagram alongside A pumping station is to be located at P on the pipeline, to pump water to the three towns Grid units are kilometres Exactly where should P be located so that the total length of the pipelines to Aden, Bracken and Caville is as short as possible? pipeline P 4m x 20 The trailing cone of a guided long range torpedo is to be conical with slant edge s cm The cone is hollow and must contain the s cm maximum possible volume of fuel r cm h cm Find the ratio of s : r such that the maximum fuel carrying capacity occurs stage cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 -b 95 x 100 a -a 50 seating 75 A company constructs rectangular seating arrangements for pop concerts on sports grounds The oval shown has equation 2 y x + = where a and b are the lengths a b of the semi-major and semi-minor axes A(x,¡y) 25 b y 21 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\660IB_HL-2_21.CDR Friday, 16 November 2007 11:57:35 AM PETERDELL IB_HL-2ed (661) 661 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) bp a ¡ x2 for A as shown a 4bx p a ¡ x2 Show that the seating area is given by A(x) = a a Show that A0 (x) = when x = p a Prove that the seating area is a maximum when x = p Given that the area of the ellipse is ¼ab, what percentage of the ground is occupied by the seats in the optimum case? a Show that y = b c d e H IMPLICIT DIFFERENTIATION For relations such as y3 + 3xy2 ¡ xy + 11 = it is often difficult or impossible to make y the subject of the formula Such relationships between x and y are called implicit relations To gain insight into how such relations can be differentiated we will examine a familiar case Consider the circle with centre (0, 0) and radius The equation of the circle is x2 + y = Suppose A(x, y) lies on the circle y A (x,¡y) y¡0 y y-step = = x-step x¡0 x x ) the tangent at A has gradient = ¡ fthe negative reciprocalg y x dy =¡ for all points (x, y) on the circle Thus dx y dy by using a circle property So, in this case we have found dx dy for a circle is to split the Another way of finding dx relation into two parts The radius OA has gradient = If x2 + y = magenta y x -~`4`-``x ) dy = ¡ 12 (4 ¡ x2 )¡ £ (¡2x) dx x = p ¡ x2 x = ¡y yellow 95 100 50 75 25 95 The question is: “Is there a better way of finding 100 50 75 25 ~`4`-``x Case 2: p y = ¡ ¡ x2 = ¡(4 ¡ x2 ) dy x =¡ dx y 95 100 50 75 25 95 100 50 75 25 cyan x tangent dy = 12 (4 ¡ x2 )¡ £ (¡2x) dx ¡x = p ¡ x2 x =¡ y So, in both cases (0,¡0) then y = ¡ x2 p and so y = § ¡ x2 Case 1: p y = ¡ x2 = (4 ¡ x2 ) ) black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_21\661IB_HL-2_21.CDR Friday, 12 March 2010 4:11:20 PM PETER dy ?” dx IB_HL-2ed (662) 662 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) IMPLICIT DIFFERENTIATION The process by which we differentiate implicit relations is called implicit differentiation dy If we are given an implicit relation between y and x and we want , we differentiate both dx sides of the equation with respect to x, applying the chain, product, quotient and any other dy rules as appropriate This will generate terms containing , which we then proceed to make dx the subject of the equation d d (x + y ) = (4) For example, if x2 + y = 4, then dx dx dy =0 ) 2x + 2y dx dy x ) =¡ as required dx y d dy (y n ) = ny n¡ dx dx A useful property is that using the chain rule Example 21 d (y ) a dx dy = 3y dx d dx b d (y ) dx a If y is a function of x find: µ ¶ y b d dx µ ¶ y c d (xy2 ) dx d (xy2 ) dx c = £ y2 + x £ 2y d ¡1 (y ) dx dy = ¡y ¡2 dx = = y2 + 2xy dy dx dy dx fproduct ruleg Example 22 dy if: i x2 + y = dx d2 y : For part a i only, find dx2 magenta 95 100 50 75 95 yellow dy =0 dx 50 95 100 50 75 25 95 100 50 75 25 cyan x + x2 y + y = 100 d(x) d d d ) + (x y) + (y ) = (100) dx dx dx dx · ¸ dy dy +3y2 ) + 2xy + x2 =0 dx dx | {z } fproduct ruleg dy = ¡1 ¡ 2xy ) (x2 + 3y ) dx dy ¡1 ¡ 2xy ) = dx x + 3y dy ¡2x = dx 3y2 ) x + x2 y + y = 100 ii 100 ) 2x + 3y 75 ) x2 + y3 = d d d (x ) + (y ) = (8) dx dx dx 25 i a b ii Find 25 a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\662IB_HL-2_21.CDR Friday, 16 November 2007 12:33:28 PM PETERDELL IB_HL-2ed (663) 663 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) d2 y d = dx2 dx b µ dy dx ¶ = d dx µ ¶ ¡2x 3y ¡2(3y ) ¡ (¡2x)6y = 9y µ ¡6y + 12xy = fquotient ruleg ¶ 9y ¡6y ¡ = = ¡2x 3y dy dx 8x2 y 9y µ ¶ y £ y ¡6y ¡ 8x2 9y Example 23 Find the gradient of the tangent to x2 + y = at the point where x = First we find dy : dx dy = fimplicit differentiationg dx dy dy ¡2x = ¡2x and so = ) 3y dx dx 3y2 2x + 3y When x = 2, + y = and ) y = dy ¡2(2) = ¡ 43 = dx 3(1)2 Consequently So, the gradient of the tangent at x = is ¡ 43 EXERCISE 21H If y is a function of x, find: d d a (2y) b (¡3y) dx dx µ ¶ d p d f g ( y) dx dx y2 Find dy if: dx c h d (y ) dx d (xy) dx x2 + y = 25 x2 ¡ y = 10 a d i d ( ) dx y d (x y) dx x2 + 3y = x2 + xy = b e Find the gradient of the tangent to: a x + y3 = 4y at y = d b x + y = 8xy d2 y for each implicit relation in question dx2 dV Given that 3V + 2q = 2V q, find a b dq d (y ) dx d (xy2 ) dx e j c f y ¡ x2 = x3 ¡ 2xy = at x = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_21\663IB_HL-2_21.CDR Friday, 12 March 2010 4:13:00 PM PETER d2 q dV IB_HL-2ed (664) 664 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) REVIEW SET 21A A particle P moves in a straight line with position relative to the origin O given by s(t) = 2t3 ¡ 9t2 + 12t ¡ cm, where t is the time in seconds, t > a Find expressions for the particle’s velocity and acceleration and draw sign diagrams for each of them b Find the initial conditions c Describe the motion of the particle at time t = seconds d Find the times and positions where the particle changes direction e Draw a diagram to illustrate the motion of P f Determine the time intervals when the particle’s speed is increasing v 9000 + dollars The cost per hour of running a freight train is given by C(v) = 30 v where v is the average speed of the train in km h¡1 a Find the cost of running the train for: ii hours at 64 km h¡1 i two hours at 45 km h¡1 b Find the rate of change in the hourly cost of running the train at speeds of: ii 66 km h¡1 i 50 km h¡1 c At what speed will the cost be a minimum? For a b c d the function f(x) = 2x3 ¡ 3x2 ¡ 36x + : find and classify all stationary points and points of inflection find intervals where the function is increasing and decreasing find intervals where the function is concave up or down sketch the graph of y = f (x) showing all important features y Rectangle ABCD is inscribed within the parabola y = k ¡ x2 and the x-axis, as shown a If OD = x, show that the rectangle ABCD has area function A(x) = 2kx ¡ 2x3 b If the area p of ABCD is a maximum when AD = 3, find k y = k - x2 C B x A D A manufacturer of open steel boxes has to make one with a square base and a open capacity of m3 The steel costs $2 per square metre ym a If the base measures x m by x m and the xm height is y m, find y in terms of x dollars b Hence, show that the total cost of the steel is C(x) = 2x2 + x c Find the dimensions of the box which would cost the least in steel to make a dy given that x2 y + 2xy3 = ¡18 dx Find the equation of the tangent to x2 y + 2xy3 = ¡18 at the point (1, ¡2) Find b y y=¦'(x) cyan magenta yellow 95 x 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Given the graph of y = f (x) drawn alongside, sketch a possible curve for y = f(x) Show clearly any turning points and points of inflection black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\664IB_HL-2_21.CDR Monday, 26 November 2007 4:04:20 PM PETERDELL IB_HL-2ed (665) 665 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) REVIEW SET 21B A triangular pen is enclosed by two fences AB and BC each of length 50 m, with the river being the third side a If AC = 2x m, show that the area of triangle ABC p is A(x) = x 2500 ¡ x2 m2 d[A(x)]2 and hence find x such that dx the area is a maximum b Find B 50 m river A C A particle P moves in a straight line with position from O given by 60 s(t) = 15t ¡ cm, where t is the time in seconds, t > (t ¡ 1)2 a Find velocity and acceleration functions for P’s motion b Describe the motion of P at t = seconds c For what values of t is the particle’s speed increasing? end view A rectangular gutter is formed by bending a 24 cm wide sheet of metal as shown in the illustration 24 cm Where must the bends be made in order to maximise the capacity of the gutter? Consider the curve with equation x2 ¡ 2xy2 + y = k where k is a constant a If (2, ¡1) lies on the curve, find k dy b Find c Find the equation of the normal to x2 ¡ 2xy2 + y = k at (2, ¡1) dx A particle moves along the x-axis with position relative to origin O given by p x(t) = 3t ¡ t cm, where t is the time in seconds, t > a Find expressions for the particle’s velocity and acceleration at any time t, and draw sign diagrams for each function b Find the initial conditions and hence describe the motion at that instant c Describe the motion of the particle at t = seconds d Find the time and position when the particle reverses direction e Determine the time interval when the particle’s speed is decreasing x2 ¡ : x2 + a find the axes intercepts b explain why f (x) has no vertical asymptotes c find the position and nature of any stationary points q d show that y = f(x) has non-stationary inflections at x = § 13 For the function f(x) = e sketch the graph of y = f (x) showing all features found in a, b, c and d above The graph of y = f(x) is given On the same axes sketch the graph of y = jf (x) j : y x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 y¡=¡¦(x) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\665IB_HL-2_21.CDR Monday, 26 November 2007 4:04:58 PM PETERDELL IB_HL-2ed (666) 666 APPLICATIONS OF DIFFERENTIAL CALCULUS (Chapter 21) REVIEW SET 21C For a b c the function f(x) = x3 ¡ 4x2 + 4x : find all axes intercepts find and classify all stationary points and points of inflection sketch the graph of y = f (x) showing features from a and b A 200 m fence is placed around a lawn which has the shape of a rectangle with a semi-circle on one of its sides a Using the dimensions shown on the figure, show that y = 100 ¡ x ¡ ¼2 x b Hence, find the area of the lawn A in terms of x only 2x m c Find the dimensions of the lawn if it has the maximum possible area ym x2 + 2x x¡2 Determine the equations of any asymptotes Find the position and nature of its turning points c Find its axes intercepts Sketch the graph of the function showing the important features of a, b and c x2 + 2x For what values of p does = p have two real distinct roots? x¡2 Consider f(x) = a b d e A machinist has a spherical ball of brass with diameter 10 cm The ball is placed in a lathe and machined into a cylinder h cm a If the cylinder has radius x cm, show that the cylinder’s volume is given by p V (x) = ¼x2 100 ¡ 4x2 cm3 x cm b Hence, find the dimensions of the cylinder of largest volume which can be made Two roads AB and BC meet at right angles A straight A pipeline LM is to be laid between the two roads with L the requirement that it must pass through point X a If PM = x km, find LQ in terms of x b Hence show that the length of the pipeline is km µ ¶ Q p km given by L(x) = x + 1 + B x d[L(x)] and hence find the shortest possible length for the c Find dx 2k cm A rectangular sheet of tin-plate is 2k cm by k cm and four squares each with sides x cm are cut from its corners The remainder is bent into the shape of an open rectangular container Find the value of x which will maximise the capacity of the container magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 x cyan km C P M pipeline k cm x cm The graph of y = f (x) is drawn On the same axes clearly draw a possible graph of y = f (x) Show all turning points and points of inflection y¡=¡¦'(x) y X black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_21\666IB_HL-2_21.CDR Monday, 26 November 2007 4:05:46 PM PETERDELL IB_HL-2ed (667) 22 Chapter Derivatives of exponential and logarithmic functions Contents: A B C D E Exponential e Natural logarithms Derivatives of logarithmic functions Applications Some special exponential functions cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 22A Review set 22B black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_22\667IB_HL-2_22.CDR Monday, 19 November 2007 1:42:33 PM PETERDELL IB_HL-2ed (668) 668 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 22) The simplest exponential functions are of the form f (x) = ax constant, a 6= The graphs of all members of the exponential family f(x) = ax have the following properties: ² pass through the point (0, 1) ² asymptotic to the x-axis at one end ² lie above the x-axis for all x ² concave up for all x ² monotone increasing for a > ² monotone decreasing for < a < where a is any positive For example, y y = (0.2)x y = (0.5) y = 5x y = 2x x y = (1.2)x x A EXPONENTIAL e THE DERIVATIVE OF y = a x INVESTIGATION This investigation could be done by using a graphics calculator or by clicking on the icon The purpose of this investigation is to observe the nature of the derivative of f(x) = ax for a = 2, 3, 4, 5, 12 and 14 What to do: For y = 2x find the gradient of the tangent at x = 0, 0:5, 1, 1:5, and 2:5 Use modelling techniques from your graphics dy calculator or the software provided to show that ¼ 0:693£2x dx Repeat for y = 3x : CALCULUS DEMO Repeat for y = 5x : Repeat for y = (0:5)x : Use 1, 2, and to help write a statement about the derivative of the general exponential y = ax for a > 0, a 6= From the previous investigation you should have discovered that: cyan yellow 95 100 50 75 25 95 100 50 75 25 95 100 k is the derivative of y = a magenta where k is a constant at x = 0, i.e., k = f (0) x 50 25 95 100 50 75 25 ² then f (x) = kax if f(x) = ax 75 ² black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_22\668IB_HL-2_22.CDR Monday, 19 November 2007 2:29:17 PM PETERDELL IB_HL-2ed (669) 669 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 22) This result is easily proved algebraically f (x + h) ¡ f (x) h If f(x) = ax , then f (x) = lim h!0 ax+h ¡ ax h = lim h!0 ax (ah ¡ 1) h!0 h µ ¶ ah ¡ x = a £ lim h!0 h ffirst principles definition of derivativeg = lim f (0 + h) ¡ f (0) h But f (0) = lim h!0 h!0 ) y y = ax ah ¡ h = lim fas ax is independent of hg gradient is ƒ'(0) x f (x) = a f (0) x So, if we can find a value of a such that f (0) = 1, then we have found a function which is its own derivative FINDING a WHEN y = ax AND INVESTIGATION Click on the icon to graph f(x) = ax and its derivative function y = f (x) Experiment with different values of a until the graphs of f (x) = ax appear the same Estimate the corresponding value of a to decimal places dy = ax dx DEMO and y = f (x) From Investigation you should have discovered that if a ¼ 2:72 and f (x) = ax f (x) = ax also then To find this value of a more accurately we return to the algebraic approach: µ x We showed that if f (x) = a then f (x) = a So if f (x) = ax lim ) we require h!0 ah ¡ lim h!0 h x ¶ : ah ¡ = 1: h a ¡1 ¼ for values of h which are close to h h cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ) ah ¼ + h for h close to black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_22\669IB_HL-2_22.CDR Friday, 12 March 2010 4:19:30 PM PETER IB_HL-2ed (670) 670 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 22) 1 1 , an ¼ + for large values of n fh = ! as n ! 1g n n n µ ¶n for large values of n ) a ¼ 1+ n µ ¶n as n ! We now examine 1+ n Letting h = n ¶n µ 1+ n n ¶n µ 1+ n 10 102 103 104 105 106 2:593 742 460 2:704 813 829 2:716 923 932 2:718 145 927 2:718 268 237 2:718 280 469 107 108 109 1010 1011 1012 2:718 281 693 2:718 281 815 2:718 281 827 2:718 281 828 2:718 281 828 2:718 281 828 µ ¶n 1+ ! 2:718 281 828 459 045 235 :::: n In fact as n ! 1, and this irrational number is denoted by the symbol e e = 2:718 281 828 459 045 235 :::: and is called exponential e If f(x) = ex then f (x) = ex ex is sometimes written as exp(x) Alternative notation: For example, exp(1 ¡ x) = e1¡x e is an important number with similarities to the number ¼ Both numbers are irrational (not surds) with non-recurring, non-terminating decimal expansions, and both are discovered naturally We also saw e in an earlier chapter when looking at continuous compound interest PROPERTIES OF y¡=¡ex dy = ex = y dx Notice that y As x ! 1, y ! very rapidly, dy ! and so dx 10 magenta yellow 95 -1 100 50 75 25 -2 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 This means that the gradient of the curve is very large for large values of x The curve increases in steepness as x gets larger cyan y=e x 15 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_22\670IB_HL-2_22.CDR Friday, 12 March 2010 4:20:06 PM PETER x IB_HL-2ed (671) DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 22) 671 dy ! dx As x ! ¡1, y ! and so This means for large negative x, the graph becomes flatter and approaches the asymptote y = ex > for all x, so the range of f : x 7! ex is R + or ] 0, [ THE DERIVATIVE OF ef(x) The functions e¡x , e2x+3 and e¡x often used in problem solving are all of the form ef (x) Such functions are In general, ef (x) > for all x, no matter what the function f (x) Consider y = ef (x) : Now y = eu where u = f (x): dy dy du But = fchain ruleg dx du dx du dy ) = eu = ef (x) £ f (x) dx dx Summary: Function ex Derivative ex ef (x) ef (x) £ f (x) Example Find the gradient function for y equal to: b x2 e¡x c If y = e2x x dy = 2xe¡x + x2 e¡x (¡1) dx = 2xe¡x ¡ x2 e¡x then then dy e2x (2)x ¡ e2x (1) = dx x2 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 = fproduct ruleg fquotient ruleg e2x (2x ¡ 1) x2 95 If y = x2 e¡x 100 b e2x x dy = 2ex + e¡3x (¡3) dx = 2ex ¡ 3e¡3x then 50 If y = 2ex + e¡3x c 75 a 25 2ex + e¡3x a black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_22\671IB_HL-2_22.CDR Friday, 12 March 2010 4:20:37 PM PETER IB_HL-2ed (672) 672 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 22) Example a (ex ¡ 1)3 Find the gradient function for y equal to: y = (ex ¡ 1)3 = u3 where u = ex ¡ a p ¡x 2e + b y = (2e¡x + 1)¡ b = u¡ where u = 2e¡x + dy dy du = dx du dx dy dy du = dx du dx du = 3u2 dx x = 3(e ¡ 1)2 £ ex = 3ex (ex ¡ 1)2 ¡3 = ¡ 12 u du dx = ¡ 12 (2e¡x + 1)¡ £ 2e¡x (¡1) = e¡x (2e¡x + 1)¡ Example Find the position and nature of any turning points of y = (x ¡ 2)e¡x dy fproduct ruleg = (1)e¡x + (x ¡ 2)e¡x (¡1) dx ¡x = e (1 ¡ (x ¡ 2)) 3¡x = where ex is positive for all x ex dy = when x = 3: dx dy is: The sign diagram of dx ) at x = we have a maximum turning point So, But when x = 3, y = (1)e¡3 = + e3 ) the maximum turning point is (3, ) e3 EXERCISE 22A i e¡x m e2x+1 cyan 4e ¡ 3e¡x h ex + e¡x k 10(1 + e2x ) l 20(1 ¡ e¡2x ) o e1¡2x p e¡0:02x exp (¡2x) f ¡ 2e¡x g j ex n e4 yellow 50 25 95 100 50 75 x 25 95 50 75 100 magenta 25 ¡x c x 95 2e e2 ex + 100 e x d b 75 e4x 95 a 100 50 75 25 Find the gradient function for f(x) equal to: black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_22\672IB_HL-2_22.CDR Friday, 12 March 2010 4:22:00 PM PETER IB_HL-2ed (673) DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 22) 673 Find the derivative of: a xex b x3 e¡x c e x2 e3x f ex p x g ex x p ¡x xe Find the gradient function for f(x) equal to: a (ex + 2)4 b ¡ e¡x (1 ¡ e3x )2 d e p ¡ e¡x d h x ex ex + e¡x + c p e2x + 10 f p x ¡ 2e¡x If y = Aekx , where A and k are constants: dy d2 y = k2 y a show that i = ky ii dx dx2 dn y and y and prove your conjecture using the dxn principle of mathematical induction b predict the connection between d2 y dy ¡7 + 12y = dx2 dx If y = 2e3x + 5e4x , show that Find dy if x3 e3y + 4x2 y = 27e¡2x dx Find the position and nature of the turning point(s) of: y = xe¡x a y = x2 ex b B c y= ex x y = e¡x (x + 2) d NATURAL LOGARITHMS In Chapter we found that: ² if ex = a then x = ln a and vice versa i.e., ex = a , x = ln a ² The graph of y = ln x is the reflection of the graph of y = ex in the mirror line y = x y = ex y = ln x y=x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 and y = ln x are inverse functions 100 50 y = ex 75 25 95 100 50 75 25 ² black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_22\673IB_HL-2_22.CDR Friday, 12 March 2010 4:23:25 PM PETER IB_HL-2ed (674) 674 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 22) From the definition ex = a , x = ln a we observe that eln a = a ² This means that any positive real number a can be written as a power of e, or alternatively, the natural logarithm of any positive number is its power of e, i.e., ln en = n Recall that the laws of logarithms in base e are identical to those for base 10 and indeed for any base These are: ² For a > 0, b > ln(ab) = ln a + ln b ³a´ ln = ln a ¡ ln b b ² Notice also that: ² ln (an ) = n ln a ² ln = and ln e = µ ¶ ln = ¡ ln a a ² ² logb a = Note: ln en = n ln a , b 6= ln b Notice that: ax = (eln a )x = e(ln a)x d(ax ) = e(ln a)x £ ln a = ax ln a dx ) if y = ax So, dy = ax ln a dx then Example Find algebraically, the exact points of intersection of y = ex ¡ and y = ¡ 3e¡x Check your solution using technology The functions meet where ex ¡ = ¡ 3e¡x ) e ¡ + 3e¡x = fmultiplying each term by ex g ) e2x ¡ 4ex + = ) (ex ¡ 1)(ex ¡ 3) = ) ex = or ) x = ln or ln ) x = or ln GRAPHING PACKAGE x TI C cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 When x = 0, y = e0 ¡ = ¡2 ) y =3¡3=0 When x = ln 3, ex = ) the functions meet at (0, ¡2) and at (ln 3, 0) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_22\674IB_HL-2_22.CDR Monday, 26 November 2007 4:13:26 PM PETERDELL IB_HL-2ed (675) 675 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 22) Example Consider the function y = ¡ e¡x a Find the x-intercept b Find the y-intercept c Show algebraically that the function is increasing for all x d Show algebraically that the function is concave down for all x e Use technology to help graph y = ¡ e¡x : f Explain why y = is a horizontal asymptote a The x-intercept occurs when y = 0, ) e¡x = ) ¡x = ln ) x = ¡ ln ) the x-intercept is ¡ ln ¼ ¡0:69 b The y-intercept occurs when x = ) y = ¡ e0 = ¡ = dy = ¡ e¡x (¡1) = e¡x = x dx e c dy > for all x dx ) the function is increasing for all x Now ex > for all x, so d d2 y ¡1 = e¡x (¡1) = x which is < for all x dx2 e ) the function is concave down for all x f As x ! 1, ex ! and e¡x ! ) y ! (below) e Hence, HA is y = 2: EXERCISE 22B Write as a natural logarithmic equation: b P = 8:69e¡0:0541t a N = 50e2t Without using a calculator, evaluate: p b ln e a ln e2 magenta yellow 50 75 95 c 10 µ ¶ e c ln g e¡ ln a d 100 b 25 95 50 a 75 25 95 100 50 75 25 95 100 50 75 25 Write as a power of e: cyan e2 ln f 100 eln e c black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_22\675IB_HL-2_22.CDR Monday, 19 November 2007 4:32:49 PM PETERDELL S = a2 e¡kt µ p e d ln h e¡2 ln ¶ ax IB_HL-2ed (676) 676 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 22) Solve for x: a ex = b ex = ¡2 c ex = d e2x = 2ex e ex = e¡x f e2x ¡ 5ex + = g ex + = 3e¡x h + 12e¡x = ex i ex + e¡x = dy for dx Find a y = 2x b d y = x3 6¡x e y = 5x 2x y= x c f y = x2x x y= x Find algebraically, the point(s) of intersection of: a c y = ex and y = e2x ¡ y = ¡ ex and y = 5e¡x ¡ b y = 2ex + and y = ¡ ex Check your answers using technology f(x) = ex ¡ and g(x) = ¡ 5e¡x a b c d Find the x and y-intercepts of both functions Discuss f (x) and g(x) as x ! and as x ! ¡1 Find algebraically the point(s) of intersection of the functions Sketch the graph of both functions on the same set of axes Show all important features on your graph The function y = ex ¡ 3e¡x cuts the x-axis at P and the y-axis at Q a Determine the coordinates of P and Q b Prove that the function is increasing for all x d2 y = y dx2 What can be deduced about the concavity of the function above and below the x-axis? c Show that d Use technology to help graph y = ex ¡ 3e¡x Show the features of a, b and c on the graph For the function y = 4x ¡ 2x : cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 find the axes intercepts discuss the graph as x ! and as x ! ¡1 find the position and nature of any turning points discuss the concavity of the function draw a graph showing the features found above 95 100 50 75 25 a b c d e black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_22\676IB_HL-2_22.CDR Monday, 19 November 2007 3:16:59 PM PETERDELL IB_HL-2ed (677) 677 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 22) C DERIVATIVES OF LOGARITHMIC FUNCTIONS THE DERIVATIVE OF ¡ln¡x INVESTIGATION If y = ln x, what is the gradient function? CALCULUS DEMO What to do: Click on the icon to see the graph of y = ln x A tangent is drawn to a point on the graph and the gradient of this tangent is given As the point moves from left to right, a graph of the gradient of the tangent is displayed What you conjecture that the equation of the gradient is? Find the gradient at x = 0:25, x = 0:5, x = 1, x = 2, x = 3, x = 4, x = Do your results confirm your conjecture from 2? From the investigation you should have observed that Proof: if y = ln x then dy = dx x If y = ln x then x = ey Using implicit differentiation with respect to x, dy = ey fchain ruleg dx dy ) 1=x fas ey = xg dx dy ) = x dx if y = ln f (x) then By use of the chain rule, we can also show that dy f (x) = dx f(x) Proof: If y = ln f (x) then y = ln u where u = f (x) Now dy du dy = dx du dx = f (x) u = f (x) f (x) Summary: Function ln x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ln f(x) black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_22\677IB_HL-2_22.CDR Friday, 12 March 2010 4:26:58 PM PETER Derivative x f (x) f(x) IB_HL-2ed (678) 678 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 22) Example a b Find the gradient function of: a y = ln(kx) where k is a constant y = ln(1 ¡ 3x) c y = x3 ln x dy k = = dx kx x Note: ln(kx) = ln k + ln x = ln x + constant If y = ln(kx) then dy ¡3 = = dx ¡ 3x 3x ¡ µ ¶ dy = 3x2 ln x + x3 fproduct ruleg dx x = 3x2 ln x + x2 = x2 (3 ln x + 1) b If y = ln(1 ¡ 3x) then c If y = x3 ln x then The laws of logarithms can help us to differentiate some logarithmic functions more easily Example Differentiate with respect to x: · ¡x b y = ln a y = ln(xe ) x2 (x + 2)(x ¡ 3) y = ln x + ln e¡x ) y = ln x ¡ x ¸ a If y = ln(xe¡x ) then b dy Differentiating with respect to x, we get = ¡1 dx x · ¸ x2 If y = ln then y = ln x2 ¡ ln[(x + 2)(x ¡ 3)] (x + 2)(x ¡ 3) = ln x ¡ [ln(x + 2) + ln(x ¡ 3)] = ln x ¡ ln(x + 2) ¡ ln(x ¡ 3) flog of a product lawg fln ea = ag 1 dy = ¡ ¡ dx x x+2 x¡3 ) EXERCISE 22C c y= cyan magenta yellow 95 100 50 l 75 25 p x ln(2x) 100 95 k 50 y = e¡x ln x 75 j 25 i y = (ln x)2 h 95 y = ex ln x 100 g 50 f 75 y = x2 ln x 25 e y = ¡ ln x 95 y = ln(2x + 1) d 100 50 75 25 Find the gradient function of: a y = ln(7x) b black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_22\678IB_HL-2_22.CDR Friday, 12 March 2010 4:28:02 PM PETER y = ln(x ¡ x2 ) ln x y= 2x p y = ln x p x y= ln x IB_HL-2ed (679) DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 22) Find a dy for: dx y = x ln y = ln(10 ¡ 5x) µ ¶ y = ln x d g b y = ln(x3 ) c e y = [ln(2x + 1)]3 f h y = ln(ln x) i Differentiate with respect to x: p a y = ln ¡ 2x b d p y = ln (x ¡ x) e g f(x) = ln ((3x ¡ 4)3 ) h a Find dy for: dx i µ ¶ 2x + µ ¶ x+3 y = ln x¡1 ii f y = log10 x for in y = 2x find dy = ax £ ln a dx c Show that if y = ax , then y = ln(x4 + x) ln(4x) y= x y= ln x p y = ln (ex x) µ f(x) = ln (x(x2 + 1)) y = log2 x b By substituting eln c y = ln 679 i ¶ x2 y = ln 3¡x µ ¶ x + 2x f (x) = ln x¡5 iii y = x log3 x dy dx Consider f (x) = ln(2x ¡ 1) ¡ a Find the x-intercept b Can f (0) be found? What is the significance of this result? c Find the gradient of the tangent to the curve at x = d For what values of x does f(x) have meaning? e Find f 00 (x) and hence explain why f (x) is concave down whenever f(x) has meaning f Graph the function Prove that ln x ln x for all x > Hint: Let f (x) = and find its greatest value x e x Consider the function f(x) = x ¡ ln x: Show that the graph of y = f (x) has a local minimum and that this is the only turning point Hence prove that ln x x ¡ for all x > Find da if e2a ln b2 ¡ a3 b + ln(ab) = 21 db D APPLICATIONS cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The applications we consider here are: ² tangents and normals ² rates of change ² curve properties ² displacement, velocity and acceleration ² optimisation (maxima and minima) black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_22\679IB_HL-2_22.CDR Friday, 12 March 2010 4:28:53 PM PETER IB_HL-2ed (680) 680 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 22) EXERCISE 22D Find the equation of the tangent to y = e¡x at the point where x = Find the equation of the tangent to y = ln(2 ¡ x) at the point where x = ¡1 The tangent to y = x2 ex at x = cuts the x and y-axes at A and B respectively Find the coordinates of A and B p Find the equation of the normal to y = ln x at the point where y = ¡1 Find the equation of the tangent to y = ex at the point where x = a Hence, find the equation of the tangent to y = ex which passes through the origin Consider f (x) = ln x a For what values of x is f (x) defined? b Find the signs of f (x) and f 00 (x) and comment on the geometrical significance of each c Sketch the graph of f (x) = ln x and find the equation of the normal at the point where y = Find, correct to decimal places, the angle between the tangents to y = 3e¡x y = + ex at their point of intersection and A radioactive substance decays according to the formula W = 20e¡kt grams where t is the time in hours a Find k given that the weight is 10 grams after 50 hours b Find the weight of radioactive substance present at: iii t = week i t = hours ii t = 24 hours c How long will it take for the weight to reach gram? d Find the rate of radioactive decay at: i t = 100 hours ii t = 1000 hours dW e Show that is proportional to the weight of substance remaining dt The temperature of a liquid after being placed in a refrigerator is given by T = + 95e¡kt o C where k is a positive constant and t is the time in minutes a Find k if the temperature of the liquid is 20o C after 15 minutes b What was the temperature of the liquid when it was first placed in the refrigerator? dT = c (T ¡ 5) for some constant c c Show that dt d At what rate is the temperature changing at: i t = mins ii t = 10 mins iii t = 20 mins? 10 The height of a certain species of shrub t years after it is planted is given by H(t) = 20 ln(3t + 2) + 30 cm, t > cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a How high was the shrub when it was planted? b How long will it take for the shrub to reach a height of m? c At what rate is the shrub’s height changing: ii 10 years after being planted? i years after being planted black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_22\680IB_HL-2_22.CDR Friday, 12 December 2008 12:39:23 PM TROY IB_HL-2ed (681) 681 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 22) 11 In the conversion of sugar solution to alcohol, the chemical reaction obeys the law A = s(1 ¡ e¡kt ), t > where t is the number of hours after the reaction commenced, s is the original sugar concentration (%), and A is the alcohol produced, in litres a Find A when t = b If s = 10 and A = after hours, find k c If s = 10, find the speed of the reaction at time hours d Show that the speed of the reaction is proportional to A ¡ s ex x Does the graph of y = f(x) have any x or y-intercepts? Discuss f (x) as x ! and as x ! ¡1: Find and classify any stationary points of y = f (x) Sketch the graph of y = f (x) showing all important features ex Find the equation of the tangent to f (x) = at the point where x = ¡1: x 12 Consider the function f(x) = a b c d e 13 A particle P moves in a straight line Its displacement from the origin O is given by ¡t s(t) = 100t + 200e cm where t is the time in seconds, t > a Find the velocity and acceleration functions b Find the initial position, velocity and acceleration of P c Discuss the velocity of P as t ! d Sketch the graph of the velocity function e Find when the velocity of P is 80 cm per second 14 A psychologist claims that the ability A to memorise simple facts during infancy years can be calculated using the formula A(t) = t ln t + where < t 5, t being the age of the child in years a At what age is the child’s memorising ability a minimum? b Sketch the graph of A(t) 15 One of the most common functions used in statistics is the normal distribution function ¡ x2 f(x) = p e 2¼ a Find the stationary points of the function and find intervals where the function is increasing and decreasing b Find all points of inflection c Discuss f (x) as x ! and as x ! ¡1 d Sketch the graph of y = f (x) showing all important features 16 A manufacturer of electric kettles performs a cost control study and discovers that to produce x kettles per day, the cost per kettle C(x) is given by ¶2 µ 30 ¡ x hundred dollars C(x) = ln x + 10 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 with a minimum production capacity of 10 kettles per day How many kettles should be manufactured to keep the cost per kettle a minimum? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_22\681IB_HL-2_22.CDR Monday, 19 November 2007 3:48:33 PM PETERDELL IB_HL-2ed (682) 682 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 22) y 17 Infinitely many rectangles which sit on the x-axis can be inscribed under the curve y = e¡x B Determine the coordinates of C such that rectangle ABCD has maximum area C A y = e- x x D 18 The revenue generated when a manufacturer sells x torches per day is given by ³ x ´ R(x) ¼ 1000 ln + + 600 dollars 400 Each torch costs the manufacturer $1:50 to produce plus fixed costs of $300 per day How many torches should be produced daily to maximise the profits made? 19 A quadratic of the form y = ax2 , a > 0, touches the logarithmic function y = ln x a If the x-coordinate of the point of contact is b, explain why ab2 = ln b and 2ab = b p b Deduce that the point of contact is ( e, ) c What is the value of a? d What is the equation of the common tangent? y = ax y y = ln x b x 20 A small population of wasps is observed After t weeks the population is modelled by 50 000 wasps, where t 25 P (t) = + 1000e¡0:5t Find when the wasp population is growing fastest 21 f(t) = atebt has a maximum value of when t = Find constants a and b 22 For the function f (x) = eax (x + 1), a R , show that: a f (x) = eax (a[x + 1] + 1) b f 00 (x) = aeax (a[x + 1] + 2) c if f (k) (x) = ak¡1 eax (a[x + 1] + k), k Z , then f (k+1) (x) = ak eax (a[x + 1] + [k + 1]) : 23 Consider the function f (x) = e¡x (x + 2) ii f 00 (x) iii f 000 (x) iv a Find i f (x) + (n) b Conjecture a formula for finding f (x), n Z c Use the principle of mathematical induction to prove your conjecture in b 24 Consider the function f (x) = xeax cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a Find f (n) (x) for n = 1, 2, and b Conjecture a formula for f (n) (x), n Z + c Use the principle of mathematical induction to prove your conjecture in b black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_22\682IB_HL-2_22.CDR Friday, 12 December 2008 12:41:58 PM TROY IB_HL-2ed (683) 683 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 22) E SOME SPECIAL EXPONENTIAL FUNCTIONS I surge functions: y = Ate¡bt , t > where A and b are positive constants y point of inflection µ ¶ 2A , b be b b t This model is used extensively when studying medicinal doses There is an initial rapid increase to a maximum and then a slower decay to zero The independent variable t is usually time, t > I logistic functions: y = C , t > where A, b and C are positive constants + Ae¡bt y y¡=¡C µ C 1+ A ¶ ln A C , b point of inflection C t The logistical model is useful for studying the growth of populations that are limited by resources or predators The independent variable t is usually time, t > EXERCISE 22E When a new pain killing injection is administered the effect is modelled by E = 750te¡1:5t units, where t > is the time in hours after the injection of the drug cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a Sketch the graph of E against t b What is the effect of the drug after i 30 minutes ii hours? c When is the drug most effective? d During the operating period, the level of the drug must be at least 100 units i When can the operation commence? ii How long has the surgeon to complete the operation if no further injection is possible? e Find t at the point of inflection of the graph What is the significance of this point? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_22\683IB_HL-2_22.CDR Monday, 19 November 2007 4:15:55 PM PETERDELL IB_HL-2ed (684) 684 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 22) a Prove that f (t) = Ate¡bt has i a local maximum at t = ii a point of inflection at t = b b b Use question to check the facts obtained in a The a b c d e The a b c d e The a b c d e f velocity of a body after t seconds, t > 0, is given by v = 25te¡2t cm s¡1 Sketch the velocity function Show that the body’s acceleration at time t is 25(1 ¡ 2t)e¡2t cm s¡2 When is the velocity increasing? Find the point of inflection of the velocity function What is its significance? Find the time interval when the acceleration is increasing 25 000 : number of ants in a colony after t months is modelled by A(t) = + 0:8e¡t Sketch the graph of A(t) What is the inital ant population? What is the ant population after months? Is there a limit to the population size? If so, what is it? At what time does the population reach 24 500? C : number of bees in a hive after t months is modelled by B(t) = + 0:5e¡1:73t What is the inital bee population? Find the percentage increase in the population after month Is there a limit to the population size? If so, what is it? If after months the bee population is 4500, what was the original population size? Find B (t) and use it to explain why the population is increasing over time Sketch the graph of B(t) C , show that: + Ae¡bt a f (t) = C is its horizontal asymptote C b it has a point of inflection with y-coordinate For the logistic function f(t) = REVIEW SET 22A Find a dy if: dx y = ex +2 b y= ex x2 c 2 Find the equation of the normal to y = e¡x ln(2y + 1) = xey at the point where x = Sketch the graphs of y = ex + and y = ¡ 5e¡x on the same set of axes Determine the exact coordinates of the points of intersection ex x¡1 a Find the x and y-intercepts b For what values of x is f (x) defined? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Consider the function f(x) = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_22\684IB_HL-2_22.CDR Monday, 19 November 2007 4:17:57 PM PETERDELL IB_HL-2ed (685) DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 22) 685 c Find the signs of f (x) and f 00 (x) and comment on the geometrical significance of each d Sketch the graph of y = f(x) and find the equation of the tangent at the point where x = The height of a tree t years after it was planted is given by H(t) = 60 + 40 ln(2t + 1) cm, t > a How high was the tree when it was planted? b How long does it take for the tree to reach: i 150 cm c At what rate is the tree’s height increasing after: i years ii ii 300 cm? 20 years? t A particle P moves in a straight line with position given by s(t) = 80e¡ 10 ¡ 40t m where t is the time in seconds, t > a Find the velocity and acceleration functions b Find the initial position, velocity, and acceleration of P c Discuss the velocity of P as t ! d Sketch the graph of the velocity function e Find when the velocity is ¡44 metres per second y Infinitely many rectangles can be inscribed under the curve y = e¡2x as shown Determine the coordinates of A such that the rectangle OBAC has maximum area C A y¡=¡e -2x B x A shirt maker sells x shirts per day with revenue function ³ x ´ + 1000 dollars R(x) = 200 ln + 100 The manufacturing costs are determined by the cost function C(x) = (x ¡ 100)2 + 200 dollars How many shirts should be sold daily to maximise profits? What is the maximum daily profit? Find where the tangent to y = ln (x2 + 3) at x = cuts the x-axis 10 Find a dy by first taking natural logarithms of both sides: dx (x2 + 2)(x ¡ 3) b y= y = x2x ¡ x3 REVIEW SET 22B dy if: dx Find a µ y = ln (x ¡ 3x) b y = ln x+3 x2 ¶ c ex+y = ln(y + 1) d y = xx cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find where the tangent to y = ln (x4 + 3) at x = cuts the y-axis black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_22\685IB_HL-2_22.CDR Tuesday, 27 November 2007 1:49:29 PM PETERDELL IB_HL-2ed (686) 686 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS (Chapter 22) Solve exactly for x: e2x = 3ex a e2x ¡ 7ex + 12 = b Consider the function f(x) = ex ¡ x a Find and classify any stationary points of y = f (x) b Discuss f (x) as x ! ¡1 and as x ! c Find f 00 (x) and draw its sign diagram Give geometrical interpretations for the signs of f 00 (x) d Sketch the graph of y = f (x) e Deduce that ex > x + for all x · ¸ (x + 2)3 x Differentiate with respect to x: a f (x) = ln(e + 3) b f (x) = ln x Find the exact roots of the following equations: µ ¶ b ln x ¡ ln a 3ex ¡ = ¡2e¡x = 10 x A particle P moves in a straight line with position given by s(t) = 25t ¡ 10 ln t cm, t > 1, where t is the time in minutes a Find the velocity and acceleration functions b Find the position, velocity, and acceleration when t = e minutes c Discuss the velocity as t ! d Sketch the graph of the velocity function e Find when the velocity of P is 12 cm per minute A manufacturer determines that the total weekly cost C of producing x clocks per ³ x ´2 dollars day is given by C(x) = 10 ln x + 20 ¡ 10 How many clocks per day should be produced to minimise the costs given that at least 50 clocks per day must be made to fill fixed daily orders? The graph of y = ae¡x for a > is shown P lies on the graph and the rectangle OAPB is drawn As P moves along the curve, the rectangle constantly changes shape Find the x-coordinate of P such that the rectangle OAPB has minimum perimeter y B P A y¡=¡ae -x x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 10 For the function f(x) = x + ln x : a find the values of x for which f (x) is defined b find the signs of f (x) and f 00 (x) and comment on the geometrical significance of each c sketch the graph of y = f(x) and find the equation of the normal at the point where x = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_22\686IB_HL-2_22.CDR Tuesday, 27 November 2007 1:50:01 PM PETERDELL IB_HL-2ed (687) Chapter 23 Derivatives of circular functions and related rates Contents: A B C D E Derivatives of circular functions The derivatives of reciprocal circular functions The derivatives of inverse circular functions Maxima and minima with trigonometry Related rates cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 23A Review set 23B black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_23\687IB_HL-2_23.CDR Tuesday, 20 November 2007 9:31:54 AM PETERDELL IB_HL-2ed (688) 688 DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES (Chapter 23) INTRODUCTION In Chapter 12 we saw that sine and cosine curves arise naturally from motion in a circle Click on the icon to observe the motion of point P around the unit circle Observe the graphs of P’s height relative to the x-axis, and then P’s displacement from the y-axis The resulting graphs are those of y = cos t and y = sin t DEMO Suppose P moves anticlockwise around the unit circle with constant linear speed of unit per second After 2¼ seconds, P will travel 2¼ units which is one full revolution y So, after t seconds P will travel through t radians, and at time t, P is at (cos t, sin t) P(cos¡q, sin¡q) q Note: ² The angular velocity of P is the time rate of change in ]AOP Angular velocity is only meaningful in motion along a circular or elliptical arc For the example above, the angular velocity dµ dµ of P is and = radian per sec dt dt x y P q ² If l is arc length AP, the linear speed of P is dl the time rate of change in l, which is dt For the example above, l = µr = µ £ = µ dµ dl = = radian per sec and dt dt l A(1, 0) x A DERIVATIVES OF CIRCULAR FUNCTIONS DERIVATIVES OF sin¡t AND cos¡t INVESTIGATION Our aim is to use a computer demonstration to investigate the derivatives of sin t and cos t What to do: Click on the icon to observe the graph of y = sin t A tangent with t-step of length unit moves across the curve, and its y-step is translated onto the slope graph Suggest the derivative of the function y = sin t DERIVATIVES DEMO cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Repeat the process in for the graph of y = cos t Hence suggest the derivative of the function y = sin t black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_23\688IB_HL-2_23.CDR Monday, 26 November 2007 4:26:01 PM PETERDELL IB_HL-2ed (689) DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES (Chapter 23) 689 From the investigation you may have deduced that d d (sin t) = cos t and (cos t) = ¡ sin t dt dt We will now show these derivatives using first principles To this we make use of the two results: sin µ =1 festablished in Chapter 19g ² If µ is in radians, then lim µ!0 µ ¢ ¡ ¢ ¡ sin S¡D festablished on page 314g ² sin S ¡ sin D = cos S+D 2 THE DERIVATIVE OF sin¡x Consider f (x) = sin x f (x + h) ¡ f(x) h Now f (x) = lim h!0 sin(x + h) ¡ sin x h!0 h ¡ ¢ ¡ x+h+x ¢ sin x+h¡x cos 2 = lim h!0 h ¢ ¡ ¢ ¡ h cos x + sin h2 = lim h!0 h ¢ ¡ ¡ ¢ cos x + h2 sin h2 = lim £ h h!0 2 = lim fidentity aboveg = cos x £ fas h ! 0, h ! 0, sin = cos x ¡h¢ h ! 1g if f (x) = sin x then f (x) = cos x, provided that x is in radians So, Alternatively, sin(x + h) ¡ sin x h sin x cos h + cos x sin h ¡ sin x = lim h!0 h f (x) = lim h!0 sin x(cos h ¡ 1) + cos x sin h h!0 h ¡ ¢ µ ¶ h sin x(¡2 sin ) sin h + lim cos x = lim h!0 h!0 h h ¡h¢ ¡h¢ sin sin sin h = lim ¡2 sin x + cos x £ lim h h!0 h!0 h = lim cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 = ¡2 sin x £ £ + cos x £ = cos x black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_23\689IB_HL-2_23.CDR Tuesday, 20 November 2007 9:46:10 AM PETERDELL IB_HL-2ed (690) 690 DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES (Chapter 23) THE DERIVATIVE OF cos¡x Consider y = cos x = sin ¡¼ ¢ ¡x ) y = sin u where u = ¼2 ¡ x dy dy du Now = fchain ruleg dx du dx = cos u £ (¡1) = ¡ cos( ¼2 ¡ x) = ¡ sin x if f (x) = cos x then f (x) = ¡ sin x, provided that x is in radians So, THE DERIVATIVE OF tan¡x Consider y = tan x = sin x cos x dy cos x cos x ¡ sin x(¡ sin x) = dx [cos x]2 ) Summary: For x in radians fquotient ruleg Function sin x cos x tan x cos2 x + sin2 x = cos2 x = which is sec2 x cos2 x Derivative cos x ¡ sin x sec2 x THE DERIVATIVES OF sin [f(x)], cos [f(x)] AND tan [f(x)] Consider y = sin[f (x)] ) y = sin u where u = f (x) Summary: For x in radians dy du dy = fchain ruleg dx du dx = cos u £ f (x) = cos[f (x)] £ f (x) Now Function Derivative sin[f (x)] cos[f (x)] f (x) cos[f (x)] ¡ sin[f (x)] f (x) tan[f(x)] sec2 [f(x)] f (x) Example a b tan2 (3x) a x sin x Differentiate with respect to x: b If y = tan2 (3x) = 4[tan(3x)]2 If y = x sin x then by the product rule then by the chain rule dy d = 8[tan(3x)]1 £ [tan(3x)] dx dx dy = (1) sin x + (x) cos x dx = sin x + x cos x = tan(3x) sec2 (3x) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 = 24 tan(3x) sec2 (3x) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_23\690IB_HL-2_23.CDR Tuesday, 20 November 2007 10:02:07 AM PETERDELL IB_HL-2ed (691) 691 DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES (Chapter 23) Example Find the equation of the tangent to y = tan x at the point where x = ¼4 Let f (x) = tan x so f( ¼4 ) = tan ¼4 = p ¡ ¢ f (x) = sec2 x so f ( ¼4 ) = [sec ¼4 ]2 = ( 2)2 = At ( ¼4 , 1), the tangent has slope y¡1 = which is y ¡ = 2x ¡ ¼2 x ¡ ¼4 ¢ ¡ or y = 2x + ¡ ¼2 ) the equation is Example B Find the rate of change in the area of triangle ABC as µ changes, at the time when µ = 60o 12 cm q A 10 cm C Area A = fArea = 12 ab sin Cg £ 10 £ 12 £ sin µ ) A = 60 sin µ dA ) = 60 cos µ dµ When µ = ¼ 3, cos µ = Note: µ must be in radians dA = 30 cm2 / radian dµ ) EXERCISE 23A Find a d dy for: dx y = sin(2x) y = sin(x + 1) ¡ ¢ y = sin x2 ¡ cos x g b y = sin x + cos x c y = cos(3x) ¡ sin x e y = cos(3 ¡ 2x) f y = tan(5x) h y = tan(¼x) i y = sin x ¡ cos(2x) cyan k sin x x l magenta yellow 95 x cos x 100 j 50 tan(2x) 75 i 25 h sin(3x) g 95 e2x tan x 100 f 50 ln(sin x) 75 e 25 d ex cos x c 95 tan x ¡ sin x 100 b 50 x2 + cos x 75 a 25 95 100 50 75 25 Differentiate with respect to x: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_23\691IB_HL-2_23.CDR Tuesday, 20 November 2007 10:05:14 AM PETERDELL e¡x sin x ³x´ cos x tan x IB_HL-2ed (692) 692 DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES (Chapter 23) Differentiate with respect to x: p b cos( x) a sin(x2 ) c p cos x d sin2 x cos3 (4x) ¡ ¢ cot3 x2 e cos3 x f cos x sin(2x) g cos(cos x) h i csc x j sec(2x) k sin (2x) l dy d2 y d3 y d4 y = 12x , = 24x, = 24 and higher = 4x3 , dx dx2 dx3 dx4 derivatives are all zero Now consider y = sin x If y = x4 a Find then dn y dy d2 y d3 y d4 y , , b Explain why can have four different values , dx dx2 dx3 dx4 dxn a If y = sin(2x + 3), show that d2 y + 4y = dx2 b If y = sin x + cos x, show that y 00 + y = where y 00 represents c Show that the curve with equation y = d2 y dx2 cos x cannot have horizontal tangents + sin x Find the equation of: a the tangent to y = sin x at the origin b the tangent to y = tan x at the origin c the normal to y = cos x at the point where x = ¼ d the normal to y = csc(2x) at the point where x = ¼4 On the Indonesian coast, the depth of water at time t hours after midnight is given by d = 9:3 + 6:8 cos(0:507t) metres a Is the tide rising or falling at 8:00 am? b What is the rate of change in the depth of water at 8:00 am? The voltage in a circuit is given by V (t) = 340 sin(100¼t) where t is the time in seconds At what rate is the voltage changing: a when t = 0:01 b when V (t) is a maximum? A piston is operated by rod AP attached to a flywheel of radius m AP = m P has coordinates (cos t, sin t) and point A is (¡x, 0) p a Show that x = ¡ sin2 t ¡ cos t b Find the rate at which x is changing at the instant when: iii i t=0s ii t = ¼2 s y piston P(cos¡t, sin¡t) 2m t xm x (1, 0) A t= 2¼ s cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 10 For each of the following functions, determine the position and nature of the stationary points on the interval x 2¼, then show them on a graph of the function b f(x) = cos(2x) c f(x) = sin2 x a f(x) = sin x black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_23\692IB_HL-2_23.CDR Thursday, 11 March 2010 10:54:36 AM PETER IB_HL-2ed (693) 693 DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES (Chapter 23) 11 Consider the function f(x) = sec x for x 2¼ a b c d For what values of x is f (x) undefined on this interval? Find the position and nature of any stationary points on this interval Prove that f(x + 2¼) = f (x), x R What is the geometrical significance of this? Sketch the graph of y = sec x, x [¡ ¼2 , 5¼ ] and show its stationary points 12 Determine the position and nature of the stationary points of y = sin(2x)+2 cos x on x 2¼ Sketch the graph of the function on this interval, and show the positions of the stationary points you found 13 A particle P moves along the x-axis with position given by x(t) = ¡ cos t cm where t is the time in seconds a State the initial position, velocity and acceleration of P b Describe the motion when t = ¼4 seconds c Find the times when the particle reverses direction on t 2¼ and find the position of the particle at these instants d When is the particle’s speed increasing on t 2¼? B THE DERIVATIVES OF RECIPROCAL CIRCULAR FUNCTIONS If y = csc x, then y = = u¡1 sin x where u = sin x du dy ¡1 cos x = ¡1u¡2 = dx dx (sin x)2 ) cos x sin x sin x = ¡ csc x cot x =¡ Likewise: ² if y = sec x, then ² if y = cot x, then Summary: dy = sec x tan x dx dy = ¡ csc2 x dx Function csc x sec x cot x Derivative ¡ csc x cot x sec x tan x ¡ csc2 x Example a y = csc(3x) cyan magenta b yellow 50 75 25 95 100 50 75 25 95 100 50 b y= dy d = ¡ csc(3x) cot(3x) (3x) dx dx = ¡3 csc(3x) cot(3x) 75 25 95 100 50 75 25 ) a y = csc(3x) for: p cot( x2 ) ¢1 ¡ y = cot( x2 ) ¡ ¢¡ dy ) = 12 cot( x2 ) £ ¡ csc2 ( x2 ) £ dx ¡ csc2 ( x2 ) = p cot( x2 ) 95 dy dx 100 Find black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_23\693IB_HL-2_23.CDR Tuesday, 20 November 2007 10:11:31 AM PETERDELL IB_HL-2ed (694) 694 DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES (Chapter 23) EXERCISE 23B d to prove that (sec x) = sec x tan x dx dy = ¡ csc2 x using the quotient rule b If y = cot x, prove that dx dy Find c y = sec(2x) for: a y = x sec x b y = ex cot x dx p x ¡x e y = x csc x f y = x csc x d y = e cot( ) cot x i y= p g y = ln(sec x) h y = x csc(x2 ) x Find the equation of the tangent to: b y = cot( x2 ) at x = ¼3 a y = sec x at x = ¼4 a Use y = (cos x)¡1 Find the equation of the normal to: a y = csc x at x = ¼6 C b y= p sec( x3 ) at x = ¼ THE DERIVATIVES OF INVERSE CIRCULAR FUNCTIONS y = sin x, x [¡ ¼2 , ¼ 2] has an inverse function f y is a 1-1 function and so ¡1 y = arcsin x or y = sin -1 x This function is called f ¡1 (x) = arcsin x or sin¡1 x Note: sin¡1 x is not -p _ y = sin x -1 or csc x sin x sin¡1 x is the inverse function of y = sin x -p _ and y = tan x, x ]¡ ¼2 , ¼2 [ has inverse function f ¡1 (x) = arctan x or tan¡1 x y p p -p _ p -1 x -p _ p x -p _ magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 y = cos x 75 25 95 100 50 75 25 y = arctan x or y = tan -1 x p -1 y = tan x p y=x cyan x -1 y y = arccos x or y = cos -1 x p csc x is the reciprocal function of y = sin x Likewise, y = cos x, x [ 0, ¼ ] has inverse function f ¡1 (x) = arccos x or cos¡1 x y=x p black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_23\694IB_HL-2_23.CDR Tuesday, 20 November 2007 10:18:31 AM PETERDELL IB_HL-2ed (695) 695 DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES (Chapter 23) Example Show that: arctan( 12 ) + arctan( 13 ) = ¼ Let arctan( 12 ) = µ and arctan( 13 ) = Á, so tan µ = Now tan(µ + Á) = 1 tan µ + tan Á + =1 = ¡ tan µ tan Á ¡ ( 12 )( 13 ) ) µ+Á= ¼ and tan Á = + k¼, k Z But < arctan( 12 ) < and < arctan( 13 ) < ¼ ¼ ) < arctan( 12 ) + arctan( 13 ) < ¼ ) arctan( 12 ) + arctan( 13 ) = µ + Á = ¼ EXERCISE 23C.1 Use a calculator to check the graphs of y = arcsin x, y = arccos x and y = arctan x Find, giving your answer in radians: a arccos(1) b arcsin(¡1) c e arcsin( 12 ) f p arccos( ¡2 ) i arctan(¡ p13 ) j sin¡1 (¡0:767) Find the exact solution of: Use tan(µ ¡ Á) = a arcsin x = tan µ ¡ tan Á + tan µ tan Á d arctan(¡1) g arctan(1) p arctan( 3) h arccos(¡ p12 ) k cos¡1 (0:327) l tan¡1 (¡50) b arctan(3x) = ¡ ¼4 ¼ to show that arctan(5) ¡ arctan( 23 ) = ¼4 Without using technology, show that: arctan( 15 ) + arctan( 23 ) = a ¼ b arctan( 43 ) = arctan( 12 ) ) Find the exact value of arctan( 15 ) ¡ arctan( 239 DERIVATIVES OF INVERSE CIRCULAR FUNCTIONS Consider the differentiation of y = arcsin x by these methods: Method 1: = arcsin x then x = sin y p = cos y = ¡ sin2 y = p ¡ x2 cyan magenta yellow so 95 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 , x ] ¡ 1, [ = p ¡ x2 dy dx dy fFrom the chain rule, = = 1, dx dy dy dy dx and are reciprocals.g dx dy 100 If y dx ) dy dx ) dy dy ) dx black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_23\695IB_HL-2_23.CDR Tuesday, 20 November 2007 10:22:47 AM PETERDELL IB_HL-2ed (696) 696 DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES (Chapter 23) Method 2: If y = arcsin x then x = sin y Using implicit differentiation, = cos y dy = dx cos y dy fy [¡ ¼2 , = p dx ¡ sin2 y dy , x ] ¡ 1, [ = p dx ¡ x2 ) ) ) If y = arcsin x, dy , x ] ¡ 1, [ =p dx ¡ x2 dy dx If y = arccos x, dy ¡1 , x ]¡1, [ =p dx ¡ x2 ¼ ] ) cos y > 0g If y = arctan x, dy , x2R = dx + x2 EXERCISE 23C.2 If y = arccos x, show that dy ¡1 , x ]¡1, [ =p dx ¡ x2 If y = arctan x, show that dy , x2R = dx + x2 a dy for: dx y = arctan(2x) b y = arccos(3x) c y = arcsin( x4 ) d y = arccos( x5 ) e y = arctan(x2 ) f y = arccos(sin x) Find Find dy for: dx a b y = x arcsin x y = ex arccos x c y = e¡x arctan x dy for x ] ¡ a, a [ =p dx a ¡ x2 dy a b Prove that if y = arctan( xa ), then for x R = dx a + x2 dy c If y = arccos( xa ), find dx a Prove that if y = arcsin( xa ), then Sonia approaches a painting which has its bottom edge m above eye level and its top edge m above eye level 1m a Given ® and µ as shown in the diagram, find tan ® and tan(® + µ) b Find µ in terms of x only Hint: µ = (® + µ) ¡ ® painting dµ = ¡ and dx x +4 x +9 dµ hence find x when = dx d Interpret the result you have found in c q cyan magenta yellow 2m a 95 xm 100 50 eye level 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 c Show that black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_23\696IB_HL-2_23.CDR Tuesday, 20 November 2007 10:25:25 AM PETERDELL IB_HL-2ed (697) 697 DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES (Chapter 23) D MAXIMA AND MINIMA WITH TRIGONOMETRY Example B Two corridors meet at right angles and are m and m wide respectively µ is the angle marked on the 2m given figure and AB is a thin metal tube which must q be kept horizontal and cannot be bent as it moves around the corner from one corridor to the other DEMO a Show that the length AB is given by A 3m L = sec µ + csc µ ³q ´ dL = when µ = arctan 23 ¼ 41:14o b Show that dµ ³q ´ c Find L when µ = arctan 23 and comment on the significance of this value a cos µ = and sin µ = a b so sec µ = a b and csc µ = ) L = a + b = sec µ + csc µ B dL = sec µ tan µ + 2(¡ csc µ cot µ) dµ sin µ cos µ ¡ = cos2 µ sin2 µ b b a sin3 µ ¡ cos3 µ = cos2 µ sin2 µ Thus c q A q dL = , sin3 µ = cos3 µ dµ ) tan3 µ = 23 q ³q ´ ) tan µ = 23 and so µ = arctan 23 ¼ 41:14o dL : dµ Sign diagram of - 30° 60° 41.14° dL ¼ ¡4:93 < 0, dµ + 90° dL ¼ 9:06 > dµ cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Thus, AB is minimised when µ ¼ 41:14o At this time L ¼ 7:023 metres, so if we ignore the width of the rod then the greatest length of rod able to be horizontally carried around the corner is 7:023 m black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_23\697IB_HL-2_23.CDR Tuesday, 20 November 2007 10:33:10 AM PETERDELL IB_HL-2ed (698) 698 DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES (Chapter 23) EXERCISE 23D A circular piece of tinplate of radius 10 cm has segments removed as illustrated If µ is the measure of angle COB, show that the remaining area is given by A = 50(µ + sin µ) of a degree when the Hence, find µ to the nearest 10 area A is a maximum B C 10 cm q A symmetrical gutter is made from a sheet of metal 30 cm wide by bending it twice as shown For µ as indicated: q q a deduce that the cross-sectional area is given by A = 100 cos µ(1 + sin µ) 10 cm dA end view = when sin µ = 12 or ¡1 b show that dµ c for what value of µ does the gutter have maximum carrying capacity? Hieu can row a boat across a circular lake of radius km at km h¡1 He can walk around the edge of the lake at km h¡1 What is the longest possible time Hieu could take to get from P to R by rowing from P to Q and then walking from Q to R? Q q km P km Fence AB is m high and is m from a house XY is a ladder which touches the ground at X, the house at Y, and the fence at B Y a If L is the length of XY, show that L = sec µ + csc µ b Show that B sin3 µ ¡ cos3 µ dL = dµ sin2 µ cos2 µ R house 2m q X A 2m c What is the length of the shortest ladder XY which touches at X, B and Y? In Example 6, suppose the corridors are those in a hospital and are m wide and m wide respectively What is the maximum length of thin metal tube that can be moved around the corner? Remember it must be kept horizontal and must not be bent 4m a cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 3m black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_23\698IB_HL-2_23.CDR Monday, 26 November 2007 4:26:58 PM PETERDELL IB_HL-2ed (699) 699 DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES (Chapter 23) A 5m q X B How far should X be from A if angle µ is to be a maximum? 4m 3m A and B are two homesteads A pump house is to be located at P on the canal to pump water to both A and B a If A and B are a km and b km from the canal respectively, show that: AP + PB = a sec µ + b sec Á = L, say b Show that A q B f a sin µ b sin Á dÁ dL = + dµ cos µ cos2 Á dµ P canal c Explain why a tan µ +b tan Á is a constant and hence show that dÁ ¡a cos2 Á = dµ b cos2 µ dL = , sin µ = sin Á dµ e What can be deduced from d? Include all reasoning and an appropriate test d Hence, show that Note: Question is solvable using geometry only The solution is to reflect point B in the line representing the canal We call the image B0 and join this point to A The location of point P which minimises AP + PB is the intersection of [AB0 ] and the edge of the canal E A B P B' canal RELATED RATES A m ladder rests against a vertical wall at point B, with its feet at point A on horizontal ground The ladder slips and slides down the wall The following diagram shows the positions of the ladder at certain instances Click on the icon to view the motion of the sliding ladder B 5m A xm y m wall O cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 DEMO black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_23\699IB_HL-2_23.CDR Tuesday, 20 November 2007 10:56:55 AM PETERDELL IB_HL-2ed (700) 700 DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES (Chapter 23) If AO = x m and OB = y m, then x2 + y2 = 52 : fPythagorasg 5m Differentiating this equation with respect to time t dx dy gives 2x + 2y =0 dt dt dy dx +y = or x dt dt Ac Av Az Ax xm Bz Bx Bc Bv O This equation is called a differential equation and describes the motion of the ladder at any instant dx is the rate of change in x with respect to time t, and is the speed of A Notice that dt relative to point O Likewise, dy is the rate at which B moves downwards dt Observe that: ² dx is positive as x is increasing dt ² dy is negative as y is decreasing dt Problems involving differential equations where one of the variables is t (time) are called related rates problems The method for solving such problems is: Step 1: Draw a large, clear diagram of the general situation Sometimes two or more diagrams are necessary Step 2: Write down the information, label the diagram(s), and make sure you distinguish between the variables and the constants Step 3: Write down an equation connecting the variables Step 4: Differentiate the equation with respect to t to obtain a differential equation Step 5: Finally, solve for the particular case which is some instant in time ² ² ² ² Checklist for finding relationships: Pythagoras’ theorem Similar triangles where corresponding sides are in proportion Right angled triangle trigonometry Sine and Cosine Rules Warning: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 We must not substitute the particular case values too early Otherwise we will incorrectly treat variables as constants The differential equation in fully generalised form must be established first black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_23\700IB_HL-2_23.CDR Tuesday, 20 November 2007 11:40:05 AM PETERDELL IB_HL-2ed (701) 701 DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES (Chapter 23) Example A m long ladder rests against a vertical wall with its feet on horizontal ground The feet on the ground slip, and at the instant when they are m from the wall, they are moving at 10 m¡s¡1 At what speed is the other end of the ladder moving at this instant? B 5m Let OA = x m and OB = y m fPythagorasg ) x2 + y2 = 52 Differentiating with respect to t gives dy dx dy dx + 2y = or x +y =0 2x dt dt dt dt ym A O xm We must perform differentiation before we substitute values for the particular case Otherwise we will incorrectly treat the variables as constants Particular case: B dx = 10 m s¡1 , dt dy ) 3(10) + =0 dt dy ¡1 ) = ¡ 15 = ¡7:5 m s dt At the instant 5m A 4m O 3m Thus OB is decreasing at 7:5 m s¡1 ) B is moving down the wall at 7:5 m¡s¡1 at that instant Example The volume of a cube increases at a constant rate of 10 cm3 per second Find the rate of change in its total surface area at the instant when its sides are 20 cm long Let x cm be the lengths of the sides of the cube, so A = 6x2 dx dA dx dV ) = 12x and = 3x2 dt dt dt dt Particular case: At the instant when x = 20, ) 10 = £ 202 £ dV = 10 dt x cm dx dt dx 10 = 120 cm s¡1 = 1200 dt dA Thus cm2 s¡1 = cm2 s¡1 = 12 £ 20 £ 120 dt ) the surface area is increasing at cm2 per second magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ) cyan and V = x3 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_23\701IB_HL-2_23.CDR Tuesday, 27 November 2007 10:13:02 AM PETERDELL x cm x cm IB_HL-2ed (702) 702 DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES (Chapter 23) Example Triangle ABC is right angled at A, and AB = 20 cm The angle ABC increases at a constant rate of 1o per minute At what rate is BC changing at the instant when angle ABC measures 30o ? b =µ Let ABC C x cm q A B 20 cm and BC = x cm 20 = 20x¡1 Now cos µ = x dx dµ ) ¡ sin µ = ¡20x¡2 dt dt Remember that dµ must dt be measured as radians/time unit 20 Particular case: cos 30o = x p dµ 20 and ) = = 1o per x dt ¼ = 180 radians per 40 ) x= p 40 cm 3 dx £ 1600 dt dx dt Thus ¡ sin 30o £ ¼ 180 = ¡20 £ ) ¡ 12 £ ¼ 180 = ¡ 80 30° 20 cm dx ¼ £ 80 = 360 cm per dt ¼ 0:2327 cm per ) BC is increasing at approximately 0:233 cm per ) EXERCISE 23E a and b are variables related by the equation ab3 = 40 At the instant when a = 5, b is increasing at unit per second What is happening to a at this instant? The area of a variable rectangle remains constant at 100 cm2 The length of the rectangle is decreasing at cm per minute At what rate is the breadth increasing at the instant when the rectangle is a square? A stone is thrown into a lake and a circular ripple moves out at a constant speed of m s¡1 Find the rate at which the circle’s area is increasing at the instant when: a t = seconds b t = seconds Air is being pumped into a spherical weather balloon at a constant rate of 6¼ m3 per minute Find the rate of change in its surface area at the instant when its radius is m cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 5 For a given mass of gas in a piston, pV¡¡1:5¡¡=¡400 where p is the pressure in N/m2 and V is the volume in m3 If the pressure increases at 3¡N/m2 per minute, find the rate at which the volume is changing at the instant when the pressure is 50¡N/m2 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_23\702IB_HL-2_23.CDR Tuesday, 20 November 2007 12:00:21 PM PETERDELL V IB_HL-2ed (703) 703 DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES (Chapter 23) Wheat runs from a hole in a silo at a constant rate and forms a conical heap whose base radius is treble its height If after minute, the height of the heap is 20 cm, find the rate at which the height is rising at this instant A trough of length m has a uniform cross-section which is an equilateral triangle with sides m Water leaks from the bottom of the trough at a constant rate of 0:1 m3 /min Find the rate at which the water level is falling at the instant when it is 20 cm deep 1m end view Two jet aeroplanes fly on parallel courses which are 12 km apart Their air speeds are 200 m s¡1 and 250 m s¡1 respectively How fast is the distance between them changing at the instant when the slower jet is km ahead of the faster one? A ground-level floodlight located 40 m from the foot of a building shines in the direction of the building A m tall person walks directly towards the building at m s¡1 How fast is L the person’s shadow on the building shortening at the instant when the person is: a 20 m from the building b 10 m from the building? 40 m 10 A right angled triangle ABC has a fixed hypotenuse AC of length 10 cm, and side AB increases at 0:1 cm per second At what rate is angle CAB increasing at the instant when the triangle is isosceles? 11 An aeroplane passes directly overhead then flies horizontally away from an observer at an altitude of 5000 m with an air speed of 200 m s¡1 At what rate is its angle of elevation to the observer changing at the instant when the angle of elevation is: b 30o ? a 60o 12 A rectangle PQRS has PQ of length 20 cm and QR increases at a constant rate of cm s¡1 At what rate is the acute angle between the diagonals of the rectangle changing at the instant when QR is 15 cm long? 13 Triangle PQR is right angled at Q and PQ is cm long If QR increases at cm per minute, find the rate of change in angle P at the instant when QR is cm 14 Two cyclists A and B leave X simultaneously at 120o to one another with constant speeds of 12 m s¡1 and 16 m s¡1 respectively Find the rate at which the distance between them is changing after minutes cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 15 AOB is a fixed diameter of a circle of radius cm A point P moves around the circle at a constant rate of revolution in 10 seconds Find the rate at which AP is changing at the instant when: a AP is cm and increasing b P is at B black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_23\703IB_HL-2_23.CDR Monday, 26 November 2007 4:30:31 PM PETERDELL P A B IB_HL-2ed (704) 704 DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES (Chapter 23) 16 Shaft AB is 30 cm long and is attached to a flywheel at A B is confined to motion along OX The radius of the wheel is 15 cm, and the wheel rotates clockwise at 100 revolutions per second Find the rate of change in angle ABO when angle AOX is: b 180o a 120o A B X 17 A farmer has a water trough of length m which has 1m a semi-circular cross-section of diameter m Water q is pumped into the trough at a constant rate of 0:1 m per minute a Show that the volume of water in the trough is given by V = µ ¡ sin µ where µ is the angle as illustrated (in radians) b Find the rate at which the water level is rising at the instant when it is 25 cm deep dµ dh Hint: First find and then find at the given instant dt dt REVIEW SET 23A Differentiate with respect to x: a sin(5x) ln(x) b c sin(x) cos(2x) e¡2x tan x Show that the equation of the tangent to y = x tan x at x = ¼2 (2 + ¼)x ¡ 2y = Find f (x) and f 00 (x) for: a f (x) = sin x ¡ cos(2x) b f (x) = ¼ is p x cos(4x) A particle moves in a straight line along the x-axis with position given by x(t) = + sin(2t) cm after t seconds a Find the initial position, velocity and acceleration of the particle b Find the times when the particle changes direction during t ¼ secs c Find the total distance travelled by the particle in the first ¼ seconds p Consider f(x) = cos x for x 2¼ a For what values of x is f (x) meaningful? b Find f (x) and hence find intervals where f (x) is increasing and decreasing c Sketch the graph of y = f (x) on x 2¼ A cork moves up and down in a bucket of water such that the distance from the centre of the cork to the bottom of the bucket is given by s(t) = 30 + cos(¼t) cm where t is the time in seconds, t > cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a Find the cork’s velocity at times t = 0, 12 , 1, 12 , sec b Find the time intervals when the cork is falling black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_23\704IB_HL-2_23.CDR Tuesday, 20 November 2007 12:12:55 PM PETERDELL s(t) IB_HL-2ed (705) 705 DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES (Chapter 23) The point (3 cos µ, sin µ) lies on a curve, µ [0, 2¼] a Find the equation of the curve in Cartesian form dy b Find in terms of µ dx c Suppose a tangent to the curve meets the x-axis at A and the y-axis at B Find the smallest area of triangle OAB and the values of µ when it occurs Four straight sticks of fixed length a, b, c and d are hinged together at P, Q, R and S a Use the cosine rule to find an equation which connects a, b, c, d, cos µ and cos Á and hence cd sin Á dµ = : show that dÁ ab sin µ Q q a b P R d c f b Hence, show that the area of quadrilateral PQRS is a maximum when it is a cyclic quadrilateral S A light bulb hangs from the ceiling at height h metres above the floor, directly above point N At any point A on the floor which is x metres from the light bulb, the illumination I is p given by cos µ units I= x2 p a If NA = metre, show that at A, I = cos µ sin2 µ b The light bulb may be lifted or lowered to change the intensity at A Assuming NA = metre, find the height the bulb should be above the floor for greatest illumination at A ceiling light bulb L q h x floor N A REVIEW SET 23B x a y=p sec x dy for: dx Find b y = ex cot(2x) Find the equation of: a the tangent to y = sec x at the point where x = b ¼ the normal to y = arctan x at the point where x = c y = arccos( x2 ) p 3 A man on a jetty pulls a boat directly towards him so the rope is coming in at a rate of 20 metres per minute The rope is attached to the boat m above water level and his hands are m above water level How fast is the boat approaching the jetty at the instant when it is 15 m from the jetty? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Two runners run in different directions, 60o apart A runs at m s¡1 and B runs at m s¡1 B passes through X seconds after A passes through X At what rate is the distance between them increasing at the time when A is 20 metres past X? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_23\705IB_HL-2_23.CDR Monday, 26 November 2007 4:31:26 PM PETERDELL A’s path X 60° B’s path IB_HL-2ed (706) 706 DERIVATIVES OF CIRCULAR FUNCTIONS AND RELATED RATES (Chapter 23) a If f (x) = arcsin x + arccos x, find f (x) What can we conclude about f (x)? µ ¶ 1¡a b Simplify arctan + arctan a 1+a c Find the value of arctan( 12 ) + arctan( 15 ) + arctan( 18 ) A and B are two houses directly opposite one A another and km from a straight road CD MC km D is km and C is a house at the roadside M A power unit is to be located on DC at P such km that PA + PB + PC is to be a minimum so B that the cost of trenching and cable will be as small as possible q q C P km a What cable length would be required if P is at i M ii C? b Show that if µ = Ab PM = Bb PM, then the length of cable will be L = csc µ + ¡ cot µ metres ¡ cos µ dL and hence show that the minimum length of cable = dµ sin2 µ p required is (3 + 3) km c Show that Water exits a conical tank at a constant rate of 0:2 m3 /minute If the surface of the water has radius r: a find V (r), the volume of the water remaining b find the rate at which the surface radius is changing at the instant when the height of water is m 6m rm 8m Consider a circle with centre O and radius r A, B and C are fixed points An ant starts at B and moves at D P constant speed v in a straight line r to point P q a The ant then moves along the arc r a C A B from P to C via D at constant speed w where w¡>¡v a Show that the total time for the journey is p r2 + (a + r)2 ¡ 2r(a + r) cos µ r(¼ ¡ µ) + T = v w µ ¶ a+r rv dT = sin ® ¡ b Show that dµ v (a + r)w cyan magenta yellow 95 100 50 rv (a + r)w 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 c Prove that T is minimised when sin ® = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_23\706IB_HL-2_23.CDR Tuesday, 20 November 2007 1:33:31 PM PETERDELL IB_HL-2ed (707) 24 Chapter Integration Contents: A B C D E F G Antidifferentiation The fundamental theorem of calculus Integration Integrating eax +b and (ax¡+¡b)n Integrating f(u)u0(x) by substitution Integrating circular functions Definite integrals cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 24A Review set 24B Review set 24C black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\707IB_HL-2_24.CDR Tuesday, 20 November 2007 2:04:02 PM PETERDELL IB_HL-2ed (708) 708 INTEGRATION (Chapter 24) In the previous chapters we used differential calculus to find the derivatives of many types of functions We also used it in problem solving, in particular to find the slopes of graphs and rates of changes, and to solve optimisation problems In this chapter we consider integral calculus This involves antidifferentiation which is the reverse process of differentiation We will see that integral calculus also has many useful applications, including: ² ² ² ² ² ² ² ² ² finding areas where curved boundaries are involved finding volumes of revolution finding distances travelled from velocity functions finding hydrostatic pressure finding work done by a force finding centres of mass and moments of inertia solving problems in economics and biology solving problems in statistics solving differential equations A ANTIDIFFERENTIATION In many problems in calculus we know the rate of change of one variable with respect to another, but we not have a formula which relates the variables In other words, we know dy , but we need to know y in terms of x dx Examples of problems we need to solve include: ² The slope function f (x) of a curve is 2x + and the curve passes through the origin What is the function y = f (x)? dT = 10e¡t o C per minute where t > What dt is the temperature function given that initially the temperature was 11o C? ² The rate of change in temperature is dy or f(x) from f (x) is the reverse process of dx differentiation We call it antidifferentiation The process of finding y from dy = x2 , what is y in terms of x? dx From our work on differentiation we know that when we differentiate power functions the index reduces by We hence know that y must involve x3 Consider the following problem: Now if y = x3 If dy = 3x2 , so if we start with y = 13 x3 dx then then cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 However, if y = 13 x3 + 2, y = 13 x3 + 100 or y = 13 x3 ¡ then black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\708IB_HL-2_24.CDR Tuesday, 20 November 2007 2:12:53 PM PETERDELL dy = x2 dx dy = x2 dx IB_HL-2ed (709) INTEGRATION (Chapter 24) 709 In fact, there are infinitely many such functions of the form y = 13 x3 + c where c is dy an arbitrary constant which will give = x2 Ignoring the arbitrary constant, we say dx that 13 x3 is the antiderivative of x2 It is the simplest function which when differentiated gives x2 If F (x) is a function where F (x) = f (x) we say that: ² ² the derivative of F (x) is f (x) and the antiderivative of f (x) is F (x): Example Find the antiderivative of: a x3 a b e2x 4x d ¡ 2x ¢ e = e2x £ 2, dx Since d ¡ 2x ¢ = e dx ) the antiderivative of e2x is c p x We know that the derivative of x4 involves x3 d ¡ 4¢ d ¡ 4¢ x = 4x3 , Since x = x3 dx dx ) the antiderivative of x3 is b c 1 p = x¡ x £ e2x £ = e2x 2x 2e : d 1 (x ) = 12 x¡ dx p ) the antiderivative of p is x x Now ) d 1 (2x ) = 2( 12 )x¡ = x¡ dx EXERCISE 24A a Find the antiderivative of: 1 iii x5 iv x¡2 v x¡4 vi x vii x¡ i x ii x2 b From your answers in a, predict a general rule for the antiderivative of xn a Find the antiderivative of: x ii e5x iii e x iv e0:01x v e¼x vi e i e2x b From your answers in a, predict a general rule for the antiderivative of ekx where k is a constant cyan magenta yellow e3x+1 by differentiating e3x+1 d (2x + 1)3 by differentiating (2x + 1)4 95 b 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find the antiderivative of: a 6x2 +4x by differentiating x3 +x2 p p c x by differentiating x x black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\709IB_HL-2_24.CDR Tuesday, 20 November 2007 4:24:17 PM PETERDELL IB_HL-2ed (710) 710 INTEGRATION (Chapter 24) B THE FUNDAMENTAL THEOREM OF CALCULUS Sir Isaac Newton and Gottfried Wilhelm Leibniz showed the link between differential calculus and the definite integral or limit of an area sum This link is called the fundamental theorem of calculus The beauty of this theorem is that it enables us to evaluate complicated summations We have already observed in Chapter 19 that: y=ƒ(x) y If f(x) is a continuous positive function on an interval [a, b] then the Z b f (x) dx area under the curve between x = a and x = b is x a a INVESTIGATION b THE AREA FUNCTION Consider the constant function f (x) = y The corresponding area function is Z t dx A(t) = y=5 a = shaded area in graph = (t ¡ a)5 = 5t ¡ 5a a t¡-a t x ) we can write A(t) in the form F (t) ¡ F (a) where F (t) = 5t or equivalently, F (x) = 5x What to do: What is the derivative F (x) of the function F (x) = 5x? How does this relate to the function f (x)? Consider the simplest linear function f (x) = x The corresponding area function is Z t x dx A(t) = y¡=¡x y t a a = shaded area in graph µ ¶ t+a = (t ¡ a) a t-a t x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a Can you write A(t) in the form F (t) ¡ F (a)? b If so, what is the derivative F (x)? How does it relate to the function f (x)? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\710IB_HL-2_24.cdr Wednesday, 21 November 2007 3:22:56 PM PETERDELL IB_HL-2ed (711) 711 INTEGRATION (Chapter 24) Consider f(x) = 2x + The corresponding area function is Z t y (2x + 3) dx A(t) = y=2x+3 a = shaded area in graph µ ¶ 2t + + 2a + = (t ¡ a) 2t+3 2a+3 a Can you write A(t) in the form F (t) ¡ F (a)? b If so, what is the derivative F (x)? How does it relate to the function f (x)? a t-a t x Repeat the procedure in and for finding the area functions of a f(x) = 12 x + b f (x) = ¡ 2x Do your results fit with your earlier observations? If f (x) = 3x2 + 4x + 5, predict what F (x) would be without performing the algebraic procedure From the investigation you should have discovered that, for f(x) > 0, Z t f (x) dx = F (t) ¡ F (a) where F (x) = f (x) F (x) is the antiderivative of f(x) a The following argument shows why this is true for all functions f (x) > Consider a function y = f(x) which has antiderivative F (x) and an area function A(t) which is the area from x = a to x = t, y Z t y=ƒ(x) f (x) dx i.e., A(t) = a A(t) is clearly an increasing function and A(t) A(a) = (1) a Now consider a narrow strip of the region between x = t and x = t + h t b x y y=ƒ(x) The area of this strip is A(t + h) ¡ A(t) Since the narrow strip is contained within two rectangles then area of smaller A(t + h) ¡ A(t) area of larger rectangle rectangle a b x t t+h ) hf(t) A(t + h) ¡ A(t) hf(t + h) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 A(t + h) ¡ A(t) f (t + h) h 95 100 50 75 25 95 100 50 75 25 ) f (t) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\711IB_HL-2_24.CDR Tuesday, 20 November 2007 2:24:46 PM PETERDELL IB_HL-2ed (712) 712 INTEGRATION (Chapter 24) Taking the limit as h ! gives y=ƒ(x) h f(t) A (t) f (t) ) A0 (t) = f(t) The area function A(t) is an antiderivative of f(t), ƒ(t+h) ƒ(t) so A(t) and F (t) differ by a constant ) A(t) = F (t) + c x Letting t = a, A(a) = F (a) + c t t+h enlarged strip But A(a) = ffrom (1)g so c = ¡F (a) ) A(t) = F (t) ¡ F (a) Z b ) letting t = b, f (x) dx = F (b) ¡ F (a) a This result is in fact true for all continuous functions f (x), and can be stated as the fundamental theorem of calculus: Z b f (x) dx = F (b) ¡ F (a) For a continuous function f (x) with antiderivative F (x), a The fundamental theorem of calculus has many applications beyond the calculation of areas ds For example, given a velocity function v(t) we know that = v dt So, s(t) is the antiderivative of v(t) and by the fundamental theorem of calculus, Z t2 v(t) dt = s(t2 ) ¡ s(t1 ) gives the displacement over the time interval [t1 , t2 ] t1 PROPERTIES OF DEFINITE INTEGRALS The following properties of definite integrals can all be deduced from the fundamental theorem of calculus: Z a a Z magenta b Z b 100 a 95 f (x) dx a [f(x) § g(x)]dx = 50 50 75 25 95 yellow c f (x) dx = a 95 50 75 25 b ² f(x) dx a 95 cyan Z b c f (x) dx = c 100 50 75 25 a 100 Z b a fc is a constantg c f (x) dx + a Z ² ² f (x) dx Z b 75 f (x) dx = ¡ b Z b 25 Z a ² c dx = c(b ¡ a) a 100 Z b ² f (x) dx = 0 ² Z black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\712IB_HL-2_24.CDR Tuesday, 20 November 2007 2:27:43 PM PETERDELL Z f (x) dx § b g(x) dx a IB_HL-2ed (713) 713 INTEGRATION (Chapter 24) Example proof: Z Z b f (x) dx + a y y=ƒ(x) c f (x) dx b = F (b) ¡ F (a) + F (c) ¡ F (b) A1 = F (c) ¡ F (a) Z c f (x) dx = a a Z Z b b c x Z c f (x) dx + a A2 c f (x) dx = A1 + A2 = b f (x) dx a Example Use the fundamental theorem of calculus to find the area: a between the x-axis and y = x2 from x = to x = p b between the x-axis and y = x from x = to x = a f (x) = x2 has antiderivative F (x) = y y = x2 ) the area = R1 x3 x2 dx = F (1) ¡ F (0) = = f (x) = y units2 p x = x2 y= x F (x) = x2 x has antiderivative p = 23 x x ) the area = ¡0 x b 3 R9 1 x dx = F (9) ¡ F (1) = = 17 13 £ 27 ¡ £1 units TI cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 C black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\713IB_HL-2_24.CDR Tuesday, 20 November 2007 2:38:01 PM PETERDELL IB_HL-2ed (714) 714 INTEGRATION (Chapter 24) EXERCISE 24B Use the fundamental theorem of calculus to show that: Z a f(x) dx = and explain the result graphically a Z a b b Z a c Z c dx = c(b ¡ a) where c is a constant a Z b f(x) dx = ¡ b f(x) dx a Z b d b c f (x) dx = c Z f (x) dx where c is a constant a a Z b e a Z b [f(x) + g(x)] dx = b f (x) dx + a g(x) dx a Use the fundamental theorem of calculus to find the area between the x-axis and: a y = x3 from x = to x = b y = x3 from x = to x = p y = x from x = to x = c y = x2 + 3x + from x = to x = d e y = ex from x = to x = 1:5 f g y = x3 + 2x2 + 7x + from x = to x = 1:25 y = p from x = to x = x Check each answer using technology Using technology, find correct to significant figures, the area between the x-axis and: a c y = ex from x = to x = 1:5 p y = ¡ x2 from x = to x = y = (ln x)2 from x = to x = b a Use the fundamental theorem of calculus to show that Z b Z b y (¡f (x)) dx = ¡ f (x) dx a a a x b b Use the result in a to show that if f (x) for all x on [a, b] then Z b f (x) dx the shaded area = ¡ y¡=¡¦(x) a Z c Calculate: i Z (¡x2 ) dx ii 0 (x2 ¡ x) dx Z d Use graphical evidence and known area facts to find cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\714IB_HL-2_24.CDR Wednesday, 21 November 2007 3:23:44 PM PETERDELL Z iii 3x dx ¡2 ¡ p ¢ ¡ ¡ x2 dx IB_HL-2ed (715) INTEGRATION (Chapter 24) C 715 INTEGRATION Earlier we showed that the antiderivative of x2 was 3x We also showed that any function of the form 13 x3 + c where c is any constant, has derivative x2 R x dx = 13 x3 + c We say that the integral of x2 is 13 x3 + c and write We read this as “the integral of x2 with respect to x” R if F (x) = f (x) then f (x) dx = F (x) + c In general, DISCOVERING INTEGRALS Since integration is the reverse process of differentiation we can sometimes discover integrals by differentiation For example: R 4x dx = x4 + c ² if F (x) = x4 , then F (x) = 4x3 ) ² if F (x) = p x = x2 , 1 then F (x) = 12 x¡ = p x Z p p dx = x + c ) x The following rules may prove useful: ² Any constant may be written in front of the integral sign R R k f(x) dx = k f(x) dx, k is a constant Proof: Consider differentiating kF (x) where F (x) = f (x) d (k F (x)) = k F (x) = k f (x) dx R ) k f (x) dx = k F (x) R = k f(x) dx ² The integral of a sum is the sum of the separate integrals This rule enables us to integrate term by term R R R [f(x) + g(x)] dx = f (x) dx + g(x) dx Example If y = x4 + 2x3 , find cyan magenta ) R 95 100 50 25 95 50 100 yellow 75 2(2x3 + 3x2 ) dx = x4 + 2x3 + c1 R ) (2x3 + 3x2 ) dx = x4 + 2x3 + c1 R ) (2x3 + 3x2 ) dx = 12 x4 + x3 + c ) 75 25 95 100 50 75 25 95 100 50 75 25 dy and hence find dx Z (2x3 + 3x2 ) dx: If y = x4 + 2x3 , dy then = 4x3 + 6x2 dx R 4x + 6x2 dx = x4 + 2x3 + c1 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\715IB_HL-2_24.CDR Tuesday, 20 November 2007 2:49:18 PM PETERDELL IB_HL-2ed (716) 716 INTEGRATION (Chapter 24) EXERCISE 24C.1 R dy and hence find x dx: We can always check dx that an integral is R dy 2 correct by If y = x + x , find and hence find (3x + 2x) dx: differentiating the dx answer.¡ It should give R 2x+1 dy us the integrand, the If y = e2x+1 , find dx: and hence find e dx function we originally integrated R dy If y = (2x + 1)4 , find and hence find (2x + 1)3 dx: dx Rp p dy x dx: and hence find If y = x x, find dx Z 1 dy p dx: If y = p , find and hence find x dx x x R R R Prove the rule [f(x) + g(x)] dx = f (x) dx + g(x) dx: Z p dy p dx: if y = ¡ 4x and hence find Find dx ¡ 4x Z 4x ¡ d By considering dx: ln(5 ¡ 3x + x ), find dx ¡ 3x + x2 R x d x By considering dx: Hint: 2x = (eln )x (2 ), find dx R d (x ln x), find ln x dx: By considering dx If y = x7 , find 10 11 RULES FOR INTEGRATION cyan magenta yellow 95 100 50 75 25 95 100 ef (x) f (x) 50 ef (x) 75 ex 25 ex chain rule dy dy du = dx du dx 95 y = f (u) where u = u(x) 100 quotient rule 50 u0 (x)v(x) ¡ u(x)v (x) [v(x)]2 75 u(x) v(x) 25 Derivative m nxn¡1 cu0 (x) u0 (x) + v (x) u0 (x)v(x) + u(x)v (x) Function c, a constant mx + c, m and c are constants xn cu(x) u(x) + v(x) u(x)v(x) 95 100 50 75 25 In earlier chapters we developed rules to help us differentiate functions more efficiently Following is a summary of these rules: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\716IB_HL-2_24.CDR Tuesday, 20 November 2007 2:53:46 PM PETERDELL Name power rule addition rule product rule IB_HL-2ed (717) 717 INTEGRATION (Chapter 24) Function Derivative ln x x ln f (x) f (x) f (x) [f(x)]n n[f (x)]n¡1 f (x) These rules or combinations of them can be used to differentiate almost all functions However, the task of finding antiderivatives is not so easy and cannot be written as a simple list of rules as we did above In fact huge books of different types of functions and their integrals have been written Fortunately our course is restricted to a few special cases SIMPLE INTEGRALS Notice that: R d ) k dx = kx + c (kx + c) = k dx ¶ µ R n xn+1 (n + 1)xn d xn+1 x dx = + c, n 6= ¡1 +c = =xn ) If n 6= ¡1, n+1 dx n + n+1 For k a constant, d x (e + c) = ex dx ) d (ln x + c) = dx x If x > 0, ) ¡1 d (ln(¡x) + c) = = dx ¡x x If x < 0, R ex dx = ex + c R dx = ln j x j + c x ² c is always an arbitrary constant called the integrating constant or constant of integration Note: ² Remember that we can check our integration by differentiating the resulting function Example R a Find: R a (x ¡ 2x2 + 5) dx ¶ Z µ p ¡ x dx b x3 = (x ¡ 2x + 5) dx x4 2x3 ¡ + 5x + c Z µ b R = (x¡3 ¡ x ) dx = x¡2 x ¡ +c ¡2 magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 =¡ cyan ¶ p ¡ x dx x3 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\717IB_HL-2_24.CDR Tuesday, 27 November 2007 1:56:26 PM PETERDELL ¡ 23 x + c 2x IB_HL-2ed (718) 718 INTEGRATION (Chapter 24) There is no product or quotient rule for integration Consequently we often have to carry out multiplication or division before we integrate Example ¶2 Z µ a 3x + dx x Find: Z µ b Z µ ¶2 Z µ dx 3x + x ¶ Z µ = 9x + 12 + dx x Z = (9x2 + 12 + 4x¡2 ) dx a = b Z µ = Z 9x 4x + 12x + +c ¡1 = x ¶ x2 ¡ p x dx ¶ dx x2 p ¡p x x ¶ dx (x ¡ 2x¡ ) dx = ¡1 = 3x3 + 12x ¡ x2 ¡ p x 5 ¡ 2x 2 +c p p = 25 x2 x ¡ x + c +c x Notice that we expanded the brackets and simplified to a form that can be integrated We can find c if we are given a point on the curve Example Find f (x) given that f (x) = x3 ¡ 2x2 + and f(0) = Since f (x) = x3 ¡ 2x2 + 3, R f(x) = (x3 ¡ 2x2 + 3) dx ) f(x) = x4 2x3 ¡ + 3x + c But f(0) = 2, so ¡ + + c = and so c = Thus f(x) = x4 2x3 ¡ + 3x + cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 If we are given the second derivative we need to integrate twice to find the function This creates two integrating constants and so we need two other facts about the curve in order to find them black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\718IB_HL-2_24.CDR Tuesday, 20 November 2007 3:00:53 PM PETERDELL IB_HL-2ed (719) 719 INTEGRATION (Chapter 24) Example Find f (x) given that f 00 (x) = 12x2 ¡ 4, f (0) = ¡1 and f(1) = If f 00 (x) = 12x2 ¡ 12x3 ¡ 4x + c ) f (x) = 4x3 ¡ 4x + c then f (x) = fintegrating with respect to xg But f (0) = ¡1 so ¡ + c = ¡1 and so c = ¡1 Thus f (x) = 4x3 ¡ 4x ¡ ) f(x) = 4x4 4x2 ¡ ¡x+d fintegrating againg ) f(x) = x4 ¡ 2x2 ¡ x + d But f(1) = so ¡ ¡ + d = and so d = Thus f(x) = x4 ¡ 2x2 ¡ x + EXERCISE 24C.2 Z b Z µ e Z µ h Find: ¡ 4x p dx x x e 2x ¡ p dx x h magenta x2 ¡ 4x + 10 p dx x2 x ¶ Z µ x i 5e + x ¡ dx x c ¶ Z µ x 2e ¡ dx x Z f i R x2 + x ¡ dx x (x + 1)3 dx c dy = 5x ¡ x2 dx e dy = 2ex ¡ dx f dy = 4x3 + 3x2 dx yellow 95 dy = 4x2 dx 100 95 ¶ x + x ¡ e dx 2x ¶ Z µ x c 3e ¡ dx x Z ³ ´ 3 x ¡ x + x dx f b 100 50 75 25 95 100 50 25 95 100 50 75 25 cyan 75 dy = dx x dx (2x + 1)2 dx Z Find y if: dy a =6 dx d R 25 Z g b Z d (x + 3x ¡ 2) dx ¶ ¶ Z µ p x¡ p dx x a p + x x x 50 ¶ Z µ p dx x x¡ x ¶ Z µ dx g x + x d R p ( x + ex ) dx 75 Find: R a (x ¡ x2 ¡ x + 2) dx black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\719IB_HL-2_24.CDR Tuesday, 20 November 2007 3:02:03 PM PETERDELL IB_HL-2ed (720) 720 INTEGRATION (Chapter 24) Find y if: dy = (1 ¡ 2x)2 dx a b dy p = x¡ p dx x c x2 + 2x ¡ dy = dx x2 b p f (x) = x(1 ¡ 3x) c f (x) = 3ex ¡ Find f (x) if: a f (x) = x3 ¡ 5x + x Find f (x) given that: a f (x) = 2x ¡ and f (0) = b f (x) = 3x2 + 2x and f(2) = c f (x) = ex + p x d f (x) = x ¡ p x and f(1) = and f (1) = Find f (x) given that: a f 00 (x) = 2x + 1, f (1) = and f (2) = p b f 00 (x) = 15 x + p , f (1) = 12 and f (0) = x c f 00 (x) = 2x and the points (1, 0) and (0, 5) lie on the curve INTEGRATING eax+b AND (ax + b)n D d dx Notice that µ ax+b e a ¶ = ax+b £ a = eax+b e a Z ax +b +c e a eax +b dx = ) Likewise if n 6= ¡1, ¶ µ d n+1 = (ax + b) (n + 1)(ax + b)n £ a, dx a(n + 1) a(n + 1) = (ax + b)n Z (ax + b)n dx = ¶ ln(ax + b) = a Z ) dx = ax + b µ cyan magenta a µ (ax + b)n+1 + c, n 6= ¡1 a n+1 a ax + b ¶ = ax + b for ax + b > ln(ax + b) + c a yellow 95 100 50 75 25 95 100 1 dx = ln j ax + b j + c ax + b a 50 95 100 50 75 25 95 100 50 75 25 In fact, 25 Z d dx Also, 75 ) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\720IB_HL-2_24.CDR Tuesday, 20 November 2007 3:04:34 PM PETERDELL IB_HL-2ed (721) 721 INTEGRATION (Chapter 24) Example R a Find: R a (2x + 3)4 dx Z p dx b ¡ 2x Z (2x + 3)4 dx (2x + 3)5 +c = = 10 (2x £ p dx ¡ 2x b R ¡1 = (1 ¡ 2x) dx + 3) + c = ¡2 £ (1 ¡ 2x) 2 +c p = ¡ ¡ 2x + c Example Find: a R a R Z (2e2x ¡ e¡3x ) dx dx ¡ 2x b Z (2e2x ¡ e¡3x ) dx dx = ¡ 2x Z dx ¡ 2x ³ ´ = ¡2 ln j ¡ 2xj + c b = 2( 12 )e2x ¡ ( ¡3 )e¡3x + c = e2x + 13 e¡3x + c = ¡2 ln j ¡ 2x j + c EXERCISE 24D Find: R a (2x + 5)3 dx Z b Z Rp e 3x ¡ dx 3(1 ¡ x) dx f b R ¡ 5x¡2 ¢ 3e dx c yellow 95 i 100 (e¡x + 2)2 dx black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_24\721IB_HL-2_24.CDR Friday, 12 December 2008 12:42:44 PM TROY R R Z f 50 c dx ¡ 3x 75 R h 95 50 (ex + e¡x )2 dx magenta e e 75 50 25 95 100 50 75 R g 25 d 75 g (x2 ¡ x)2 dx R p ¡ x dx Z dx 2x ¡ Z 25 R dx (2x ¡ 1)4 R 25 ¢ R ¡ x 2e + 5e2x dx a (1 ¡ x2 )2 dx 95 Find: b R d 3(2x ¡ 1)2 dx 100 R 100 a c d R (4x ¡ 3)7 dx Z h p dx ¡ 4x a Find y = f(x) given Find: 10 p dx ¡ 5x Z dy p = 2x ¡ and that y = 11 when x = dx and passes through the point (¡3, ¡11) b Function f (x) has slope function p 1¡x Find the point on the graph of y = f (x) with x-coordinate ¡8 cyan f dx (3 ¡ 2x)2 (1 ¡ 3x)3 dx (x2 + 1)3 dx ¡ 7¡3x ¢ e dx Z µ e¡x ¡ ¶ dx 2x + ¶ Z µ x¡ dx 1¡x IB_HL-2ed (722) 722 INTEGRATION (Chapter 24) Find y given that: dy = (1 ¡ ex )2 dx a dy = ¡ 2x + dx x+2 b c dy = e¡2x + dx 2x ¡ Z Z 1 dx, Tracy’s answer was dx = 14 ln j 4x j + c 4x 4x Z Z 1 and Nadine’s answer was dx = dx = 14 ln j x j + c 4x x To find Which of them has found the correct answer? Prove your statement a If f (x) = 2e¡2x b If f (x) = 2x ¡ and f(0) = 3, find f(x) 1¡x and f (¡1) = 3, find f(x) p x + 12 e¡4x c If a curve has slope function equation of the function and passes through (1, 0), find the Z Show that 2x ¡ ¡ = , and hence find x+2 x¡2 x ¡4 Show that 1 ¡ = , and hence find 2x ¡ 2x + 4x ¡ 2x ¡ dx: x2 ¡ Z dx: ¡1 4x2 INTEGRATING f(u)u0(x) BY SUBSTITUTION E R R (x2 + 3x)4 (2x + 3) dx is of the form f (u) u0 (x) dx where f(u) = u4 , u = x2 + 3x and u0 (x) = 2x + Z ex Likewise, ¡x (2x ¡ 1) dx is of the form R f (u) u0 (x) dx where f(u) = eu , u = x2 ¡ x and u0 (x) = 2x ¡ 1, Z R 3x2 + dx is of the form f(u) u0 (x) dx x3 + 2x where f (u) = , u = x3 + 2x and u0 (x) = 3x2 + u and We can integrate funtions of this form using the theorem cyan magenta yellow Z 95 100 50 75 f (u) du 25 95 100 50 75 du dx = f (u) dx 25 95 100 50 75 25 95 100 50 75 25 Z black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\722IB_HL-2_24.CDR Tuesday, 20 November 2007 3:20:46 PM PETERDELL IB_HL-2ed (723) 723 INTEGRATION (Chapter 24) Suppose F (u) is the antiderivative of f (u), so F (u) = f (u) R ) f(u) du = F (u) + c (1) Proof: d d du F (u) = F (u) fchain ruleg dx du dx du = F (u) dx du = f (u) dx R du ) f (u) dx = F (u) + c dx R = f (u) du ffrom (1)g So, for the first example: Z R du du (x + 3x) (2x + 3) dx = u4 dx fu = x2 + 3x, = 2x + 3g dx dx R du freplacing dx by dug = u4 du dx u5 + c which is 15 (x2 + 3x)5 + c = But Example 10 Rp x3 + 2x (3x2 + 2) dx Z p du u = dx where u = x3 + 2x dx R = u du Use substitution to find: Rp x3 + 2x(3x2 + 2) dx = u2 +c = 23 (x3 + 2x) + c Example 11 Z a Use substitution to find: a Z = Z Z 3x2 + dx x3 + 2x x3 b (3x2 + 2) dx + 2x du dx fu = x3 + 2xg u dx R = du u = ln j u j +c = yellow xe1¡x dx Z = ¡ 12 (¡2x) e1¡x dx Z du = ¡ eu dx fu = ¡ x2 , dx du R u = ¡2xg = ¡ e du dx 95 50 75 25 95 100 50 xe1¡x dx = ¡ 12 e1¡x + c 75 25 95 50 75 25 95 100 50 75 25 100 magenta R = ¡ 12 eu + c = ln j x3 + 2x j +c cyan b 100 Z 3x2 + dx x3 + 2x black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\723IB_HL-2_24.CDR Tuesday, 20 November 2007 3:27:18 PM PETERDELL IB_HL-2ed (724) 724 INTEGRATION (Chapter 24) EXERCISE 24E Integrate with respect to x: p x3 + x(3x2 + 1) a 3x2 (x3 + 1)4 b 2x p x2 + c d 4x3 (2 + x4 )3 e (x3 + 2x + 1)4 (3x2 + 2) f x2 (3x3 ¡ 1)4 g x (1 ¡ x2 )5 h (x2 i x4 (x + 1)4 (2x + 1) Find: Z a ¡2e1¡2x dx Z d Find: Z a Z d x+2 + 4x ¡ 3)2 Z Z 2xex dx b Z p e x p dx x Z b 6x2 ¡ dx x3 ¡ x e x¡x2 (2x ¡ 1)e e 2x dx x2 + c Z Z f dx c 4x ¡ 10 dx 5x ¡ x2 f x2 (3 ¡ x3 )2 d xe1¡x F e Z x¡1 x x2 Z x dx ¡ x2 Find f (x) if f (x) is: a x2 ex +1 dx dx 2x ¡ dx x2 ¡ 3x ¡ x2 dx x3 ¡ 3x b x ln x c p x ¡ x2 e ¡ 3x2 x3 ¡ x f (ln x)3 x INTEGRATING CIRCULAR FUNCTIONS Observe the following: d (sin x + c) = cos x + = cos x dx R ) d (¡ cos x + c) = ¡(¡ sin x) + = sin x dx R ) d (tan x + c) = sec2 x dx R ) cos x dx = sin x + c sin x dx = ¡ cos x + c sec2 x dx = tan x + c We can now complete our list of basic integrals: sin x + c sin x ¡ cos x + c ln j x j + c sec2 x tan x + c cyan magenta 25 95 100 50 75 25 95 100 50 75 25 x yellow 95 cos x 100 xn+1 +c n+1 50 (n 6= ¡1) xn 75 ex + c 25 ex kx + c (a constant) k 95 Integral 100 Function 50 Integral 75 Function black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\724IB_HL-2_24.CDR Tuesday, 20 November 2007 3:30:29 PM PETERDELL IB_HL-2ed (725) INTEGRATION (Chapter 24) 725 Example 12 Integrate with respect to x: a sin x ¡ cos x R a b sec2 x ¡ [2 sin x ¡ cos x]dx ¸ Z · sec2 x ¡ + x dx x b = 2(¡ cos x) ¡ sin x + c = ¡2 cos x ¡ sin x + c p + x x = tan x ¡ ln j x j + x2 +c 3 = tan x ¡ ln j x j + 23 x + c Example 13 Find R d (x sin x) and hence deduce x cos x dx dx d (x sin x) = (1) sin x + (x) cos x fproduct rule of differentiationg dx = sin x + x cos x R fantidifferentiationg Thus (sin x + x cos x)dx = x sin x + c1 R R ) sin x dx + x cos x dx = x sin x + c1 R ) ¡ cos x + c2 + x cos x dx = x sin x + c1 R ) x cos x dx = x sin x + cos x + c Example 14 Z determine d d (csc x) = [sin x]¡1 dx dx d (csc x), dx By considering = ¡[sin x]¡2 £ cos x dx: sin2 x £ cos x sin2 x cos x =¡ sin x =¡ Z Hence ¡ Z magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ) cyan d (sin x) dx black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\725IB_HL-2_24.CDR Wednesday, 21 November 2007 3:24:15 PM PETERDELL cos x dx = csc x + c1 sin2 x cos x dx = ¡ csc x + c sin2 x IB_HL-2ed (726) 726 INTEGRATION (Chapter 24) Z h i f (x) = sin x ¡ x dx Example 15 Find f(x) given that p f (x) = sin x ¡ x and f (0) = ) f (x) = £ (¡ cos x) ¡ x2 +c ) f (x) = ¡2 cos x ¡ 23 x + c But f (0) = ¡2 cos ¡ + c ) = ¡2 + c and so c = Thus f (x) = ¡2 cos x ¡ 23 x + EXERCISE 24F.1 Integrate with respect to x: a sin x ¡ b d sec2 x + sin x e g p x2 x ¡ 10 sin x h R ¡p x+ b Find: a d R ¢ cos x dx (2et ¡ sin t) dt 4x ¡ cos x x ¡ sec2 x x(x ¡ 1) + cos x R (µ ¡ sin µ) dµ ¶ µ R dt cos t ¡ t e c p x + sec2 x f sin x ¡ cos x + ex i p sec2 x ¡ sin x + x c f ¢ R¡p t t + sec2 t dt ¶ µ R ¡ + sec2 µ dµ µ R x d x e (sin x + cos x) dx (e sin x) and hence find dx Z cos x ¡ sin x d ¡x b By considering dx (e sin x), determine dx ex R d c Find (x cos x) and hence find x sin x dx dx Z d d By considering (sec x), determine tan x sec x dx dx a Find Find f (x) given that: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a f (x) = x2 ¡ cos x and f (0) = ¡ ¢ b f (x) = cos x ¡ sin x and f ¼4 = p12 p c f (x) = x ¡ sec2 x and f(¼) = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\726IB_HL-2_24.CDR Tuesday, 20 November 2007 3:34:18 PM PETERDELL IB_HL-2ed (727) 727 INTEGRATION (Chapter 24) INTEGRALS OF CIRCULAR FUNCTIONS OF THE FORM f(ax¡+¡b) Observe the following: As R ) d (sin(ax + b)) = cos(ax + b) £ a, dx a cos(ax + b) dx = sin(ax + b) + c1 R ) a cos(ax + b) dx = sin(ax + b) + c1 R sin(ax + b) + c a R sin(ax + b) dx = ¡ cos(ax + b) + c a Z sec2 (ax + b) dx = tan(ax + b) + c a ) Likewise we can show and cos(ax + b) dx = SUMMARY OF INTEGRALS FOR FUNCTIONS OF THE FORM f(ax¡+¡b) Function Integral Function Integral cos(ax + b) sin(ax + b) + c a eax+b ax+b +c e a sin(ax + b) ¡ cos(ax + b) + c a (ax + b)n (ax + b)n+1 + c, n 6= ¡1 a n+1 sec2 (ax + b) tan(ax + b) + c a ax + b ln j ax + b j + c a Example 16 a e¡2x ¡ sec2 (2x) Integrate with respect to x: Z a = ¡ ¡2x ¢ e ¡ sec2 (2x) dx ¡2 e¡2x ¡ £ ¡ 12 e¡2x R b b sin(3x) + cos(4x + ¼) (2 sin(3x) + cos(4x + ¼)) dx = £ 13 (¡ cos(3x)) + tan(2x) + c = ¡ 23 cos(3x) + ¡ tan(2x) + c 4 sin(4x + ¼) + c sin(4x + ¼) + c INTEGRALS OF POWERS OF CIRCULAR FUNCTIONS Integrals involving sin2 (ax + b) and cos2 (ax + b) can be found by first using sin2 µ = ¡ cos(2µ) or cos2 µ = + cos(2µ) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 These formulae are simply rearrangements of cos(2µ) formulae black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\727IB_HL-2_24.CDR Tuesday, 20 November 2007 3:38:46 PM PETERDELL IB_HL-2ed (728) 728 INTEGRATION (Chapter 24) sin2 (3x ¡ ¼2 ) becomes 12 ¡ 12 cos(6x ¡ ¼ ) ³x´ ³x´ cos2 becomes 12 + 12 cos = 12 + 2 ² For example, ² R Example 17 R Integrate (2 ¡ sin x) cos x (2 ¡ sin x)2 dx = (4 ¡ sin x + sin2 x)dx ¢ R¡ = ¡ sin x + 12 ¡ 12 cos(2x) dx ¢ R ¡9 = ¡ sin x ¡ cos(2x) dx = 92 x + cos x ¡ £ = 92 x + cos x ¡ sin(2x) + c sin(2x) + c INTEGRATION BY SUBSTITUTION Example 18 a cos3 x sin x Integrate with respect to x: R a R cos3 x sin x dx b cot x Z b cot x dx = = [cos x]3 sin x dx du We let u = cos x, = ¡ sin x dx R ) cos3 x sin x dx µ ¶ Z du = u ¡ dx dx R = ¡ u3 du =¡ = Z We let u = sin x, Z ) Z cot x dx = Z = du = cos x dx du dx u dx du u = ln j u j + c u4 +c ¡ 14 cos x dx sin x = ln j sin x j + c cos x + c Note: The substitutions we make need to be chosen with care For example, in Example 18 part b, if we let u = cos x, Z Z u cos x dx dx = du sin x ¡ dx du = ¡ sin x then dx cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 This substitution leads nowhere as we cannot perform this integration black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_24\728IB_HL-2_24.cdr Thursday, 11 March 2010 4:38:20 PM PETER IB_HL-2ed (729) 729 INTEGRATION (Chapter 24) Example 19 R Find R sin3 x dx = R R sin3 x dx sin2 x sin x dx = (1 ¡ cos2 x) sin x dx µ ¶ R du = (1 ¡ u2 ) ¡ dx dx R = (u2 ¡ 1) du = u3 ¡u+c = du = ¡ sin xg dx fu = cos x, cos3 x ¡ cos x + c EXERCISE 24F.2 Integrate with respect to x: a sec2 (2x) b cos(4x) c e h sin(2x) ¡ e¡x ¡ ¢ ¡3 cos ¼4 ¡ x f g sin(3x) ¡ ¢ cos x2 ¡ ¢ sin 2x + ¼6 i ¡ ¢ e2x ¡ sec2 x2 ¡ ¢ sec2 ¼3 ¡ 2x j cos(2x) + sin(2x) k sin(3x) + cos(4x) l d cos(8x) ¡ sin x Integrate with respect to x: a cos2 x b sin2 x c + cos2 (2x) d ¡ sin2 (3x) e f (1 + cos x)2 cos2 (4x) Use the identity cos2 µ = + cos4 x = cos(4x) + cos(2µ) to show that cos(2x) + and hence find Integrate by substitution: b sin x p cos x c tan x d e cos x (2 + sin x)2 f sin x cos3 x g sin x ¡ cos x h cos(2x) sin(2x) ¡ i x sin(x2 ) j sin3 x cos5 x k csc3 (2x) cot(2x) l cos3 x magenta yellow 95 100 50 75 sin4 x cos3 x dx 25 R 95 100 50 75 b 25 95 sin5 x dx 100 50 R 75 25 95 100 50 75 25 p sin x cos x sin4 x cos x a cos4 x dx: a Find: cyan R black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\729IB_HL-2_24.CDR Tuesday, 20 November 2007 3:45:24 PM PETERDELL IB_HL-2ed (730) 730 INTEGRATION (Chapter 24) Find f (x) if f (x) is: sin x ecos x a Find: a R R d R g b sin3 (2x) cos(2x) cot x dx b sec x tan x dx e csc ¡x¢ ¡x¢ cot 2 h dx R R R c sin x + cos x sin x ¡ cos x cot(3x) dx c csc x cot x dx f sec3 x sin x dx i G R R Z d etan x cos2 x csc2 x dx tan(3x) sec(3x) dx csc2 x p dx cot x DEFINITE INTEGRALS If F (x) is the antiderivative of f (x) where f (x) is continuous on the interval a x b Z b then the definite integral of f(x) on this interval is f (x) dx = F (b) ¡ F (a) a Z b f(x) dx reads “the integral from x = a to x = b of f (x) with respect to x” a We write F (b) ¡ F (a) = [F (x)]ba : Notation: R3 Example 20 R3 Find (x2 + 2) dx ¸3 · x + 2x = ³ ´ ³ ´ = + 2(3) ¡ 13 + 2(1) (x2 + 2) dx = (9 + 6) ¡ ( 13 + 2) = 12 23 Example 21 Z ¼ a Evaluate: Z b sin x dx Z ¼ a Z = (¡ cos ¼3 ) ¡ (¡ cos 0) = ¡ 12 + = cyan = magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 = 25 95 100 50 75 sec2 (2x) dx ¤¼ tan(2x) 08 ¢ ¡ ¡ ¢ = 12 tan ¼4 ¡ 12 tan = 25 ¼ b sin x dx £ ¤¼ = ¡ cos x 03 sec2 (2x) dx 0 ¼ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\730IB_HL-2_24.CDR Wednesday, 21 November 2007 3:24:56 PM PETERDELL £1 2 £1¡ £0 IB_HL-2ed (731) INTEGRATION (Chapter 24) 731 When we solve a definite integral by substitution, we need to make sure the endpoints are converted to the new variable Example 22 Z a Evaluate: Z 3 = Z (x2 6x dx + 1)3 du = 2x dx When x = 2, u = 22 ¡ = When x = 3, u = 32 ¡ = Let u = x2 ¡ ) du u = = [ln j u j]3 b x dx x2 ¡ µ ¶ 1 du dx u dx a Z Z x dx x ¡1 = 12 (ln ¡ ln 3) = ln( 83 ) Z 6x dx (x2 + 1)3 ¶ µ du dx u3 dx b Z = Z =3 Let u = x2 + ) du = 2x dx When x = 0, u = When x = 1, u = u¡3 du · ¸2 u¡2 ¡2 µ ¡2 ¶ 1¡2 =3 ¡ ¡2 ¡2 =3 Z Example 23 Evaluate: cyan magenta ¼ p sin x cos x dx Let u = sin x ) du = cos x dx When x = ¼2 , u = sin ¼2 = yellow 95 100 50 75 95 When x = ¼6 , u = sin ¼6 = 100 50 95 ¼ 75 p du u dx dx 25 ¼ ¼ 100 50 75 25 95 100 50 75 25 = p sin x cos x dx Z ¼ Z ¼ 25 = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\731IB_HL-2_24.CDR Tuesday, 20 November 2007 3:59:39 PM PETERDELL IB_HL-2ed (732) 732 INTEGRATION (Chapter 24) Z = " = u du u2 #1 2 = 23 (1) ¡ = ¡ ¡ ¢ 32 p EXERCISE 24G.1 Evaluate the following and check with your graphics calculator: Z Z Z x3 dx b (x2 ¡ x) dx c a Z d Z g µ ¶ x¡ p dx x e (e¡x + 1)2 dx h Z Z Z b ¼ ¼ ¼ d 2 Evaluate: Z ¼6 cos x dx a Z Z ¼ e sin(3x) dx ex dx Z x¡3 p dx x f p dx 2x ¡ i Z 1 dx x e1¡x dx Z c sin x dx ¼ ¼ Z ¼ f cos x dx sec2 x dx sin2 x dx Evaluate the following and check with your graphics calculator: Z Z Z p x x3 +1 a dx b x e dx c x x2 + 16 dx + 2)2 (x 0 Z Z Z x ln x xe¡2x dx e dx f d dx 2 ¡ x x Z Z Z 1 ¡ 3x2 6x ¡ 4x + (x2 + 2x)n (x + 1) dx dx h dx i g 3 1¡x +x x ¡ x + 2x (Careful!) Evaluate: Z ¼3 sin x p dx a cos x Z ¼2 cot x dx d Z b 95 ¼ f tan x dx sec2 x tan3 x dx 100 50 75 25 95 yellow Z cos x dx ¡ sin x ¼ 100 50 75 25 95 100 50 75 25 95 100 50 75 25 c sin x cos x dx Z e ¼ 0 magenta Z ¼ cyan ¼ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\732IB_HL-2_24.CDR Tuesday, 20 November 2007 4:02:17 PM PETERDELL IB_HL-2ed (733) 733 INTEGRATION (Chapter 24) PROPERTIES OF DEFINITE INTEGRALS Earlier in the chapter we proved the following properties of definite integrals using the fundamental theorem of calculus: Z b ² Z Z a f(x) dx a Z b ² b cf(x)dx = c Z a f (x)dx, c is any constant a Z b ² Z c f(x) dx + Z b [¡f(x)] dx = ¡ a b Z b ² c f(x) dx = f(x)dx a Z b [f (x) + g(x)]dx = b f (x)dx + a g(x)dx a a EXERCISE 24G.2 Use questions to to check the properties of definite integrals Z Z Z Z p p Find: a x dx and (¡ x) dx b x dx and Z Find: 1 a Z x2 dx Find: Find: Z x2 dx a b Z b x dx Z c Z p x) dx y y¡=¡¦(x) 2 x -2 y y¡=¡¦(x) 2 8 x -2 f (x) dx magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 (x2 + d f (x) dx 75 25 95 100 50 75 25 c (x3 ¡ 4x) dx f (x) dx 0 c cyan c Z Evaluate the following integrals using area interpretation: Z Z f (x) dx b f (x) dx a Z 3x2 dx p x dx d f (x) dx d Z 0 Z x2 dx (x3 ¡ 4x) dx Evaluate the following integrals using area interpretation: Z Z f (x) dx b f (x) dx a Z (¡x7 ) dx 0 Z (x3 ¡ 4x) dx c a Z b Z black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\733IB_HL-2_24.CDR Tuesday, 20 November 2007 4:22:37 PM PETERDELL IB_HL-2ed (734) 734 INTEGRATION (Chapter 24) Write as a single integral: Z Z f (x) dx + f (x) dx a Z Z Z f(x) dx = ¡3, find Z b If Z f(x) dx = 5, f (x) dx = ¡2 and g(x) dx f(x) dx = and Z g(x) dx + a If Z g(x) dx + Z Z b f (x) dx: Z f (x) dx = 7, find f (x) dx: REVIEW SET 24A Integrate with respect to x: p x a ¡ 2x b Integrate with respect to x: sin7 x cos x a Z Find the exact value of: ¡1 a ¡5 By differentiating y = xe1¡x c b p x2 ¡ 4, find Z e4¡3x c esin x cos x tan(2x) Z p ¡ 3x dx d b 4x2 dx (x3 + 2)3 x p dx: x2 ¡ A curve y = f (x) has f 00 (x) = 18x+10 Find f (x) if f(0) = ¡1 and f (1) = 13 Z Evaluate: ¼ a cos ¡x¢ 4x ¡ 2x + a Z ¼ b dx tan x dx can be written in the form A + B 2x + b Find the value of A and B Z Hence find Z Find the exact value of: a Z p dx 2x + b 4x ¡ dx: 2x + x2 ex +1 dx Differentiate ln sec x, given that sec x > What integral can be deduced from this derivative? Z a 10 If e1¡2x dx = 4e , find a in the form ln k magenta yellow 95 100 50 £ ¡2x ¤ e (cos x ¡ sin x) dx 75 ¼ 25 Z 95 100 50 75 25 95 100 50 75 95 100 50 75 25 cyan 25 d ¡2x sin x) and hence find (e dx 11 Find black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_24\734IB_HL-2_24.CDR Friday, 12 March 2010 1:48:30 PM PETER IB_HL-2ed (735) 735 INTEGRATION (Chapter 24) REVIEW SET 24B Find: ¶ Z µ ¡x a 2e ¡ + dx x Z Evaluate: a Integrate: Z (x2 ¡ 1)2 dx sin2 a ¶2 Z µ p x¡ p dx b x ¡x¢ c ¡ ¢2 + e2x¡1 dx x(x2 ¡ 1)2 dx (2 ¡ cos x)2 b 2 b Z By differentiating (3x2 + x)3 , find R (3x2 + x)2 (6x + 1) dx: Given that f (x) = x2 ¡ 3x + and f(1) = 3, find f (x) Differentiate sin(x2 ) and hence find Z Find the exact value of R x cos(x2 )dx: p dx 3x ¡ f 00 (x) = 3x2 + 2x and f (0) = f(2) = Find: a f (x) Z Evaluate ¼ ¼ b the equation of the normal to y = f (x) at x = cot µ dµ 10 Find A, B, C and D using the division algorithms, if: Z x3 ¡ 3x + D = Ax2 + Bx + C + Hence find x¡2 x¡2 Z dx ¡ x+2 Z dx b Hence find x¡1 11 a Find 12 a Find constants A, B and C given that Z b Hence find dx x(1 ¡ x2 ) Z yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 magenta x+5 dx (x + 2)(x ¡ 1) A B C = + + x(1 ¡ x2 ) x x+1 x¡1 c Find in simplest form, the exact value of cyan Z x3 ¡ 3x + dx: x¡2 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\735IB_HL-2_24.CDR Tuesday, 20 November 2007 4:09:41 PM PETERDELL dx x(1 ¡ x2 ) IB_HL-2ed (736) 736 INTEGRATION (Chapter 24) REVIEW SET 24C Find y if: a Evaluate: a dy = (x2 ¡ 1)2 dx Z Z b 10x p dx 3x2 + sin x dx: cos4 x Find Z d (ln x)2 dx Find Z dx 2x ¡ ¡2 ¡x dy = 400 ¡ 20e dx b ln x dx: x and hence find Given that f 00 (x) = 4x2 ¡ 3, f (0) = and f(2) = 3, find f(3) R Find the derivative of x tan x and hence determine R Find x sec2 x dx (2x + 3)n dx for all integers n a Find (ex + 2)3 using the binomial expansion Z (ex + 2)3 dx b Hence find the exact value of c Check b using technology Z Evaluate: ¼ a sin ¡x¢ Z b dx ¼ ¼ sec2 x dx tan x p a and passes through the points (0, 2) 10 A function has slope function x + p x and (1, 4) Find a and hence explain why the function y = f(x) has no stationary points (x2 + ax + 2) dx = a cyan e1¡x x2 ¶ Z and hence find the exact value of e1¡x (x + 2) dx x3 magenta yellow 95 100 50 75 25 95 100 50 95 sin x p dx Comment on the existence of this integral cosn x 100 25 95 100 50 75 25 13 Find 50 Z µ 75 d dx 12 Find 73a Find a 75 2a 11 25 Z black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_24\736IB_HL-2_24.CDR Wednesday, 21 November 2007 3:25:27 PM PETERDELL IB_HL-2ed (737) 25 Chapter Applications of integration Finding areas between curves Motion problems Problem solving by integration A B C Contents: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 25A Review set 25B Review set 25C black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_25\737IB_HL-2_25.CDR Tuesday, 29 January 2008 1:56:31 PM PETERDELL IB_HL-2ed (738) 738 APPLICATIONS OF INTEGRATION (Chapter 25) We have already seen how definite integrals can be related to the areas between functions and the x-axis In this chapter we explore this relationship further, and consider other applications of integral calculus such as motion problems Z b INVESTIGATION a Z f(x)dx AND AREAS b Does f (x) dx always give us an area? a What to do: Z Z Find x dx and x3 dx ¡1 Explain why the first integral in gives an area whereas the second integral does not Graphical evidence is essential Z Find x3 dx and explain why the answer is negative ¡1 Z Check that Z x dx + ¡1 A Z x dx = x3 dx: ¡1 FINDING AREAS BETWEEN CURVES We have already established in Chapter 24 that: If f(x) is positive and continuous on the interval a x b, then the area bounded by y = f (x), the x-axis, and the vertical lines x = a and x = b Z b f (x) dx is given by y=ƒ(x) y a a y Notice also that the area bounded by x = f (y), the y-axis, and the horizontal lines y = a and y = b Z b f (y) dy: is given by b x x=ƒ(y) b a a cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 x black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_25\738IB_HL-2_25.CDR Wednesday, 21 November 2007 9:28:09 AM PETERDELL IB_HL-2ed (739) APPLICATIONS OF INTEGRATION (Chapter 25) Example y Find the area of the region enclosed by y = x2 + 1, the x-axis, x = and x = R2 (x2 + 1) dx ¸2 · x +x = ¡8 ¢ ¡1 ¢ = +2 ¡ +1 y¡=¡xX¡+¡1 Area = x 739 = 13 units2 We can check this result using a graphics calculator or graphing package GRAPHING PACKAGE TI C AREA BETWEEN TWO FUNCTIONS If two functions f (x) and g(x) intersect at x = a and x = b and f(x) > g(x) for all x [a, b], then the area of the shaded region between their points of intersection is given by Rb a [f (x) ¡ g(x)] dx y y¡=¡ƒ(x)¡¡or¡¡y¡=¡yU Alternatively, if we describe the upper and lower functions as y = yU and y = yL respectively, then the area is Rb [yU ¡ yL ] dx a a b x y¡=¡g(x)¡¡or¡¡y¡=¡yL Proof: If we translate each curve vertically through [0, k] until it is completely above the x-axis, the area does not change Area of shaded region Rb Rb = a [f (x) + k] dx ¡ a [g(x) + k] dx Rb = a [f (x) ¡ g(x)] dx y¡=¡g(x)¡+¡k y¡=¡ƒ(x)¡+¡k a b y x cyan magenta yellow x 95 a 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 We can see immediately that if y = f (x) = Rb then the enclosed area is a [¡g(x)] dx y¡=¡g(x) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_25\739IB_HL-2_25.CDR Wednesday, 21 November 2007 10:01:42 AM PETERDELL b IB_HL-2ed (740) 740 APPLICATIONS OF INTEGRATION (Chapter 25) Example Rb Use a [yU ¡ yL ] dx to find the area bounded by the x-axis and y = x2 ¡ 2x The curve cuts the x-axis when y = ) x2 ¡ 2x = ) x(x ¡ 2) = ) x = or ) the x-intercepts are and R2 Area = [yU ¡ yL ] dx R2 = [0 ¡ (x2 ¡ 2x)] dx R2 = (2x ¡ x2 ) dx ¸2 · x3 = x2 ¡ ¢ ¡ = ¡ 83 ¡ (0) ) the area is y yU¡=¡0 x yL = x - x units2 Example Find the area of the region enclosed by y = x + and y = x2 + x ¡ y = x + meets y = x2 + x ¡ where x2 + x ¡ = x + ) x2 ¡ = ) (x + 2)(x ¡ 2) = ) x = §2 R2 Area = ¡2 [yU ¡ yL ] dx R2 = ¡2 [(x + 2) ¡ (x2 + x ¡ 2)] dx R2 = ¡2 (4 ¡ x2 ) dx ¸2 · x3 = 4x ¡ ¡2 ¢ ¡ ¢ ¡ = ¡ 83 ¡ ¡8 + 83 y y = x2 + x - 2 -2 y = x+2 x -2 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 = 10 23 units2 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_25\740IB_HL-2_25.CDR Wednesday, 21 November 2007 10:10:41 AM PETERDELL IB_HL-2ed (741) APPLICATIONS OF INTEGRATION (Chapter 25) 741 Example Find the total area of the regions contained by y = f (x) and the x-axis for f(x) = x3 + 2x2 ¡ 3x f(x) = x3 + 2x2 ¡ 3x = x(x2 + 2x ¡ 3) = x(x ¡ 1)(x + 3) y -3 ) y = f (x) cuts the x-axis at 0, 1, ¡3 x Total area R1 R0 = ¡3 (x3 + 2x2 ¡ 3x) dx ¡ (x3 + 2x2 ¡ 3x) dx · ¸0 ¸1 · x x 2x3 3x2 2x3 3x2 ¡ + ¡ + ¡ = ¡3 ¢ ¡ ¡ ¢ = ¡ ¡11 14 ¡ ¡ 12 ¡ = 11 56 units2 The area in Example may also be found using technology ¯ R1 ¯ ¯ x + 2x2 ¡ 3x¯ dx as total area = ¡3 In general, the area between the functions f (x) and g (x) on the interval [a, b] is Rb a jf (x) ¡ g(x)j dx EXERCISE 25A Find the exact value of the area of the region bounded by: a y = x2 , the x-axis and x = b y = x3 , the x-axis, x = and x = c y = ex , the x-axis, the y-axis and x = d the x-axis and the part of y = + x ¡ x2 above the x-axis e x = y2 + 1, the y-axis, and the lines y = and y = p f x = y + 5, the y-axis, and the lines y = ¡1 and y = Rb [f (x) ¡ g(x)] dx or a [yU ¡ yL ] dx to find the exact value of the area between: p a the axes and y = ¡ x b y = , the x-axis, x = and x = x c y = , the x-axis, x = ¡1 and x = ¡3 x d y = ¡ p , the x-axis and x = x Use Rb a e y = ex + e¡x , the x-axis, x = ¡1 and x = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 f y = ex , the y-axis, the lines y = and y = using black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_25\741IB_HL-2_25.CDR Friday, January 2008 9:57:29 AM DAVID3 R ln y dy = y ln y ¡ y + c IB_HL-2ed (742) 742 APPLICATIONS OF INTEGRATION (Chapter 25) a Find the area enclosed by one arch of y = sin(2x) b Show that the area enclosed by y = sin x and the x-axis from x = to x = ¼ is units2 c Find the area enclosed by y = sin2 x and the x-axis from x = to x = ¼ Rb Use a b c d a [yU ¡ yL ] dx to find the exact value of the area bounded by: the x-axis and y = x2 + x ¡ the x-axis, y = e¡x ¡ and x = the x-axis and the part of y = 3x2 ¡ 8x + below the x-axis y = x3 ¡ 4x, the x-axis, x = 1, and x = A region with x > has boundaries defined by y = sin x, y = cos x and the y-axis Find the area of the region a Find the area of the region enclosed by y = x2 ¡ 2x and y = b Consider the graphs of y = x ¡ and y = x2 ¡ 3x i Sketch the graphs on the same set of axes ii Find the coordinates of the points where the graphs meet iii Find the area of the region enclosed by the two graphs p c Determine the area of the region enclosed by y = x and y = x2 d On the same set of axes, graph y = ex ¡ and y = ¡ 2e¡x , showing axes intercepts and asymptotes Find algebraically, the points of intersection of y = ex ¡ and y = ¡ 2e¡x Find the area of the region enclosed by the two curves e Determine the exact value of the area of the region bounded by y = 2ex , y = e2x and x = On the same set of axes, draw the graphs of the relations y = 2x and y2 = 4x Determine the area of the region enclosed by these relations y The graph alongside shows a small portion of the graph of y = tan x A is a point on the graph with a y-coordinate of y¡=¡tan¡x A a Find the coordinates of A b Find the shaded area x x= Sketch the circle with equation x2 + y = p p a Explain why the upper half of the circle has equation y = ¡ x2 R3p b Hence, determine ¡ x2 dx without actually integrating the function c Check your answer using technology cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 10 Find the area enclosed by the function y = f (x) and the x-axis for: a f(x) = x3 ¡ 9x b f (x) = ¡x(x¡2)(x¡4) c f (x) = x4 ¡ 5x2 + 4: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_25\742IB_HL-2_25.CDR Wednesday, 21 November 2007 10:21:21 AM PETERDELL IB_HL-2ed (743) 743 APPLICATIONS OF INTEGRATION (Chapter 25) 11 The illustrated curves are those of y = sin x and y = sin(2x) y A a Identify each curve b Find algebraically the coordinates of A c Find the total area enclosed by C1 and C2 for x ¼ 12 y y = x3 - x y = 3x + Cx 2p p x For the given graphs of y = x3 ¡ 4x and y = 3x + 6: a write the shaded area as i the sum of two definite integrals ii a single definite integral involving modulus b Find the total shaded area -2 Cz x 13 Find the areas enclosed by: a y = x3 ¡ 5x and y = 2x2 ¡ b y = ¡x3 + 3x2 + 6x ¡ and y = 5x ¡ c y = 2x3 ¡ 3x2 + 18 and y = x3 + 10x ¡ 14 y a Explain why the total area shaded is not R7 equal to f (x) dx x b What is the total shaded area equal to in terms of integrals? 15 A E Cz B C x D c Cx y = f(x) The and a b y illustrated curves are those of y = cos(2x) y = cos2 x Identify each curve Determine the coordinates of A, B, C, D and E Show that the area of the shaded region is ¼ 2 units 16 Find, correct to significant figures, the areas of the regions enclosed by the curves: b y = xx and y = 4x ¡ 10 x a y = e¡x and y = x2 ¡ 17 The shaded area is 0:2 units2 : Find k, correct to decimal places 18 The shaded area is unit2 : Find b, correct to decimal places y y y= cyan magenta 95 100 50 75 25 50 75 25 95 yellow y= x b x k 95 100 50 75 25 95 100 50 75 25 1 + 2x 100 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_25\743IB_HL-2_25.CDR Thursday, 22 November 2007 10:34:09 AM PETERDELL x IB_HL-2ed (744) 744 APPLICATIONS OF INTEGRATION (Chapter 25) The shaded area is 2:4 units2 : Find k, correct to decimal places 19 y 20 The shaded area is 6a units2 : Find the exact value of a y y = x2 y = x2 + y=k -a x B a x MOTION PROBLEMS DISTANCES FROM VELOCITY GRAPHS Suppose a car travels at a constant positive velocity of 60 km h¡1 for 15 minutes We know the distance travelled = speed £ time = 60 km h¡1 £ 14 h = 15 km speed ¡(km¡h-1) When we graph speed against time, the graph is a horizontal line and it is clear that the distance travelled is the area shaded 60 So, the distance travelled can also be found by Z 14 60 dt = 15 the definite integral Qr_ time (t hours) Now suppose the speed decreases at a constant rate so that the car, initially travelling at 60 km h¡1 , stops in minutes speed (km¡h-1) In this case the average speed must be 30 km h¡1 , h so the distance travelled = 30 km h¡1 £ 10 = km But the triangle has area = = 2 60 v(t)¡=¡60-600t time (t hours) £ base £ altitude £ 10 v(t)¡=¡60 qA_p_ £ 60 = So, once again the shaded area gives us the distance travelled, and we can find it using the Z 10 (60 ¡ 600t) dt = definite integral Z t2 These results suggest that: distance travelled = v(t) dt provided we not change direction t1 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 If we have a change of direction within the time interval then the velocity will change sign In such cases we need to either add the components of area above and below the t-axis, or alternatively integrate the speed function j v(t) j black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_25\744IB_HL-2_25.CDR Wednesday, 21 November 2007 10:32:37 AM PETERDELL IB_HL-2ed (745) 745 APPLICATIONS OF INTEGRATION (Chapter 25) Z In general, t2 distance travelled = j v(t) j dt t1 Example The velocity-time graph for a train journey is illustrated in the graph alongside Find the total distance travelled by the train 60 v¡(km¡h-1) 30 t (h) 0.1 0.2 0.3 0.4 0.5 0.6 Total distance travelled = total area under the graph = area A + area B + area C + area D + area E ¢ ¡ (0:1) + (0:1)30 = 12 (0:1)50 + (0:2)50 + 50+30 + (0:1)30 50 = 2:5 + 10 + + + 1:5 = 21 km 50 30 30 A B C D E 0.1 0.2 0.1 0.1 b area = a c 0.1 a+b ´c EXERCISE 25B.1 velocity (m¡s-1) A runner has the velocity-time graph shown Find the total distance travelled by the runner 80 60 40 20 -20 velocity (km¡h-1) 10 15 20 time (s) A car travels along a straight road with the velocity-time function illustrated a What is the significance of the graph: i above the t-axis ii below the t-axis? b Find the total distance travelled by the car c Find the final displacement of the car t¡¡(h) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A cyclist rides off from rest, accelerating at a constant rate for minutes until she reaches 40 km h¡1 She then maintains a constant speed for minutes until reaching a hill She slows down at a constant rate to 30 km h¡1 in one minute, then continues at this rate for 10 minutes At the top of the hill she reduces her speed uniformly and is stationary minutes later After drawing a graph, find how far she has travelled black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_25\745IB_HL-2_25.CDR Thursday, 22 November 2007 10:40:55 AM PETERDELL IB_HL-2ed (746) 746 APPLICATIONS OF INTEGRATION (Chapter 25) DISPLACEMENT AND VELOCITY FUNCTIONS In this section we are concerned with motion in a straight line, or linear motion Recall that for some displacement function s(t) the velocity function is s0 (t) and that t > in all situations So, given a velocity function we can determine the displacement function by the integral s(t) = R v(t) dt Using the displacement function we can quickly determine the displacement in a time interval [a, b] Displacement = s(b) ¡ s(a) = Rb a v(t) dt We can also determine the total distance travelled in some time interval a t b Consider the following example: A particle moves in a straight line with velocity function v(t) = t ¡ cm s¡1 How far does it travel in the first seconds of motion? - + We notice that v(t) has sign diagram: v(t) t Since the velocity function changes sign at t = seconds, the particle reverses direction at this time R t2 Now s(t) = (t ¡ 3) dt = ¡ 3t + c but we have no information to find c Clearly, the displacement of the particle in the first seconds is s(4) ¡ s(0) = c ¡ ¡ c = ¡4 cm However, when calculating the distance travelled we need to remember the reversal of direction at t = We find the positions of the particle at t = 0, t = and t = 4: s(4) = c ¡ s(0) = c, s(3) = c ¡ 12 , Hence, we can draw a diagram of the motion: c-4\Qw_ c-4 c Thus the total distance travelled is (4 12 + 12 ) cm = cm Summary: To find the total distance travelled given a velocity function v(t) = s0 (t) on a t b: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 Draw a sign diagram for v(t) so that we can determine any directional changes Determine s(t) by integration, with integrating constant c, say Find s(a) and s(b) Also find s(t) at every point where there is a direction reversal Draw a motion diagram Determine the total distance travelled from the motion diagram 100 50 75 25 ² ² ² ² ² black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_25\746IB_HL-2_25.CDR Wednesday, 21 November 2007 10:51:19 AM PETERDELL IB_HL-2ed (747) 747 APPLICATIONS OF INTEGRATION (Chapter 25) Z Z b total distance travelled = jv(t)j dt b displacement = v(t) dt a a Example A particle P moves in a straight line with velocity function v(t) = t2 ¡ 3t + m s¡1 a How far does P travel in the first seconds of motion? b Find the displacement of P after seconds a v(t) = s0 (t) = t2 ¡ 3t + = (t ¡ 1)(t ¡ 2) ) sign diagram of v is: - + + t Since the signs change, P reverses direction at t = and t = secs R t3 3t2 Now s(t) = (t2 ¡ 3t + 2) dt = ¡ + 2t + c Now s(0) = c s(2) = ¡6+4+c=c+ s(1) = s(4) = 64 ¡ +2+c=c+ ¡ 24 + + c = c + 13 Motion diagram: c ¡ ) total distance = c + = + c+5\Qe_ c+\We_ c+\Ty_ ¢ ¡ 5 ¡c + c+ 2 ¡ + 53 ¡ ¢ ¡ ¢ ¡ [c + 23 ] + c + 13 ¡ [c + 23 ] = 23 m b Displacement = final position ¡ original position = s(4) ¡ s(0) = c + 13 ¡ c i.e., 13 m to the right = 13 m EXERCISE 25B.2 A particle has velocity function v(t) = ¡ 2t cm s¡1 as it moves in a straight line a Find the total distance travelled in the first second of motion b Find the displacement of the particle at the end of one second cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Particle P has velocity v(t) = t2 ¡ t ¡ cm s¡1 a Find the total distance travelled in the first seconds of motion b Find the displacement of the particle at the end of three seconds black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_25\747IB_HL-2_25.CDR Wednesday, 21 November 2007 2:07:20 PM PETERDELL IB_HL-2ed (748) 748 APPLICATIONS OF INTEGRATION (Chapter 25) A particle moves along the x-axis with velocity function x0 (t) = 16t ¡ 4t3 units/s Find the total distance travelled in the time interval: a t seconds b t seconds A particle moves in a straight line with velocity function v(t) = cos t m s¡1 a Show that the particle oscillates between two points b Find the distance between the two points in a VELOCITY AND ACCELERATION FUNCTIONS We know that the acceleration function is the derivative of velocity, so a(t) = v (t) So, given an acceleration function, we can determine the velocity function by the integral v(t) = R a(t) dt EXERCISE 25B.3 The velocity of a particle travelling in a straight line is given by v(t) = 50 ¡ 10e¡0:5t m s¡1 , where t > 0, t in seconds a b c d e f g State the initial velocity of the particle Find the velocity of the particle after seconds How long will it take for the particle’s velocity to increase to 45 m s¡1 ? Discuss v(t) as t ! Show that the particle’s acceleration is always positive Draw the graph of v(t) against t Find the total distance travelled by the particle in the first seconds of motion t ¡ m s¡2 If the initial A train moves along a straight track with acceleration 10 ¡1 velocity of the train is 45 m s , determine the total distance travelled in the first minute An object has initial velocity 20 m s¡1 as it moves in a straight line with acceleration ¡ t 20 function 4e m s¡2 a Show that as t increases the object approaches a limiting velocity b Find the total distance travelled in the first 10 seconds of motion C PROBLEM SOLVING BY INTEGRATION When we studied differential calculus, we saw how to find the rate of change of a function by differentiation cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 In practical situations it is sometimes easier to measure the rate of change of a variable, for example, the rate of water flow through a pipe In such situations we can use integration to find a function for the quantity concerned black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_25\748IB_HL-2_25.CDR Wednesday, 21 November 2007 11:35:05 AM PETERDELL IB_HL-2ed (749) APPLICATIONS OF INTEGRATION (Chapter 25) 749 Example The marginal cost of producing x urns per week is given by 2:15 ¡ 0:02x + 0:000 36x2 dollars per urn provided x 120: The initial costs before production starts are $185 Find the total cost of producing 100 urns per day dC = 2:15 ¡ 0:02x + 0:000 36x2 $=urn dx R ) C(x) = (2:15 ¡ 0:02x + 0:000 36x2 ) dx The marginal cost is = 2:15x ¡ 0:02 x2 x3 + 0:000 36 + c = 2:15x ¡ 0:01x2 + 0:000 12x3 + c ) But C(0) = 185 c = 185 ) C(x) = 2:15x ¡ 0:01x2 + 0:000 12x3 + 185 ) C(100) = 2:15(100) ¡ 0:01(100)2 + 0:000 12(100)3 + 185 = 420 ) the total cost is $420 Example A metal tube has an annulus cross-section as shown The outer radius is cm and the inner radius is cm Within the tube, water is maintained at a temperature of 100o C Within the metal the temperature drops off from inside dT 10 to outside according to =¡ where x is the dx x distance from the central axis and x Find the temperature of the outer surface of the tube x water at 100°C metal tube cross-section Z ¡10 dx x ) T = ¡10 ln j x j + c ¡10 dT = , dx x so T = But when x = 2, T = 100 ) 100 = ¡10 ln + c ) c = 100 + 10 ln Thus T = ¡10 ln x + 100 + 10 ln ¡ ¢ T = 100 + 10 ln x2 ¡ ¢ When x = 4, T = 100 + 10 ln 12 ¼ 93:1 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ) the outer surface temperature is 93:1o C black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_25\749IB_HL-2_25.CDR Thursday, 22 November 2007 10:41:59 AM PETERDELL IB_HL-2ed (750) 750 APPLICATIONS OF INTEGRATION (Chapter 25) EXERCISE 25C The marginal cost per day of producing x gadgets is C (x) = 3:15 + 0:004x euros per gadget What is the total cost of daily production of 800 gadgets given that the fixed costs before production commences are E450 per day? The marginal profit for producing x dinner plates per week is given by P (x) = 15 ¡ 0:03x dollars per plate If no plates are made a loss of $650 each week occurs a Find the profit function b What is the maximum profit and when does it occur? c What production levels enable a profit to be made? Jon needs to bulk-up for the football season His energy needs t days after starting his weight gain program are given by E (t) = 350(80 + 0:15t)0:8 ¡ 120(80 + 0:15t) calories per day Find Jon’s total energy needs over the first week of the program The tube cross-section shown has inner radius of cm and outer radius cm Within the tube, water is maintained at a temperature of 100o C Within the metal the temperature falls off at the rate dT ¡20 = 0:63 where x is the distance from the central dx x axis and x 6 Find the temperature of the outer surface of the tube A thin horizontal metal strip of length metre has a deflection of y metres at a distance of x m from the fixed end d2 y It is known that = ¡ 10 (1 ¡ x)2 dx2 x metal x deflection y metal strip a Find the function y(x) which measures the deflection from the horizontal at any dy point on the metal strip Hint: When x = 0, what are y and ? dx b Determine the greatest deflection of the metal strip cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 x A m plank of wood is supported only at its ends, O P and P The plank sags under its own weight by y metres at a distance x metres from end O sag y µ ¶ x2 d2 y = 0:01 2x ¡ The differential equation relates the variables x and y dx2 a Find the function y(x) which measures the sag from the horizontal at any point along the plank b Find the maximum sag from the horizontal c Find the sag at a distance m from P d Find the angle the plank makes with the horizontal at the point m from P black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_25\750IB_HL-2_25.CDR Thursday, 22 November 2007 10:43:19 AM PETERDELL IB_HL-2ed (751) APPLICATIONS OF INTEGRATION (Chapter 25) A contractor digs roughly cylindrical wells to a depth of h metres He estimates that the cost of digging at the depth x metres is 12 x2 +4 dollars per m3 of earth and rock extracted 751 x h If a well is to have a radius r m, show that the total cost of digging a well is given by µ ¶ h + 24h dC dC dV C(h) = ¼r2 + C0 dollars Hint: = dx dV dx The length of a continuous function y = f (x) on a x b is found using µ ¶2 Z bs dy L= 1+ dx dx a Find, correct to decimal places, the length of y = sin x on x ¼ A farmer with a large property plans a rectangular irrigation canal fruit orchard with one boundary being an irrigation canal He has km of fencing to fence the orchard x km The farmer knows that the yield per unit of area p km changes the further you are away from the canal in proportion to p x+4 p km where x is as shown in the figure If the yield from the field is denoted Y, and the area of the orchard is denoted A: a explain why k dY =p dA x+4 where k is a constant k(4 ¡ 2p) dY by using the chain rule = p dx x+4 Z p k(4 ¡ 2p) p dx c explain why Y = x+4 p d show that Y = 4k(2 ¡ p)[ p + ¡ 2] b show that cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 e What dimensions should the orchard be to maximise the yield? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_25\751IB_HL-2_25.CDR Tuesday, 27 November 2007 10:16:16 AM PETERDELL IB_HL-2ed (752) 752 APPLICATIONS OF INTEGRATION (Chapter 25) REVIEW SET 25A The function y = f (x) is graphed Find: R4 a f (x) dx R6 b f (x) dx R6 c f (x) dx y x -2 y=ƒ(x) Write the total shaded area: a as the sum of three definite integrals b as one definite integral involving a modulus a y=g(x) b c d At time t = a particle passes through the origin with velocity 27 cm s¡1 Its acceleration t seconds later is 6t ¡ 30 cm s¡2 Find the total distance that the particle has travelled when it momentarily comes to rest for the second time Draw the graphs of y = x ¡ and y = x ¡ a Find the coordinates where the graphs meet b Find the enclosed area Determine k if the enclosed region has area 13 units2 y y=k y = x2 x Z Z e x By appealing only to geometrical evidence, explain why: e dx+ ln x dx = e: A boat travelling in a straight line has its engine turned off at time t = Its velocity 100 m s¡1 at time t seconds thereafter is given by v(t) = (t + 2)2 a b c d e Find the initial velocity of the boat, and its velocity after seconds Discuss v(t) as t ! Sketch the graph of v(t) against t Find how long it takes for the boat to travel 30 metres Find the acceleration of the boat at any time t dv = ¡kv , and find the value of the constant k f Show that dt cyan magenta yellow 95 100 50 and y = 7x2 ¡ 10x: 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find the total finite area enclosed by y = x3 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_25\752IB_HL-2_25.CDR Thursday, 22 November 2007 10:50:28 AM PETERDELL IB_HL-2ed (753) 753 APPLICATIONS OF INTEGRATION (Chapter 25) Find a given that the area of the region between y = ex and the x-axis from x = to x = a is units2 y = ex y Hence determine b, given that the area of the region between x = a and x = b is also units2 x a b 10 A cantilever of length L m has a deflection of y m at a distance x m from the fixed end The variables are connected by d2 y = k(L ¡ x)2 where k is the dx2 proportionality constant x y Find the greatest deflection of the cantilever in terms of k and L 11 Determine the area enclosed by y = ¼2 x and y = sin x 12 The figure shows the graphs of y = cos(2x) and y = e3x for x [¡¼, ¼2 ] Find correct to decimal places: y y = cos x a the value of b b the area of the shaded region a Find 13 y = e3 x b x R d [ ln(tan x + sec x) ] and hence find sec x dx: dx b Consider x 7! sec(2x) for x [ 0, ¼ ] i Sketch the graph of the function on the given domain ii Find the area of the region bounded by y = sec(2x), the y-axis, and the line y = REVIEW SET 25B A particle moves in a straight line with velocity v(t) = 2t ¡ 3t2 m s¡1 Find the distance travelled in the first second of motion Find the area of the region enclosed by y = x2 + 4x + and y = 3x + R2 p Determine ¡ x2 dx using graphical evidence only cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ¡100 dI = The current I(t) milliamps in a circuit falls off in accordance with dt t2 where t is the time in seconds, t > 0:2 It is known that when t = 2, the current is 150 milliamps Find a formula for the current at any time t > 0:2, and hence find: a the current after 20 seconds b what happens to the current as t ! black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_25\753IB_HL-2_25.CDR Tuesday, 27 November 2007 1:59:03 PM PETERDELL IB_HL-2ed (754) 754 APPLICATIONS OF INTEGRATION (Chapter 25) Is it true that R3 ¡1 y f (x) dx represents the area of the shaded region? Explain your answer briefly -1 x x + x2 Find the position and nature of all turning points of y = f(x) Discuss f (x) as x ! and as x ! ¡1 Sketch the graph of y = f (x) y Find the area enclosed by y = f (x), y = x2 + k the x-axis, and the vertical line x = ¡2 A Consider f(x) = a b c d B OABC is a rectangle and the two shaded regions are equal in area Find k k a Sketch the region bounded by y = x3 + 2, the y-axis, and the horizontal lines y = and y = b Write x in the form f(y): c Find the area of the region graphed in a C x Find the area enclosed by y = 2x3 ¡ 9x and y = 3x2 ¡ 10 10 Consider f(x) = ¡ sec2 x on [¡4, ] a b c d 11 Use technology to help sketch the graph of the function Find the equations of the function’s vertical asymptotes Find the axes intercepts Find the area of the region bounded by one arch of the function and the x-axis A metal tube has an annulus cross-section with radii r1 and r2 as shown Within the tube a liquid is maintained at rz temperature T0 o C Within the metal, the temperature drops from x dT k rx inside to outside according to = where dx x metal k is a negative constant and x is the distance from the central axis µ ¶ r2 Show that the outer surface has temperature T0 + k ln r1 liquid at T/°C 12 Find the area of the region enclosed by y = tan x, the x-axis, and the vertical line x = ¼3 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 13 A particle moves in a straight line with velocity given by v(t) = sin t metres per second Find the total distance travelled by the particle in the first seconds of motion black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_25\754IB_HL-2_25.CDR Tuesday, 27 November 2007 10:21:42 AM PETERDELL IB_HL-2ed (755) 755 APPLICATIONS OF INTEGRATION (Chapter 25) REVIEW SET 25C A particle moves in a straight line with velocity v(t) = t2 ¡ 6t + m s¡1 , t > a Draw a sign diagram for v(t) b Explain exactly what happens to the particle in the first seconds of motion c After seconds, how far is the particle from its original position? d Find the total distance travelled in the first seconds of motion Determine the area enclosed by the y-axis, the line y = and the curve ¶ µ y+3 x = ln a Find a given that the shaded area is units2 b Find the x-coordinate of A if OA divides the shaded region into equal areas y A y = ax(x - 2) x a The graph of y = sin x is drawn alongside y Z Use the graph to explain why R¼ ¼ < sin x dx < ¼: ´ ³ ´ ³ p1 , b A is ¼4 , p12 and C is 3¼ B A w_ A x B y = mx + c Hint: You may need your graphics calculator to the algebra y = -x + 2x + yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 -1 magenta y = sin x y Determine m and c if the enclosed region has area 12 units2 cyan x For the given function y = f (x), x 6: p a Show that A has equation yA = 4x ¡ x2 : b Show that B has equation p yB = ¡ 10x ¡ x2 ¡ 24: R4 R6 c Find yA dx and yB dx: R6 d Hence, find f (x) dx y X p p Use the diagram to show that the area under one arch of y = sin x (as illustrated) is p close to ¼4 (1 + 2) units2 , but more than it c Find exactly the area under one arch of y = sin x Y C black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_25\755IB_HL-2_25.CDR Wednesday, 21 November 2007 1:48:10 PM PETERDELL x IB_HL-2ed (756) 756 APPLICATIONS OF INTEGRATION (Chapter 25) y x2 y2 + = 16 -4 a Sketch the graph again and mark on it the area R4 p -2 represented by 12 16 ¡ x2 dx: R4 p b Explain from the graph why we can say < 16 ¡ x2 dx < 16 The ellipse shown has equation x Find the area of the region enclosed by y = x3 +x2 +2x+6 and y = 7x2 ¡x¡4: R¼ Without actually integrating sin3 x, prove that sin3 x dx < Hint: Graph y = sin3 x for x ¼ a Sketch the graphs of y = sin2 x and y = sin x on the same set of axes for x [ 0, ¼ ] b Find the exact value of the area enclosed by these curves for x [ 0, ¼2 ] 10 11 Determine the area of the region enclosed by y = x, y = sin x and x = ¼: 12 The shaded region has area Find the value of m unit2 y y¡=¡sin¡x p m 13 Find, correct to decimal places: a the value of a b the area of the shaded region x y y¡=¡xX y¡=¡sin¡x a x 14 A bank may compound interest over various lengths of time: yearly, half-yearly, quarterly, monthly, daily, and so on RT If interest is compounded instantaneously, we can show that I = P0 rert dt where I is the interest accrued, P0 is the initial investment, r is the rate of interest per annum as a decimal, and T is the period of the loan in years a Show that the amount of money in an account at time T is given by PT = P0 erT cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b How long will it take for an amount to double at a rate of 8% p.a.? c A block of land was bought for $55 in 1940 and sold for $196 000 in 2007 at the same time of the year What rate of interest, compounded instantaneously, would produce this increase in the same time? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_25\756IB_HL-2_25.CDR Tuesday, 27 November 2007 10:25:23 AM PETERDELL IB_HL-2ed (757) 26 Chapter Volumes of revolution Contents: A B Solids of revolution Volumes for two defining functions cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 26 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_26\757IB_HL-2_26.cdr Wednesday, 21 November 2007 3:43:29 PM PETERDELL IB_HL-2ed (758) 758 VOLUMES OF REVOLUTION (Chapter 26) A SOLIDS OF REVOLUTION Consider the curve y = f(x) for a x b If the shaded part is rotated about the x-axis through 360o, a 3-dimensional solid will be formed This solid is called a solid of revolution y y¡=ƒ(x) y y¡=¡ƒ(x) DEMO x x a b a b A solid of revolution will also be formed if the part of the curve is rotated about the y-axis through 360o y y y¡=ƒ(x) DEMO x a x b VOLUME OF REVOLUTION y=ƒ(x) y We can use integration to find volumes of revolution between x = a and x = b dx The solid can be thought to be made up of an infinite number of thin cylindrical discs x Since the volume of a cylinder = ¼r2 h, the left-most disc has approximate volume ¼[f(a)]2 ±x, and the right-most disc has approximate volume ¼[f(b)]2 ±x a x b In general, ¼[f (x)]2 ±x is the approximate volume for the illustrated disc As there are infinitely many discs, we let ±x ! cyan magenta ¼[f (x)]2 ±x = Z b ¼[f (x)]2 dx = ¼ Z b y2 dx 95 100 50 a 75 95 50 75 25 100 yellow 25 a x=a 95 100 50 75 25 95 100 50 75 25 ±x!0 x=b X ) V = lim black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_26\758IB_HL-2_26.CDR Wednesday, 23 January 2008 11:39:18 AM DAVID3 IB_HL-2ed (759) 759 VOLUMES OF REVOLUTION (Chapter 26) When the region enclosed by y = f (x), the x-axis, and the vertical lines x = a, x = b is rotated about the x-axis to generate a solid, the volume of the solid is given by Z b Volume of revolution = ¼ y y¡=¡ƒ(x) a x b y2 dx a Example y=x y Use integration to find the volume of the solid generated when the line y = x for x is revolved around the x-axis x Z b Z a Volume of revolution = ¼ =¼ · y dx Note: The volume of a cone can be calculated using Vcone = 13 ¼r2 h So, in this example V = 13 ¼42 (4) ¡ 13 ¼12 (1) x2 dx ¸4 x3 =¼ ¡ 64 ¢ =¼ ¡3 = 21¼ = 64¼ = 21¼ cubic units ¡ ¼ which checks X Example Find the volume of the solid formed when the graph of the function y = x2 for x is revolved about the x-axis Volume of revolution Z b =¼ y dx y (5, 25) Z a =¼ Z x 5 =¼ · =¼ (x2 )2 dx 0 x5 x4 dx ¸5 Note: Entering Y1 = X then fnInt(¼ ¤ Y12 , X, 0, 5) gives this volume TI C cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 = ¼(625 ¡ 0) = 625¼ cubic units black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_26\759IB_HL-2_26.CDR Wednesday, 21 November 2007 3:58:15 PM PETERDELL IB_HL-2ed (760) 760 VOLUMES OF REVOLUTION (Chapter 26) Example Z One arch of y = sin x is rotated about the x-axis b Volume = ¼ a Z What is the volume of revolution? [f(x)]2 dx ¼ sin2 x dx =¼ Z ¼ =¼ £1 y ¡ ¤ cos(2x) dx ¤¼ sin(2x) £¡ ¢ ¡ = ¼ ¼2 ¡ 14 sin(2¼) ¡ ¡ =¼ y = sin x p £x =¼£ x = ¼2 ¡ ¡1¢ đô sin ¼ units3 Using the same limit method we can derive a similar formula for a solid of revolution which has been rotated about the y-axis When the region enclosed by x = f(y), the x-axis, and the horizontal lines y = a, y = b is rotated about the y-axis to generate a solid, the volume of the solid is given by Z y b x = f (y) x b Volume of revolution = ¼ a x2 dy a Example When x = 1, y = When x = e, y = ) we rotate the function for y [0, 1] The graph of y = ln x, x [ 1, e ] is rotated about the y-axis What is the volume of revolution? Volume Z b x2 dy =¼ Z a =¼ y Z (ey )2 dy =¼ x e e2y dy =¼ £1 =¼ £1 ¤ 2y 2e 2e ¡ 12 e0 ¤ cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 = ¼2 (e2 ¡ 1) units3 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_26\760IB_HL-2_26.CDR Wednesday, 21 November 2007 4:08:53 PM PETERDELL IB_HL-2ed (761) 761 VOLUMES OF REVOLUTION (Chapter 26) EXERCISE 26A Find the volume of the solid formed when the following are revolved about the x-axis: p a y = 2x for x b y = x for x c y = x3 for x d e y = x2 for x f y = x for x p y = 25 ¡ x2 for x g y= h y =x+ for x x¡1 for x x Use technology to find, correct to significant figures, the volume of the solid of revolution formed when these functions are rotated through 360o about the x-axis: a y= x3 for x [ 1, ] x2 + y = esin x for x [ 0, ] b Find the volume of revolution when the shaded region is revolved about the x-axis a b c y y y y = x +3 y= +4 x x2 + y = 64 x x a What is the name of the solid of revolution when the shaded region is revolved about the x-axis? b Find the equation of the line segment AB in the form y = ax + b c Find a formula for the volume of the solid using Z b y dx: ¼ x The shaded region is rotated about the x-axis a Find the volume of revolution b A hemispherical bowl of radius cm contains water to a depth of cm What is the volume of water? y y = ex x y (0, r) A (h, 0) B x a y cyan magenta yellow 95 100 25 95 100 50 75 25 95 x (r, 0) 100 50 75 25 95 100 50 75 25 (-r, 0) 50 A circle with centre (0, 0) and radius r units has equation x2 + y = r2 a If the shaded region is revolved about the x-axis, what solid is formed? b Use integration to show that the volume of revolution is 43 ¼r3 (0, r) 75 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_26\761IB_HL-2_26.CDR Wednesday, 21 November 2007 4:11:57 PM PETERDELL IB_HL-2ed (762) 762 VOLUMES OF REVOLUTION (Chapter 26) Find the volumes of the solids formed when the following are revolved about the y-axis: a y = x2 between y = and y = p b y = x between y = and y = c y = ln x between y = and y = p d y = x ¡ between x = and x = 11 e y = (x ¡ 1)3 between x = and x = Find the exact value of the volume of the solid of revolution formed by rotating the x2 y2 relation + = 1, x > through 360o about the y-axis 16 Find the volume of revolution when these regions are rotated about the x-axis: c y = cos x for x ¼2 p y = sin x for x ¼ e y = sec(3x) for x [0, a ¼ 12 b y = cos(2x) for x d y= for x ¼3 cos x ¡ ¢ y = tan x2 for x [0, ¼2 ] f ] ¼ a Sketch the graph of y = sin x + cos x for x ¼2 10 b Hence, find the volume of revolution of the shape bounded by y = sin x + cos x, the x-axis, x = and x = ¼4 when it is rotated about the x-axis a Sketch the graph of y = sin(2x) from x = to x = ¼4 11 b Hence, find the volume of revolution of the shape bounded by y = sin(2x), the x-axis, x = and x = ¼4 when it is rotated about the x-axis B VOLUMES FOR TWO DEFINING FUNCTIONS Consider the circle with centre (0, 3) and radius unit y y x x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 When this circle is revolved about the x-axis, we obtain a doughnut or torus black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_26\762IB_HL-2_26.CDR Wednesday, 28 November 2007 9:11:48 AM PETERDELL IB_HL-2ed (763) 763 VOLUMES OF REVOLUTION (Chapter 26) In general, if the region bounded by y = f (x) (on top) and y = g(x) and the lines x = a, x = b is revolved about the x-axis, then its volume of revolution is given by: Z b Z b [f(x)] dx ¡ ¼ [g(x)]2 dx V = ¼ a Z a b V =¼ So, Z ³ ´ [f (x)]2 ¡ [g(x)]2 dx b V =¼ or a a Rotating ¡ ¢ yU ¡ yL2 dx y=ƒ(x) or yU y y about the x-axis gives y=g(x) or yL a b x x yU reads ‘y upper’ yL reads ‘y lower’ Example y y = x2 Find the volume of revolution generated by revolving the region between y = x2 p and y = x about the x-axis y= x (1,¡1) x Z ¡ ¢ yU ¡ yL2 dx ³p ´ ( x) ¡ (x2 )2 dx Volume = ¼ Z =¼ Z =¼ · y y= x (x ¡ x4 ) dx ¸1 x2 x5 =¼ ¡ ¢ ¡¡ 1 ¢ = ¼ ¡ ¡ (0) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 x units3 95 3¼ 10 100 50 75 25 95 100 50 75 25 = y = x2 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_26\763IB_HL-2_26.CDR Friday, 30 November 2007 10:01:14 AM PETERDELL IB_HL-2ed (764) 764 VOLUMES OF REVOLUTION (Chapter 26) EXERCISE 26B y The shaded region between y = ¡ x y = is revolved about the x-axis and x a What are the coordinates of A and B? b Find the volume of revolution y y¡=¡4¡-¡xX The shaded region is revolved about the x-axis a Find the coordinates of A b Find the volume of revolution x y = e2 y¡=¡e A y¡=¡3 B A x The shaded region (between y = x, y = y x y=x and x = 2) is revolved about the x-axis a Find the coordinates of A b Find the volume of revolution A y= x x x=2 Find exactly the volume of the solid of revolution generated by rotating the region enclosed by y = x2 ¡ 4x + and x + y = through 360o about the x-axis p The shaded region (between y = x ¡ 4, y = and x = 8) is revolved about the x-axis a What are the coordinates of A? b Find the volume of revolution y y = x-4 y=1 A x y cyan magenta yellow 95 (0, 3) x 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The illustrated circle has equation x2 + (y ¡ 3)2 = p a Show that y = § ¡ x2 b Draw a diagram and show on it what part of the p circle is represented by y = + ¡ x2 and p what part by y = ¡ ¡ x2 c Find the volume of revolution of the shaded region about the x-axis Hint: Substitute x = sin u to evaluate the integral Use your calculator to check your answer black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_26\764IB_HL-2_26.CDR Thursday, 22 November 2007 9:30:36 AM PETERDELL IB_HL-2ed (765) VOLUMES OF REVOLUTION (Chapter 26) 765 The length of a chord of the circle with equation x2 + y = r2 is equal to the radius of the circle The chord is parallel to the y-axis and a solid of revolution is generated by rotating the minor segment cut off by the chord through 360o about the y-axis ¼r3 Prove that the volume of the solid formed is given by V = A circle has equation x2 + y = r2 where r > A minor segment is cut off by a chord of length units drawn parallel to the y-axis Show that the volume of the solid of revolution formed by rotating the segment through 360o about the y-axis is independent of the value of r y y= Prove that the shaded area from x = to infinity is infinite whereas its volume of revolution is finite x (1, 1) x REVIEW SET 26 Find the volume of the solid of revolution formed when the following are rotated about the x-axis: a y = x between x = and x = 10 b y = x + between x = and x = 10 c y = sin x between x = and x = ¼ p d y = ¡ x2 between x = and x = Find the volume of the solid of revolution formed when the shaded region is rotated through 360o about the x-axis: a b y y y = e-x + y = cos(2x) p 16 x x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find the volume of revolution when y = csc x is rotated through 360o about the x-axis for ¼4 x 3¼ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_26\765IB_HL-2_26.CDR Thursday, 22 November 2007 9:35:29 AM PETERDELL IB_HL-2ed (766) 766 VOLUMES OF REVOLUTION (Chapter 26) Find the volume of the solid of revolution formed when the following are rotated about the y-axis: a x = y2 between y = and y = p b y = x2 between y = and y = c y = x3 between x = and x = y Find the volume of revolution generated by rotating the shaded region through 360o about the x-axis: y = x2 y=4 x Find the volume of revolution if the shaded region is rotated through 360o about the x-axis: y y¡=¡sin¡x x y¡=¡cos¡x a b Use V = 13 ¼r2 h to find the volume of this cone Check your answer to a by integration cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find the volume enclosed when y = x3 , from the x-axis to y = 8, is revolved about the y-axis black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_26\766IB_HL-2_26.CDR Wednesday, 28 November 2007 9:41:25 AM PETERDELL IB_HL-2ed (767) 27 Chapter Further integration and differential equations Contents: The integrals of p a2 ¡ x2 and x + a2 A Further integration by substitution Integration by parts Miscellaneous integration Separable differential equations B C D E cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 27A Review set 27B black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_27\767IB_HL-2_27.CDR Tuesday, 29 January 2008 2:11:46 PM PETERDELL IB_HL-2ed (768) 768 FURTHER INTEGRATION AND DIFFERENTIAL EQUATIONS (Chapter 27) In the previous chapters we have seen several techniques for finding integrals, including: ² ² ² integrating term by term integrating by using the reverse process of differentiation integration by substitution In this chapter we consider some more special integrals and techniques for integration, including integration by parts We conclude the chapter with a study of separable differential equations which can be solved by integration A THE INTEGRALS OF p a ¡ x2 AND x2 + a2 1 and can be obtained by considering the derivatives The integrals of p 2 x + a2 a ¡x ¡ ¢ ¡ ¢ and y = arctan xa of y = arcsin xa ¡ ¢ ¡ ¢ Consider y = arcsin xa Consider y = arctan xa ) x = a sin y ) x = a tan y dx dx ) ) = a cos y = a sec2 y dy dy p = a(1 + tan2 y) = a ¡ sin2 y µ ¶ r x2 x2 = a 1+ = a 1¡ a a 2 p a +x = a2 ¡ x2 = a dy dy a ) = p ) = dx a2 ¡ x2 dx x + a2 Z Z ¡x¢ ¡ ¢ 1 p ) dx = arcsin a + c ) dx = a1 arctan xa + c 2 2 x +a a ¡x Z p dx ¡ x2 b Z Z p dx ¡ x2 ¡ ¢ = arcsin x3 + c dx +8 Z p dx = 54 x2 + ( 2)2 ³ ´ = 54 £ p12 arctan px2 + c b 4x2 cyan magenta yellow 50 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 = p 95 a dx +8 4x2 ³ arctan 100 Z a Find: 75 Example black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_27\768IB_HL-2_27.CDR Friday, 23 November 2007 10:13:50 AM PETERDELL x p ´ +c IB_HL-2ed (769) 769 FURTHER INTEGRATION AND DIFFERENTIAL EQUATIONS (Chapter 27) EXERCISE 27A Find: Z p a dx ¡ x2 Z p dx e ¡ 4x2 Z b Z f Z p dx ¡ x2 c p dx ¡ 9x2 g Z Z dx x + 16 d dx + 2x2 h Z dx +1 4x2 dx + 4x2 a Sketch the graph of y = p ¡ x2 b Explain algebraically why the function i is symmetrical about the y-axis ii has domain x ] ¡1, [ c Find the exact area enclosed by the function and the x-axis, the y-axis, and the line x = 12 B FURTHER INTEGRATION BY SUBSTITUTION Here are some suggestions of possible substitutions to help integrate more difficult functions Note that these substitutions may not always lead to success, so sometimes other substitutions will be needed With practice you will develop a feeling for which substitution is best in a given situation Example Find When a function contains p f (x) Try substituting ln x p a2 ¡ x2 p x2 + a2 or x2 + a2 p x2 ¡ a2 u = ln x u = f(x) x = a sin µ x = a tan µ x = a sec µ R p x x + dx R p x x + dx R p du = (u ¡ 2) u dx dx R = (u ¡ 2u ) du ) Let u = x + du ) =1 dx = u2 ¡ 2u +c = 25 (x + 2) ¡ 43 (x + 2) + c EXERCISE 27B cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find using a suitable substitution: R p R 2p a x x ¡ dx b x x + dx Z p R 3p x¡1 e d t t2 + dt dx x black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_27\769IB_HL-2_27.CDR Friday, 23 November 2007 10:16:47 AM PETERDELL c R p x3 ¡ x2 dx Hint: in e let u = p x ¡ IB_HL-2ed (770) 770 FURTHER INTEGRATION AND DIFFERENTIAL EQUATIONS (Chapter 27) Example Z p x x + dx Find the exact value of ¡4 Z du =1 dx Let u = x + ) ) p x x + dx = ¡4 Z Z When x = ¡4, u = When x = 6, u = 10 10 p (u ¡ 4) u du 10 h = (u ¡ 4u ) du = 52 5u ¡ 3 2 i10 u2 3 £ 10 ¡ £ 10 p p = 40 10 ¡ 80 10 p 40 = 10 = Z p x ¡9 Find dx x Example dx = sec µ tan µ dµ Z p Z p x ¡9 sec2 µ ¡ So, dx = sec µ tan µ dµ x sec µ R p = sec2 µ ¡ tan µ dµ R = tan2 µ dµ R = (3 sec2 µ ¡ 3) dµ ) Let x = sec µ ) Z b p x2 ¡9 x µ = tan µ ¡ 3µ + c p ¡ ¢ = x2 ¡ ¡ arccos x3 + c Find the exact value of: Z p x x ¡ dx a x 3 cos µ = x sec µ = ) tan µ = Z p x x + dx c p x2 ¡ p x2 x ¡ dx Check each answer using technology Integrate with respect to x: cyan magenta yellow 95 100 x3 + x2 50 h 75 p ¡ 4x2 25 g sin x cos 2x f 95 ln x ³ ´ x + [ln x]2 100 d 50 2x x2 + 75 c 25 x2 p ¡ x2 b 95 100 50 75 25 95 100 50 75 25 e x2 + x2 p x2 ¡ x a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_27\770IB_HL-2_27.CDR Friday, 23 November 2007 10:20:11 AM PETERDELL IB_HL-2ed (771) 771 FURTHER INTEGRATION AND DIFFERENTIAL EQUATIONS (Chapter 27) ³ ´ j x + [ln x] i x(x2 + 16) C k x2 p 16 ¡ x2 l p x2 ¡ x2 INTEGRATION BY PARTS Some functions can only be integrated using integration by parts, which is a method we derive from the product rule of differentiation R d (u v + uv0 ) dx = uv Since (uv) = u0 v + uv0 then dx R R ) u v dx + uv0 dx = uv R R ) uv dx = uv ¡ u0 v dx So, providing R R u0 v dx can be easily found, we can find uv0 dx = uv ¡ R Z u0 v dx u or R uv0 dx using dv dx = uv ¡ dx Z v du dx dx Example Find: a R xe¡x dx R b x cos x dx a v = e¡x u=x v = ¡e¡x u =1 R ¡x R ) xe dx = ¡xe¡x ¡ (¡e¡x ) dx = ¡xe¡x + (¡e¡x ) + c = ¡e¡x (x + 1) + c Check: d (¡e¡x (x + 1) + c) dx = e¡x (x + 1) + ¡e¡x (1) + = xe¡x + e¡x ¡ e¡x = xe¡x X b u=x v = cos x u =1 v = sin x R R ) x cos x dx = x sin x ¡ sin x dx = x sin x ¡ (¡ cos x) + c = x sin x + cos x + c Check: d (x sin x + cos x + c) dx = £ sin x + x cos x ¡ sin x = sin x + x cos x ¡ sin x = x cos x X EXERCISE 27C Use integration by parts to find the integral of the following functions with respect to x: b x sin x c x2 ln x a xex d x sin 3x e x cos 2x f x sec2 x i arctan x g ln x h (ln x)2 When using ‘integration by parts’ the function u should be easy to differentiate and v should be easy to integrate cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Hint: In g write ln x as ln x In i write arctan x as arctan x black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_27\771IB_HL-2_27.CDR Friday, 23 November 2007 10:53:56 AM PETERDELL IB_HL-2ed (772) 772 FURTHER INTEGRATION AND DIFFERENTIAL EQUATIONS (Chapter 27) Sometimes it is necessary to use integration by parts twice in order to find an integral Example Find R ex sin x dx R ex sin x dx R = ex (¡ cos x) ¡ ex (¡ cos x) dx R = ¡ex cos x + ex cos x dx R = ¡ex cos x + ex sin x ¡ ex sin x dx R x ) e sin x dx = ¡ex cos x + ex sin x R x ) e sin x dx = 12 ex (sin x ¡ cos x) + c Find these integrals: a x2 e¡x ex cos x b u = ex v = sin x u0 = ex v = ¡ cos x u = ex v = cos x u0 = ex v = sin x e¡x sin x c d x2 sin x R u a Use integration by parts to find u e du R b Hence find (ln x) dx using the substitution u = ln x R a Use integration by parts to find u sin u du p R b Hence find sin 2x dx using the substitution u2 = 2x Find R p cos 3x dx using the substitution u2 = 3x INVESTIGATION NUMERICAL INTEGRATION There are many functions that not have indefinite integrals In other words, we cannot write the indefinite integral as a function However, definite integrals can still be determined by numerical methods An example of such a method is the upper and lower rectangles we used in Chapter 19 y A slightly more accurate method is the midpoint rule in which we take the height of each rectangle to be the value of the function at the midpoint of the subinterval For example, consider finding the area between f (x) = sin(x2 ) and the x-axis from x = to x = The graph of f (x) = sin(x2 ) is: f (x) is an even function and does not have an indefinite integral, so a numerical method is essential for Z sin(x2 ) dx to be evaluated upper rectangle midpoint lower rectangle x y -2 -1 x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_27\772IB_HL-2_27.CDR Monday, 26 November 2007 3:18:31 PM PETERDELL IB_HL-2ed (773) 773 FURTHER INTEGRATION AND DIFFERENTIAL EQUATIONS (Chapter 27) What to do: Suppose the interval from to is divided into 10 equal subintervals of width 0:1 Using the midpoint rule, the shaded area is Z sin(x2 )dx ¼ [f (0:05) + f (0:15) + f (0:25) + :::::: + f (0:95)] ±x where ±x = 0:1 is the subinterval width Use this formula to estimate the integral to decimal places AREA FINDER Click on the area finder icon and find the area estimate for n = 10, 100, 1000, 10 000 Use your graphics calculator’s definite integral function to find the area R2 Now find sin(x2 ) dx: What is the area enclosed between y = sin(x2 ), the x-axis, and the vertical line x = 2? D MISCELLANEOUS INTEGRATION We have now practised several integration techniques with clues given as to what method to use In this section we attempt to find integrals without clues EXERCISE 27D Integrate with respect to x: a ex + e¡x ex ¡ e¡x b 7x c (3x + 5)5 d e x sec2 x f cot 2x g x(x + 3)3 h i x2 e¡x j p x 1¡x k p x2 ¡ x2 l m p x2 x ¡ n tan3 x o cyan (x + 2) (x + 1)3 x p x x2 ¡ p x2 + 2x + c ln (2x) d e¡x cos x n ¡ 2x p ¡ x2 o x3 (2 ¡ x)3 p sin5 x cos5 x magenta yellow 95 x+4 x2 + 100 m 50 cos3 x 75 l 25 2x + x2 ¡ 2x + k sin 4x cos x 95 j 100 x p x¡3 50 i 75 (ln x)2 x2 25 h p ¡ x2 g 95 arctan x + x2 100 f 50 x(1 + x2 ) 75 e 25 95 100 50 75 25 Integrate with respect to x: a b p p x2 + x 1¡x ln (x + 2) sin x ¡ cos x black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_27\773IB_HL-2_27.CDR Thursday, 29 November 2007 9:52:40 AM PETERDELL IB_HL-2ed (774) 774 FURTHER INTEGRATION AND DIFFERENTIAL EQUATIONS (Chapter 27) E SEPARABLE DIFFERENTIAL EQUATIONS dy Consider a function y = f(x) We know the derivative is the rate at which the dx function changes with respect to its independent variable x A differential equation is an equation which connects the derivative(s) of a function to the function itself and variables in which the function is defined x2 dy = dx y Examples of differential equations are: dy d2 y ¡3 + 4y = dx2 dx and These are some examples of situations where differential equations are observed: A falling object A parachutist Object on a spring y m v d2 y = 9:8 dt2 m dv = mg ¡ av2 dt m d2 y = ¡ky dt2 SEPARABLE DIFFERENTIAL EQUATIONS dy f(x) = dx g(y) Differential equations which can be written in the form separable differential equations are known as Z Z dy dy = f (x) and so g(y) dx = f (x) dx when we integrate dx dx both sides with respect to x R R Consequently, g(y) dy = f(x) dx which enables us to solve the original differential equation Notice that g(y) When we integrate, the solution involves unknown constants, and this is called a general solution of the differential equation The constants are evaluated using initial conditions to give a particular solution Example Find the general solutions to the following separable differential equations: dy dy a = ky b = k(A ¡ y) dx dx cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 dy = k(A ¡ y) = ¡k(y ¡ A) dx b 75 25 dy = ky dx 95 100 50 75 25 a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_27\774IB_HL-2_27.CDR Friday, 23 November 2007 12:26:45 PM PETERDELL IB_HL-2ed (775) FURTHER INTEGRATION AND DIFFERENTIAL EQUATIONS (Chapter 27) dy =k y dx ) Z ) Z R dy dx = k dx y dx Z R ) dy = k dx y ) 775 dy = ¡k y ¡ A dx R dy dx = ¡k dx y ¡ A dx Z R ) dy = ¡k dx y¡A ) ) ln j y j = kx + c ) y = §ekx+c ) y = §ec ekx ) y = Aekx for some constant A ) ln j y ¡ A j = ¡kx + c ) j y ¡ A j = e¡kx+c ) y ¡ A = §ec e¡kx ) y ¡ A = Be¡kx fB a constantg ) y = A + Be¡kx Example Solve p dV = k V given that V (9) = 1, V (13) = 4, and k is a constant dh Don’t forget to check your solution by differentiation p dV =k V dh dV ) p =k V dh R ¡ dV R ) V dh = k dh dh R R ¡1 ) V dV = k dh V ) 2 This is the general solution = kh + c p V = kh + c ) These are the initial conditions But V (9) = and V (13) = p p ) = k(9) + c and = k(13) + c ) 9k + c = and 13k + c = Solving these equations simultaneously gives k = cyan magenta yellow and c = ¡ 52 h¡5 95 100 50 75 This is the particular solution 25 95 100 50 75 25 95 = p h¡5 V = µ ¶2 h¡5 ) V = 100 50 75 25 95 100 50 75 25 ) 5 p V = 12 h ¡ ) black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_27\775IB_HL-2_27.CDR Thursday, 11 March 2010 10:56:38 AM PETER IB_HL-2ed (776) 776 FURTHER INTEGRATION AND DIFFERENTIAL EQUATIONS (Chapter 27) Example The curve y = f (x) has a y-intercept of The tangent to this curve at the point (x, y) has an x-intercept of x + Find the equation of this curve in the form y = f (x) dy y¡0 y = =¡ dx x ¡ (x + 2) dy ) = ¡ 12 y dx Z Z dy ) dx = ¡ 12 dx y dx Z ) dy = ¡ 12 x + c y Gradient of tangent at (x, y) is y (x, y) y (x+2, 0) x ln j y j = ¡ 12 x + c ) x ) y = Ae¡ x When x = 0, y = ) 3=A ) y = 3e¡ x Example 10 The number of bacteria present in a culture increases at a rate proportional to the number present If the number increases by 10% in one hour, what percentage increase occurs after a further hours? If N is the number of bacteria present at time t hours, then dN = kN dt ) dN /N dt where k is a constant dN =k N dt Z R dN and dt = k dt N dt Z R ) dN = k dt N ) ) ln N = kt + c fj N j is not necessary as N > 0g Suppose the number of bacteria when t = was N = N0 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 ln N0 = c and so ln N = kt + ln N0 (1) 100 50 75 25 95 100 50 75 25 ) black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_27\776IB_HL-2_27.CDR Friday, 12 March 2010 4:31:52 PM PETER IB_HL-2ed (777) 777 FURTHER INTEGRATION AND DIFFERENTIAL EQUATIONS (Chapter 27) fsince 110% of N0 is 1:1N0 g When t = 1, N = 1:1N0 ) in (1), ln(1:1N0 ) = k + ln N0 ) ln(1:1) + ln N0 = k + ln N0 ) k = ln (1:1) So, (1) becomes ln N = t ln (1:1) + ln N0 ) N = N0 £ (1:1)t After a further hours, t = ) N = N0 (1:1)6 ¼ 1:7716N0 ¼ 177:16% of N0 ) N has increased by 77:16% Example 11 A raindrop falls with acceleration 9:8 ¡ v m s¡2 , where v is its velocity Show t that the raindrop’s velocity is v = 29:4(1 ¡ e¡ ) m s¡1 limiting value of 29:4 m s¡1 Acceleration is dv , dt µ and it approaches a dv v 29:4 ¡ v = 9:8 ¡ = m s¡2 dt 3 so ¶ dv ) = ¡ 13 v ¡ 29:4 dt ¶ Z µ Z dv ) dt = ¡ 13 dt v ¡ 29:4 dt ¶ Z µ Z ) dv = ¡ 13 dt v ¡ 29:4 ) ln j v ¡ 29:4 j = ¡ 13 t + c t ) v ¡ 29:4 = Ae¡ t ) v = 29:4 + Ae¡ Now when t = 0, v = and so A = ¡29:4 t ) v = 29:4 ¡ 29:4e¡ t ) v = 29:4(1 ¡ e¡ ) ¡t cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Now as t ! 1, e ! ) v ¡ 29:4 ! and so v ! 29:4 m s¡1 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_27\777IB_HL-2_27.CDR Friday, 23 November 2007 12:45:19 PM PETERDELL IB_HL-2ed (778) 778 FURTHER INTEGRATION AND DIFFERENTIAL EQUATIONS (Chapter 27) Example 12 A large cylindrical tank has radius m Water flows out of a tap at the bottom of the tank at a rate proportional to the square root of the depth of the water within it Initially the tank is full to a depth of m After 15 minutes the depth of water is m How long will it take for the tank to empty? 9m hm 2m p dV / h where h is the depth of water dt p dV ) = k h where k is a constant dt dV dV dh Now = fchain ruleg dt dh dt dV = 4¼ The volume of water in the tank is V = ¼r2 h = 4¼h ) dh p dh ) k h = 4¼ dt 4¼ dh ) p =k h dt We are given that ) R dh dt = k dt dt R R ¡1 ) 4¼ h dh = k dt R 4¼h¡ ) 4¼ h2 = kt + c p ) 8¼ h = kt + c (1) Now when t = 0, h = p ) in (1), 8¼ = c and so c = 24¼ p So, 8¼ h = kt + 24¼ (2) And when t = 15, h = ) in (2), 16¼ = 15k + 24¼ p ) 8¼ h = ¡ 8¼ 15 t + 24¼ p t ) h = ¡ 15 +3 and so k = ¡ 8¼ 15 Now when it is empty h = t +3 ) = ¡ 15 ) t = 45 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 So, the tank empties in 45 minutes black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_27\778IB_HL-2_27.CDR Friday, 23 November 2007 12:48:06 PM PETERDELL IB_HL-2ed (779) FURTHER INTEGRATION AND DIFFERENTIAL EQUATIONS (Chapter 27) 779 EXERCISE 27E.1 Find the general solution of the following differential equations: dy dM dy a = 5y b = ¡2M c = dx dt dx y p dP dQ dQ d e =3 P = 2Q + f = dt dt dt 2Q + Find the particular solution of the following differential equations: dy dM a = 4y and when x = 0, y = 10 b = ¡3M and M (0) = 20 dx dt p y dP dy = and when t = 24, y = d = 2P + and P (0) = c dt dn dy p e = k y where k is a constant, y(4) = and y(5) = dx If the gradient of a curve at any point is equal to twice the y-coordinate at that point, show that the curve is an exponential function a If dp = ¡ 12 p, and p = 10 when t = 0, find p when t = dt b If dM = ¡ 2M, and M = when r = 0, find r when M = 3:5 dr ds + ks = where k is a constant, and s = 50 when t = dt t If it is also known that s = 20 when t = 3, show that s = 50(0:4) : Find the general solution of: b xy = 4y0 a xy0 = 3y c Solve dz = z + zr2 , z(0) = dr Solve dy = ¡2xy if y = when x = dx Solve yex y = yex d y = xey dy = x if y = when x = dx 10 Solve (1 + x) dy = 2xy if y = e2 when x = dx 11 Solve (1 + x2 ) dy = 2xy if y = 10 when x = dx dy = 4x + xy2 if y = when x = dx 12 Solve cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 13 The tangent to the curve y = f (x) at the point (x, y) has an x-intercept of x + If the curve has an y-intercept of 2, find f (x) black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_27\779IB_HL-2_27.CDR Friday, 12 March 2010 4:32:26 PM PETER IB_HL-2ed (780) 780 FURTHER INTEGRATION AND DIFFERENTIAL EQUATIONS (Chapter 27) dy x = Find the general solution to this differential dx y equation and hence find the curve that passes through the point (5, ¡4) Find a if (a, 3) also lies on this curve 14 A curve is known to have dy = y2 (1 + x) and passes through the point (1, 2) dx a Find the equation of this curve b Find the equations for the curve’s asymptotes 15 A curve has dy x =¡ dx y 16 Compare the differential equations: dy y = dx x with a Prove that the solution curves for these differential equations intersect at right angles b Solve the differential equations analytically and give a geometrical interpretation of the situation 17 A body moves with velocity v metres per second, and its acceleration is proportional to v If v = when t = and v = when t = 4, find the formula for v in terms of t Hence, find v when t = seconds 18 In the ‘inversion’ of raw sugar, the rate of change in the weight w kg of raw sugar is directly proportional to w If, after 10 hours, 80% reduction has occurred, how much raw sugar remains after 30 hours? 19 When a transistor ratio is switched off, the current falls away according to the differential dI equation = ¡kI where k is a constant If the current drops to 10% in the first dt second, how long will it take to drop to 0:1% of its original value? 20 A lump of metal of mass kg is released from rest in water After t seconds its velocity is v m s¡1 and the resistance due to the water is 4v Newtons dv The equation for the motion is = g ¡ 4v where g is the gravitational constant dt g a Prove that v = (1 ¡ e¡4t ) and hence show that there is a limiting velocity g b When is the metal falling at m s¡1 ? 10 21 Water evaporates from a lake at a rate proportional to the volume of water remaining dV = k(V0 ¡ V ) a Explaining the symbols used, why does dt represent this situation? b If 50% of the water evaporates in 20 days, find the percentage of water remaining after 50 days without rain cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 22 Water flows out of a tap at the bottom of a cylindrical tank of height m and radius m The tank is initially full and the water escapes at a rate proportional to the square root of the depth of the water remaining After hours the depth of water is m How long will it take for the tank to empty? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_27\780IB_HL-2_27.CDR Friday, 23 November 2007 12:49:58 PM PETERDELL IB_HL-2ed (781) 781 FURTHER INTEGRATION AND DIFFERENTIAL EQUATIONS (Chapter 27) 23 Water evaporates from a hemispherical bowl of radius dV r cm such that = ¡r2 , where t is the time in dt hours r cm h cm If the water has depth h cm then its volume is given by V = 13 ¼h2 (3r ¡ h) dV dV dh a Assuming r is a constant, use = to set up a differential equation dt dh dt between h and t b Suppose the bowl’s radius is 10 cm and that initially it is full of water ¼ (h3 ¡ 30h2 + 2000) and hence find the time taken for the Show that t = 300 depth of water to reach the cm mark Newton’s law of cooling is: “The rate at which an object changes temperature is proportional to the difference between its temperature and that of the surrounding medium, dT / (T ¡ Tm )” i.e., dt Use Newton’s law of cooling to solve questions 24 and 25 24 The temperature inside a refrigerator is maintained at 5o C An object at 100o C is placed in the refrigerator to cool After minute its temperature drops to 80o C How long will it take for the temperature to drop to 10o C? 25 At am the temperature of a corpse is 13o C and hours later it falls to 9o C Given that living body temperature is 37o C and the temperature of the corpse’s surroundings is constant at 5o C, estimate the time of death eiµ = cos µ + i sin µ INVESTIGATION In the 18th century Leonhard Euler made enormous contributions to mathematics and physics Euler was responsible for introducing the symbols e and i, and for the famous identity eiµ = cos µ + i sin µ In this investigation we explore two methods which show this result METHOD 1: POWER SERIES EXPANSION + x + x2 + x3 + ::::: = cyan for j x j < fsum of an infinite geometric seriesg magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 = + x + x2 + x3 + ::::: where the RHS is called the power series expansion 1¡x of 1¡x 25 95 100 50 75 25 ) 1¡x black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_27\781IB_HL-2_27.CDR Friday, 23 November 2007 1:32:56 PM PETERDELL IB_HL-2ed (782) 782 FURTHER INTEGRATION AND DIFFERENTIAL EQUATIONS (Chapter 27) Other functions also have power series expansions of the form X aj xj where faj g j=0 is a set of constant coefficients What to do: Suppose ex has a power series expansion, so ex = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + :::::: (1) a b c d By letting x = 0, find a0 Differentiate both sides of (1) and by letting x = 0, find a1 Repeat the above process to find a2 , a3 , a4 and a5 Conjecture the power series expansion of ex by writing an expression for the general term an Repeat the procedure of to obtain power series expansions for sin x and cos x In your power series for ex , replace x by iµ Hence, find eiµ in the form A + iB Compare the results of and to find A and B METHOD 2: DIFFERENTIAL EQUATION In this method we obtain a separable differential equation involving the complex number i We can solve this in the usual method by treating i as a constant, though in later years of mathematics you may learn this is not always such a wise tactic! What to do: Given the complex number z = r cis µ = r cos µ + ir sin µ where r = j z j is a dz constant and µ = arg z is variable, show that = iz dµ dz = iz, showing that z = reiµ where r is a Solve the differential equation dµ constant Compare your results in and to obtain the identity eiµ = cos µ + i sin µ RESULTS USING THE IDENTITY Use the identity eiµ = cos µ + i sin µ to show that: i j zw j = j z jj w j and arg(zw) = arg(z) + arg(w) for complex numbers z, w ii cis µ = cis (µ ¡ Á) cis Á iii ei¼ + = fcalled Euler’s identityg DIFFERENTIAL EQUATIONS WITH UNUSUAL SUBSTITUTIONS cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 In this final section we consider some interesting inseparable differential equations which can be converted to separable form using clever substitutions black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_27\782IB_HL-2_27.CDR Friday, 23 November 2007 1:38:20 PM PETERDELL IB_HL-2ed (783) FURTHER INTEGRATION AND DIFFERENTIAL EQUATIONS (Chapter 27) Example 13 Solve x dy =x+y dx du dy = x+u dx dx Substituting into the original DE, µ ¶ du x x + u = x + ux dx 783 by letting y = ux fproduct ruleg Let y = ux ) du + ux = x + ux dx du =x ) x2 dx du = ) dx x Z Z du ) dx = dx dx x ) x2 ) u = ln j x j + c But y = ux, so y = x ln j x j + cx is the general solution EXERCISE 27E.2 dy ¡ 2xex = y by letting y = uex dx µ ¶2 dy = y + 2ex y + e2x by letting y = uex Solve dx dy Solve 4xy = ¡x2 ¡ y for x > by letting y = ux dx dy Solve x ¡ y = 4x2 y by letting y = ux or otherwise dx Solve REVIEW SET 27A Find R p x2 ¡ x dx R Use integration by parts to find differentiation Find integrals of: a arctan x dx Check your answer using e¡x cos x Find the solution to y0 = ¡ 2ex y x2 ex b c x3 p ¡ x2 given that y(0) = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 5 The current I which flows through an electrical circuit with resistance R and dI inductance L can be determined from the differential equation L = E ¡ RI dt Both R and L are constants, and so is the electromotive force E black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_27\783IB_HL-2_27.CDR Friday, 23 November 2007 1:59:30 PM PETERDELL IB_HL-2ed (784) 784 FURTHER INTEGRATION AND DIFFERENTIAL EQUATIONS (Chapter 27) Given that R = 4, L = 0:2 and E = 20, find how long it will take for the current, initially amps, to reach 0:5 amps The graph of y = f (x) has a y-intercept of and the tangent at the point (x, y) has an x-intercept of x ¡ Find the function f (x) dy = x2 + y by letting y = ux: dx Hence show that if y = when x = 1, a particular solution is y = x2 + 3x Solve the differential equation 2xy REVIEW SET 27B Z Find: a Find: a R Z b x cos x dx Given dy = dx y+2 Given dy 2x = dx cos y Z dx + 4x2 Z p x ¡4 b dx x p dx ¡ x2 10 c and y(0) = 0, deduce that y = p x x ¡ dx p 2x + ¡ with initial condition y(1) = ¼2 , show that y = arcsin(x2 ) OBLH is a seal slide at the zoo At any point P, the gradient of the slide is equal to y the gradient of a uniformly inclined plane with H highest point P and with a m long horizontal base along OB a Show that P(x, y) satisfies the equation 10 m x y = 10e¡ b Find the height of L above OB c Find the gradient of the slide at H and at L O P(x, y) L 5m 30 m x B Bacteria grow in culture at a rate proportional to the number of bacteria present a Write down a differential equation connecting the number of bacteria N(t) at time t, to its growth rate b If the population of bacteria doubles every 37 minutes and there are 105 bacteria initially, how many bacteria are present in the culture after hours? dy = (y ¡ 1)2 (2 + x) and passes through the point (¡1, 2): dx a find the equation of the curve b find the equations of the asymptotes to the curve If a curve has cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A cylindrical rainwater tank is initially full, and the water runs out of it at a rate proportional to the volume of water left in it If the tank is half full after 20 minutes, find the fraction of water remaining in the tank after one hour black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_27\784IB_HL-2_27.CDR Friday, 12 March 2010 4:34:11 PM PETER IB_HL-2ed (785) 28 Chapter Statistical distributions of discrete random variables Contents: Discrete random variables Discrete probability distributions Expectation The measures of a discrete random variable The binomial distribution The Poisson distribution A B C D E F cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 28A Review set 28B black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\785IB_HL-2_28.CDR Friday, 23 November 2007 2:15:03 PM PETERDELL IB_HL-2ed (786) 786 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) A DISCRETE RANDOM VARIABLES RANDOM VARIABLES In previous work we have described events mainly by using words Where possible, it is far more convenient to use numbers A random variable represents in number form the possible outcomes which could occur for some random experiment A discrete random variable X has possible values x1 , x2 , x3 , For example: ² the number of houses in your suburb which have a ‘power safety switch’ ² the number of new bicycles sold each year by a bicycle store ² the number of defective light bulbs in the purchase order of a city store A continuous random variable X has all possible values in some interval on the number line For example: ² the heights of men could all lie in the interval 50 < x < 250 cm ² the volume of water in a rainwater tank during a given month could lie in the interval < x < 100 m3 To determine the value of a discrete random variable we need to count To determine the value of a continuous random variable we need to measure For any random variable there is a corresponding probability distribution The probability that the variable X takes value x is written as P(X = x) or px We can also sometimes write a probability distribution as a function P (x) For example, when tossing two coins, the random variable X could be heads, head, or heads, i.e., X = 0, or The associated probability distribution is p0 = 14 , p1 = 12 , and p2 = 14 with graph: probability probability Qw_ Qw_ or Qr_ Qr_ number of heads number of heads Example cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A supermarket has three checkout points A, B and C A government inspector checks for accuracy of the weighing scales at each checkout If a weighing scale is accurate then yes (Y) is recorded, and if not, no (N) Suppose the random variable X is the number of accurate weighing scales at the supermarket a List the possible outcomes black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\786IB_HL-2_28.CDR Monday, 26 November 2007 12:02:41 PM PETERDELL IB_HL-2ed (787) STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) b Describe using x the events of there being: i one accurate scale ii at least one accurate scale a Possible outcomes: A N Y N N N Y Y Y B N N Y N Y N Y Y b x 1 2 C N N N Y Y Y N Y i ii 787 x=1 x = 1, or EXERCISE 28A Classify the following random variables as continuous or discrete a b c d e f g h The The The The The The The The quantity of fat in a lamb chop mark out of 50 for a Geography test weight of a seventeen year old student volume of water in a cup of coffee number of trout in a lake number of hairs on a cat length of hairs on a horse height of a sky-scraper For each of the following: i identify the random variable being considered ii give possible values for the random variable iii indicate whether the variable is continuous or discrete a To measure the rainfall over a 24-hour period in Singapore, the height of water collected in a rain gauge (up to 200 mm) is used b To investigate the stopping distance for a tyre with a new tread pattern, a braking experiment is carried out c To check the reliability of a new type of light switch, switches are repeatedly turned off and on until they fail cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A supermarket has four checkouts A, B, C and D Management checks the weighing devices at each checkout If a weighing device is accurate a yes (Y) is recorded; otherwise, no (N) is recorded The random variable being considered is the number of weighing devices which are accurate a Suppose X is the random variable What values can x have? b Tabulate the possible outcomes and the corresponding values for x c Describe, using x, the events of: i devices being accurate ii at least two devices being accurate black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\787IB_HL-2_28.CDR Friday, 23 November 2007 2:29:01 PM PETERDELL IB_HL-2ed (788) 788 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) Consider tossing three coins simultaneously The random variable under consideration is the number of heads that could result a List the possible values of x b Tabulate the possible outcomes and the corresponding values of x c Find the values of P(X = x), the probability of each x value occurring d Graph the probability distribution P(X = x) against x as a probability histogram B DISCRETE PROBABILITY DISTRIBUTIONS For each random variable there is a probability distribution The probability pi of any given outcome lies between and (inclusive), i.e., pi If there are n possible outcomes then n X pi = or p1 + p2 + p3 + ::::: + pn = i=1 The probability distribution of a discrete random variable can be given ² in table form ² in graphical form ² in functional form as a probability mass function It provides us with all possible values of the variable and the probability of the occurrence of each value Example A magazine store recorded the number of magazines purchased by its customers in one week 23% purchased one magazine, 38% purchased two, 21% purchased three, 13% purchased four, and 5% purchased five a What is the random variable? b Make a probability table for the random variable c Graph the probability distribution using a spike graph 0.4 0.3 0.2 0.1 magenta yellow probability 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 xi pi 0:00 0:23 0:38 0:21 0:13 0:05 95 100 50 75 25 b cyan c The random variable X is the number of magazines sold So, x = 0, 1, 2, 3, or 5 a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\788IB_HL-2_28.CDR Friday, January 2008 9:59:36 AM DAVID3 x IB_HL-2ed (789) STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) 789 Example Show that the following are probability distribution functions: x2 + , x = 1, 2, 3, 34 a P (x) = b P (x) = Cx3 (0:6)x (0:4)3¡x , x = 0, 1, 2, a P (1) = 34 P (2) = 34 P (3) = 10 34 17 34 P (4) = P all of which obey P (xi ) 1, and P (xi ) = 34 + 34 + 10 34 + 17 34 =1 ) P (x) is a probability distribution function For P (x) = Cx3 (0:6)x (0:4)3¡x b P (0) = C03 (0:6)0 (0:4)3 = £ £ (0:4)3 P (1) = C13 (0:6)1 (0:4)2 = £ (0:6) £ (0:4)2 = 0:288 P (2) = C23 (0:6)2 (0:4)1 = £ (0:6)2 £ (0:4) = 0:432 = 0:064 P (3) = C33 (0:6)3 (0:4)0 = £ (0:6)3 £ = 0:216 Total 1:000 P All probabilities lie between and and P (xi ) = So, P (x) is a probability distribution function Example A bag contains red and blue marbles.¡ Two marbles are randomly selected without replacement.¡ If X denotes the number of reds selected, find the probability distribution of X st selection nd selection Et_ R Ry_ ´ Et_ RR X=2 Wt_ B Ry_ ´ Wt_ RB X=1 Rt_ R Wy_ ´ Rt_ BR X=1 Qt_ B Wy_ ´ Qt_ BB X=0 R Ry_ Wy_ B x P(X = x) 30 16 30 12 30 0.6 0.4 0.2 P(X=x) x EXERCISE 28B Find k in these probability distributions: cyan magenta yellow 95 x P (x) 100 50 75 25 95 100 50 b 0:5 75 k 25 0 0:3 95 100 50 x P (x) 75 25 95 100 50 75 25 a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\789IB_HL-2_28.CDR Friday, 23 November 2007 2:46:40 PM PETERDELL k 2k 3k k IB_HL-2ed (790) 790 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) The probabilities of Jason scoring home runs in each game during his baseball career are given in the following table X is the number of home runs per game x P (x) a 0:3333 0:1088 0:0084 0:0007 0:0000 a What is the value of P (2)? b What is the value of a? Explain what this number means c What is the value of P (1) + P (2) + P (3) + P (4) + P (5)? Explain what this represents d Draw a probability distribution spike graph of P (x) against x Explain why the following are not valid probability distributions: a x P (x) 0:2 0:3 0:4 b 0:2 Sally’s number of hits each softball match has the following probability distribution: x P (x) 0:3 x P (x) 0:07 0:14 0:4 k 0:5 ¡0:2 0:46 0:08 0:02 a State clearly what the random variable represents b Find k c Find: i P(x > 2) ii P(1 x 3) A die is rolled twice a Draw a grid which shows the sample space b Suppose X denotes the sum of the results for the two rolls Find the probability distribution of X c Draw a probability distribution histogram for this situation Find k for the following probability distributions: a b P (x) = k(x + 2) for x = 1, 2, P (x) = k x+1 for x = 0, 1, 2, A discrete random variable X has probability distribution given by: ¡ ¢x ¡ ¢4¡x P (x) = k 13 where x = 0, 1, 2, 3, a Find P (x) for x = 0, 1, 2, and b Find k and hence find P(x > 2) Electrical components are produced and packed into boxes of 10 It is known that 4% of the components may be faulty The random variable X denotes the number of faulty items in the box and has a probability distribution P (x) = Cx10 (0:04)x (0:96)10¡x , x = 0, 1, 2, , 10 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a Find the probability that a randomly selected box will contain no faulty component b Find the probability that a randomly selected box will contain at least one faulty component black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\790IB_HL-2_28.CDR Friday, 30 November 2007 10:03:59 AM PETERDELL IB_HL-2ed (791) STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) 791 A bag contains blue and green tickets Two tickets are randomly selected without replacement X denotes the number of blue tickets selected a Find the probability distribution of X b Suppose instead that three tickets are randomly selected without replacement Find the probability distribution of X for X = 0, 1, 2, 10 When a pair of dice is rolled, D denotes the sum of the top faces a b c d Display the possible results in a table Find P(D = 7): Find the probability distribution of D: Find P(D > j D > 6): 11 The number of cars X that pass a shop during the period from 3:00 pm to 3:03 pm is given by P(X = x) = (0:2)x e¡0:2 x! where x = 0, 1, 2, 3, a Find i P(X = 0) ii P(X = 1) iii P(X = 2): b Find the probability that at least three cars will pass the shop in the given period 12 When a pair of dice is rolled, N denotes the difference between the numbers on the top faces a Display the possible results in a table b Construct a probability distribution table for the possible values of N c Find P(N = 3) d Find P(N > j N > 1) C EXPECTATION Consider the following problem: A die is to be rolled 120 times On how many occasions would you expect the result to be a “six”? In order to answer this question we must first consider all possible outcomes of rolling the die The possibilities are 1, 2, 3, 4, and 6, and each of these is equally likely to occur Therefore, we would expect 6 of them to be a “six” of 120 is 20, so we expect 20 of the 120 rolls of the die to yield a “six” However, this does not mean that you will get 20 sixes if you roll a die 120 times cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 If there are n outcomes in an event and the probability of each outcome in the event occurring is p, then the expectation that the event will occur is np 100 50 75 25 95 100 50 75 25 In general: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\791IB_HL-2_28.CDR Monday, 26 November 2007 9:04:27 AM PETERDELL IB_HL-2ed (792) 792 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) Example Each time a footballer kicks for goal he has a 34 chance of being successful In a particular game he has 12 kicks for goal How many goals would you expect him to kick? p = P(goal) = ) the expected number of goals is np = 12 £ =9 Example In a game of chance, a player spins a square Number spinner labelled 1, 2, 3, The player wins the Winnings $1 $2 $5 $8 amount of money shown in the table alongside, depending on which number comes up Determine: a the expected return for one spin of the spinner b whether you would recommend playing this game if it costs $5 for one game a As each number is equally likely, the probability for each number is ) expected return = 14 £ + 14 £ + 14 £ + 14 £ = $4 b As the expected return of $4 is less than the cost of $5 to play the game, you would not recommend that a person play the game EXPECTATION BY FORMULAE For examples like Example part a we can define the expectation E(X) of a random variable E(X) = to be n P pi xi i=1 EXERCISE 28C In a particular region, the probability that it will rain on any one day is 0:28 On how many days of the year would you expect it to rain? a If coins are tossed what is the chance that they all fall heads? b If the coins are tossed 200 times, on how many occasions would you expect them all to fall heads? If two dice are rolled simultaneously 180 times, on how many occasions would you expect to get a double? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 A single coin is tossed once If a head appears you win $2 and if a tail appears you lose $1 How much would you expect to win when playing this game three times? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\792IB_HL-2_28.CDR Monday, 26 November 2007 9:25:37 AM PETERDELL IB_HL-2ed (793) STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) 793 During the snow season there is a 37 probability of snow falling on any particular day If Udo skis for five weeks, on how many days could he expect to see snow falling? In a random survey of her electorate, politician A discovered the residents’ voting intentions in relation to herself and her two opponents B and C The results are indicated alongside: A 165 B 87 C 48 a Estimate the probability that a randomly chosen voter in the electorate will vote for: i A ii B iii C b If there are 7500 people in the electorate, how many of these would you expect to vote for: i A ii B iii C? A person rolls a normal six-sided die and wins the number of dollars shown on the face a How much does the person expect to win for one roll of the die? b If it costs $4 to play the game, would you advise the person to play several games? A charity fundraiser gets a licence to run the following gambling game: A die is rolled and the returns to the player are given in the ‘pay table’ alongside To play the game $4 is needed A result of getting a wins $10, so in fact you are ahead by $6 if you get a on the first roll Result 4, 1, 2, Wins $10 $4 $1 a What are your chances of playing one game and winning: i $10 ii $4 iii $1? b Your expected return from throwing a is from throwing: i a or ii a 1, or £$10 What is your expected return iii a 1, 2, 3, 4, or 6? c What is your overall expected result at the end of one game? d What is your overall expected result at the end of 100 games? A person plays a game with a pair of coins If two heads appear then $10 is won If a head and a tail appear then $3 is won If two tails appear then $5 is lost cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a How much would a person expect to win playing this game once? b If the organiser of the game is allowed to make an average of $1 per game, how much should be charged to play the game once? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\793IB_HL-2_28.CDR Monday, 26 November 2007 9:25:15 AM PETERDELL IB_HL-2ed (794) 794 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) D THE MEASURES OF A DISCRETE RANDOM VARIABLE Suppose xi are the possible values of the random variable X, and fi are the frequencies with which these values occur P fi xi We calculate the population mean as ¹ = P , fi sP fi (xi ¡ ¹)2 , the population standard deviation as ¾ = P fi and the population variance is ¾ Suppose we have 10 counters, one with a written on it, four with a written on them, three with a 3, and two with a One counter is to be randomly selected from a hat We can summarise the possible results in a table: P fi xi ¹= P fi Now = outcome xi frequency fi 1 3 probability pi 10 10 10 10 1£1 + 2£4 + 3£3 + 4£2 1+4+3+2 = £ 10 + 2£ P so ¹ = xi pi P fi (xi ¡ ¹)2 P Also, ¾ = fi 10 + 3£ 10 + 4£ 10 = 1(x1 ¡ ¹)2 4(x2 ¡ ¹)2 3(x3 ¡ ¹)2 2(x4 ¡ ¹)2 + + + 10 10 10 10 = 10 (x1 so ¾ = ¡ ¹)2 + P (xi ¡ ¹)2 pi 10 (x2 ¡ ¹)2 + 10 (x3 ¡ ¹)2 + 10 (x4 ¡ ¹)2 We can show that these formulae are also true in the general case Suppose a random variable has n possible values with frequencies and probabilities P fi xi The population mean ¹ = P fmean for fi cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 tabled valuesg f1 x1 + f2 x2 + f3 x3 + ::::: + fn xn N 95 100 50 75 25 95 100 50 75 25 = x1 , x2 , x3 , , xn f1 , f2 , f3 , , fn p1 , p2 , p3 , , pn black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\794IB_HL-2_28.CDR Monday, 26 November 2007 9:27:53 AM PETERDELL fletting P fi = Ng IB_HL-2ed (795) 795 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) µ = x1 f1 N µ ¶ + x2 f2 N µ ¶ + x3 f3 N µ ¶ + ::::: + xn fn N ¶ = x1 p1 + x2 p2 + x3 p3 + ::::: + xn pn P = xi pi P The variance ¾ = = fi (xi ¡ ¹)2 P fi f1 (x1 ¡ ¹)2 f2 (x2 ¡ ¹)2 f3 (x3 ¡ ¹)2 fn (xn ¡ ¹)2 + + + :::::: + N N N N = p1 (x1 ¡ ¹)2 + p2 (x2 ¡ ¹)2 + p3 (x3 ¡ ¹)2 + :::::: + pn (xn ¡ ¹)2 P = (xi ¡ ¹)2 pi If a discrete random variable has n possible values x1 , x2 , x3 , ., xn with probabilities p1 , p2 , p3 , ., pn of occurring, P then ² the population mean is ¹ = xi pi P ² the population variance is ¾2 = (xi ¡ ¹)2 pi pP (xi ¡ ¹)2 pi ² the population standard deviation is ¾ = The population mean of a discrete random variable is often referred to as the ‘expected value of x’ or sometimes as the ‘average value of x in the long run’ In practice, we can define: ² the mean as E(X) = ¹ = ² the variance as P xi pi and P Var(X) = ¾2 = (xi ¡ ¹)2 pi = E(X ¡ ¹)2 Example Find the mean and standard deviation of the data of Example The probability table is: Now ¹ = P xi pi 0:00 0:23 0:38 0:21 0:13 0:05 xi pi = 0(0:00) + (0:23) + 2(0:38) + 3(0:21) + 4(0:13) + 5(0:05) = 2:39 so in the long run, the average number of magazines purchased per customer is 2:39 The standard deviation p ¾ = (xi ¡ ¹)2 pi p = (1 ¡ 2:39)2 £ 0:23 + (2 ¡ 2:39)2 £ 0:38 + ::::: + (5 ¡ 2:39)2 £ 0:05 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ¼ 1:122 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\795IB_HL-2_28.CDR Wednesday, 28 November 2007 10:00:52 AM PETERDELL IB_HL-2ed (796) 796 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) An alternative formula for the population standard deviation is ¾= pP x2i pi ¡ ¹2 This formula is often easier to use than the first one For example, for a die: P ¹= xi pi = 1( 16 ) + 2( 16 ) + 3( 16 ) + 4( 16 ) + 5( 16 ) + 6( 16 ) = 3:5 P and ¾ = xi pi ¡ ¹2 = 12 ( 16 ) + 22 ( 16 ) + 32 ( 16 ) + 42 ( 16 ) + 52 ( 16 ) + 62 ( 16 ) ¡ (3:5)2 ¼ 2:92 Consequently, ¾ ¼ 1:71 These results can be checked using your calculator by generating 800 random digits from to Then find the mean and standard deviation You should get a good approximation to the theoretical values obtained above TI C THE MEDIAN AND MODE In Chapter 17 we also found two other measures for the centre of a distribution or data set: ² the median of a data set is the middle score ² the mode is the most frequently occurring score Example Find the median and mode of the data from Example The probability table is: xi pi 0:00 0:23 0:38 0:21 0:13 0:05 So, in 100 trials the expected frequencies fi are: ) middle score = xi fi 0 23 38 21 13 5 x50 + x51 2+2 = =2 2 ) median = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 The mode is the most frequently occurring score, which should be the score with highest probability ) mode = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\796IB_HL-2_28.CDR Friday, January 2008 10:00:34 AM DAVID3 IB_HL-2ed (797) STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) 797 EXERCISE 28D.1 A country exports crayfish to overseas markets The buyers are prepared to pay high prices when the crayfish arrive still alive If X is the number of deaths per dozen crayfish, the probability distribution for X is given by: xi P (xi ) 0:54 0:26 0:15 k >5 0:01 0:01 0:00 a Find k b Over a long period, what is the mean number of deaths per dozen crayfish? c Find ¾, the standard deviation for the probability distribution A random variable X has probability distribution given by x2 + x P (x) = for x = 1, 2, Calculate ¹ and ¾ for this distribution 20 A random variable X has probability distribution given by P (x) = Cx3 (0:4)x (0:6)3¡x for x = 0, 1, 2, a Find P (x) for x = 0, 1, and and display the results in table form b Find the mean and standard deviation for the distribution P P Using ¾ = (xi ¡ ¹)2 pi show that ¾ = x2i pi ¡ ¹2 A random variable X has the probability distribution shown a Copy and complete: xi P (xi ) 0.4 probability 0.3 0.2 0.1 b Find the mean ¹ and standard deviation ¾ for the distribution c Determine i P(¹ ¡ ¾ < x < ¹ + ¾) ii x P(¹ ¡ 2¾ < x < ¹ + 2¾) An insurance policy covers a $20 000 sapphire ring against theft and loss If it is stolen then the insurance company will pay the policy owner in full If it is lost then they will pay the owner $8000 From past experience, the insurance company knows that the probability of theft is 0:0025 and of being lost is 0:03 How much should the company charge to cover the ring if they want a $100 expected return? A pair of dice is rolled and the random variable M is the larger of the two numbers that are shown uppermost a In table form, obtain the probability distribution of M b Find the mean and standard deviation of the M-distribution A uniform distribution has P (x1 ) = P (x2 ) = P (x3 ) = :::::: Give two examples of a uniform distribution cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find the median and mode of the discrete random variables given in questions 1, 2, 3, 5, above black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\797IB_HL-2_28.CDR Thursday, 29 November 2007 10:04:43 AM PETERDELL IB_HL-2ed (798) 798 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) PROPERTIES OF E(X) If E(X) is the expected value of random variable X then: ² E(k) = k ² E(kX) = kE(X) for any constant k ² E(A(X) + B(X)) = E(A(X)) + E(B(X)) for functions A and B i.e., the expectation of a sum is the sum of the individual expectations for any constant k These properties enable us to deduce that: E(5) = 5, E(3X) = 3E(X) E(X + 2X + 3) = E(X ) + 2E(X) + and PROPERTY OF Var(X) Var(X) = E(X ) ¡ fE(X)g2 Var(X) = = = = = Proof: Var(X) = E(X ) ¡ ¹2 or E(X ¡ ¹)2 E(X ¡ 2¹X + ¹2 ) E(X ) ¡ 2¹E(X) + ¹2 E(X ) ¡ 2¹2 + ¹2 E(X ) ¡ ¹2 fproperties of E(X)g Example X has probability distribution Find: a the mean of X P x px 0:1 b the variance of X 0:3 c 0:4 0:2 the standard deviation of X a E(X) = b ) E(X) = 2:7 so ¹ = 2:7 P E(X ) = x2 px = 12 (0:1) + 22 (0:3) + 32 (0:4) + 42 (0:2) = 8:1 c ) Var(X) = E(X ) ¡ fE(X)g2 = 8:1 ¡ 2:72 = 0:81 p ¾ = Var(X) = 0:9 xpx = 1(0:1) + 2(0:3) + 3(0:4) + 4(0:2) EXERCISE 28D.2 cyan magenta 0:3 c 0:1 0:1 the standard deviation of X 95 50 25 95 50 75 100 yellow 0:2 100 0:3 the variance of X 25 b 95 100 50 75 25 95 100 50 75 25 Find: a the mean of X x px 75 X has probability distribution: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\798IB_HL-2_28.CDR Monday, 26 November 2007 10:00:23 AM PETERDELL IB_HL-2ed (799) STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) X has probability distribution: Find: a the value of k b x px c the mean of X X has probability distribution: Find: a E(X) e E(X ) b f E(X + 1) X has probability distribution: a Find a and b given that E(X) = 2:8 b Hence show that Var(X) = 1:26 k 0:4 0:1 the variance of X x px 0:4 c Var(X) 0:3 0:2 0:1 d ¾ E(2X + 3X ¡ 7) g Var(X + 1) 0:2 799 x px 0:2 a 0:3 b Suppose X is the number of marsupials entering a park at night It is suspected that X has a probability distribution of the form P(X = x) = a(x2 ¡ 8x) where X = 0, 1, 2, 3, a Find the constant a b Find the expected number of marsupials entering the park on a given night c Find the standard deviation of X An unbiased coin is tossed four times X is the number of heads which could appear a Find the probability distribution of X b Find: i the mean of X ii the standard deviation of X A box contains 10 almonds, two of which are bitter and the remainder are normal Brit randomly selects three almonds without replacement Let X be the random variable for the number of bitter almonds Brit selects a Find the probability distribution of X b Find: i the mean of X ii the standard deviation of X The probability distribution of a discrete random variable Y is illustrated in the table below Y ¡1 P(Y = y) 0:1 a 0:3 b a Given E(Y ) = 0:9, find the values of a and b b Calculate Var(Y ): The score X obtained by rolling a biased pentagonal die has the following probability distribution: X P(X = x) 12 a cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Find a and hence find E(X) and Var(X) black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_28\799IB_HL-2_28.CDR Thursday, 11 March 2010 4:38:47 PM PETER IB_HL-2ed (800) 800 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) E(aX + b) AND Var(aX + b) INVESTIGATION The purpose of this investigation is to discover a relationship between E(aX + b) and E(X) and also Var(aX + b) and Var(X) What to do: Consider the X-distribution: 1, 2, 3, 4, 5, each occurring with equal probability a Find E(X) and Var(X) b If Y = 2X+3, find the Y -distribution Hence find E(2X+3) and Var(2X+ 3): c Repeat b for Y = 3X ¡ 2: X +1 : e Repeat b for Y = d Repeat b for Y = ¡2X + 5: Make up your own sample distribution for a random variable X and repeat Record all your results in table form for both distributions ² ² From 3, what is the relationship between: E(X) and E(aX + b) Var(X) and Var(aX + b)? From the investigation you should have discovered that E(aX + b) = aE(X) + b and Var(aX + b) = a2 Var(X) These results will be formally proved in the exercise which follows Example 10 X is distributed with mean 8:1 and standard deviation 2:37 If Y = 4X ¡ 7, find the mean and standard deviation of the Y -distribution E(X) = 8:1 and Var(X) = 2:372 E(4X ¡ 7) = 4E(X) ¡ = 4(8:1) ¡ = 25:4 Var(4X ¡ 7) = 42 Var(X) = 42 £ 2:372 For the Y -distribution, the mean is 25:4 and the standard deviation is £ 2:37 = 9:48 EXERCISE 28D.3 X is distributed with mean and standard deviation If Y = 2X + 5, find the mean and standard deviation of the Y -distribution a Use the properties of E(X) to prove that E(aX + b) = aE(X) + b cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b The mean of an X-distribution is Find the mean of the Y -distribution where: 4X ¡ i Y = 3X + ii Y = ¡2X + iii Y = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\800IB_HL-2_28.CDR Monday, 26 November 2007 10:13:15 AM PETERDELL IB_HL-2ed (801) 801 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) X is a random variable with mean and standard deviation Find i E(Y ) ii Var(Y ) for: a Y = ¡2X + b Y = 2X + Suppose Y = 2X + where X is a random variable Find in terms of E(X) and E(X ) : a E(Y ) b E(Y ) X ¡5 c Y = c Var(Y ) Using Var(X) = E(X ) ¡ fE(X)g2 , prove that Var(aX + b) = a2 Var(X) E THE BINOMIAL DISTRIBUTION Thus far in the chapter we have considered the general properties of discrete random variables We now examine a special discrete random variable which is applied to sampling with replacement The probability distribution associated with this variable is the binomial probability distribution Note that for sampling without replacement the hypergeometric probability distribution is the model used It is not part of this core course, but questions involving sampling of this type can be tackled using combinations BINOMIAL EXPERIMENTS Consider an experiment for which there are two possible results: success if some event occurs, or failure if the event does not occur If we repeat this experiment in a number of independent trials, we call it a binomial experiment The probability of a success p must be constant for all trials If q is the probability of a failure, then q = ¡ p (since p + q = 1) The random variable X is the total number of successes in n trials THE BINOMIAL PROBABILITY DISTRIBUTION Suppose a spinner has three blue edges and one white edge Clearly, for each spin we will get either a blue or a white The chance of finishing on blue is and on white is 14 If we call a blue result a ‘success’ and a white result a ‘failure’, then we have a binomial experiment We let p be the probability of getting a blue and q be the probability of getting a white ) p= and q = 14 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Consider twirling the spinner n = times Let the random variable X be the number of ‘successes’ or blue results, so x = 0, 1, or black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\801IB_HL-2_28.CDR Monday, 26 November 2007 10:15:01 AM PETERDELL IB_HL-2ed (802) 802 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) P (0) = P(all are white) = 14 £ 14 £ 14 ¡ ¢3 = 14 nd spin st spin rd spin Er_ P (1) = P(1 blue and white) = P(BWW or WBW or WWB) ¡ ¢ ¡ ¢2 = 34 14 £ fthe branches X g B Er_ B W Qr_ B Er_ P (2) = P(2 blue and white) = P(BBW or BWB or WBB) ¡ ¢2 ¡ ¢ = 34 £3 W Qr_ W P (3) = P(3 blues) ¡ ¢3 = 34 B Qr_ WX Er_ B Qr_ WX Er_ B X B Er_ Qr_ Er_ W Qr_ W Qr_ The coloured factor is the number of ways of getting one success in three trials, which is ¡¢ combination C13 or 31 We note that ¡ ¢3 P (0) = 14 ¡ ¢2 ¡ ¢1 P (1) = 14 ¡ ¢1 ¡ ¢2 P (2) = 4 ¡ ¢3 P (3) = = C03 = C13 = C23 = C33 ¡ ¢0 ¡ ¢3 4 ¼ 0:0156 4 ¼ 0:1406 4 ¼ 0:4219 4 ¼ 0:4219 ¡ ¢1 ¡ ¢2 ¡ ¢2 ¡ ¢1 ¡ ¢3 ¡ ¢0 This suggests that: P (x) = Cx3 ¡ ¢x ¡ ¢3¡x 4 probability 0.4 0.3 0.2 0.1 number of blues where x = 0, 1, 2, In general: Consider a binomial experiment for which p is the probability of a success and q is the probability of a failure If there are n trials then the probability that there are r successes and n ¡ r failures is given by P(X = r) = Crn pr qn¡r where q = ¡ p and r = 0, 1, 2, 3, 4, , n P(X = r) is the binomial probability distribution function cyan magenta yellow 95 100 50 75 25 B(n, p) is a useful notation It indicates that the distribution is binomial and gives the values of n, the number of independent trials, and p, the probability of success in each trial 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Note: ² If X is the random variable of a binomial experiment with parameters n and p, then we write X » B(n, p): » reads ‘is distributed as’ ¡ ¢ ² Cxn = nx black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\802IB_HL-2_28.CDR Friday, 30 November 2007 10:04:39 AM PETERDELL IB_HL-2ed (803) 803 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) ² We can quickly calculate binomial probabilities using a graphics calculator I To find the probability P(X = r) that the variable takes the value r, we use the function binompdf(n, p, r) binompdf stands for ‘binomial probability distribution function’ I To find the probability P(X r) that the variable takes a value which is at most r, we use the function binomcdf(n, p, r) binomcdf stands for ‘binomial cumulative distribution function’ Example 11 72% of union members are in favour of a certain change to their conditions of employment A random sample of five members is taken Find the probability that: a three members are in favour of the change in conditions b at least three members are in favour of the changed conditions Let X denote the number of members in favour of the changes n = 5, so X = 0, 1, 2, 3, or 5, and p = 72% = 0:72 ) X » B(5, 0:72) a C35 C P(x > 3) = 1¡ P(x 2) = 1¡ binomcdf(5, 0:72, 2) ¼ 0:862 b P(x = 3) TI = (0:72) (0:28) or binompdf(5, 0:72, 3) ¼ 0:293 SUMMARY OF BINOMIAL DISTRIBUTIONS: ² ² ² The probability distribution is discrete There are two outcomes which we usually call success and failure The trials are independent, so the probability of success in a particular trial is not affected by the success or failure of previous trials In other words, the probability of success is a constant for each trial of the experiment EXERCISE 28E.1 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 For which of these probability experiments does the binomial distribution apply? Justify your answers, using a full sentence a A coin is thrown 100 times The variable is the number of heads b One hundred coins are each thrown once The variable is the number of heads c A box contains blue and red marbles I draw out marbles, replacing the marble each time The variable is the number of red marbles drawn d A box contains blue and red marbles I draw marbles without replacement The variable is the number of red marbles drawn e A large bin contains ten thousand bolts, 1% of which are faulty I draw a sample of 10 bolts from the bin The variable is the number of faulty bolts black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\803IB_HL-2_28.CDR Friday, 25 January 2008 11:10:07 AM PETERDELL IB_HL-2ed (804) 804 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) At a manufacturing plant, 35% of the employees work night-shift If employees were selected at random, find the probability that: a exactly of them work night-shift b less than of them work night-shift c at least of them work night-shift Records show that 6% of the items assembled on a production line are faulty A random sample of 12 items is selected at random (with replacement) Find the probability that: a none will be faulty b at most one will be faulty c at least two will be faulty d less than will be faulty The local bus service does not have a good reputation It is known that the am bus will run late on average two days out of every five For any week of the year taken at random, find the probability of the am bus being on time: a all days b only on Monday c on any days d on at least days An infectious flu virus is spreading through a school The probability of a randomly selected student having the flu next week is 0:3 a Calculate the probability that out of a class of 25 students, or more will have the flu next week b If more than 20% of the students are away with the flu next week, a class test will have to be cancelled What is the probability that the test will be cancelled? MEAN AND STANDARD DEVIATION OF A BINOMIAL RANDOM VARIABLE Suppose we toss a coin n = 20 times For each toss, the probability of it falling ‘heads’ is p = 12 , so for the 20 trials we expect it to fall ‘heads’ np = 10 times Now suppose we roll a die n = 30 times For each roll, the probability of it finishing as a is p = 16 , so for the 30 trials we expect to obtain a on np = occasions So, in the general case: If a binomial experiment is repeated n times and a particular variable has probability p of occurring each time, then our expectation is that the mean ¹ will be ¹ = np Finding the standard deviation is not so simple We will start with a theoretical approach, after which we verify the generalised result by simulation ONE TRIAL (n¡=¡1) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 In the case of n = where p is the probability of success and q is the probability of failure, the number of successes x could be or black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\804IB_HL-2_28.CDR Monday, 26 November 2007 10:45:20 AM PETERDELL IB_HL-2ed (805) 805 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) xi pi q Now ¹ = p P and ¾ = pi xi = p ¡ p2 = p(1 ¡ p) = pq fas q = ¡ pg p ) ¾ = pq TWO TRIALS (n¡=¡2) P (0) = C02 p0 q = q P (1) = C12 p1 q = 2pq P (2) = C22 p2 q = p2 So, the table of probabilities is ¹= P xi pi q2 and ¾ = pi xi 2pq P x2i pi ¡ ¹2 = [(0)2 q + (1)2 p] ¡ p2 = q(0) + p(1) =p In the case where n = 2, P > > = > > ; as x = 0, or 2 p2 x2i pi ¡ ¹2 = q (0) + 2pq(1) + p2 (2) = 2pq + 2p2 = 2p(q + p) = 2p fas p + q = 1g = [02 £ q + 12 £ 2pq + 22 £ p2 ] ¡ (2p)2 = 2pq + 4p2 ¡ 4p2 = 2pq p ) ¾ = 2pq p fsee the following exerciseg The case n = produces ¹ = 3p and ¾ = 3pq p The case n = produces ¹ = 4p and ¾ = 4pq: These results suggest that in general: If X is a binomial random variable with parameters n and p i.e., X » B(n, p) p then the mean of X is ¹ = np and the standard deviation of x is ¾ = npq A general proof of this statement is beyond the scope of this course However, the following investigation should help you appreciate the truth of the statement INVESTIGATION THE MEAN AND STANDARD DEVIATION OF A BINOMIAL RANDOM VARIABLE In this investigation we will examine binomial distributions randomly generated by a sorting simulation STATISTICS PACKAGE SIMULATION What to do: Obtain experimental binomial distribution results for 1000 repetitions with a n = 4, p = 0:5 b n = 5, p = 0:6 c n = 6, p = 0:75 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 For each of the distributions obtained in 1, find the mean ¹ and standard deviation ¾ from the statistics package black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\805IB_HL-2_28.CDR Monday, 26 November 2007 10:52:59 AM PETERDELL IB_HL-2ed (806) 806 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) p Use ¹ = np and ¾ = npq to see how your experimental values for ¹ and ¾ in agree with the theoretical expectation Finally, comment on the shape of a distribution for different values of p For the fixed value of n = 50, consider the distribution with p = 0:2, 0:35, 0:5, 0:68, 0:85 Example 12 5% of a batch of batteries are defective A random sample of 80 is taken with replacement Find the mean and standard deviation of the number of defectives in the sample This is a binomial sampling situation with n = 80, p = 5% = 20 : If X is the random variable for the number of defectives then X is B(80, q p 1 So, ¹ = np = 80 £ 20 = and ¾ = npq = 80 £ 20 £ 19 20 ¼ 1:95 20 ) We expect to find defective batteries with standard deviation 1:95 EXERCISE 28E.2 Suppose X is B(6, p) For each of the following cases: i find the mean and standard deviation of the X-distribution ii graph the distribution using a histogram iii comment on the shape of the distribution a when p = 0:5 b when p = 0:2 c when p = 0:8 A coin is tossed 10 times and X is the number of heads which occur Find the mean and standard deviation of the X-distribution Suppose X is B(3, p) a Find P (0), P (1), P (2) and P (3) using xi pi P (x) = Cx3 px q3¡x and display your results in a table: P b Use ¹ = pi xi to show that ¹ = 3p P p c Use ¾ = x2i pi ¡ ¹2 to show that ¾ = 3pq: Bolts produced by a machine vary in quality The probability that a given bolt is defective is 0:04 A random sample of 30 bolts is taken from the week’s production If X denotes the number of defectives in the sample, find the mean and standard deviation of the X-distribution cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 5 A city restaurant knows that 13% of reservations are not honoured, which means the group does not come Suppose the restaurant receives five reservations and X is the random variable on the number of groups that not come Find the mean and standard deviation of the X-distribution black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\806IB_HL-2_28.CDR Friday, 30 November 2007 10:06:44 AM PETERDELL IB_HL-2ed (807) 807 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) F THE POISSON DISTRIBUTION The Poisson random variable was first introduced by the French mathematician Siméon-Denis Poisson (1781 - 1840) He discovered it as a limit of the binomial distribution as the number of trials n ! PROOF Whereas the binomial distribution B(n, p) is used to determine the probability of obtaining a certain number of successes in a given number of independent trials, the Poisson distribution is used to determine the probability of obtaining a certain number of successes that can take place in a certain interval (of time or space) ² ² ² ² Examples are: the the the the number number number number of of of of incoming telephone calls to a given phone per hour misprints on a typical page of a book fish caught in a large lake per day car accidents on a given road per month The probability distribution function for the discrete Poisson random variable is: px = P(X = x) = mx e¡ m x! for x = 0, 1, 2, 3, 4, 5, where m is called the parameter of the distribution INVESTIGATION POISSON MEAN AND VARIANCE In this investigation you should discover the mean and variance of the Poisson distribution What to do: Prove by solving the differential equation that there is only one solution of the differential equation f (x) = f (x) where f(0) = 1, and that this solution is f (x) = ex x x2 x3 xn Consider f(x) = + + + + ::: + + ::: which is an infinite power series 1! 2! 3! n! a Check that f(0) = and find f (x) What you notice? b From the result of question 1, what can be deduced about f (x)? c Let x = in your discovered result in b, and calculate the sum of the first 12 terms of the power series x x 3x2 x2 x2 2x = + , = + , etc., By observing that 1! 1! 1! 2! 2! 1! show that + 2x 3x2 4x3 + + + ::::: = ex (1 + x): 1! 2! 3! If px = P(X = x) = a X mx e¡m x! b px = for x = 0, 1, 2, 3, ., show that: c E(X) = m Var(X) = m cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 x=0 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\807IB_HL-2_28.CDR Friday, January 2008 10:04:26 AM DAVID3 IB_HL-2ed (808) 808 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) From the investigation, you should have discovered that the Poisson distribution has mean m and variance m Thus ¹ = m and ¾ = m Since ¹ = ¾ = m, we can describe a Poisson distribution using the single parameter m We can therefore denote the distribution simply by Po(m) If X is a Poisson random variable then we write X » Po(m) where m = ¹ = ¾2 Conditions for a distribution to be Poisson: The average number of occurrences ¹ is constant for each interval This means it should be equally likely that the event occurs in one specific interval as in any other The probability of more than one occurrence in a given interval is very small The typical number of occurrences in a given interval should be much less than is theoretically possible, say about 10% or less The number of occurrences in disjoint intervals are independent of each other Consider the following example: When Sandra proof read 80 pages of a text book she observed the following distribution for X, the number of errors per page: X frequency 11 16 18 15 9 10 P fi xi 257 The mean for this data is ¹ = P = = 3:2125 fi 80 Using the Poisson model with m = 3:2125 the expected frequencies are: xi fi 3:22 10:3 16:6 17:8 14:3 9:18 4:92 2:26 0:906 0:323 10 0:104 The expected frequencies are close to the observed frequencies, which suggests the Poisson model is a good model for representing this distribution Further evidence is that X ¾2 = xi pi2 ¡ ¹2 16 2 = 02 ( 80 ) + 12 ( 11 80 ) + ( 80 ) + :::::: + 10 ( 80 ) ¡ 3:2125 ¼ 3:367 which is fairly close to m = 3:2125 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Note that there is a formal statistic test for establishing whether a Poisson distribution is appropriate This test is covered in the Statistics Option Topic black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\808IB_HL-2_28.CDR Friday, January 2008 10:05:06 AM DAVID3 IB_HL-2ed (809) 809 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) EXERCISE 28F Sven’s Florist Shop receives the X X distribution of phone calls shown between frequency 12 18 12 9:00 am and 9:15 am on Fridays a Find the mean of the X-distribution TI b Compare the actual data with that generated by a Poisson model C a A Poisson distribution has a standard deviation of 2:67 i What is its mean? ii What is its probability generating function? b For the distribution in a, find: ii P(X 3) i P(X = 2) P(X > 5) iii P(X > 3) j X > 1) iv One gram of radioactive substance is positioned so that each emission of an alpha-particle will flash on a screen The emissions over 500 periods of 10 second duration are given in the following table: number per period frequency 91 156 132 75 33 a Find the mean of the distribution b Fit a Poisson model to the data and compare the actual data to that from the model p c Find the standard deviation of the distribution How close is it to m found in a? Top Cars rent cars to tourists They have four cars which are hired out on a daily basis The number of requests each day is distributed according to the Poisson model with a mean of Determine the probability that: a none of its cars are rented c some requests will have to be refused d all are hired out given that at least two are b at least of its cars are rented Consider a random variable X » Po(m) a Find m given that P(X = 1) + P(X = 2) = P(X = 3) P(X > 3) b If m = 2:7, find i ii P(X j X > 2) Wind tunnel experiments on a new aerofoil produced data showing there was a 98% chance of the aerofoil not disintegrating at maximum airspeed In a sample of 100 aerofoils, use the Poisson distribution to determine the probability that: a only one b only two c at most aerofoils will disintegrate cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Road safety figures for a large city show that any driver has a 0:02% chance of being killed each time he or she drives a car a Use the Poisson approximation to the binomial distribution to find the probability that a driver can use a car 10 times a week for a year and survive b If this data does not change, how many years can you drive in Los Angeles and still have a better than even chance of surviving? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\809IB_HL-2_28.CDR Wednesday, 28 November 2007 10:17:57 AM PETERDELL IB_HL-2ed (810) 810 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) A supplier of clothing materials looks for flaws before selling it to customers The number of flaws follows a Poisson distribution with a mean of 1:7 flaws per metre a Find the probability that there are exactly flaws in metre of material b Determine the probability that there is at least one flaw in metres of material c Find the modal value of this Poisson distribution The random variable Y is Poisson with mean m and satisfies P(Y = 3) = P(Y = 1) + 2P(Y = 2) a Find the value of m correct to decimal places b Using the value of m found in a above, find P(1 < Y < 5) c Calculate P(2 Y 6 j Y > 4) 10 The random variable U has a Poisson distribution with mean x Let y be the probability that U takes one of the values 0, or a Write down an expression for y as a function of x b Sketch the graph of y for x c Use calculus to show that as the mean increases, P(U 2) decreases REVIEW SET 28A f (x) = a a , x = 0, 1, 2, is a probability distribution function x2 + b Find a Hence, find P(x > 1) A random variable X has probability distribution function P (x) = Cx4 for x = 0, 1, 2, 3, a b Find P (x) for x = 0, 1, 2, 3, ¡ ¢x ¡ ¢4¡x 2 Find ¹ and ¾ for this distribution A manufacturer finds that 18% of the items produced from its assembly lines are defective During a floor inspection, the manufacturer randomly selects ten items Find the probability that the manufacturer finds: a one defective b two defective c at least two defective items A random sample of 120 toothbrushes is made with replacement from a very large batch where 4% are known to be defective Find: a the mean b the standard deviation of the number of defectives in the sample cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 5 At a social club function, a dice game is played where on a single roll of a six-sided die the following payouts are made: $2 for an odd number, $3 for a 2, $6 for a 4, and $9 for a a What is the expected return for a single roll of the die? b If the club charges $5 for each roll, how much money would the club expect to make if 75 people played the game once each? black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\810IB_HL-2_28.CDR Monday, 26 November 2007 11:46:48 AM PETERDELL IB_HL-2ed (811) 811 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) A biased tetrahedral die has the numbers 6, 12 and 24 clearly indicated on of its faces The fourth number is unknown The table below indicates the probability of each of these numbers occurring if the die is rolled once If the die is rolled once: Number 12 x 24 a Find the probability of obtaining the Probability y number 24 b Find the fourth number if the average result when rolling the die once is 14 c Find the median and modal score for this die The probability distribution for the discrete random variable X is given by the ¡ ¢x for x = 0, 1, 2, 3, probability distribution function: P(X = x) = a 56 Find the value of a A hot water unit relies on 20 solar components for its power and will operate provided at least one of its 20 components is working The probability that an individual solar component will fail in a year is 0:85, and the components’ failure or otherwise are independent of each other a Find the probability that this hot water unit will fail within one year b Find the smallest number of solar components required to ensure that the hot water service is operating at the end of one year with a probability of at least 0:98 For a given binomial random variable X with independent trials, we know that P(X = 3) = 0:226 89 a Find the smallest possible value of p, the probability of obtaining a success in one trial b Hence calculate the probability of getting at most successes in 10 trials 10 During peak period, customers arrive at random at a fish and chip shop at the rate of 20 customers every 15 minutes a Find the probability that during peak period, 15 customers will arrive in the next quarter of an hour b If the probability that more than 10 customers will arrive at the fish and chip shop in a 10 minute period during peak period is greater than 80%, the manager will employ an extra shop assistant Will the manager hire an extra shop assistant? 11 A Poisson random variable X is such that P(X = 1) = P(2 X 4) cyan magenta yellow 95 i 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a Find the mean and standard deviation of: b Find P(X > 2) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\811IB_HL-2_28.CDR Thursday, 29 November 2007 10:07:43 AM PETERDELL X ii Y = X +1 IB_HL-2ed (812) 812 STATISTICAL DISTRIBUTIONS OF DISCRETE RANDOM VARIABLES (Chapter 28) REVIEW SET 28B A discrete random variable X has probability distribution function ¡ ¢x ¡ ¢3¡x where x = 0, 1, 2, and k is a constant P (x) = k 34 b Find P(x > 1) a Find k An X-ray has probability of 0:96 of showing a fracture in the arm If four different X-rays are taken of a particular fracture, find the probability that: a all four show the fracture b the fracture does not show up c at least three X-rays show the fracture d only one X-ray shows the fracture A random variable X has probability x distribution function given by: P (x) 0:10 0:30 0:45 0:10 a Find k b Find the mean ¹ and standard deviation ¾ for the distribution of X k From data over the last fifteen years it is known that the chance of a netballer with a knee injury needing major knee surgery in any one season is 0:0132 In 2007 there were 487 cases of knee injuries Find the mean and standard deviation of the number of major knee surgeries An author was known to make one mistake in 200 pages of her work A second author was known to make mistakes in 200 pages of her work During one writing period, the first author produced 20 pages and the second author produced 40 pages Assuming independent work of the authors, use the Poisson distribution to find the probability that the authors made or more mistakes between them A die is biased such that the probability of obtaining a is 25 The die is rolled 1200 times Let X be the number of sixes obtained Find: a the mean of X b the standard deviation of X A discrete random variable has its probability distribution given by ¢ ¡ P(X = x) = k x + x¡1 where x = 1, 2, 3, Find: a the exact value of k b E(X) and Var(X) c the median and mode of X A Poisson random variable X satisfies the rule 5Var(X) = [E(X)]2 ¡ 12 a Find the mean of X b Find P(X < 3) The random variable X has a binomial distribution for which P(X > 2) ¼ 0:070 198 for 10 independent trials Find P(X < 2) 10 The random variable Y has a Poisson distribution with P(Y > 3) ¼ 0:033 768 97 Find P(Y < 3): cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 11 The random variable X has mean ¹ and standard deviation ¾ Prove that the random variable Y = aX+b has mean a¹+b and standard deviation j a j ¾ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_28\812IB_HL-2_28.CDR Friday, 30 November 2007 10:08:32 AM PETERDELL IB_HL-2ed (813) 29 Chapter Statistical distributions of continuous random variables Contents: Continuous probability density functions Normal distributions The standard normal distribution (Z-distribution) Applications of the normal distribution A B C D cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Review set 29A Review set 29B black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_29\813IB_HL-2_29.CDR Tuesday, 27 November 2007 10:43:31 AM PETERDELL IB_HL-2ed (814) 814 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chapter 29) A CONTINUOUS PROBABILITY DENSITY FUNCTIONS In the previous chapter we looked at discrete random variables and examined some probability distributions where the random variable X could take the non-negative integer values x = 0, 1, 2, 3, 4, For a continuous random variable X, x can take any real value Consequently, a function is used to specify the probability distribution, and that function is called the probability density function Probabilities are found by finding areas under the probability density function ¦(x) x a b A continuous probability density function (pdf) is a function f (x) such that Z b f (x) dx = f(x) > on a given interval [ a, b ] and a For a continuous probability density function: ² The mode is the value of x at the maximum value of f(x) on [ a, b ] Z m The median m is the solution for m of the equation f(x) dx = 12 a Z b The mean ¹ or E(X) is defined as ¹ = x f(x) dx: ² ² a The variance Var(X) = E(X ) ¡ fE(X)g2 = ² Z b x2 f (x) dx ¡ ¹2 a f(x) = on [ 0, ] elsewhere Area = y (2,¡1) Z magenta yellow 95 50 75 0 50 75 x 25 95 100 50 75 25 95 100 50 75 25 2 or 95 y¡=¡Qw_ x the mean of the distribution 25 a cyan is a probability density function Check that the above statement is true Find i the mode ii the median iii Find Var(X) and ¾ 100 a b c 2x £2£1 =1 X x dx 100 ( Example black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_29\814IB_HL-2_29.CDR Tuesday, 27 November 2007 10:59:31 AM PETERDELL = ¤ 2 4x £1 =1 X IB_HL-2ed (815) STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chapter 29) b i ii f (x) is a maximum when x = ) the mode = Z iii The median is the solution of Z m 1 x dx = ) ¤ m 4x £1 = E(X ) = Z Z = Z = = x2 f(x) dx = 2x dx ¤ 6x £1 = 13 ) Var(X) = ¡ (1 13 )2 = 2x x f (x) dx 0 2 ¹= m ) = 12 ) m2 = p ) m = fas m [ 0, ] g c 815 dx and ¾ = ¤ 8x £1 = p Var(X) p =2 EXERCISE 29A ( f(x) = ax(x ¡ 4), x elsewhere a Find a: c Find: ( f(x) = b i f(x) = Sketch the graph of y = f (x): ii the mean b i Find: ke¡x , x elsewhere b ii the mean iv the variance the variance is a probability density function Find k to decimal places ( kx2 (x ¡ 6), x f(x) = elsewhere k the median is a probability density function b a a iii the mode ¡0:2x(x ¡ b), x b elsewhere a Find b: ( is a continuous probability density function c the mode Find the median is a probability density function Find: the median d the mean e the variance cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 5 The probability density function of the random variable Y is given by ( ¡ 12y, y k f(y) = elsewhere black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_29\815IB_HL-2_29.CDR Tuesday, 27 November 2007 11:06:23 AM PETERDELL IB_HL-2ed (816) 816 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chapter 29) a Under what conditions can Y be a continuous random variable? b Without using calculus, find k Z 12 (5 ¡ 12y) dy = Explain why k 6= 12 despite this result c Notice that d Find the mean and median value of Y The probability density function for the random variable X is f(x) = k, a x b a Find k in terms of a and b b Calculate the mean, median and mode of X: c Calculate Var(X) and the standard deviation of X The continuous random variable X has the probability density function f (x) = 2e¡2x , x > a Calculate the median of X b Calculate the mode of X A continuous random variable X has probability density function f (x) = cos 3x for x a a Find a b Find the mean of X d Find the standard deviation of X c Find the 20th percentile of X The continuous random variable X has the probability density function f (x) = ax4 , x k ¢ ¡ , find a and k Given that P X 23 = 243 > > > < 10 The time taken in hours to perform a particular task has the probability density function: f(x) = 125 18 x > 10x > > : 0 x < 0:6 0:6 x 0:9 otherwise a Sketch the graph of this function b Show that f(x) is a well defined probability density function for the random variable X, the time taken to perform the task c Find the mean, median and mode of X d Find the variance and standard deviation of X e Find P(0:3 < X < 0:7) and interpret your answer cyan magenta yellow 95 100 50 75 25 = E(X ) ¡ fE(X)g2 95 100 50 75 25 95 E(X ¡ ¹)2 P (x ¡ ¹)2 px P x px ¡ ¹2 E(X ) ¡ fE(X)g2 100 50 75 ² ¾ = Var(X) = E(X ¡ ¹)2 R = (x ¡ ¹)2 f (x) dx R = x2 f (x) dx ¡ ¹2 25 ² ¾ = Var(X) = = = = Continuous random variable R ² ¹ = E(X) = xf(x) dx 95 Discrete random variable P ² ¹ = E(X) = xpx 100 50 75 25 SUMMARY black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_29\816IB_HL-2_29.CDR Tuesday, 27 November 2007 11:07:43 AM PETERDELL IB_HL-2ed (817) STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chapter 29) B 817 NORMAL DISTRIBUTIONS The normal distribution is the most important distribution for a continuous random variable Many naturally occurring phenomena have a distribution that is normal, or approximately normal Some examples are: ² physical attributes of a population such as height, weight, and arm length ² crop yields ² scores for tests taken by a large population If X is normally distributed then its probability density function is given by f (x) = x¡ ¹ ¡ 1( ¾ ) e p ¾ 2¼ for ¡1 < x < where ¹ is the mean and ¾ is the variance of the distribution Each member of the family is specified by the parameters ¹ and ¾ , and we can write X » N(¹, ¾2 ) This probability density function for the normal distribution represents a family of bell-shaped normal curves These curves are all symmetrical about the vertical line x = ¹ A typical normal curve is illustrated alongside Notice that f(¹) = p : ¾ 2¼ & m, s p* f (x) m x HOW THE NORMAL DISTRIBUTION ARISES Consider the oranges picked from an orange tree They not all have the same weight The variation may be due to several factors, including: ² genetics ² different times when the flowers were fertilised ² different amounts of sunlight reaching the leaves and fruit ² different weather conditions such as the prevailing winds The result is that most of the fruit will have weights close to the mean, while there are far fewer oranges that are much heavier or much lighter This results in the bell-shaped distribution cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Once a normal model has been established, we can use it to make predictions about a distribution and to answer other relevant questions black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_29\817IB_HL-2_29.CDR Thursday, 29 November 2007 10:13:46 AM PETERDELL IB_HL-2ed (818) 818 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chapter 29) CHARACTERISTICS OF THE NORMAL PROBABILITY DENSITY FUNCTION ² The curve is symmetrical about the vertical line x = ¹: ² As j x j ! the normal curve approaches its asymptote, the x-axis ² f(x) > for all x Z ² The area under the curve is one unit2 , and so f (x) dx = ¡1 ² More scores are distributed closer to the mean than further away This results in the typical bell shape A TYPICAL NORMAL DISTRIBUTION VIDEO CLIP A large sample of cockle shells was collected and the maximum distance across each shell was measured Click on the video clip icon to see how a histogram of the data is built up Now click on the demo icon to observe the effect of changing the class interval lengths for normally distributed data DEMO THE GEOMETRICAL SIGNIFICANCE OF ¹ AND ¾ ¡1 ³ For f (x) = p e ¾ 2¼ x¡¹ ¾ ´2 ¡1 we can obtain f (x) = p ¾ 2¼ µ x¡¹ ¾ ¶ ¡ 12 ³ e x¡¹ ¾ ´2 ) f (x) = only when x = ¹ This corresponds to the point on the graph when f(x) is a maximum ³ ´2 · ¸ x¡¹ ¡ (x ¡ ¹)2 ¡1 ¾ Differentiating again, we can obtain f 00 (x) = p e ¡ ¾ ¾3 ¾ 2¼ ) f 00 (x) = when (x ¡ ¹)2 = ¾3 ¾ ) (x ¡ ¹)2 = ¾ ) x ¡ ¹ = §¾ ) x = ¹§¾ So, the points of inflection are at x = ¹ + ¾ and x = ¹ ¡ ¾: point of inflection s s m-s m point of inflection m+s x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 For a given normal curve, the standard deviation is uniquely determined as the horizontal distance from the vertical line x = ¹ to a point of inflection black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_29\818IB_HL-2_29.CDR Thursday, 29 November 2007 4:34:09 PM PETERDELL IB_HL-2ed (819) STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chapter 29) 819 For a normal distribution with mean ¹ and standard deviation ¾, the proportional breakdown of where the random variable could lie is given below Normal distribution curve 34.13% 2.15% 0.13% 34.13% 2.15% 13.59% m-3s 13.59% m-2s ² ² ² Notice that: 0.13% m-s m m+s m+2s m+3s ¼ 68:26% of values lie between ¹ ¡ ¾ and ¹ + ¾ ¼ 95:44% of values lie between ¹ ¡ 2¾ and ¹ + 2¾ ¼ 99:74% of values lie between ¹ ¡ 3¾ and ¹ + 3¾ INVESTIGATION STANDARD DEVIATION SIGNIFICANCE The purpose of this investigation is to check the proportions of normal distribution data which lie within ¾, 2¾ and 3¾ of the mean DEMO What to do: Click on the icon to start the demonstration in Microsoft® Excel Take a random sample of size n = 1000 from a normal distribution Find: a x and s b x ¡ s, x + s c x ¡ 2s, x + 2s d x ¡ 3s, x + 3s Count all values between: b x ¡ 2s and x + 2s c x ¡ 3s and x + 3s a x ¡ s and x + s Determine the percentage of data values in these intervals Do these confirm the theoretical percentages given above? Repeat the procedure several times Example The chest measurements of 18 year old male footballers is normally distributed with a mean of 95 cm and a standard deviation of cm a Find the percentage of footballers with chest measurements between: ii 103 cm and 111 cm i 87 cm and 103 cm b Find the probability that the chest measurement of a randomly chosen footballer is between 87 cm and 111 cm a i We need the percentage between ¹ ¡ ¾ and ¹ + ¾ This is ¼ 68:3% 34.13% ii We need the percentage between ¹ + ¾ and ¹ + 2¾ 13.59% s magenta s s yellow 95 100 50 75 25 95 87 95 103 111 m-s m m+s m+2s 100 50 75 25 95 100 50 75 25 95 100 50 75 25 This is ¼ 13:6%: cyan 34.13% black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_29\819IB_HL-2_29.CDR Tuesday, 27 November 2007 12:08:18 PM PETERDELL IB_HL-2ed (820) 820 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chapter 29) b This is between ¹ ¡ ¾ and ¹ + 2¾ The percentage is 68:26% + 13:59% ¼ 81:9%: So, the probability is ¼ 0:819 m-s m m+2s EXERCISE 29B.1 Draw each of the following normal distributions accurately on one set of axes Distribution A B C mean (mL) 25 30 21 standard deviation (mL) 10 Explain why it is likely that the distributions of the following variables will be normal: a the volume of soft drink in cans b the diameter of bolts immediately after manufacture It is known that when a specific type of radish is grown without fertiliser, the weights of the radishes produced are normally distributed with a mean of 40 g and a standard deviation of 10 g When the same type of radish is grown in the same way except for the inclusion of fertiliser, the weights of the radishes produced are also normally distributed, but with a mean of 140 g and a standard deviation of 40 g Determine the proportion of radishes grown: a without fertiliser with weights less than 50 grams b with fertiliser with weights less than 60 grams c i with and ii without fertiliser with weights between 20 and 60 g inclusive d i with and ii without fertiliser with weights greater than or equal to 60 g The height of male students is normally distributed with a mean of 170 cm and a standard deviation of cm a Find the percentage of male students whose height is: ii between 170 cm and 186 cm i between 162 cm and 170 cm b Find the probability that a randomly chosen student from this group has a height: ii less than 162 cm i between 178 cm and 186 cm iii less than 154 cm iv greater than 162 cm A bottle filling machine fills an average of 20 000 bottles a day with a standard deviation of 2000 Assuming that production is normally distributed and the year comprises 260 working days, calculate the approximate number of working days that: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a under 18 000 bottles are filled b over 16 000 bottles are filled c between 18 000 and 24 000 bottles (inclusive) are filled black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_29\820IB_HL-2_29.CDR Tuesday, 27 November 2007 12:14:08 PM PETERDELL IB_HL-2ed (821) 821 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chapter 29) PROBABILITIES BY GRAPHICS CALCULATOR We can use a graphics calculator to quickly find probabilities for a normal distribution Suppose X » N(10, 22 ), so X is normally distributed with mean 10 and standard deviation How we find P(8 X 11) ? 10 11 How we find a if P(X > a) = 0:479? Click on the icon for your graphics calculator to obtain instructions for answering these questions TI C EXERCISE 29B.2 Use a calculator to find these probabilities: X is a random variable that is distributed normally with mean 70 and standard deviation Find: P(70 X 74) a P(68 X 72) b c P(X 65) X is a random variable that is distributed normally with mean 60 and standard deviation Find: P(60 X 65) P(X 68) a d P(62 X 67) P(X 61) b e c f P(X > 64) P(57:5 X 62:5) Given that X » N(23, 52 ), find a if: a c b P(X < a) = 0:378 P(23 ¡ a < X < 23 + a) = 0:427 C P(X > a) = 0:592 THE STANDARD NORMAL DISTRIBUTION (Z-DISTRIBUTION) Every normal X-distribution can be transformed into the standard normal distribution or X ¡¹ Z-distribution using the transformation Z = ¾ In the following investigation we determine the mean and standard deviation of this Z-distribution INVESTIGATION MEAN AND STANDARD DEVIATION OF Z= x ¡¹ ¾ Suppose a random variable X is normally distributed with mean ¹ and standard deviation ¾ For each value of X we can calculate a Z-value using the algebraic cyan magenta yellow 95 100 50 75 25 X ¡¹ ¾ 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 transformation Z = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_29\821IB_HL-2_29.CDR Tuesday, 27 November 2007 12:19:06 PM PETERDELL IB_HL-2ed (822) 822 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chapter 29) What to do: Consider the X-values: 1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, a Draw a graph of the distribution to check that it is approximately normal b Find the mean ¹ and standard deviation ¾ for the distribution of X-values X ¡¹ to convert each X-value into a Z-value c Use the transformation Z = ¾ d Find the mean and standard deviation for the distribution of Z-values Click on the icon to load a large sample drawn from a normal population By clicking appropriately we can repeat the four steps of question Write a brief report of your findings DEMO You should have discovered that for a Z-distribution the mean is and the standard deviation is This is true for all Z-distributions generated by transformation of a normal distribution, and this is why we call it the standard normal distribution x¡¹ For a normal X-distribution we know f (x) = p e¡ ( ¾ ) the probability density function is: ¾ 2¼ Substituting x = z, ¹ = and ¾ = 1, we find the probability density function for the Z-distribution is f (z) = p1 2¼ e¡ z , ¡1< z < Notice that the normal distribution function f (x) has two parameters ¹ and ¾, whereas the standard normal distribution function f (z) has no parameters This means that a unique table of values can be constructed for f(z), and we can use this table to compare normal distributions Before graphics calculators and computer packages the standard normal distribution was used exclusively for normal probability calculations such as those which follow CALCULATING PROBABILITIES USING THE Z-DISTRIBUTION Since z is continuous, y P(Z a) = P(Z < a) Z a 1 p e¡ z dz and P(Z a) = 2¼ ¡1 this area is P(Z a) y=ƒ(z) z a The table of curve areas on page 824 enables us to find P(Z a) USING A GRAPHICS CALCULATOR TO FIND PROBABILITIES cyan yellow 95 use normalcdf(¡E99, a) use normalcdf(a, E99) use normalcdf(a, b) 100 50 75 25 95 100 50 75 25 95 50 75 100 magenta P(Z a) or P(Z < a) P(Z > a) or P(Z > a) P(a Z b) or P(a < Z < b) To find To find To find 25 95 100 50 75 25 For a TI-83: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_29\822IB_HL-2_29.CDR Tuesday, 27 November 2007 12:26:16 PM PETERDELL IB_HL-2ed (823) 823 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chapter 29) Example If Z is a standard normal variable, find: a P(Z 1:5) b P(Z > 0:84) c P(¡0:41 Z 0:67) P(Z 1:5) ¼ 0:933 a b P(Z > 0:84) = ¡ P(Z 0:84) ¼ ¡ 0:79954 ¼ 0:200 P(Z 1:5) = normalcdf(¡E99, 1:5) ¼ 0:933 or P(Z > 0:84) = normalcdf(0:84, E99) ¼ 0:200 or P(¡0:41 Z 0:67) = normalcdf(¡0:41, 0:67) ¼ 0:408 1.5 0.84 P(¡0:41 Z 0:67) = P(Z 0:67) ¡ P(Z ¡0:41) ¼ 0:7486 ¡ 0:3409 ¼ 0:408 c or -0.41 0.67 WHAT DO Z-VALUES TELL US? ² ² If z1 = 1:84 then z1 is 1:84 standard deviations to the right of the mean If z2 = ¡0:273 then z2 is 0:273 standard deviations to the left of the mean So, Z-values are useful when comparing results from two or more different distributions Example Kelly scored 73% in History where the class mean was 68% and the standard deviation was 10:2% In Mathematics she scored 66%, the class mean was 62%, and the standard deviation was 6:8% In which subject did Kelly perform better compared with the rest of her class? Assume the scores for both subjects were normally distributed Kelly’s Z-score for History = Kelly’s Z-score for Maths = 73 ¡ 68 ¼ 0:490 10:2 66 ¡ 62 ¼ 0:588 6:8 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 So, Kelly’s result in Maths was 0:588 standard deviations above the mean, whereas her result in History was 0:490 standard deviations above the mean ) Kelly’s result in Maths was better, even though it was a lower score black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_29\823IB_HL-2_29.CDR Tuesday, 27 November 2007 12:30:39 PM PETERDELL IB_HL-2ed (824) cyan magenta yellow 100 95 75 50 25 100 95 75 50 25 100 95 75 50 25 100 95 75 50 25 z -3.4 -3.3 -3.2 -3.1 -3.0 -2.9 -2.8 -2.7 -2.6 -2.5 -2.4 -2.3 -2.2 -2.1 -2.0 -1.9 -1.8 -1.7 -1.6 -1.5 -1.4 -1.3 -1.2 -1.1 -1.0 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0.0 .00 0.0003 0.0005 0.0007 0.0010 0.0013 0.0019 0.0026 0.0035 0.0047 0.0062 0.0082 0.0107 0.0139 0.0179 0.0228 0.0287 0.0359 0.0446 0.0548 0.0668 0.0808 0.0968 0.1151 0.1357 0.1587 0.1841 0.2119 0.2420 0.2743 0.3085 0.3446 0.3821 0.4207 0.4602 0.5000 .01 0.0003 0.0005 0.0007 0.0009 0.0013 0.0018 0.0025 0.0034 0.0045 0.0060 0.0080 0.0104 0.0136 0.0174 0.0222 0.0281 0.0351 0.0436 0.0537 0.0655 0.0793 0.0951 0.1131 0.1335 0.1562 0.1814 0.2090 0.2389 0.2709 0.3050 0.3409 0.3783 0.4168 0.4562 0.4960 .02 0.0003 0.0005 0.0006 0.0009 0.0013 0.0018 0.0024 0.0033 0.0044 0.0059 0.0078 0.0102 0.0132 0.0170 0.0217 0.0274 0.0344 0.0427 0.0526 0.0643 0.0778 0.0934 0.1112 0.1314 0.1539 0.1788 0.2061 0.2358 0.2676 0.3015 0.3372 0.3745 0.4129 0.4522 0.4920 Each table value is the area to the left of the specified Z-value the second decimal digit of z 03 04 05 06 0.0003 0.0003 0.0003 0.0003 0.0004 0.0004 0.0004 0.0004 0.0006 0.0006 0.0006 0.0006 0.0009 0.0008 0.0008 0.0008 0.0012 0.0012 0.0011 0.0011 0.0017 0.0016 0.0016 0.0015 0.0023 0.0023 0.0022 0.0021 0.0032 0.0031 0.0030 0.0029 0.0043 0.0041 0.0040 0.0039 0.0057 0.0055 0.0054 0.0052 0.0075 0.0073 0.0071 0.0069 0.0099 0.0096 0.0094 0.0091 0.0129 0.0125 0.0122 0.0119 0.0166 0.0162 0.0158 0.0154 0.0212 0.0207 0.0202 0.0197 0.0268 0.0262 0.0256 0.0250 0.0336 0.0329 0.0322 0.0314 0.0418 0.0409 0.0401 0.0392 0.0516 0.0505 0.0495 0.0485 0.0630 0.0618 0.0606 0.0594 0.0764 0.0749 0.0735 0.0721 0.0918 0.0901 0.0885 0.0869 0.1093 0.1075 0.1056 0.1038 0.1292 0.1271 0.1251 0.1230 0.1515 0.1492 0.1469 0.1446 0.1762 0.1736 0.1711 0.1685 0.2033 0.2005 0.1977 0.1949 0.2327 0.2296 0.2266 0.2236 0.2643 0.2611 0.2578 0.2546 0.2981 0.2946 0.2912 0.2877 0.3336 0.3300 0.3264 0.3228 0.3707 0.3669 0.3632 0.3594 0.4090 0.4052 0.4013 0.3974 0.4483 0.4443 0.4404 0.4364 0.4880 0.4840 0.4801 0.4761 area to the left of a ƒ(z) .07 0.0003 0.0004 0.0005 0.0008 0.0011 0.0015 0.0021 0.0028 0.0038 0.0051 0.0068 0.0089 0.0116 0.0150 0.0192 0.0244 0.0307 0.0384 0.0475 0.0582 0.0708 0.0853 0.1020 0.1210 0.1423 0.1660 0.1922 0.2206 0.2514 0.2843 0.3192 0.3557 0.3936 0.4325 0.4721 a STANDARD NORMAL CURVE AREAS (Z 0) .08 0.0003 0.0004 0.0005 0.0007 0.0010 0.0014 0.0020 0.0027 0.0037 0.0049 0.0066 0.0087 0.0113 0.0146 0.0188 0.0239 0.0301 0.0375 0.0465 0.0571 0.0694 0.0838 0.1003 0.1190 0.1401 0.1635 0.1894 0.2177 0.2483 0.2810 0.3156 0.3520 0.3897 0.4286 0.4681 09 0.0002 0.0003 0.0005 0.0007 0.0010 0.0014 0.0019 0.0026 0.0036 0.0048 0.0064 0.0084 0.0110 0.0143 0.0183 0.0233 0.0294 0.0367 0.0455 0.0559 0.0681 0.0823 0.0985 0.1170 0.1379 0.1611 0.1867 0.2148 0.2451 0.2776 0.3121 0.3483 0.3859 0.4247 0.4641 z z 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 3.1 3.2 3.3 3.4 .00 0.5000 0.5398 0.5793 0.6179 0.6554 0.6915 0.7257 0.7580 0.7881 0.8159 0.8413 0.8643 0.8849 0.9032 0.9192 0.9332 0.9452 0.9554 0.9641 0.9713 0.9772 0.9821 0.9861 0.9893 0.9918 0.9938 0.9953 0.9965 0.9974 0.9981 0.9987 0.9990 0.9993 0.9995 0.9997 01 0.5040 0.5438 0.5832 0.6217 0.6591 0.6950 0.7291 0.7611 0.7910 0.8186 0.8438 0.8665 0.8869 0.9049 0.9207 0.9345 0.9463 0.9564 0.9649 0.9719 0.9778 0.9826 0.9864 0.9896 0.9920 0.9940 0.9955 0.9966 0.9975 0.9982 0.9987 0.9991 0.9993 0.9995 0.9997 02 0.5080 0.5478 0.5871 0.6255 0.6628 0.6985 0.7324 0.7642 0.7939 0.8212 0.8461 0.8686 0.8888 0.9066 0.9222 0.9357 0.9474 0.9573 0.9656 0.9726 0.9783 0.9830 0.9868 0.9898 0.9922 0.9941 0.9956 0.9967 0.9976 0.9982 0.9987 0.9991 0.9994 0.9995 0.9997 Each table value is the area to the left of the specified Z-value the second decimal digit of z 03 04 05 06 0.5120 0.5160 0.5199 0.5239 0.5517 0.5557 0.5596 0.5636 0.5910 0.5948 0.5987 0.6026 0.6293 0.6331 0.6368 0.6406 0.6664 0.6700 0.6736 0.6772 0.7019 0.7054 0.7088 0.7123 0.7357 0.7389 0.7422 0.7454 0.7673 0.7704 0.7734 0.7764 0.7967 0.7995 0.8023 0.8051 0.8238 0.8264 0.8289 0.8315 0.8485 0.8508 0.8531 0.8554 0.8708 0.8729 0.8749 0.8770 0.8907 0.8925 0.8944 0.8962 0.9082 0.9099 0.9115 0.9131 0.9236 0.9251 0.9265 0.9279 0.9370 0.9382 0.9394 0.9406 0.9484 0.9495 0.9505 0.9515 0.9582 0.9591 0.9599 0.9608 0.9664 0.9671 0.9678 0.9686 0.9732 0.9738 0.9744 0.9750 0.9788 0.9793 0.9798 0.9803 0.9834 0.9838 0.9842 0.9846 0.9871 0.9875 0.9878 0.9881 0.9901 0.9904 0.9906 0.9909 0.9925 0.9927 0.9929 0.9931 0.9943 0.9945 0.9946 0.9948 0.9957 0.9959 0.9960 0.9961 0.9968 0.9969 0.9970 0.9971 0.9977 0.9977 0.9978 0.9979 0.9983 0.9984 0.9984 0.9985 0.9988 0.9988 0.9989 0.9989 0.9991 0.9992 0.9992 0.9992 0.9994 0.9994 0.9994 0.9994 0.9996 0.9996 0.9996 0.9996 0.9997 0.9997 0.9997 0.9997 area to the left of a STANDARD NORMAL CURVE AREAS (Z > 0) .07 0.5279 0.5675 0.6064 0.6443 0.6808 0.7157 0.7486 0.7794 0.8078 0.8340 0.8577 0.8790 0.8980 0.9147 0.9292 0.9418 0.9525 0.9616 0.9693 0.9756 0.9808 0.9850 0.9884 0.9911 0.9932 0.9949 0.9962 0.9972 0.9979 0.9985 0.9989 0.9992 0.9995 0.9996 0.9997 a ƒ(z) .08 0.5319 0.5714 0.6103 0.6480 0.6844 0.7190 0.7517 0.7823 0.8106 0.8365 0.8599 0.8810 0.8997 0.9162 0.9306 0.9429 0.9535 0.9625 0.9699 0.9761 0.9812 0.9854 0.9887 0.9913 0.9934 0.9951 0.9963 0.9973 0.9980 0.9986 0.9990 0.9993 0.9995 0.9996 0.9997 .09 0.5359 0.5753 0.6141 0.6517 0.6879 0.7224 0.7549 0.7852 0.8133 0.8389 0.8621 0.8830 0.9015 0.9177 0.9319 0.9441 0.9545 0.9633 0.9706 0.9767 0.9817 0.9857 0.9890 0.9916 0.9936 0.9952 0.9964 0.9974 0.9981 0.9986 0.9990 0.9993 0.9995 0.9997 0.9998 z 824 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chapter 29) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_29\824IB_HL-2_29.CDR Tuesday, 27 November 2007 12:33:26 PM PETERDELL IB_HL-2ed (825) STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chapter 29) 825 EXERCISE 29C.1 For a random variable X the mean is ¹ and standard deviation is ¾ µ ¶ µ ¶ X ¡¹ X ¡¹ Using properties of E(X) and Var(X) find: a E b Var ¾ ¾ If Z has standard normal distribution, find using tables and a sketch: a P(Z 1:2) b P(Z > 0:86) c P(Z ¡0:52) d P(Z > ¡1:62) e P(¡0:86 Z 0:32) If Z has standard normal distribution, find using technology: a P(Z > 0:837) b P(Z 0:0614) c P(¡0:3862 Z 0:2506) d e P(Z > ¡0:876) P(¡2:367 Z ¡0:6503) If Z has standard normal distribution, find: a P(¡0:5 < Z < 0:5) b P(¡1:960 < Z < 1:960) Find a if Z has standard normal distribution and: a P(Z a) = 0:95 b P(Z > a) = 0:90 The table alongside shows Sergio’s results in his mid-year examinations, along with the class means and standard deviations a Find Sergio’s Z-value for each subject b Arrange Sergios performances in each subject in order from ‘best’ to ‘worst’ Physics Chemistry Mathematics German Biology Sergio 83% 77% 84% 91% 72% ¹ 78% 72% 74% 86% 62% ¾ 10:8% 11:6% 10:1% 9:6% 12:2% Pedro is studying Algebra and Geometry He sits for the mid-year exams in each subject He is told that his Algebra mark is 56%, whereas the class mean and standard deviation are 50:2% and 15:8% respectively In Geometry he is told that the class mean and standard deviation are 58:7% and 18:7% respectively What percentage does Pedro need to have scored in Geometry to have an equivalent result to his Algebra mark? STANDARDISING ANY NORMAL DISTRIBUTION To find probabilities for a normally distributed random variable X: Step 2: X ¡¹ : ¾ Sketch a standard normal curve and shade the required region Step 3: Use the standard normal tables or a graphics calculator to find the probability cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 Convert X-values to Z using Z = 75 25 95 100 50 75 25 Step 1: black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_29\825IB_HL-2_29.CDR Tuesday, 27 November 2007 12:39:21 PM PETERDELL IB_HL-2ed (826) 826 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chapter 29) Example Given that X is a normal variable with mean 62 and standard deviation 7, find: a P(X 69) b P(58:5 X 71:8) a P(X 69) ¶ µ X ¡ 62 69 ¡ 62 =P 7 = P(Z 1) ¼ 0:841 This means that there is an 84:1% chance that a randomly selected X-value is 69 or less b P(58:5 X 71:8) ¶ µ X ¡ 62 71:8 ¡ 62 58:5 ¡ 62 6 =P 7 = P(¡0:5 Z 1:4) ¼ 0:9192 ¡ 0:3085 -0.5 1.4 ¼ 0:611 This means that there is a 61:1% chance that a randomly selected X-value is between 58:5 and 71:8 inclusive These probabilities can also be found using a graphics calculator without actually converting to standard normal Z-scores Click on the icon for your calculator for instructions, and hence check the answers above TI C EXERCISE 29C.2 A random variable X is normally distributed with mean 70 and standard deviation By converting to the standard variable Z and then using the tabled probability values for Z, find: a P(X > 74) b P(X 68) c P(60:6 X 68:4) A random variable X is normally distributed with mean 58:3 and standard deviation 8:96 By converting to the standard variable Z and then using your graphics calculator, find: a P(X > 61:8) b P(X 54:2) c P(50:67 X 68:92) The length L of a nail is normally distributed with mean 50:2 mm and standard deviation 0:93 mm Find, by first converting to Z-values: cyan magenta yellow 95 100 50 75 c 25 95 100 50 P(L 51) 75 25 95 b 100 50 75 25 P(L > 50) 95 100 50 75 25 a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_29\826IB_HL-2_29.CDR Tuesday, 27 November 2007 12:48:38 PM PETERDELL P(49 L 50:5) IB_HL-2ed (827) 827 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chapter 29) FINDING QUANTILES OR k VALUES Consider a population of crabs where the length of a shell, X mm, is normally distributed with mean 70 mm and standard deviation 10 mm A biologist wants to protect the population by allowing only the largest 5% of crabs to be harvested He therefore asks the question: “95% of the crabs have lengths less than what?” To answer this question we need to find the value of k such that P(X k) = 0:95 Example Find k for which P(X k) = 0:95 given that X » N(70, 102 ) and X is measured in mm P(X k) = 0:95 ¶ µ X ¡ 70 k ¡ 70 = 0:95 ) P 10 10 µ ¶ k ¡ 70 ) P Z6 = 0:95 10 or Using technology: If P(X k) = 0:95 then k = invNorm(0:95, 70, 10) ) k ¼ 86:5 Searching amongst the standard normal tables or using your graphics calculator: k ¡ 70 ¼ 1:645 10 ) k ¼ 86:5 So, approximately 95% of the values are expected to be 86:5 mm or less EXERCISE 29C.3 Z has a standard normal distribution Find k using tabled values if: a P(Z k) = 0:81 b P(Z k) = 0:58 c P(Z k) = 0:17 Z has a standard normal distribution Find k using technology if: a P(Z k) = 0:384 b P(Z k) = 0:878 c P(Z k) = 0:1384 a Show that P(¡k Z k) = P(Z k) ¡ 1: b Suppose Z has a standard normal distribution Find k if: i P(¡k Z k) = 0:238 ii P(¡k Z k) = 0:7004 a Find k if P(X k) = 0:9 and X » N(56, 182 ) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b Find k if P(X > k) = 0:8 and X » N(38:7, 8:82 ) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_29\827IB_HL-2_29.CDR Monday, December 2007 12:12:36 PM PETERDELL IB_HL-2ed (828) 828 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chapter 29) D APPLICATIONS OF THE NORMAL DISTRIBUTION Example In 1972 the heights of rugby players were found to be normally distributed with mean 179 cm and standard deviation cm Find the probability that a randomly selected player in 1972 was: b between 170 cm and 190 cm a at least 175 cm tall If X is the height of a player then X is normally distributed with ¹ = 179, ¾ = 7: a P(X > 175) = normalcdf (175, E99, 179, 7) ¼ 0:716 fgraphics calculatorg b P(170 < X < 190) = normalcdf (170, 190, 179, 7) ¼ 0:843 fgraphics calculatorg Example A university professor determines that 80% of this year’s History candidates should pass the final examination The examination results are expected to be normally distributed with mean 62 and standard deviation 13 Find the expected lowest score necessary to pass the examination Let the random variable X denote the final examination result, so X » N(62, 132 ) We need to find k such that P(X > k) = 0:8 ) P(X k) = 0:2 ) k = invNorm(0:2, 62, 13) ) k ¼ 51:1 So, the minimum pass mark is more than 51% If the final marks are given as integer percentages then the pass mark will be 52% Example Find the mean and standard deviation of a normally distributed random variable X if P(X > 50) = 0:2 and P(X 20) = 0:3 P(X 20) = 0:3 P(X > 50) = 0:2 ) P(X 50) = 0:8 20 ¡ ¹ ) = 0:3 ¾ 50 ¡ ¹ ) = 0:8 ) P(Z ¾ 20 ¡ ¹ ) = invNorm(0:3) 50 ¡ ¹ ¾ = 0:8416 ) ¼ ¡0:5244 ¾ ) 20 ¡ ¹ ¼ ¡0:5244¾ (1) ) 50 ¡ ¹ ¼ 0:8416¾ (2) ) P(Z cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Solving (1) and (2) simultaneously we get ¹ ¼ 31:5, ¾ ¼ 22:0 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_29\828IB_HL-2_29.CDR Tuesday, 27 November 2007 2:08:49 PM PETERDELL IB_HL-2ed (829) STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chapter 29) 829 Note: ² In Example we must convert to Z-scores to answer the question We always need to convert to Z-scores if we are trying to find an unknown mean ¹ or standard deviation ¾ ² Z-scores are also useful when trying to compare two scores from different normal distributions ² If possible, it is good practice to verify your results using an alternative method EXERCISE 29D A machine produces metal bolts The lengths of these bolts have a normal distribution with mean 19:8 cm and standard deviation 0:3 cm If a bolt is selected at random from the machine, find the probability that it will have a length between 19:7 cm and 20 cm Max’s customers put money for charity in a collection box on the front counter of his shop Assume that the average weekly collection is approximately normally distributed with a mean of $40 and a standard deviation of $6 What proportion of weeks would he b at least $50:00? expect to collect: a between $30:00 and $50:00 The students of Class X sat a Physics test The average score was 46 with a standard deviation of 25 The teacher decided to award an A to the top 7% of the students in the class Assuming that the scores were normally distributed, find the lowest score that a student needed to obtain in order to achieve an A Eels are washed onto a beach after a storm Their lengths have a normal distribution with a mean of 41 cm and a variance of 11 cm2 a If an eel is randomly selected, find the probability that it is at least 50 cm long b Find the proportion of eels measuring between 40 cm and 50 cm long c How many eels from a sample of 200 would you expect to measure at least 45 cm in length? Find the mean and standard deviation of a normally distributed random variable X if P(X > 35) = 0:32 and P(X 8) = 0:26 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a A random variable X is normally distributed Find the mean and the standard deviation of X, given that P(X > 80) = 0:1 and P(X 30) = 0:15 b In the Mathematics examination at the end of the year, it was found that 10% of the students scored at least 80 marks, and no more than 15% scored less than 30 marks Assuming the marks are normally distributed, what proportion of students scored more than 50 marks? 95 100 50 75 25 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_29\829IB_HL-2_29.CDR Tuesday, December 2007 2:34:28 PM PETERDELL IB_HL-2ed (830) 830 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chapter 29) Circular metal tokens are used to operate a washing machine in a laundromat The diameters of the tokens are normally distributed, and only tokens with diameters between 1:94 and 2:06 cm will operate the machine a Find the mean and standard deviation of the distribution given that 2% of the tokens are too small, and 3% are too large b Find the probability that at most one token out of a randomly selected sample of 20 will not operate the machine REVIEW SET 29A The arm lengths of 18 year old females are normally distributed with mean 64 cm and standard deviation cm a Find the percentage of 18 year old females whose arm lengths are: i between 60 cm and 72 cm ii greater than 60 cm b Find the probability that a randomly chosen 18 year old female has an arm length in the range 56 cm to 68 cm The length of steel rods produced by a machine is normally distributed with a standard deviation of mm It is found that 2% of all rods are less than 25 mm long Find the mean length of rods produced by the machine ( ax(x ¡ 3), x f (x) = is a continuous probability distribution function elsewhere a Find a c Find: i the mean d Find P(1 x 2) b Sketch the graph of y = f(x): ii the mode iii the median iv the variance A factory has a machine designed to fill bottles of drink with a volume of 375 mL It is found that the average amount of drink in each bottle is 376 mL, and that 2:3% of the drink bottles have a volume smaller than 375 mL Assuming that the amount of drink in each bottle is distributed normally, find the standard deviation The continuous random variable Z is distributed such that Z » N(0, 1) Find the value of k if P (j Z j > k) = 0:376 X is a continuous random variable where X » N(¹, 22 ) Find P(j X ¡ ¹ j < 0:524) The marks of 2376 candidates in an IB examination are normally distributed with a mean of 49 marks and variance 225 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a If the pass mark is 45, estimate the number of candidates who passed the examination b If 7% of the candidates scored scored a ‘7’, find the minimum mark required to obtained a ‘7’ c Find the interquartile range of the distribution of marks obtained black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_29\830IB_HL-2_29.CDR Tuesday, December 2007 2:35:48 PM PETERDELL IB_HL-2ed (831) STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chapter 29) 831 The lengths of metal rods produced in a manufacturing process are distributed normally with mean ¹ cm and standard deviation cm It is known that 5:63% of the rods have length greater than 89:52 cm Find the mean, median, and modal length of these metal rods ½ ¡1x The continuous random variable X e for x k: f (x) = has probability density function otherwise a Find the exact value of k, writing your answer in the form ln a where a Z : b What is the probability that X lies between 14 and 78 ? c Find the exact values of the mean and variance of X 10 The random variable X is distributed normally with mean 50 and P(X < 90) ¼ 0:975: Find the shaded area in the given diagram which illustrates the probability density function for the random variable X 80 x REVIEW SET 29B The contents of a certain brand of soft drink can is normally distributed with mean 377 mL and standard deviation 4:2 mL a Find the percentage of cans with contents: i less than 368:6 mL ii between 372:8 mL and 389:6 mL b Find the probability that a randomly selected can has contents between 364:4 mL and 381:2 mL The life of a Xenon battery is known to be normally distributed with a mean of 33:2 weeks and a standard deviation of 2:8 weeks a Find the probability that a randomly selected battery will last at least 35 weeks b Find the maximum number of weeks for which the manufacturer can expect that not more than 8% of batteries will fail The edible part of a batch of Coffin Bay oysters is normally distributed with mean 38:6 grams and standard deviation 6:3 grams If the random variable X is the mass of a Coffin Bay oyster: a find a if P(38:6 ¡ a X 38:6 + a) = 0:6826 b find b if P(X > b) = 0:8413 cyan magenta yellow 95 100 50 75 25 95 100 50 Find P(0:6 < X < 1:2) 75 d 25 Find the median of X c Find the mode of X 95 b 100 Show that a = 34 50 a 75 25 95 100 50 75 25 A random variable X has probability density function f (x) = ax2 (2 ¡ x) for < x < black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_29\831IB_HL-2_29.CDR Tuesday, 27 November 2007 2:28:49 PM PETERDELL IB_HL-2ed (832) 832 STATISTICAL DISTRIBUTIONS OF CONTINUOUS RANDOM VARIABLES (Chapter 29) The random variable T represents the lifetime in years of a component of a solar cell Its probability density function is F (t) = 0:4e¡0:4t , t > a Find the probability that this component of the solar cell fails within year Give your answer correct to decimal places b Each solar cell has of these components which operate independently of each other The cell will work provided at least of the components continue to work Find the probability that a solar cell will still operate after year < It is claimed that the continuous random variable X has probability density function f (x) = + x2 for x : otherwise a Show that this is not possible b Use your working from a to find an exact value of k for which F (x) = kf (x) would be a well-defined probability density function c Hence, find the exact values of the mean and variance of X Hint: x2 =1¡ + x2 + x2 A continuous random variable X has probability density function ( ax(4 ¡ x2 ), x f (x) = elsewhere a b c d Find the exact value of a Find the exact value of the mode of X Calculate the median value of X Find the exact value of the mean of X The heights of 18 year old boys are normally distributed with a mean of 187 cm Fifteen percent of all these boys have heights greater than 193 cm Find the probability that two 18 year old boys chosen at random will have heights greater than 185 cm The random variable X is normally distributed with P(X 30) = 0:0832 and P(X > 90) = 0:101 a Find the mean ¹ and standard deviation ¾ for X, correct to decimal places b Hence find P(j X ¡ ¹ j > 7) : 10 The continuous random variable X has a interval [ 0, k ] by > < f(x) = > : probability density function defined on the x 5x2 for x 2 x k: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a Find the value of k b Find the exact value of the median of X c Find the mean and variance of X black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_29\832IB_HL-2_29.CDR Tuesday, 27 November 2007 2:39:25 PM PETERDELL IB_HL-2ed (833) 30 Chapter cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Miscellaneous questions black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_30\833IB_HL-2_30.CDR Thursday, 24 January 2008 10:39:36 AM PETERDELL IB_HL-2ed (834) 834 MISCELLANEOUS QUESTIONS (Chapter 30) EXERCISE 30 p a Simplify (¡1 + i 2)3 : p b Write + i in the form a3 cis µ stating the exact values of a and µ p c Find the exact solutions of ³ z ´= + i ³ ´ p ¡1 d Hence, show that arctan 52 + 2¼ = arccos p a b c d Find the cube roots of ¡2 ¡ 2i Display the cube roots of ¡2 ¡ 2i on an Argand diagram If the cube roots are ®1 , ®2 and ®3 , show that ®1 + ®2 + ®3 = 0: Using an algebraic argument, prove that if ¯ is any complex number then the sum of the zeros of z n = ¯ is a Evaluate (1 ¡ i)2 and simplify (1 ¡ i)4n b Hence, evaluate (1 ¡ i)16 c Use your answers above to find two solutions of z 16 = 256 Give clear reasons for your answers p p p 2+i ¡1 + i Let z = and w = 4 a Write z and w in the form r(cos µ + i sin µ) where µ ¼ ¡ ¢ 11¼ b Show that zw = 14 cos 11¼ 12 + i sin 12 c Evaluate zw in the form a + ib and hence find the exact values of 11¼ cos 11¼ 12 and sin 12 The sum of the first n terms of a series is given by Sn = n3 + 2n ¡ Find un , the nth term of the series The tangent to the curve y = f (x) at the point A(x, y) meets the x-axis at the point B(x ¡ 12 , 0) The curve meets the y-axis at the point C(0, 1e ) Find the equation of the curve The diagram shows a sector POR of a circle of radius unit and centre O The b = µ, and the line segments angle POR [PQ], [P1 Q1 ], [P2 Q2 ], [P3 Q3 ], are all perpendicular to [OR] Calculate, in terms of µ, the sum to infinity of the lengths PQ + P1 Q1 + P2 Q2 + P3 Q3 + P P1 P2 q O Q2 Q1 Q R R Use the method of integration by parts to find x arctan x dx: Check that your answer is correct using differentiation cyan magenta yellow 95 100 50 75 b 32x+1 = 5(3x ) + 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Solve the following equations: ¡ ¢ a log2 x2 ¡ 2x + = + log2 (x ¡ 1) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_30\834IB_HL-2_30.CDR Monday, 21 January 2008 9:44:57 AM PETERDELL IB_HL-2ed (835) MISCELLANEOUS QUESTIONS (Chapter 30) 10 Solve exactly for x: 835 3x ¡ > jx + 1j 11 Find the exact values of x for which sin2 x + sin x ¡ = and ¡2¼ x 2¼ 12 If f : x 7! ln x and g : x 7! + x find: a f ¡1 (2) £ g ¡1 (2) b (f ± g)¡1 (2) 13 Given an angle µ where sin µ = ¡ 25 and ¡ ¼2 < µ < 0, find the exact values of: a cos µ b tan µ c sin 2µ d sec 2µ: p cos x csc x + = for x 2¼ 14 Solve 15 The number of snails in a garden plot follows a Poisson distribution with standard deviation d Find d if the chance of finding exactly snails is half that of finding exactly snails in this plot 16 Find the coordinates of the point on the line L that is nearest to the origin if the equation of L is r = 2i ¡ 3j + k + ¸(¡i + j ¡ k), ¸ R 17 The function f(x) satisfies the following criteria: f (x) > and f 00 (x) < for all x, f(2) = 1, and f (2) = a Find the equation of the tangent to f (x) at x = and sketch it on a graph b Hence, sketch a graph of f (x) on the same axes c Explain why f(x) has exactly one zero d Estimate an interval in which the zero of f(x) lies 18 If n Z , n > ¡2, prove by induction that 2n3 ¡ 3n2 + n + 31 > n X 19 Prove by induction that r3r = [(2n ¡ 1)3n + 1] for all n Z + r=1 20 Prove by induction that for all n Z + , 1 1 n + + + :::::: + = a(a + 1) (a + 1)(a + 2) (a + 2)(a + 3) (a + n ¡ 1)(a + n) a(a + n) 21 Prove by induction that xn ¡ y n has a factor of x ¡ y for all n Z + ¢ ¡ 22 Prove that 52n+1 + 23n+1 is divisible by 17 for all n Z + ¡ n ¢ ³ n ´ ³ n+1 ´ 23 Assuming Pascal’s rule r + r+1 = r+1 , ¡ ¢ ¡ ¢ ¡ ¢ n a prove that (1 + x) = + n1 x + n2 x2 + :::::: + nn xn for all n Z + b b Establish the binomial expansion for (a + b)n by letting x = in a a 1 24 Prove that + + :::::: + = cot x ¡ cot(2n x) for all n Z + sin 2x sin 4x sin(2n x) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 25 Show that if y = mx + c is a tangent to y = 4x then c = and the coordinates m ¶ µ , : of the point of contact are m2 m black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_30\835IB_HL-2_30.CDR Monday, 21 January 2008 9:45:45 AM PETERDELL IB_HL-2ed (836) 836 MISCELLANEOUS QUESTIONS (Chapter 30) x2 y2 + = 1: a2 b a Show that the shaded region has area given by Z b ap a ¡ x2 dx a b 26 The illustrated ellipse has equation -a a -b b Find the area of the ellipse in terms of a and b c An ellipsoid is obtained by rotating the ellipse about the x-axis through 360o Prove that the volume of the ellipsoid is given by V = 43 ¼ab2 27 Consider the following quadratic function where , bi R : f (x) = (a1 x ¡ b1 )2 + (a2 x ¡ b2 )2 + (a3 x ¡ b3 )2 + :::::: + (an x ¡ bn )2 Use quadratic theory to prove the ‘Cauchy-Schwartz inequality’: ¶µ n ¶ µn ¶2 µn P P P bi > bi i=1 i=1 i=1 28 Show that the equation of the tangent to the ellipse with equation ³x ´ ³y ´ x2 y2 1 + = at the point P(x , y ) is x + y = 1 a2 b2 a2 b2 29 Prove that in any triangle with angles A, B and C: a sin 2A + sin 2B + sin 2C = sin A sin B sin C b tan A + tan B + tan C = tan A tan B tan C 30 a A circle has radius r and the acute angled triangle ABC has vertices on the circle abc Show that the area of the triangle is given by 4r b In triangle ABC it is known that sin A = cos B + cos C Show that the triangle is right angled 31 [AB] is a thin metal rod of fixed length and P is its centre A is free to move on the x-axis and B is free to move on the y-axis What path or locus is traced out by point P as the rod moves to all possible places? y B P x A p p p a Show that 14 ¡ cannot be written in the form a + b where a, b Z p p p p b Can 14 ¡ be written in the form a m + b n where a, b, m, n Z ? ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ 33 (1 + x)n = n0 + n1 x + n2 x2 + n3 x3 + :::::: + nn xn for all n Z + Prove that: ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ a n1 + n2 + n3 + :::::: + n nn = n2n¡1 ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ b n0 + n1 + n2 + :::::: + (n + 1) nn = (n + 2)2n¡1 cyan magenta yellow 95 100 50 75 25 95 ¡ n ¢ 2n+1 ¡ = n+1 n n+1 100 50 + :::::: + 75 ¡n¢ 25 95 + ¡n¢ 100 50 0 + 75 ¡n¢ 95 100 50 75 25 c 25 32 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_30\836IB_HL-2_30.CDR Monday, 21 January 2008 9:47:30 AM PETERDELL IB_HL-2ed (837) 837 MISCELLANEOUS QUESTIONS (Chapter 30) 34 x+5 Ax + B C = + , find A, B and C + 5)(1 ¡ x) x +5 x¡1 Z x+5 b Hence, find the exact value of dx: + 5)(1 ¡ x) (x a If (x2 35 Consider the series 1 1 + + + :::::: + : 1£3 2£4 3£5 n(n + 2) n(n + 2) A B + , find the values of A and B n n+2 1 b Use a to show that the sum of the series is 34 ¡ ¡ 2n + 2n + X d Check b using mathematical induction c Find r(r + 2) r=1 a By writing Z 36 Find: a in the form Z x p dx ¡ x2 Z 1+x dx + x2 b c p dx ¡ x2 37 A flagpole is erected at A and its top is B At C, due west of A, the angle of elevation to B is ® At D, due south of A, the angle of elevation to B is ¯ Point E is due³p south of C and due ´ west of D Show that at E, the angle of elevation to cot2 ® + cot2 ¯ B is arccot 38 tan µ ¡ tan3 µ ¡ tan2 µ + tan4 µ b Hence, find the roots of the equation x4 + 4x3 ¡ 6x2 ¡ 4x + = a Use complex number methods to show that tan 4µ = 39 Find the sum of the series: a + a cos µ + a2 cos 2µ + a3 cos 3µ + :::::: + an cos nµ b a sin µ + a2 sin 2µ + a3 sin 3µ + :::::: + an sin nµ for n Z + 40 Suppose ex can be written as the infinite series ex = a0 +a1 x+a2 x2 +::::::+an xn +:::::: a Show that a0 = 1, a1 = 1, a2 = 2! , a3 = 3! , b Hence conjecture an infinite geometric series representation for ex c Check your answer to b using the substitution x = 41 P Q = + , find P and Q ¡x a¡x a+x ¯ ¯ Z ¯ a+x¯ 1 ¯ ¯ + c b Use a to show that dx = ln a2 ¡ x2 2a ¯ a ¡ x ¯ a By considering a2 c Check b using differentiation cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 a Assuming that ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ cos S¡D and sin S+sin D = sin S+D cos S¡D , cos S+cos D = cos S+D 2 2 ³ ´ ³ ´ prove that cis µ + cis Á = cos µ¡Á cis µ+Á 2 25 95 100 50 75 25 42 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_30\837IB_HL-2_30.CDR Thursday, 24 January 2008 10:42:19 AM PETERDELL IB_HL-2ed (838) 838 MISCELLANEOUS QUESTIONS (Chapter 30) b From a, what is the modulus and argument of cis µ + cis Á? c Show that your answers in b are correct using a geometrical argument µ ¶ µ ¶5 k¼ z+1 = are z = ¡i cot d Prove that the solutions of , k = 1, 2, 3, z¡1 is real, prove that either jzj = or z is real z b If jz + wj = jz ¡ wj prove that arg z and arg w differ by c If z = r cis µ, write z , and iz ¤ in a similar form z a If z + 43 ¼ 44 x2 + ax + bc = and x2 + bx + ca = where a 6= 0, b 6= 0, c 6= 0, have a single common root Prove that the other roots satisfy x2 + cx + ab = 1 45 If x = a + b , show that x3 = 3(ab) x + (a + b) Hence, find all real solutions of the equation x3 = 6x + 46 Solve simultaneously: and logy x ¡ logx y = 83 x = 16y 47 Find all values of m for which the quartic equation x4 ¡ (3m + 2)x2 + m2 = has real roots in arithmetic progression 48 ® and ¯ are two of the roots of x3 + ax2 + bx + c = Prove that ®¯ is a root of x3 ¡ bx2 + acx ¡ c2 = 49 x and y satisfy the equations x2 + 3xy + = and y2 + x ¡ = Solve these equations simultaneously for x given that x is real 1 1 a Find the value p +p p +p p + + p p of the sum: 1+ 2+ 3+ 99 + 100 b Can you make any generalizations from a? 50 51 The three numbers x, y and z are such that x > y > z > Show that if 1 x, y and z are in arithmetic progression, then x ¡ z, y and x ¡ y + z are the lengths of the sides of a right angled triangle 52 Each summer, 10% of the trees on a certain plantation die out, and each winter, workmen plant 100 new trees At the end of the winter in 1980 there were 1200 trees in the plantation a How many living trees were there at the end of winter in 1970? b What will happen to the number of trees in the plantation during the 21st century providing the conditions remain unchanged? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 a I wish to borrow $20000 for 10 years at 12% p.a where the interest is compounded quarterly I intend to pay off the loan in quarterly instalments How much I need to pay back each quarter? b Find a formula for calculating the repayments R if the total amount borrowed is $P , for n years, at r% p.a., and there are to be m equal payments at equal intervals each year 100 50 75 25 53 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_30\838IB_HL-2_30.CDR Monday, 21 January 2008 9:48:56 AM PETERDELL IB_HL-2ed (839) MISCELLANEOUS QUESTIONS (Chapter 30) 839 54 A rectangle is divided by m lines parallel to one pair of opposite sides and n lines parallel to the other pair How many rectangles are there in the figure obtained? 55 a Schools A and B each preselect 11 members for a team to be sent interstate However, circumstances allow only a combined team of 11 to be sent away In how many ways can a team of 11 be selected and a captain be chosen if the captain must come from A? ³ ´ ¡ ¢2 ¡ ¢2 ¡ ¢2 ¡ ¢2 b Use a to show that: n1 + n2 + n3 + + n nn = n 2n¡1 n¡1 56 Two different numbers are randomly chosen out of the set f1; 2; 3; 4; 5; , ng, where n is a multiple of four Determine the probability that one of the numbers is four times larger than the other 57 A hundred seeds are planted in ten rows of ten seeds per row Assuming that each seed independently germinates with probability 12 , find the probability that the row with the maximum number of germinations contains at least seedlings 58 Consider a randomly chosen n child family, where n > Let A be the event that the family has at most one boy, and B be the event that every child in the family is of the same sex For what values of n are the events A and B independent? 59 Two marksmen, A and B, fire simultaneously at a target If A is twice as likely to hit the target as B, and if the probability that the target does get hit is 12 , find the probability of A hitting the target 60 A quadratic equation ax2 + bx + c = is copied by a typist However, the numbers standing for a, b and c are blurred and she can only see that they are integers of one digit What is the probability that the equation she types has real roots? 61 Two people agree to meet each other at the corner of two city streets between pm and pm, but neither will wait for the other for more than 30 minutes If each person is equally likely to arrive at any time during the one hour period, determine the probability that they will in fact meet p 1+ o 62 Use the figure alongside to show that cos 36 = a 10° A D 60° 63 Find ®: a° F a C E 30° a 2a 40° B 64 For A, B, C not necessarily the angles of a triangle, what can be deduced about A + B + C if tan A + tan B + tan C = tan A tan B tan C ? cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 65 Without using a calculator, show how to find arctan( 17 ) + arctan( 13 ): black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_30\839IB_HL-2_30.CDR Monday, 21 January 2008 9:56:23 AM PETERDELL IB_HL-2ed (840) 840 MISCELLANEOUS QUESTIONS (Chapter 30) 66 A mountain is perfectly conical in shape The base is a circle of radius km, and the steepest slopes leading up to the top are km long From the southernmost point A on the base, a path leads up on the side of the mountain to B, a point on the northern slope which is 1:5 km up the slope from C A and C are diametrically opposite If the path leading from A to B is the shortest possible distance from A to B along the mountainside, find the length of this path 67 H cm wall A C km An A cm by B cm rectangular refrigerator leans at an angle of µ to the floor against a wall a Find H in terms of A, B and µ: b Explain how the figure can be used to prove p that A sin µ + B cos µ A2 + B , with A equality when tan µ = B X B cm northern slope km B A cm q floor 68 Over 2000 years ago, Heron or Hero discovered a formula for finding the area of a p triangle with sides a, b and c It is A = s(s ¡ a)(s ¡ b)(s ¡ c) where 2s = a+b+c Prove that this formula is correct dy = k csc(2x) for some 69 a Given that y = ln(tan x), x ] 0, ¼2 [ , show that dx constant k y b The graph of y = csc(2x) is illustrated on the interval ] 0, ¼2 [ Find the area of the shaded region x Give your answer in the form a ln b where a Q and b Z + p 70 P is a point and line l, with direction vector v, passes through points A and Q ¡! j AP £ v j a Prove that PQ = jvj x y z = ¡1 +¸ ¡1 y 71 x= p P l b Hence, find the shortest distance from (2, ¡1, 3) à ! à ! à ! to the line p Q v A Find a given that the shaded region has area 16 units2 y¡=¡xX, ¡x¡>¡0 x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a a+2 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_30\840IB_HL-2_30.CDR Monday, 21 January 2008 4:44:31 PM PETERDELL IB_HL-2ed (841) MISCELLANEOUS QUESTIONS (Chapter 30) 72 If dy = x csc y dx 841 and y(2) = 0, find y as a function of x R d sec4 x dx: (tan3 x), find dx 73 By considering 74 What can be deduced if A \ B and A [ B are independent events? 75 For a continuous function defined on the interval [ a, b ], the length of the curve can be Z bq + [f (x)]2 dx Find the length of: found using L = a b y = sin x on the interval [ 0, ¼ ] a y = x on the interval [ 0, ] ii (A \ B) [ (A0 \ B) a Simplify: i (A [ B) \ A0 b Verify that (A \ B) [ C = (A [ C) \ (B [ C) c Prove that if A and B are independent events then so are: ii A and B i A0 and B p p 77 Write (3 ¡ i 2)4 in the form x + y 2i where x, y Z 76 78 Solve the equation sin µ cos µ = for the interval µ [ ¡¼, ¼ ] 79 z and w are two complex numbers such that 2z + w = i and z ¡ 3w = ¡ 10i Find z + w in the form a + bi, where a and b Z 80 Solve the differential equation (x + 1)2 dy = 2xy, x > ¡1 given that y(1) = dx 81 f is defined by x 7! ln (x(x ¡ 2)) a State the domain of f b Find f (x) c Find the equation of the tangent to f at the point where x = 82 Hat contains three green and four blue tickets Hat contains four green and three blue tickets One ticket is randomly selected from each hat a What is the probability that the tickets are the same colour? b Given that the tickets are different colours, what is the probability that the green ticket came from Hat 2? 83 If A3 = A, what can be said about: a jAj b A¡1 ? 84 If P (x) is divided by (x¡a)2 , prove that the remainder is P (a)(x¡a)+P (a) where P (x) is the derivative of P (x): 85 A lampshade is a truncated cone open at the bottom 20 cm Find the pattern needed to make this lampshade from a flat sheet of material 15 cm cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 32 cm black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_30\841IB_HL-2_30.CDR Monday, 21 January 2008 4:44:46 PM PETERDELL IB_HL-2ed (842) 842 MISCELLANEOUS QUESTIONS (Chapter 30) 86 [AB] represents a painting on a wall AB = m and BC = m The angle of view observed by a girl between the top and bottom of the painting is 30o How far is the girl from the wall? A side view 2m B 30° 1m G 87 C eye level A circle is centred at the origin O A second circle has half the diameter of the original circle and touches it internally P is a fixed point on the smaller circle as shown, and lies on the x-axis The smaller circle now rolls around the inside of the larger one without slipping Show that for all positions of the smaller circle, P remains on the x-axis P Write ¡8i in polar form Hence find the three cube roots of ¡8i, calling them z1 , z2 and z3 Illustrate the roots from b on an Argand diagram Show that z12 = z2 z3 where z1 is any one of the three cube roots Find the product of the three cube roots 88 a b c d e 89 a Complex number z has an argument of µ Show that iz has an argument of µ + ¼2 b In an Argand plane, points P, Q and R represent the complex numbers z1 , z2 and z3 respectively If i(z3 ¡ z2 ) = z1 ¡ z2 , what can be deduced about triangle PQR? 90 z = reiµ , r > 0, is a non-zero complex number such that z + a Find expressions for a and b in terms of r and µ b Hence, find all complex numbers z such that z + = a + bi, a, b R z is real z 91 The diagram shows a simple electrical network Each symbol represents a switch A B All four switches operate independently, and the probability of each one of them being closed is p a In terms of p, find the probability that the current flows from A to B b Find the least value of p for which the probability of current flow is more than 0:5 ³ ´ 92 If A = : a find A2 and A3 ³ n b prove using mathematical induction that An = n2n¡1 2n ´ for all n Z + cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 93 By considering the identity (1 + x)n = (1 + x)2 (1 + x)n¡2 , deduce that ¡n¢ ¡n¡2¢ ¡ ¢ ¡n¡2¢ + n¡2 r = r r¡1 + r¡2 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_30\842IB_HL-2_30.CDR Thursday, 11 March 2010 11:00:27 AM PETER IB_HL-2ed (843) MISCELLANEOUS QUESTIONS (Chapter 30) 843 94 While driving, Bernard passes through n intersections which are independently controlled by traffic lights Each set of lights has probability p of stopping him a What is the probability that Bernard will be stopped at least once b Suppose Ak is the event that Bernard is stopped at exactly k intersections and Bk is the event that Bernard is stopped at at least k intersections Write down the conditional probability P(Ak j Bk ) c If A1 and B1 are independent, find p d Find p if P(A2 j B2 ) = P(A1 ) and n = 95 A club has n female members and n male members A committee of three members is to be randomly chosen, and must contain more females than males a How many committees consist of females and male? b How many committees consist of females? ¡ ¢ ¡ ¢ ¡ ¢ c Use a and b to deduce that n n2 + n3 = 12 2n d Suppose the club consists of 12 people, and that Mr and Mrs Jones are both members Find the probability that a randomly selected committee contains: i Mrs Jones ii Mr Jones given that it contains Mrs Jones 96 In triangle ABC, the angle at A is double the angle at B If AC = cm and BC = cm, find: a the cosine of the angle at B b the length of [AB] using the cosine rule c Are both solutions in b valid? 97 x2 + b1 x + c1 = and x2 + b2 x + c2 = are two quadratic equations where b1 b2 = 2(c1 + c2 ) Prove that at least one of the equations has real roots p p 98 Suppose that for all n Z + , (2 ¡ 3)n = an ¡ bn where an and bn are integers a Show that an+1 = 2an + 3bn and bn+1 = an + 2bn b Calculate an2 ¡ 3bn2 for n = 1, and c What you propose from b? d Prove your proposition from c 99 A sequence un is defined by u1 = u2 = and un+2 = un+1 + un for all n Z + Prove by induction that un 2n for all n Z + cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 a Use complex number methods to prove that cos3 µ = 34 cos µ + b Solve the equation x3 ¡ 3x + = by letting y = mx 25 102 a Graph y = x3 ¡ 12x2 + 45x and on the graph mark the coordinates of its turning points b If x3 ¡ 12x2 + 45x = k has three real roots, what values can k have? 95 101 100 50 75 25 100 Use the Principle of mathematical induction to prove that, if n > 2, n Z + , 1 1 n+1 then (1 ¡ )(1 ¡ )(1 ¡ ) :::::: (1 ¡ ) = n 2n black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_30\843IB_HL-2_30.CDR Friday, 12 December 2008 12:43:18 PM TROY cos 3µ IB_HL-2ed (844) 844 MISCELLANEOUS QUESTIONS (Chapter 30) B 103 Triangle ABC has perimeter 20 cm x cm a Find y in terms of x and µ and hence y cm find cos µ in terms of x only q C b If the triangle has area A, show that A 2 cm A = ¡20(x ¡ 12x + 20) c Hence, without calculus, find the maximum area of the triangle and comment on the triangle’s shape when its area is a maximum ´ ³ 104 a If A = , predict the form of An b Use mathematical induction to prove your conjecture in a correct c If Sn = A + A2 + A3 + + An , find Sn in simplest form and hence find S20 105 If m X f(n) = m3 + 3m, find f(n) n=1 106 ABC is an equilateral triangle with sides 10 cm long P is a point within the triangle which is cm from A and cm from B How far is it from C? 107 A normally distributed random variable X has a mean of 90 Given that the probability P(X < 85) ¼ 0:16 : a find the proportion of scores between 90 and 95, i.e., find P(90 < X < 95) b find an estimate of the standard deviation for the random variable X 108 A normally distributed random variable X has a mean of 90 Given that the probability P(X < 88) ¼ 0:28925, find the: a standard deviation of X to decimal places b probability that a randomly chosen score is either greater than 91 or less than 89 109 In an International school there are 78 students preparing for the IB Diploma Of these students, 38 are male and 17 of these males are studying Mathematics at the higher level Of the female students, 25 are not studying Mathematics at the higher level A student is selected at random and found to be studying Mathematics at the higher level Find the probability that this student is male 110 A company manufactures computer chips, and it is known that 3% of them are faulty In a batch of 500 such chips, find the probability that between and percent (inclusive) of the chips are faulty 111 A factory manufactures rope, and the rope has an average of 0:7 flaws per metre It is known that the number of flaws produced in the rope follows a Poisson distribution a Determine the probability that there will be exactly flaws in metres of rope b Find the probability that there will be at least flaws in metres of rope 112 A random variable X is known to be distributed normally with standard deviation 2:83 Find the probability that a randomly selected score from X will differ from the mean by less than cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 113 A discrete random variable X has a probability function given by the rule ¡ ¢x P(X = x) = a 25 , x = 0, 1, 2, 3, Find the value of a black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_30\844IB_HL-2_30.CDR Monday, 21 January 2008 10:13:05 AM PETERDELL IB_HL-2ed (845) 845 MISCELLANEOUS QUESTIONS (Chapter 30) 114 Given that events A and B are independent with P(A j B) = find P(A [ B ) and P(B j A) = 25 , 115 In a game, a player rolls a biased tetrahedral Score (four-faced) die The probability of each possible Probability 12 k 14 13 score is shown alongside in the table a Find the value of k b Let the random variable X denote the number of 2s that occur when the die is rolled 2400 times Calculate the exact mean and standard deviation of X 116 The lifetime n (in years) of a particular component of a solar cell is given by the ½ 0:6e¡0:6n , n > probability density function f (n) = 0, otherwise a What is the chance that a randomly chosen component will last for at least one year? b A solar cell has components, each of which operates independently of each other The solar cell will continue to operate provided at least one of the components are operating Find the probability that a randomly chosen solar cell fails within one year 117 The random variable X has a Poisson distribution with standard deviation ¾ such that P(X = 2) ¡ P(X = 1) = 3P(X = 0) Find the exact value of ¾ in surd form 118 A machine produces soft drink in bottles The volumes in millilitres (mL) of a sample of drinks chosen at random are shown below Volume (mL) Frequency 374:7 374:8 12 374:9 15 375:0 16 375:1 375:2 11 375:3 375:4 Find unbiased estimates of: a the mean of the population from which this sample is taken b the variance of the population from which this sample is taken 119 In a particular year, a randomly chosen Year 12 group completed a calculus test with the following results: 25 25 X X xi = 1650 and xi2 = 115 492, where xi denotes the percentage result of the i=1 i=1 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 i th student in the class Calculate an unbiased estimate of: a the mean percentage result of all Year 12 students in the calculus test b the variance of the percentage result of all Year 12 students in the calculus test R 120 a Using integration by parts, find ln x dx Show how to check that your answer is correct b The continuous random variable X has probability density function defined by ½ ln x, x k f (x) = Find the exact value of k 0, otherwise c Write down an equation that you would need to solve to find the median value of the random variable X Do not attempt to solve this equation black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_30\845IB_HL-2_30.CDR Monday, 21 January 2008 10:13:55 AM PETERDELL IB_HL-2ed (846) 846 MISCELLANEOUS QUESTIONS (Chapter 30) 121 Use the cosine rule and the given kite to show that sin2 µ = 12 ¡ 12 cos 2µ and cos2 µ = 12 + 12 cos 2µ 122 Show that tan µ = tan ® D q q a q A B 123 Points P and Q are free to move on the coordinate axes N is the foot of the perpendicular from the origin to the line segment [PQ] [PQ] makes an angle of µ with the y-axis a Show that N is at (3 sin µ cos2 µ, sin2 µ cos µ): b Use technology to sketch the graph of the curve defined by: x = sin µ cos2 µ, y = sin2 µ cos µ: 124 125 C y P q 3m N a Find the general term un of the sequence: 1 ¡ sin µ, cos µ, sin µ, ¡ cos µ, sin µ cos µ b Find an equation connecting consecutive terms of the sequence: 1, cos µ, cos3 µ, cos7 µ, cos15 µ, k, k, k + where k Q respectively a Find k b x Q are the 3rd, 4th and 6th terms of an arithmetic sequence Find the general term un 126 ABC is an equilateral triangle with sides of length 2k P is any point within the triangle [PX], [PY] and [PZ] are altitudes from P to the sides [AB], [BC] and [CA] respectively b be µ, find PX + PY + PZ in terms of µ, and hence show that a By letting PCZ PX + PY + PZ is constant for all positions of P b Check that your solution to a is correct when P is at A c Prove that the result in a is true using areas of triangles only 127 R and Q are two fixed points on either side of line segment [AB] P is free to move on the line segment so that the angles µ and Á vary a and b are the distances of Q and R respectively from [AB] R b a Show that for all positions of P, A dÁ ¡b cos2 Á = dµ a cos2 µ b A particle moves from R to P with constant speed v1 and from P to Q with constant speed v2 q M a f P Q magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 Deduce that the time taken to go from R to P to Q is a minimum when cyan B N black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_30\846IB_HL-2_30.CDR Friday, 12 December 2008 12:45:08 PM TROY sin µ v1 = sin Á v2 IB_HL-2ed (847) MISCELLANEOUS QUESTIONS (Chapter 30) 847 R 128 At A on the surface of the Earth, a rocket is launched vertically upwards After t hours it is h km at R, h km above the surface B is the horizon A y km seen from R b Suppose BOR is µ and arc AB is y km long B q a If the Earth’s radius is r km, show that r km dy cos2 µ dh = dt sin µ dt b If the velocity of the rocket after t hours is given by r sin t for any t [ 0, ¼ ], find the height of the rocket at t = ¼2 hours c If r ¼ 6000, find the rate at which arc AB is changing at the instant when t = ¼2 129 Prove that the roots of (m ¡ 1)x2 + x ¡ m = 0 < m < a Show that sin 15o = 130 p p 6¡ using sin 45o = are always real and positive for p1 and sin 30o = 12 , together with a suitable trigonometric formula b Find the exact value of cos2 165o + cos2 285o 131 For ¡¼ x ¼, find the exact solutions to sec 2x = cot 2x + tan 2x: p p 132 Solve exactly for x if sin x = csc x + ¡ where x 2¼ 133 The first terms of a geometric sequence have a sum of 39 If the middle term is increased by 66 23 %, the first three terms now form an arithmetic sequence Find the smallest possible value of the first term 134 Show algebraically that the equation log3 (x ¡ k) + log3 (x + 2) = has a real solution for every real value of k 135 Solve the following equations, giving exact answers: p b 32x+1 + 8(3x ) = a 82x+3 = c ln (ln x) = d log 19 x = log9 136 Solve the following inequalities, giving exact answers: b ( 23 )x > ( 32 )x¡1 c 4x + 2x+3 < 48 a (0:5)x+1 > 0:125 µ ¶ x¡y 137 If x2 + y2 = 52xy, show that log = 12 (log x + log 2y) 138 If z = cos µ + i sin µ where < µ < ¼4 , find the modulus and argument of ¡ z : 139 Find z in the form a + bi if z = + i + 58 9(3 ¡ 7i) 140 Solve the following equations simultaneously: 4x = 8y and 9y = 243 3x z¡1 where z = a+bi and z ¤ is the complex conjugate of z, write w z¤ + in the form x + yi Hence determine the conditions under which w is purely imaginary cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 141 Given w = black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_30\847IB_HL-2_30.CDR Thursday, 11 March 2010 11:01:42 AM PETER IB_HL-2ed (848) 848 MISCELLANEOUS QUESTIONS (Chapter 30) 142 Given that x = log3 y2 , express logy 81 in terms of x 143 An infinite number of circles are drawn in a sector of a circle of radius 10 cm and angle ® = ¼3 as shown a What is the total area of this infinite series of circles? b Find an expression for the total area of all circles for a general angle ® such that ® ¼2 : a 144 The ratio of the zeros of x2 + ax + b is : Find a relationship between a and b 145 Find real numbers a and b if the polynomial z +az +bz +15 = has a root +i 146 If xn + ax2 ¡ leaves a remainder of ¡3 when divided by (x ¡ 1) and a remainder of ¡15 when divided by (x + 3), find the values of a and n 147 When a cubic polynomial P (x) is divided by x(2x ¡ 3), the remainder is ax + b where a and b are real a If the quotient is the same as the remainder, write down an expression for P (x) b Prove that (2x ¡ 1) and (x ¡ 1) are both factors of P (x) c Find the equation of P (x) given that it has a y-intercept (0, 7) and passes through the point (2, 39) 148 Factorise f (x) = 2x3 ¡x2 ¡8x¡5, and hence find the values of x for which f(x) > 149 The graph of a quartic polynomial y = f(x) cuts the x-axis at x = ¡3 and at x = ¡ 14 , and touches it at x = 32 The y-intercept is Find f(x) 150 The polynomial p(x) = x3 + (5 + 4a)x + 5a where a is real, has a zero ¡2 + i a Find a real quadratic factor of p(x) b Hence, find the value of a and the real zero of p(x) 151 Let a b c h(x) = x3 ¡ 6tx2 + 11t2 x ¡ 6t3 where t is real Show that t is a zero of h(x) Factorise h(x) as a product of linear factors Hence or otherwise, find the coordinates of the points where the graphs of y = x3 + 6x2 and y = ¡6 ¡ 11x meet cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 152 A real polynomial P (x) = x4 + ax3 + bx2 + cx ¡ 10 has two integer zeros p and q a If P (x) also has a complex zero + ki, where k is an integer: i use this zero to write an expression for a real quadratic factor of P (x) ii state all possible values of k black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_30\848IB_HL-2_30.CDR Monday, 21 January 2008 1:28:56 PM PETERDELL IB_HL-2ed (849) 849 MISCELLANEOUS QUESTIONS (Chapter 30) b Using p and q, write another expression for a real quadratic factor of P (x) Hence list all possible values of pq c Given that p + q = ¡1, show that there is only one possible value for pq Hence find all zeros of P (x) 153 The real polynomial P (z) of degree has one complex zero of the form ¡ 2i, and another of the form ai, where a 6= and a is real Find P (z) if P (0) = 10 and the coefficient of z is Leave the answer in factorised form 154 The point A(¡2, 3) lies on the graph of y = f (x) Give the coordinates of the point that A moves to under the following transformations: a y = f (x ¡ 2) + b y = 2f (x ¡ 2) c y = ¡ j f (x) j ¡2 d y = f (2x ¡ 3) e y= f (x) f y = f ¡1 (x) 155 The points A(¡1, 0), B(1, 0) and C(0, ¡0:5) are the x- and y-intercepts of y = f (x) On the same set of axes, sketch the following graphs For each case, explain what happens to the points A, B and C a y = f (x + 1) ¡ b c y = jf(x)j d y = ¡2f (x ¡ 1) y= f (x) y -2 A -1 C B x -1 156 The real quadratic function f (x) has a zero of + 2i, and a y-intercept of ¡13 Write the function in the form: b f (x) = a(x ¡ h)2 + k a f (x) = ax2 + bx + c 157 Find a trigonometric equation of the form y = a sin(b(x + c)) + d that represents the following graph with the information given below You may assume that (3, ¡5) is a minimum point and (6, ¡1) lies on the principal axis (6,-1) (3,-5) < x + 3y ¡ z = 15 2x + y + z = 158 Solve the system using an inverse matrix: : x ¡ y ¡ 2z = cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 159 The Ferris wheel at the Royal Show turns one full circle every minute The lowest point is metre from the ground, whilst the highest point is 25 metres above the ground a The height of the Ferris wheel above ground level after t seconds is given by the model h(t) = a + b sin(c(t ¡ d)) Find the values of a, b, c and d given that you start your ride after entering your seat at the lowest point b If the motor driving the Ferris wheel breaks down after 91 seconds, how high up would you be while waiting to be rescued? black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_30\849IB_HL-2_30.CDR Thursday, 11 March 2010 11:29:34 AM PETER IB_HL-2ed (850) 850 MISCELLANEOUS QUESTIONS (Chapter 30) 160 The equations of two lines are: à ! à ! l1 : r = ¡4 ¡1 +¸ , ¸ R , and l2 : x = y¡5 ¡z ¡ = 2 Determine the point of intersection of l1 and the plane 2x + y ¡ z = Clearly explain why l1 and l2 are not parallel Find the point of intersection of l1 and l2 Find the equation of the plane that contains l1 and l2 a b c d 161 Find the acute angle between the plane 2x + 2y ¡ z = and the line x = ¸ ¡ 1, y = ¡2¸ + 4, z = ¡¸ + x ¡ 2y + 3z = 162 Consider the following system of < x + py + 2z = linear equations in which p and q : ¡2x + p2 y ¡ 4z = q are constants: a Write this system of equations in augmented matrix form à b Show, using clearly defined row operations, that this augmented matrix can be reduced to: ¡2 p+2 ¯ ¯ ¯ ¡1 ¯ ¡1 p ¯p+q ! c What values can p and q take when the system has i a unique solution ii no solutions iii infinite solutions? d Specify the infinite solutions in parametric form 163 a Show that the plane 2x + y + z = contains the line l1 : x = ¡2t + 2, y = t, z = 3t + 1, t R b For what values of k does the plane x + ky + z = contain l1 ? c Without using row operations, find the values of p and q < 2x + y + z = x¡y+z =3 for which the following system of equations has an infinite : 2x + py + 2z = q number of solutions Clearly explain your reasoning d Check your result using row operations ! à ! à 164 For A = ¡1 ¡1 1 and B = ¡1 ¡2 12 ¡3 ¡5 , calculate AB and < 4a + 7b ¡ 3c = ¡8 ¡a ¡ 2b + c = hence solve the system of equations : 6a + 12b ¡ 5c = ¡15 165 Use vector methods to prove that joining the midpoints of the sides of a rhombus gives a rectangle 166 a Given a = i + j ¡ 3k and b = j + 2k, find a £ b b Find a vector of length units which is perpendicular to both a and b cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 167 Let r = 2i ¡ 2j + k, s = 3i + j + 2k and t = i + 2j ¡ k be the position vectors of the points R, S, and T respectively Find the area of the triangle RST black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_30\850IB_HL-2_30.CDR Monday, 21 January 2008 11:37:20 AM PETERDELL IB_HL-2ed (851) 851 MISCELLANEOUS QUESTIONS (Chapter 30) 168 In the given figure, ABCD is a parallelogram X is the midpoint of [BC], and Y is on [AX] such that AY : YX = : The coordinates of A, B and C are (1, 3, ¡4), (4, 4, ¡2) and (10, 2, 0) respectively a Find the coordinates of D, X and Y b Prove that B, D and Y are collinear C X B Y D A 169 Let a = 3i + 2j ¡ k, b = i + j ¡ k and c = 2i ¡ j + k a Show that b £ c = ¡3j – 3k b Verify for the given vectors that a £ (b £ c) = b (a ² c) ¡ c (a ² b) à 170 Given the vectors p = ¡2 ! à and q = a p and q are perpendicular ¡t 1+t 2t ! , find t such that: b p and q are parallel 171 Suppose A and B are events such P(A) = 0:3 + x, P(B) = 0:2 + x and P(A \ B) = x a Find x if A and B are mutually exclusive events b Calculate the possible values of x if A and B are independent events 172 Find exact solutions for the following: j ¡ 4x j > a j 2x ¡ j b x¡2 60 ¡ 5x ¡ x2 173 The average number of amoebas in 50 mL of pond water is 20 a Assuming that the number of amoebas in pond water follows a Poisson distribution, find the probability that no more than amoebas are present in 10 mL of randomly sampled pond water b If a researcher collected 10 mL of pond water each weekday over weeks (20 days in all), find the probability that the researcher collected no more than amoebas on more than 10 occasions in that week period 174 Solve dy = cos2 x given that y(0) = dx 175 Solve the differential equation xy dy = + y2 given that y = when x = dx 176 A current of I amperes flows through a coil of inductance L henrys and resistance dI R ohms with electromotive force E = L + RI volts dt Assuming that E , L and R are constants, show by separating the variables that ´ E³ R I= ¡ e¡ L t , given that I = when t = R cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 177 A pair of guinea pigs was released onto an island in early January Infrared scans of the island in early May showed the guinea pig population to be 180 Given that the rate of increase in such a population is proportional to the population at that time, estimate the island’s guinea pig population in early October black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_30\851IB_HL-2_30.CDR Monday, 21 January 2008 4:51:26 PM PETERDELL IB_HL-2ed (852) 852 MISCELLANEOUS QUESTIONS (Chapter 30) 178 a Show algebraically that 1 P + = y P ¡y y(P ¡ y) ³ y´ b Using part a, solve the differential equation y = k y ¡ P given that P = 624, y = when t = 0, and that y = 12 when t = 179 c The differential equation above describes how a rumour is spread at Beijing College by people starting at 12 noon 12 people have heard the rumour by pm i Find the number of people at Beijing College, giving a reason for your answer ii How many people have heard the rumour by pm? iii At what time have 90% of the people heard the rumour? p ¡ i in the form reiµ a Express + i and ¡1 ¡ i Hence write z = p in the form re iµ 3¡i b What is the smallest positive integer n such that z n is a real number? 180 There are 12 students in a school’s Hungarian class Being well-mannered, they line up in a single file to enter the class a How many orders are possible? b How many orders are there if: i Irena and Eva are among the last four in the line ii Istvan is between Paul and Laszlo and they are all together iii Istvan is between Paul and Laszlo but they are not necessarily together iv there are exactly three students between Annabelle and Holly? c Once inside, the class is split into groups of four students each for a vocabulary quiz How many ways can this be done: i if there are no restrictions ii if Ben and Marton must be in the same group? 181 The velocity of a particle travelling in a straight line is given by v = cos( 13 t) cm s¡1 Find the distance travelled by this particle in the first 10¼ seconds of motion 182 Year 12 students at a government school can choose from 16 subjects for their Certificate Seven of these subjects are in group I, six are in group II, and the other three are in group III Students must study six subjects to qualify for the Certificate How many combinations of subjects are possible if: a there are no restrictions b students must choose subjects from groups I and II and the remaining subjects could be from any group c French (a group I subject) is compulsory, and they must choose at least one subject from group III? ¡ ¢ ¡n¡1¢ ¡ ¢ ¡ 183 Solve the equation: n3 = n¡1 184 Find the coefficient of: µ ¶8 2x3 ¡ b x2 in the expansion of (1+2x)5 (2¡x)6 2x c x3 in the expansion of (1 + 2x ¡ 3x2 )4 : cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a x12 in the expansion of black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_30\852IB_HL-2_30.CDR Monday, 21 January 2008 1:30:20 PM PETERDELL IB_HL-2ed (853) 853 MISCELLANEOUS QUESTIONS (Chapter 30) 185 The function f is defined by f : x 7! esin x , x [ 0, ¼ ] a Use calculus to find the exact value(s) of x for which f (x) has a maximum value b Find f 00 (x) and write down an equation that will enable you to find any points of inflection in the given domain c Find the point(s) of inflection in the given domain 186 Solve for x: logx + log2 x = Z a 187 Find the exact value of a if a > and x dx = x2 + 188 Given that A is an acute angle and tan 2A = 32 , find the exact value of tan A 189 The scores a, b, 6, 13 and where b > a have a mean and variance of Find the values of a and b 190 The graph of y = f (x) for ¡9 x is shown alongside The function has vertical asymptotes at x = and x = ¡3 and a horizontal asymptote at y = Copy and sketch the graph of y -8 , indicating clearly the axes y= f (x) intercepts and all asymptotes 191 For what values of x is the matrix A = ³ a b -4 x -4 à 192 Find a and b if the matrix A = y¡=¡¦(x) ¡1 x¡1 ¡4 ¡2 ´ ¡2 3¡x 5 ¡2 ¡8 ! singular? is its own inverse Hence find A11 193 If z = x + 2i and u = + iy where x, y R , find the smallest positive value of x z+u for which is purely imaginary z¡u 194 If ¡ 2i is a zero of P (x) = x4 + 11x2 ¡ 10x + 50, find all the other zeros 195 Find the exact value of the volume of the solid formed when the region enclosed by y = xex , the x-axis, and the line x = 1, is rotated through 360o about the x-axis 196 Determine the sequence of transformations which transform the function f(x) = 3x2 ¡ 12x + to g(x) = ¡3x2 + 18x ¡ 10 197 Find the area of the region bounded by the curve y = tan2 x + sin2 x, the x-axis, and the line x = ¼4 Z sin(2 arcsin x) dx 198 Simplify sin(2 arcsin x) and hence find cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_30\853IB_HL-2_30.CDR Monday, 21 January 2008 4:52:42 PM PETERDELL IB_HL-2ed (854) 854 MISCELLANEOUS QUESTIONS (Chapter 30) < 2x ¡ y + 3z = 2x + y + (a + 3)z = 10 ¡ a : 4x + 6y + (a2 + 6)z = a2 , 199 For the system of linear equations: find a b c the value(s) of a for which the system has: no solutions infinitely many solutions, and find the form of these solutions a unique solution, and find the solution in the case where a = 2: x2 + (x + 1)2 Write down the equations of the asymptotes of the graph of y = f (x): Find f (x) and hence find the position and nature of any stationary points Find f 00 (x) and hence find the coordinates of all points of inflection Sketch the graph of y = f(x) showing all the above features 200 A function f is defined by f (x) = a b c d 201 A particle moves in a straight line such that its displacement from point O is s The acceleration of the particle is a and its velocity is v where a = 12 v a Find v(t) given that v(0) = ¡1 b Find the distance travelled in the first seconds of motion 202 Determine the domain of f (x) = arccos(1 + x ¡ x2 ), and find f (x) 203 Find the area of the region enclosed by the graph of y = the line x = ¼3 tan x , the x-axis, and cos(2x) + tan x 204 Find the equations of all asymptotes of the graph of the function y = sin(2x) +1 where ¡¼ x ¼ sin x 205 Find the exact coordinates of the stationary points on the curve y = tan x + where ¡¼ x ¼2 206 The sum of an infinite geometric series is 49 and the second term of the series is 10 Find the possible values for the sum of the first three terms of the series x+1 , find: x¡2 207 If f : x 7! 2x + and g : x 7! 208 (f ± g)(x) a dy if x2 ¡ 3xy + y2 = 7: dx b Hence find the coordinates of all points on the curve for which the gradient is 23 and P(B) = 27 a Find P(A [ B) if A and B are: i mutually exclusive ii cyan magenta yellow 95 100 50 75 25 95 independent 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b Find P(A j B) if P(A [ B) = g ¡1 (x) a Find 209 A and B are two events such that P(A) = b black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_30\854IB_HL-2_30.CDR Monday, 21 January 2008 4:53:07 PM PETERDELL IB_HL-2ed (855) MISCELLANEOUS QUESTIONS (Chapter 30) 210 855 a Find the coordinates of A, the point of intersection of l1 and l2 , where l1 is given ! à ! à x + 10 y¡7 z ¡ 11 and l2 is given by = = by r = ¡13 + ¸ ¡5 ¡5 ¡5 ¡3 ¡2 b Find the coordinates of B, where l1 meets the plane 3x + 2y ¡ z = ¡2 c The point C(p, 0, q) lies on the plane in b Find the possible values of p if the area of triangle ABC is p units2 211 Find x in terms of a if a > and loga (x + 2) = loga x + 212 Solve for y: (x2 + 1) R 213 Find dy = y + given that y = when x = dx x2 sin x dx 214 If f(2x + 3) = 5x ¡ 7, find f ¡1 (x) 215 Find, to significant figures, the area of the region enclosed by the graphs of y = xesin x and y = x2 ¡ 4x + 6: dy if exy + xy2 ¡ sin y = dx 216 Find j 2x ¡ j +3 < ¡x j x + j ¡2 217 Solve for x: Z 218 Find x p dx 1+ x+2 dy dx 219 Find if sin(xy) + y2 = x à 220 The lines r = ¡2 ! à +¸ a ¡1 ! and x¡4 z+2 =1¡y = intersect at point P a Find the value of a and hence find the coordinates of P b Find the acute angle between the two lines c Find the equation of the plane which contains the two lines 221 For what values of a does the graph of y = ax+2 cut the graph of y = 3x2 ¡2x+5 in two distinct points? 222 The height of a cone is always twice the radius of its base The volume of the cone is increasing at a constant rate of cm3 s¡1 Find the rate of change in the radius when the height is 20 cm 223 Find the value of a if the line passing through the points A(0, 5, 6) and B(4, 1, ¡2) à ! à ! cyan magenta yellow 95 100 50 75 are coplanar 25 ¡1 95 +s 100 50 a 75 25 95 100 50 75 25 95 100 50 75 25 and the line r = black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_30\855IB_HL-2_30.CDR Monday, 21 January 2008 12:00:11 PM PETERDELL IB_HL-2ed (856) 856 MISCELLANEOUS QUESTIONS (Chapter 30) p 224 Let f (x) = x tan ¡ x2 , ¡1 x a Sketch the graph of y = f(x) b Write down an expression for V , the volume of the solid formed by rotating the region bounded by y = f (x), x = and x = through 360o about the x-axis 225 Let f (x) = xe1¡2x a Find f (x) and f 00 (x) b Find the exact coordinates of the stationary points of the function and determine their nature c Find the exact values of the x-coordinates of the points of inflexion of the function d Discuss the behaviour of the function as x ! §1: e Sketch the graph of the function f Find the exact value of k if k > and the region bounded by y = f (x), the x-axis, and the line x = k has area equal to 14 (e ¡ 1) units2 à ! à ! 226 lies on the plane P1 a Find a and k if the line l1 given by r = ¡1 + ¸ a ¡1 with equation 3x ¡ ky + z = b Show that the plane P2 with equation 2x ¡ y ¡ 4z = is perpendicular to P1 c Find the equation of l2 , the line of intersection of P1 and P2 d Find the point of intersection of l1 and l2 e Find the angle between the lines l1 and l2 227 Construct a quartic polynomial f (x) with integer coefficients such that f(x) < for all x R Write your polynomial in expanded form 228 P (z) = z + az + bz + c where a, b and c R Two of the roots of P (z) are ¡2 and ¡3 + 2i Find a, b and c and also find possible values of z when P (z) > 229 f(x) = tan(3(x ¡ 1)) + for x [ ¡1, ] Find: cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 period of y = f (x) equations of any asymptotes transformations that transform y = tan x into y = f (x) domain and range of y = f (x) 25 the the the the 95 100 50 75 25 a b c d black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_30\856IB_HL-2_30.CDR Thursday, 11 March 2010 11:29:47 AM PETER IB_HL-2ed (857) ANSWERS 857 EXERCISE 1C ANSWERS a ¡ 2x b ¡2x ¡ c 11 f (g(x)) = (2 ¡ x)2 , Domain fx j x is in R g, Range fy j y > 0g g(f (x)) = ¡ x2 , Domain fx j x is in R g, Range fy j y 2g a x2 ¡ 6x + 10 b ¡ x2 c x = § p12 a f ± g = f(0, 1), (1, 0), (2, 3), (3, 2)g b g ± f = f(0, 1), (1, 0), (2, 3), (3, 2)g c f ± f = f(0, 0), (1, 1), (2, 2), (3, 3)g f ± g = f(0, 0), (1, 1), (2, 2), (3, 3)g a f ± g = f(2, 7), (5, 2), (7, 5), (9, 9)g b g ± f = f(0, 2), (1, 0), (2, 1), (3, 3)g EXERCISE 1A p a, d, e a, b, c, e, g No, e.g., x = y = § ¡ x2 EXERCISE 1B.1 a b c ¡1 d ¡13 e a b c ¡16 d ¡68 e 17 a ¡ 3a b + 3a c ¡3a ¡ d 10 ¡ 3b e ¡ 3x f ¡ 3x ¡ 3h a 2x2 + 19x + 43 b 2x2 ¡ 11x + 13 c 2x2 ¡ 3x ¡ d 2x4 + 3x2 ¡ e 2x4 ¡ x2 ¡ f 2x2 + 4xh + 2h2 + 3x + 3h ¡ 2x + a i ¡ 72 ii ¡ 34 iii ¡ 49 b x = c d x = 95 4x ¡ , x 6= 1, Domain fx j x 6= 13 or 1g 3x ¡ b (f ± g)(x) = 2x + 5, x 6= ¡2, Domain fx j x 6= ¡2g c (g ± g)(x) = x, x 6= 1, Domain fx j x 6= 1g a Let x = 0, ) b = d and so ax + b = cx + b ) ax = cx for all x Let x = 1, ) a = c b (f ± g)(x) = [2a]x + [2b + 3] = 1x + for all x ) 2a = and 2b + = ) a = 12 and b = ¡ 32 a (f ± g)(x) = x¡2 f is the function which converts x into f (x) whereas f (x) is the value of the function at any value of x a 6210 Euros value after years b t = 4:5, the time for the photocopier to reach a value of 5780 Euros c 9650 Euros y 10 (5' 3) (2' 1) x f (x) = ¡2x + a = 3, b = ¡1, c = ¡4, T (x) = 3x2 ¡ x ¡ c Yes, f(g ± f )(x) = [2a]x + [3a + b]g EXERCISE 1D a - EXERCISE 1B.2 a b c d a b c d e f a b c d e f g h i a b c d e Domain f1, 2, 3g, Range f3, 5, 7g Domain f¡1, 0, 2g, Range f3, 5g Domain f¡3, ¡2, ¡1, 3g, Range f1g p Domain f¡2, ¡1, 0, 1, 2g, Range f0, 3, 2g Domain fxj ¡1 < x 5g, Range fy j < y 3g Domain fx j x 6= 2g, Range fy j y 6= ¡1g Domain fx j x R g, Range fy j < y 2g Domain fx j x R g, Range fy j y > ¡1g Domain fx j x > ¡4g, Range fy: y > ¡3g Domain fx j x 6= §2g, Range fy: y ¡1 or y > 0g Domain fx j x is in R g, Range fy j y is in R g Domain fx j x is in R g, Range f3g Domain fx j x is in R g, Range fy j y > 2g Domain fx j x is in R g, Range fy j y > 2g Domain fx j x ¡2, x > 2g, Range fy j y > 0g Domain fx j x 6= 2g, Range fy j y 6= 0g Domain fx j x 2g, Range fy j y > 0g Domain fx j x > 52 g, Range fy j y > 0g c + e - g - i k + + -1 m - black 95 Y:\HAESE\IB_HL-2ed\IB_HL-2ed_an\857IB_HL-2_AN.CDR Thursday, 24 January 2008 11:30:08 AM PETERDELL h j + - l - - Qe_ n + + + - + -\Qw_ + - -\We_ - \Qw_ + 100 50 75 25 100 yellow + -2 - + - - + Qw_ + + -3 o f + - -\Qw_ - -1 - + - d k + -3 - + - + -2 -1 - - b + + + + i + - - g 95 h Qw_ 50 + - -2 e 75 25 95 100 50 75 25 95 100 50 75 25 + - + -2 l -\Wt_ f - + - + -1 + + - j -4 c + - + + -3 + a f Domain fx j x 6= 0g, Range fy j y ¡2 or y > 2g g Domain fx j x 6= 2g, Range fy j y 6= 1g h Domain fx j x R g, Range fy j y R g i Domain fx j x 6= ¡1 or 2g, Range fy j y 13 or y > 3g j Domain fx j x 6= 0g, Range fy jy > 2g k Domain fx j x 6= 0g, Range fy j y ¡2 or y > 2g l Domain fx j x R g, Range fy j y > ¡8g d Domain fx j x R g, Range fy j y 12 g magenta + + -1 -2 Domain fx j x 6= 5g, Range fy j y 6= 2g Domain fx j x > 0g, Range fy j y > 0g Domain fx j x 6= 0g, Range fy j y > 0g Domain fx j x 4g, Range fy j y > 0g Domain fx j x R g, Range fy j y > ¡2 14 g cyan b + + \Qw_ + Qe_ IB_PD (858) 858 ANSWERS + a + + b -2 - c - - d -2 e + g + i - + EXERCISE 1F.1 - a b c d e f ¡1 g h i j k l 2 a b c 54 d a No b No - 4 + + f - h - -3 Qw_ - a b j ab j j a jj b j 6 ¡6 ¡6 ¡2 ¡2 12 12 12 12 12 12 12 12 Ew_ a + c - - + -2 + -\Ew_ + e - g + - + - 1 -1 0 l -~`3 -1 ~`3 o - -1 + - - + s + - - r - + t - Qw_ -8 -3 y= n ¡x, x > x, x<0 d y= x 2x, x > 0, x < y y x y¡=|¡x¡|+ x y¡=¡-|¡x¡| x - ½1, + e y= + n x>0 undefined, x = ¡1, x<0 - y y= -3 -\Qd_R_ + - + - + u c -1 x n + + -1 y =j x + j n x + 1, x > ¡1 y = ¡x ¡ 1, x < ¡1 y + 2 + -3 q -2 -\Ew_ p + b y¡=¡|¡x¡+¡1¡| + - + - n - + - + y =j x ¡ j n x ¡ 2, x > y = ¡ x, x < (b 6= 0) y¡=¡|¡x-2¡| - jaj a j= b jbj jaj jbj 3 3 a j b 3 3 y + + - + - + - + -2 - - a - j + 0 + + j ab j=j a jj b j, j + -3 f + m - -2 k d - h - i + Qr_ + b j f y= ¡x, 3x, x>0 x<0 y x x x y = x -2| x| x -1 EXERCISE 1E a x [ ¡3, ] b no solutions c x ] ¡ 12 , [ g d x ] ¡1, ] or [ 1, [ e x ] ¡1, ] or [ 3, [ f x ] ¡ 23 , [ g x ] ¡2, [ p p y =j x j + j x ¡ j ½2x ¡ 2, x > 06x<2 y = 2, ¡ 2x, x < h x ] ¡1, ¡ ] or [ 2, [ i x 6= ¡2 j x ] ¡1, ¡1 ] or [ 32 , [ k no solutions x [ ¡1, ] or [ 1, [ q x ] ¡2, 11 y = | x|+| x -2| x -1 x i y = j x2 + j y = x2 + for all x j y = j x2 ¡ j n y = x2 ¡ 1, x > 1, x ¡1 ¡ x2 , ¡1 < x < y [ or ] 2, [ y cyan y = | x2 - | y = | x2 + | [ or [ 1, ] magenta yellow 95 100 50 75 25 95 100 50 75 25 x 95 95 50 75 25 r x ] ¡1, y = | x | - | x -1| x ] ¡3, [ or ] 2, [ l x [ ¡5, ¡2 [ or [ 0, [ x ] ¡1, ¡2 [ or ] ¡1, [ or ] 2, [ x ] ¡2, [ or ] 2, [ o x [ ¡1, [ or [ 1, [ 100 k m n p 50 h 75 e 25 c q a y m x ] ¡1, ¡ 43 [ or ] 4, [ n x 6= p x ] ¡1, ¡ 16 [ or ] 1, [ x ] ¡1, ¡ ] or [ 23 , [ r x ] ¡1, 32 [ or ] 3, [ x ] ¡1, ¡4 [ or ] 12 , [ b x ] ¡1, ¡1 [ or ] 4, [ x ] ¡1, ¡3 ] or ] ¡ 32 , [ d x ] ¡1, ¡3 ] or ] 3, [ x ] ¡1, 14 ] or ] 1, [ f x ] ¡1, 12 [ g x ] 0, 100 [ x ] , ] i x ] ¡1, ¡1 [ or ] ¡ , [ j x ] ¡7, 52 [ o y =j x j ¡ j x ¡ j ½1, x>1 y = 2x ¡ 1, x < ¡1, x<0 y x ] ¡ 32 , 13 [ x [ 13 , 12 ] 100 l h black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_an\858IB_HL-2_AN.CDR Friday, 12 December 2008 12:49:02 PM TROY -1 x IB_PD (859) ANSWERS y = j x2 ¡ 2x j n x ¡ 2x, x > 2, x y = 2x ¡ x2 , 0<x<2 k b ii Anywhere between O and Q, length of cable is 10 km iii At O, minimum length of cable is 17 km a True b True y EXERCISE 1G y =| x - x | y g (x ) = x f (x ) = y g (x) = - 2x x h (x ) = y = j x2 + 3x + j n x + 3x + 2, x > ¡1, x ¡2 y = ¡x2 ¡ 3x ¡ 2, ¡2 < x < ¡1 l x f (x) = - 1x x x x y y =| x + x + | h(x) = - 4x x -2 -1 EXERCISE 1H a i vertical asymptote x = 2, horizontal asymptote y = ii as x ! 2¡ , y ! ¡1 as x ! 1, y ! 0+ EXERCISE 1F.2 as x ! 2+ , y ! 1 a x = §3 b no solution c x = d x = or ¡2 e x = ¡1 or f no solution g x = or 13 h x = or i x = ¡2 or 14 a x= a x= e x= a x= 2 or b x = ¡2 or ¡ 47 c b x = 52 c x = ¡ 14 or § f x = ¡6 or 25 b x = ¡ 45 or c x = iii as x ! ¡1, y ! 0¡ y f (x ) = x = ¡1 or d x = ¡6 or ¡ 43 or x x¡=¡2 x2] , [ b x ] ¡1, ¡3 ] or [ 3, [ x [ ¡4, ¡2 ] d x ] ¡1, ¡6 ] or [ ¡2, [ x ] ¡1, [ f x ] ¡1, 14 [ or ] 54 , [ x ] ¡ 52 , x ] ¡1, iv Function does not cross its asymptotes b i vertical asymptote x = ¡1, horizontal asymptote y = ii as x ! ¡1¡ , y ! as x ! 1, y ! 2¡ as x ! ¡1+ , y ! ¡1 [ h x [ ¡1, ] i x ] ¡ 17 , [ ¡ 37 , ] or [ 1, [ k x [ x [ ¡1, ] b x [ ¡1, ] x ] ¡1, ¡1 [ or x ] , as x ! ¡1, y ! 2+ iii ] l x [ ¡2, ] f (x ) = - 1[ y x +1 d x ] ¡1, ¡1] or x [6, [ e x [1, [ f x [¡1, 15 ] g x [ 32 , [ or x ] 2, 3] y¡=¡2 x h x [¡ 14 , [ or x ] 1, [ x¡=¡-1 a x ] 1, [ b x ] ¡1, [ c x ] ¡1, ¡1 [ or x ] 2, [ d x [3, [ iv Function does not cross its asymptotes c i vertical asymptotes x = ¡1, x = 2, y ¦(x) = horizontal asymptote y = x x-2 ii as x ! ¡1¡ , y ! as x ! 1, y ! 0+ as x ! ¡1, y ! 0¡ as x ! ¡1+ , y ! ¡1 as x ! 2¡ , y ! ¡1 as x ! 2+ , y ! y =1 x y = -1 iii y f (x ) = x=2 x [¡2, x-2 y¡=¡0 EXERCISE 1F.3 a c e g j a c ] x+3 (x + 1)(x - 2) or x ] 2, [ y¡=¡0 a x y y = -4 x - y = 4x + 16 y = x + 10 x¡=-1 y = 10 cyan magenta yellow 95 100 25 95 100 50 75 25 as x ! 3+ , y ! 95 100 50 x 75 25 -2 95 100 50 75 25 -5 x¡=¡2 iv Function crosses the horizontal asymptote at (¡3, 0) d i vertical asymptote x = 3, oblique asymptote y = x ii as x ! 3¡ , y ! ¡1 as x ! 1, y ! x+ 50 10 75 y = -2 x + 859 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\859IB_HL-2_AN.CDR Friday, 12 December 2008 12:50:53 PM TROY as x ! ¡1, y ! x¡ IB_PD (860) 860 ANSWERS iii y i i vert asymptotes x = ¡1, x = 5, horiz asymptote y = as x ! 1, y ! 0+ ii as x ! ¡1¡ , y ! ¡1 y¡=¡x as x ! ¡1+ , y ! as x ! 5¡ , y ! ¡1 as x ! 5+ , y ! f (x ) = x + x-3 iii x x=-1 as x ! ¡1, y ! 0¡ y y=0 x¡=¡3 x iv Function does not cross its asymptotes e i horizontal asymptote y = ii as x ! 1, y ! 1¡ as x ! ¡1, y ! 1¡ iii x=5 y iv Function crosses the horizontal asymptote at (0, 0) j i vertical asymptote x = 0, parabolic asymptote y = x2 ii as x ! 0¡ , y ! as x ! 1, y ! (x2 )¡ y¡=¡1 as x ! 0+ , y ! ¡1 x x2 - y= x +4 as x ! ¡1, y ! (x2 )+ iii y y=xX y = x2 - iv Function does not cross its asymptote f i vert asymptotes x = ¡2, x = 2, horiz asymptote y = as x ! 1, y ! 2+ ii as x ! ¡2¡ , y ! as x ! ¡2+ , y ! ¡1 as x ! 2¡ , y ! ¡1 as x ! 2+ , y ! x x as x ! ¡1, y ! 2+ x=0 iii y iv Function does not cross its asymptotes x2 + y= x -4 EXERCISE 1I a i b i y¡=¡¡2 y y x Qw_ x¡=¡-2 -2 -\Qe_ x¡=¡2 ƒ -1 as x ! ¡1+ , y ! ¡1 iii f (x ) = x + + x-2 x +1 a ƒ -1 f ¡1 (x) = x¡1 f ¡1 (x) = x¡5 i ii, iii f ¡1 (x) = 4x ¡ b f ¡1 (x) = ¡2x + i ii ii x ƒ y y ƒ -1 y=x y=x ƒ -1 Ew_ x¡=¡-1 -\Tw_ iv Function does not cross its asymptotes h i oblique asymptote y = 2x ii as x ! 1, y ! (2x)¡ as x ! ¡1, y ! Qw_ y=x ƒ ii, iii y -2 -\Qe_ y=x as x ! ¡1, y ! (x + 3)+ x ƒ x iv Function does not cross its asymptotes g i vertical asymptote x = ¡1, oblique asymptote y = x + ii as x ! ¡1¡ , y ! as x ! 1, y ! (x + 3)¡ y¡=¡x¡+¡3 4x x2 - 4x - y= x x Ew_ Er_ Er_ -\Tw_ ƒ (2x)+ iii c i y 2x g (x ) = x - x +1 y=2x f ¡1 (x) =x¡3 ii y ƒ y=x ƒ -1 x -3 x -3 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 iv Function crosses the oblique asymptote at (0, 0) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\860IB_HL-2_AN.CDR Thursday, 24 January 2008 11:30:43 AM PETERDELL IB_PD (861) ANSWERS y a b y=x y ƒ ƒ -2 -2 c 4 ƒ -1 x that they both equal x p a f ¡1 (x) = x ¡ ¡ 1, x > b c i y ƒ -1 ƒ -1 y c Hint: Inverse is x = y2 ¡ 4y + for y > d i Domain is fx j x > 2g, Range is fy j y > ¡1g ii Domain is fx j x > ¡1g, Range is fy j y > 2g e Hint: Find (f ± f ¡1 )(x) and (f ¡1 ± f )(x) and show y=x x ƒ d (-1, 3) 1 -3 -3 y=x e y=x f y x+3 x+3 and (g ± f )¡1 (x) = 8 11 a Is not b Is c Is d Is e Is 13 a B is (f (x), x) b x = f ¡1 (f (x)) = (f ¡1 ± f )(x) ƒ=ƒ -1 x a a b ¡15 c ¡ 54 a i Range = fy j y > ¡5g, Domain = fx j x is in R g ii x-int ¡1, 5; y-int ¡ 25 iii is a function iv no b i Range = fy j y = or ¡3g Domain = fx j x is in R g ii no x-intercepts; y-intercept iii is a function iv no a = 1, b = ¡6, c = a b + - + y ƒ ƒ x -4 x ƒ -1 inverse -4 REVIEW SET 1A y=x y c Start with B first and repeat the process used in a and b x ƒ=ƒ -1 y=x y=x -\We_ p No c Yes, it is y = x + b a a f(2, 1), (4, 2), (5, 3)g b not invertible c f(0, ¡1), (1, 2), (2, 0), (3, 1)g d f(¡1, ¡1), (0, 0), (1, 1)g -2 b f ¡1 (x) = y x+7 3\Qw_ , x 6= satisfies both the vertical and x horizontal line tests and ) has an inverse function 1 and f (x) = i.e., f = f ¡1 b f ¡1 (x) = x x ) f is a self-inverse function -7 a f : x 7! ) a 2x2 + b 4x2 ¡ 12x + 11 a x ] ¡1, ¡5 ] or ] ¡2, ] b x ] ¡8, ¡ 12 [ or ] 1, [ EXERCISE 1J a y= Domain fx j x > 3g Range fy j y > ¡1g 10 (f ¡1 ± g¡1 )(x) = y (2,¡2) Domain fx j x > ¡1g Range fy j y > 3g a 10 b x = a i 25 ii 16 b x = y=x (3,-1) x x ii ƒ -1 x ƒ -1 ƒ y=x y ƒ 3\Qw_ ƒ -1 x -7 ƒ a x = or b x ] ¡1, ¡ 15 ] or [ 5, [ a vertical asymptotes x = ¡3, x = 3x ¡ is symmetrical about y = x, x¡3 f is a self-inverse function horizontal asymptote y = b as x ! ¡3¡ , y ! ¡1 as x ! ¡3+ , y ! as x ! 2¡ , y ! ¡1 as x ! 2+ , y ! 3x ¡ 3x ¡ and f (x) = x¡3 x¡3 i.e., f = f ¡1 ) f is a self-inverse function b f ¡1 (x) = c x=-3 b i is the only one c ii Domain fx j x 1g iii Domain fx j x > 1g p a f ¡1 (x) = ¡ x b y y=x as x ! 1, y ! 0+ as x ! ¡1, y ! 0¡ y f (x ) = y=0 4x + x2 + x - x ƒ x=2 x 10 (f ¡1 ± h¡1 )(x) = x ¡ and (h ± f )¡1 (x) = x ¡ ƒ -1 cyan magenta yellow REVIEW SET 1B a x2 ¡ x ¡ b x4 ¡ 7x2 + 10 7¡x 5x ¡ b f ¡1 (x) = a Domain fx j x R g, Range fy j y > ¡4g b Domain fx j x 6= 0, 2g, Range fy j y ¡1 or y > 0g 95 100 50 75 25 a f ¡1 (x) = 95 100 50 95 50 25 (2, 1) 95 100 50 75 25 b 75 x 75 25 ƒ x=2 A horizontal line above the vertex cuts the graph twice So, it does not have an inverse For x > 2, all horizontal lines cut or once only, ) has an inverse y 100 a 861 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_an\861IB_HL-2_AN.CDR Thursday, 11 March 2010 11:30:50 AM PETER IB_PD (862) 862 ANSWERS a y b y y=x ƒ -1 c y=x ƒ y 2 ƒ - + a - + -2 [ ¡ 52 , a x2 a f (x) = p - b y=x-2 x=1 + -5 + -3 10 a, d ] b x ] ¡2, ¡1 [ or ] 4, [ x, g(x) = ¡ x2 b g(x) = x2 , f (x) = c f (x ) = + a 4, 13, 22, 31, b 45, 39, 33, 27, c 2, 6, 18, 54, d 96, 48, 24, 12, a Starts at and each term is more than the previous term b c d e f g h i a b c d Function crosses horizontal asymptote at ( 23 , 3) p 10 a h¡1 (x) = + x ¡ REVIEW SET 1C a 10 ¡ 6x b x = 2 a Domain fx j x > ¡3g, Range fy j ¡3 < y < 5g b Domain fx j x 6= 1g, Range fy j y ¡3, y > 5g p p a i ¡ x ii ¡ 2x b For f ± g, Domain fx j x > 0g, Range fy j y 1g Starts at 243, each term is Starts at 50 000, each term is of the previous term; 3, 1: of the previous term; 80, 16: Each term is the square of the term number; 25, 36, 49: Each term is the cube of the term number; 125, 216, 343: Each term is n(n + 1) where n is the term number; 30, 42, 56: a 2, 4, 6, 8, 10 b 4, 6, 8, 10, 12 c 1, 3, 5, 7, d ¡1, 1, 3, 5, e 5, 7, 9, 11, 13 f 13, 15, 17, 19, 21 g 4, 7, 10, 13, 16 h 1, 5, 9, 13, 17 a 2, 4, 8, 16, 32 b 6, 12, 24, 48, 96 c 3, 12 , 34 , 38 , 16 d ¡2, 4, ¡8, 16, ¡32 17, 11, 23, ¡1, 47 or ] 1, [ y y= Range = fy j y > 0g Next two terms 40, 48: Starts at 2, each term is more than the previous term; 14, 17: Starts at 36, each term is less than the previous term; 16, 11: Starts at 96, each term is less than the previous term; 68, 61: Starts at 1, each term is times the previous term; 256, 1024: Starts at 2, each term is times the previous term; 162, 486: Starts at 480, each term is half the previous term; 30, 15: EXERCISE 2B Range fy j y > 0g c Domain = fx j x 6= 0g, c EXERCISE 2A 3x - x2 - x=2 a x ] ¡1, [ b x ] ¡2, a x=0 b g A -1 If x ¡3, we have the graph to the left of x = ¡3 and any horizontal line cuts it at most once p y = ¡3 ¡ x + Range of g fy j y > ¡2g, Domain of g ¡1 fx j x > ¡2g Range of g ¡1 fy j y ¡3g e x g, ¡ 54 ] (-3,-2) y=3 For g ± f , Domain fx j x b y=x x as x ! ¡1, y ! 3¡ y x=-2 y x=-3 g x¡2 x+1 a x = ¡1 or b x ] ¡1, ¡ 13 ] or [ 5, [ a vertical asymptotes x = §2, horizontal asymptote y = as x ! 1, y ! 3+ b as x ! ¡2¡ , y ! ¡1 as x ! ¡2+ , y ! as x ! 2¡ , y ! ¡1 as x ! 2+ , y ! b (x - 2)2 x¡2 ¡ 4x f ¡1 (x) = f ¡1 (x) = x ƒ -1 x f (x ) = x - + x a x2 EXERCISE 2C a d a b b a a d x a x ] ¡1, [ or ] 9, [ b y= y x | x | +1 y=\Qe_ (Qw_ ' Qe_) x u1 = 6, d = 11 b yes, u30 e no u1 = 87, d = ¡4, u1 = 1, d = c u1 = 32, d = ¡ 72 k = 17 12 b k = c k = 3, k = ¡1 11 c un = ¡5n + 36 1[ a b = 18, c = 54 b b = 12 , c = 14 c b = 3, c = ¡1 12 a u1 = 5, r = b un = £ 2n¡1 , u15 = 81 920 3 a u1 = 12, r = ¡ 12 b un = 12 £ (¡ 12 )n¡1 , u13 = 1024 a vertical asymptote x = 1, oblique asymptote y = x ¡ b as x ! 1¡ , y ! as x ! 1, y ! (x ¡ 2)+ cyan u1 = 8, r = ¡ 34 , u10 = ¡0:600 677 49 magenta yellow 95 p1 , n un = 2 ¡ 100 50 75 95 u1 = 8, r = 100 50 75 25 95 100 50 75 as x ! ¡1, y ! (x ¡ 2)+ 25 95 100 50 75 25 as x ! 1+ , y ! b 100 100 006 EXERCISE 2D.1 b x ] ¡1, ¡2 [ or ] 1, [ - + - + -2 b 57 , 37 , 13 17 , 17 67 , 22 47 , 27 27 a u1 = 36, d = ¡ 23 25 a for x [ , ) b un = 91 ¡ 4n c ¡69 d no 169 d u151 = 451 c ¡227 d n > 68 un = 6n ¡ b un = ¡ 32 n + un = ¡ 32 n + 12 a 14 , 12 , 34 x > j x j +1 un = 11n ¡ c 545 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\862IB_HL-2_AN.CDR Friday, 12 December 2008 12:52:46 PM TROY IB_PD (863) ANSWERS a k = §14 b k = c k = ¡2 or a un = £ 2n¡1 b un = 32 £ (¡ 12 )n¡1 p a 420 b 2231:868 211 a n = 37 b n = 11 x = 12 a u1 = 7, u2 = 10, b 64 10 a A3 = $8000(1:03)3 ¡ (1:03)2 R ¡ 1:03R ¡ R b A8 = $8000(1:03)8 ¡ (1:03)7 R ¡ (1:03)6 R ¡ (1:03)5 R p c un = £ (§ 2)n¡1 d un = 10 £ (§ 2)1¡n p = 2916 ¼ 5050:66 a u9 = 13 122 b u14 c u18 ¼ 0:000 091 55 ¡(1:03)4 R ¡ (1:03)3 R ¡ (1:03)2 R ¡ (1:03)R ¡ R =0 R = $1139:65 EXERCISE 2D.2 REVIEW SET 2A a $3993:00 b $993:00 E 11 470:39 a 43 923 Yen b 13 923 Yen $23 602:32 148 024:43 Yen $ 51 249:06 $14 976:01 $ 11 477:02 E 19 712:33 10 19 522:47 Yen a i 1550 ants ii 4820 ants b 12:2 weeks a 278 animals b Year 2044 EXERCISE 2E.1 un = EXERCISE 2D.3 a i Sn = + 11 + 19 + 27 + :::: + (8n ¡ 5) ii 95 b i Sn = 42 + 37 + 32 + :::: + (47 ¡ 5n) ii 160 c i Sn = 12 + + + 12 + :::: + 12( 12 )n¡1 ii 23 14 d i Sn = + + e i Sn = + + 12 + + 34 + :::: + + :::: + 2( 32 )n¡1 ii 2n¡1 a 10 b 25 c 168 d 310 For u1 = 54, 13 14 (3n ¡ 1) = 610 k=1 n(n2 +6n+11) a 81 b ¡1 12 a 11 x = 12 d as n ! 1, Sn ! 1 a b c a b c during the 86th month its value is $500 54 or 23 70 cm The 20th terms are: arithmetic 39, geometric 319 magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 or arithmetic 13 , geometric ( 43 )19 50 + (7k ¡ 3) b 64 1875 + n P + + ( 12 )k+1 k=1 10 + 11 10 a 1587 b 47 253 ¼ 47:99 256 ¡ ¢ x = ¡ 67 gives a divergent series 14 S = 2¡2 n+1 n+1 ¡ 21 = 2, 22 = 4, 23 = 8, 24 = 16, 25 = 32, 26 = 64 31 = 3, 32 = 9, 33 = 27, 34 = 81, 35 = 243, 36 = 729 41 = 4, 42 = 16, 43 = 64, 44 = 256, 45 = 1024, 46 = 4096 51 = 5, 52 = 25, 53 = 125, 54 = 625 61 = 6, 62 = 36, 63 = 216, 64 = 1296 71 = 7, 72 = 49, 73 = 343, 74 = 2401 95 34th week (total sold = 2057) After 85 months its value is $501:88 and after 86 months its 75 EXERCISE 3B a ¡1 b c d ¡1 e f ¡1 g ¡1 h ¡32 i ¡32 j ¡64 k 625 l -625 a 16 784 b 2401 c ¡3125 d ¡3125 e 262 144 f 262 144 g ¡262 144 h 902:436 039 i ¡902:436 039 j ¡902:436 039 a 0:1 b 0:1 c 0:027 d 0:027 e 0:012 345 679 f 0:012 345 679 g h 50 ii r = 0:1 b S1 = EXERCISE 2F 25 + EXERCISE 3A b Sn = + 18(1 ¡ (0:9)n¡1 ) c 19 seconds 5 c un = 100(0:9)n¡1 a $18 726:65 b $18 885:74 $13 972:28 a 3470 b Year 2014 18 metres a < x < b 35 57 10 a un = 3n + 2 a 49 b 16 c 104 a 54 b 14:175 a b 27 99 333 a convergent, sum = 12 b not convergent, n = 10 u1 = 9, r = 23 u1 = 8, r = 15 and u1 = 2, r = 45 cyan c ¡486 21, 19, 17, 15, 13, 11 un = ( 34 )2n¡1 a 49 152 b 24 575:25 12 u11 = 19 8683 ¼ 0:000 406 a 17 b 255 511 ¼ 256:0 512 EXERCISE 2E.4 value is $491:84, ) S = 162 REVIEW SET 2C a 3069 b 4095 ¼ 3:999 c ¡134 217 732 1024 c $26 361:59 2n ¡ a 12 , 34 , 78 , 15 , 31 b Sn = 16 32 n 2n ¡ 2n , a 70 b 241:2 u12 = 10 240 a E 8415:31 b E 8488:67 c E 8505:75 p 10 a 10 45 b 16 + p ¢ 3+ ¡ p n a Sn = ( 3) ¡ b Sn = 24(1 ¡ ( 12 )n ) c Sn = ¡ (0:1)n d Sn = 40 (1 ¡ (¡ 12 )n ) 3 10 + n P k=1 a 23:9766 ¼ 24:0 b ¼ 189 134 c ¼ 4:000 d ¼ 0:5852 a i u1 = b 2, 5, 8, 11, 14 or 14, 11, 8, 5, c ¡ ( 12 )n = r= 2n + n+3 a + + + 16 + 25 + 36 + 49 a 820 b 3087:5 c ¡1460 d ¡740 a 1749 b 2115 c 1410 12 a 160 b ¡630 c 135 203 ¡115:5 18 a 65 b 1914 c 47 850 a 14 025 b 71 071 c 3367 a un = 2n ¡ c S1 = 1, S2 = 4, S3 = 9, S4 = 16 56, 49 12 10, 4, ¡2 or ¡2, 4, 10 EXERCISE 2E.3 and u1 = 150, r = ¡ 25 a un = 89 ¡ 3n b un = EXERCISE 2E.2 10 11 13 x > ¡ 12 REVIEW SET 2B n=1 (k + 1) (k + 2) = 11 For u1 = 150, r = ¡ 25 , S = 107 17 a=b=c x = 3, y = ¡1, z = 13 or x = 13 , y = ¡1, z = 75 b 15 ¼ 0:634 b b j r j < in both cases 15 16 25 n P 20 P or ¡ 16 £ (¡2)n¡1 £ 2n¡1 12 a u1 = 54, r = 26 38 ii 1331 2100 10 a f i Sn = + + 27 + 64 + :::: + n3 ii 225 a 13 , 1, 3, b 54 , 85 , 11 , c 5, ¡5, 35, ¡65 b u1 = 63, d = ¡5 c ¡117 d u54 = ¡202 a u1 = 3, r = b un = £ 4n¡1 , u9 = 196 608 k = ¡ 11 un = 73 ¡ 6n, u34 = ¡131 b u1 = 6, r = 12 c 0:000 183 p un = 33 ¡ 5n, Sn = n (61 ¡ 5n) k = § 3 2 100 863 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\863IB_HL-2_AN.CDR Friday, 12 December 2008 1:11:22 PM TROY IB_PD (864) 864 ANSWERS Yes, n = 2s¡1 239 5:5 £ 1011 ¼ 264 ¡ 25 2¡5 ¼ f k p a 1:8 £ 1019 EXERCISE 3C.1 a h a h a g a g a 22 2¡2 b 23 2¡3 c d e 2¡1 i 26 j 2¡6 k 27 l 2¡7 32 b 3¡2 c 33 d 3¡3 e 31 3¡4 i 30 j 35 k 3¡5 2a+1 b 2b+2 c 2t+3 d 22x+2 22m h 2n+1 i 21 j 23x¡1 3p+2 b 33a c 32n+1 d 3d+3 31¡y h 32¡3t i 33a¡1 j 33 a b2 a2 b2 b b3 g a3 h i a2 4a2 b2 c a a¡n b bn 9b2 a4 d f g g x = ¡ 13 h x = 53 i x = 14 j x = 72 x = ¡2 l x = ¡4 m x = n x = 52 o x = ¡2 x = ¡6 x= a EXERCISE 3F e 2n¡1 f 2c¡2 a 1:4 b 1:7 c 2:8 d 0:3 e 2:7 f 0:4 a b y y y = 2- x ! d 33t+2 f 3y¡1 @\=\2 ! @\=\2 ! a2 bc2 a2 c2 b f 1 -1 c e a¡2n¡2 i ¡4 c d y 1 d ¡1 e f g ! !-2 c 34 d 32 x e 3¡ 2 f 7¡ e 77 g 3¡ 4 a b y 27 g h ! 16 c x x !-1 e¡2x @\=\3 l 25 ¡ 10(2¡x ) + 4¡x Qe_ x2 a en (1 e2n ) @\=\2 x c d y y -! @\=\3+2 @\=\3 x @\=\3 x -! @\=\3-2 a y ¼ 3:67 b y ¼ ¡0:665 c y ¼ 3:38 d y ¼ 2:62 a as x ! 1, y ! as x ! ¡1, y ! (above) HA is y = b as x ! 1, y ! ¡1 as x ! ¡1, y ! (below) EXERCISE 3E as x ! ¡1, y ! HA is y = d as x ! 1, y ! (below) 95 100 50 75 25 as x ! ¡1, y ! ¡1 95 50 75 25 95 100 50 75 25 100 100 yellow HA is y = c as x ! 1, y ! (above) a x = b x = c x = d x = e x = ¡1 f x = ¡1 g x = ¡3 h x = i x = j x = ¡4 k x = l x = a x = 12 b x = ¡ 23 c x = ¡ 12 d x = ¡ 12 e x = ¡1 12 magenta x @\=\2-2 @\=\1 a n 2n+1 b ¡3n¡1 a x = or b x = c x = or d x=1 e x=2 f x=0 95 @\=\2 +1 5(5n (2x + 3)(2x + 6) b (2x + 4)(2x ¡ 5) c (3x + 2)(3x + 7) (3x + 5)(3x ¡ 1) e (5x + 2)(5x ¡ 1) f (7x ¡ 4)(7x ¡ 3) a 2n b 10a c 3b d n e 5x f ( 34 )a g h 5n 5 a 3m + b + 6n c 4n + 2n d 6n e 5n f g 12 h 12 i 12 50 y ! 10(3n ) (3x + 2)(3x ¡ 2) b (2x + 5)(2x ¡ 5) c (4 + 3x )(4 ¡ 3x ) (5 + 2x )(5 ¡ 2x ) e (3x + 2x )(3x ¡ 2x ) f (2x + 3)2 (3x + 5)2 h (2x ¡ 7)2 i (5x ¡ 2)2 75 b y + 1) b c + d ¡ 1) 6(6n+1 ¡1) f 16(4n ¡1) g 5(2n ) h 7(2n ) i 16(3n¡1 ) 25 x ! @\=\-3 EXERCISE 3D.2 -1 ! cyan ! @\=\3 1 g x ¡ h 4x ¡ i x ¡ x¡1 j x2 + + y ! @\=\3 a 4x + 22+x + b 9x + 7(3x ) + 10 c 25x ¡ 6(5x ) + d 4x + 6(2x ) + e 9x ¡ 2(3x ) + f 16x + 14(4x ) + 49 a e a d g a d @\=\1 x d y i x2 + x2 + 5x (5x ! @\=\3 +1 @\=\3 a x5 + 2x4 + x2 b 22x + 2x c x + d e2x + 2ex e 2(3x ) ¡ f x2 + 2x + g + 5(2¡x ) h 5x + ¡2+ y @\=\3 y = 3- x x ! EXERCISE 3D.1 k @\=\2 Qr_ a 2:28 b 1:83 c 0:794 d 0:435 a b 1:68 c 1:93 d 0:523 a b 32 c d 125 e f 1 i 81 j 25 e2x @\=\2´2 @\=\2 ¡3 a 73 b 34 c 25 d 23 h 2¡ i 2¡ j 7¡ y j b 3¡ a 33 x @\=\-2 @\=\2 b @\=\2 -2 x ! ¡1 h 22 b has no solutions c x = 12 f 3¡1 g 34 EXERCISE 3C.2 a x = b x = c x = d x = e x = ¡2 f x = ¡2 e j 12am3 ad2 c 3n¡2 d an bm 21 x = ¡ 12 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\864IB_HL-2_AN.CDR Friday, 12 December 2008 1:14:25 PM TROY HA is y = IB_PD ! (865) ANSWERS a ¼ 7:39 b ¼ 20:1 c ¼ 2:01 d ¼ 1:65 e ¼ 0:368 EXERCISE 3G.1 a b c 100 grams i 132 g ii 200 g iii 528 g Wt (grams) a e2 b a e0:18t a 10:074 e 41:914 10 (24' 528) Wt\=\100´2 0.1t (10' 200) 100 a b c 50 865 ¡1 d e¡2 c e e0:004t b c e¡0:005t d ¼ e¡0:167t b 0:099 261 c 125:09 d 0:007 994 f 42:429 g 3540:3 h 0:006 342 y = ex y t (hours) (4' 132"0) Pn i 76 ii 141 iii 400 e2 y = ex-2 y = ex + (10' 400) y = Pn\=\50´2 0.3n x (5' 141) Domain of f , g and h is fx j x R g Range of f is fy j y > 0g Range of g is fy j y > 0g Range of h is fy j y > 3g y = ex 11 y y = 10 n (years) (2' 76) 50 a V0 b 2V0 c 100% d 183% increase, percentage increase at 50o C compared with 20o C a 12 bears b 146 bears c 248% increase EXERCISE 3G.2 a 250 g b i 112 g ii 50:4 g iii 22:6 g c d ¼ 346 years 250 200 150 100 50 W (t ) = 250 ´ (0.998)t (400, 112) i ii iii c t Tt (°C) 81:2o C 12 a i g ii 2:57 g iii 4:23 g iv 40:2 g b W Tt\=\100´2 -0.02t 100 75:8o C 33:9o C x Domain of f , g and h is fx j x R g Range of f is fy j y > 0g Range of g is fy j y < 0g Range of h is fy j y < 10g (800, 50.4) (1200, 22.6) 100o C y = 10 - e y = -e x 1000 a b x W(t) t (20' 75"8) W (t) = 2e 2 (78' 33"9) (15' 81"23) t t (min) 13 a b 1000 g c 1000 i 812 g ii 125 g iii 9:31 £ 10¡7 g Wt (grams) a i ii b (10' 812) f ¡1 (x) = loge x ƒ y y¡=¡x I 75 Wt\=\1000´2 -0.03t 14 a b 64:6 amps 16:7 amps I (t) = 75e-0.15t ƒ -1 I¡=¡1 (100' 125) t (years) t a W0 b 12:9% c EXERCISE 3H e1 ¼ 2:718 281 828 :::: a ¡1 b 27 c y = 3x x The graph of y = lies between y = One is the other 2x and y = a x = ¡2 b x = 3x reflected in the y-axis y y = ex yellow 95 100 50 75 25 -3 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a a ex > for all x b i 0:000 000 004 12 ii 970 000 000 magenta b3 a b y=2 x a x y = 2x -4 x cyan y2 c 3x a b c ab b a a6 b7 b a 2:28 b 0:517 c 3:16 a b 24 c y y = e -x 3 a 2¡3 b 27 c 212 a y = 2x ex 28:8 sec REVIEW SET 3A y = ex y x black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_an\865IB_HL-2_AN.CDR Monday, 12 January 2009 4:36:35 PM TROY y = -4 b 2x y= has yintercept and horizontal asymptote y = y = 2x ¡ has yintercept ¡3 and horizontal asymptote y = ¡4 IB_PD (866) ANSWERS 80o C c 1 a 104 = 10 000 b 10¡1 = 0:1 c 10 = p d 23 = e 2¡2 = 14 f 31:5 = 27 T = 80´ ( 0.913) t ¼ 12:8 10 20 a 2x+2 a b c a x = b x = ¡ 25 b 22x¡3 c a x5 a2 b2 b 2a b2 c a ¼ 2:18 b ¼ 1:40 c ¼ 1:87 d ¼ ¡0:0969 a x = b x = c x = d x = 14 a Let loga an = x ) ax = an ) x = n 5¡9x a 34 b 30 c 3¡3 d 3¡5 a 33¡2a b x ¡2 ¡1 y ¡4 89 ¡4 23 ¡4 ¡2 b i ii iii ¡1 iv y =3 -5 a x= 10 b x = ¡ 45 log x c log y¡=¡-5 d log 10 ^ ( e log 100 ^ ( f log 10 g log h log ¦(x)¡=¡ex y g(x)¡=¡ex-1 x a f a a h(x)¡=¡3-ex Domain of f , g and h is fx j x R g Range of f is fy j y > 0g Range of g is fy j y > 0g Range of h is fy j y < 3g a 2n+2 b ¡ 67 c 38 d ¡1 a 50 b a2 b4 4a6 c ¡ a g l a f b3 27 400 x ¡2 ¡1 y 15:8 6:44 1:74 1:27 y¡=¡1 800 ) ) ENTER , 0:¹6 ) ) ENTER , 1:5 10 × 2nd ) ENTER , ¡0:5 ÷ 2nd 10 ) ÷ 10 ^ 0:25 ) ENTER , ¡0:25 100:7782 b 101:7782 c 103:7782 d 10¡0:2218 e 10¡2:2218 101:1761 g 103:1761 h 100:1761 i 10¡0:8239 j 10¡3:8239 i 0:477 ii 2:477 b log 300 = log(3 £ 102 ) i 0:699 ii ¡1:301 b log 0:05 = log(5 £ 10¡2 ) cyan magenta a c e g i k e x = 10 g x ¼ 6:84 h x ¼ 0:000 631 log 16 b log c log d log 20 e log f log 24 log 30 h log 0:4 i log 10 j log 200 k log 0:4 log or m log 0:005 n log 20 o log 28 log 96 b log 72 c log d log log ¡ 25 ¢ e log g log 20 h log 25 i log y = x log b log y ¼ 1:301 + log b log M = log a + log d d log T ¼ 0:6990 + 12 log d log R = log b + 12 log l f log Q = log a ¡ n log b log y = log a + x log b h log F ¼ 1:301 ¡ 12 log n log L = log a + log b ¡ log c j log N = 12 log a ¡ 12 log b log S ¼ 2:301 + 0:301t l log y = m log a ¡ n log b p a D = 2e b F = c P = x d M = b2 c t m3 100 e B= f N= p g P = 10x3 h Q = p n x yellow 95 100 50 75 25 95 100 50 75 25 10 EXERCISE 4C.2 y¡=¡2e-x¡+¡1 95 100 50 75 ÷3 ÷ ) ENTER , 0:5 ) ) ENTER , 0:¹3 a b 32 c d 12 e ¡2 f ¡ 32 a p+q b 2p+3q c 2q+r d r+ 12 q¡p e r¡5p f p¡2q a x + z b z + 2y c x + z ¡ y d 2x + 12 y e 3y ¡ 12 z f 2z + 12 y ¡ 3x a 0:86 b 2:15 c 1:075 x 25 ) EXERCISE 4C.1 d 52a+6 c 10 2nd f x = 10 a 288 = 25 £ 32 b 22x b as x ! 1, y ! (above); as x ! ¡1, y ! c d y=1 y 10 000 ¡1 a + e2x b 22x + 10(2x ) + 25 c x ¡ 49 a ¡ 6(2a ) + 22a b x ¡ c 2x + a x = b x = ¡4 a 1500 g c W (grams) 1500 b i 90:3 g ii 5:4 g W = 1500 ´ (0.993)t d 386 years t (years) 64 10 a vii ¡ 12 viii ¡ 34 a x = 100 b x = 10 c x = d x = REVIEW SET 3C ENTER , b log 0:001 ENTER , ¡3 a a ¡4 b v ¡ 12 vi a b ¡3 c d e 12 f 13 g ¡ 14 h 12 i 23 j 12 k 13 l 12 m n n a + o ¡ m p a ¡ b b as x ! 1, y ! 1; as x ! ¡1, y ! ¡5 (above) c d y = ¡5 y x -4 EXERCISE 4B 95 a 100 2¡4 a b ¡2 c 12 d e f g h i ¡3 j 12 k l 12 m n 13 o n p 13 q ¡1 r 32 s t 4b a3 p4 q 50 a b a21 10 d log7 49 = e log2 64 = f log3 ( 27 ) = ¡3 REVIEW SET 3B ¡ 45 p a log2 = b log2 ( 18 ) = ¡3 c log10 (0:01) = ¡2 30 40 t (minutes) 75 d EXERCISE 4A T (°C) 80 60 40 20 26:8o C 9:00o C 3:02o C i ii iii 25 10 a b 866 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\866IB_HL-2_AN.CDR Thursday, 24 January 2008 11:40:10 AM PETERDELL IB_PD (867) ANSWERS p EXERCISE 4D x ] 2, [, y2R ii VA is x = 2, x-intercept 7, no y-intercept a b c d ¡2 x does not exist such that ex = ¡2 or a a b a + c a + b d ab e a ¡ b a f a e a g e1:7918 e4:0943 e8:6995 e¡0:5108 iv v e¡5:1160 b c d e g e7:3132 h e0:4055 i e¡1:8971 j e¡8:8049 x ¼ 20:1 b x ¼ 2:72 c x = d x ¼ 0:368 x ¼ 0:006 74 f x ¼ 2:30 g x ¼ 8:54 h x ¼ 0:0370 ln 45 b ln c ln d ln 24 e ln = f ln 30 ¡ ¢ ¡ ¢ ln 4e h ln 6e i ln 20 j ln 4e2 k ln e202 l ln = e2:7081 e.g., for a, ln 27 = ln 33 f ¡1 (x) = 51¡x + x R , x 6= 0, y2R ii VA is x = 0, p x-intercepts § 2, no y-intercept iv x = §2 VA x = 4, x = ¡1 x-ints 4:19, ¡1:19, no y-intercept x ¼ 3:32 b x ¼ 2:73 c x ¼ 3:32 d x ¼ 37:9 x ¼ ¡3:64 f x ¼ ¡7:55 g x ¼ 7:64 h x ¼ 32:0 x ¼ 1150 t ¼ 6:340 b t ¼ 74:86 c t ¼ 8:384 d t ¼ 132:9 t ¼ 121:5 f t ¼ 347:4 x ¼ 2:303 b x ¼ 6:908 c x ¼ ¡4:754 d x ¼ 3:219 x ¼ 15:18 f x ¼ ¡40:85 g x ¼ ¡14:63 x ¼ 137:2 i x ¼ 4:868 iv v f x d M = e3 y EXERCISE 4E a i EXERCISE 4F ¼ 2:26 b x ¼ ¡4:29 x ¼ 0:683 x = 16 b ¼ ¡10:3 c ¼ ¡2:46 d ¼ 5:42 b x ¼ 3:87 c x ¼ 0:139 b x ¼ ¡1:89 log or log25 x ¼ 1:71 x = log 25 i x ] ¡1, [, y2R ii VA is x = ¡1, x and y-intercepts x ] ¡1, [, y2R ii VA is x = ¡1, x-intercept 2, y-intercept ii VA is x = 2, x-intercept 27, no y-intercept -1 4.19 f ¡1 (x) = ln(x + 3) ¡ x=5 f ( x) = e + x fx j x R g, range is fy j y > 5g domain of f ¡1 is fx j x > 5g, range is fy j y R g f has a HA y = 5, f ¡1 has a VA x = y=5 x f -1 y=x y ii x=-3 f -1 x -1 iv c i iii f ¡1 (x) = ex+4 ii fx j x > 0g, range of f is fy j y R g domain of f ¡1 is fx j x R g, range is fy j y > 0g 27 x x=7 f ¡1 (x) = 52+x + y=-3 y iii domain of f is y x fx j x R g, f ( x) = e x +1 - range is fy j y > ¡3g domain of f ¡1 is fx j x > ¡3g, y=x range is fy j y R g ¡1 f has a HA y = ¡3, f has a VA x = ¡3 y f ¡1 (x) = 31¡x ¡ x ] 2, [, y2R iv v iii x=8 i -1.19 x iii domain of f is f ¡1 (x) = 3x ¡ iv v c b i y if f (x) = log2 (x2 ¡ 3x ¡ 4), x < ¡1, p ¡ 25 + 2x+2 f ¡1 (x) = f ¡1 (x) ii y = ln(x ¡ 5) iv -1 1¡x if f (x) = log2 (x2 ¡ 3x ¡ 4), x > 4, p + 25 + 2x+2 f ¡1 (x) = y x i 1¡x iii domain of f is x = ¡ 23 iv v b iii x ~`2 iii x = ¡1:10 and 4:10 EXERCISE 4G.1 a -~`2 if x > 0, f ¡1 (x) = v ii g f N= p y iii g Q ¼ 8:66x3 h D ¼ 0:518n0:4 t3 e p x x = 27 x ] ¡1, ¡1 [ or x ] 4, [ y2R e B= a a a a i c P= y if x < 0, f ¡1 (x) = ¡2 = ln e2 10 a D = ex b F = p a e i a e a e h iii i e a ln ¡972¢ b ln 200 ¡ ¢ c ln = d ln 16 e ln f ln 13 g ln 12 h ln i ln 16 i d a x = b x = or c x = 25 d x = 200 e x=5 f x=3 f -1 e4 x f y=x magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 iv f has a VA x = 0, f ¡1 has a HA y = cyan 867 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\867IB_HL-2_AN.CDR Tuesday, 22 January 2008 11:11:15 AM PETERDELL IB_PD (868) 868 ANSWERS f ¡1 (x) = + ex¡2 d i ii y f ( x) = ln(x - 1) + iii domain of f is fx j x > 1g, range is fy j y R g domain of f ¡1 is fx j x R g, range is fy j y > 1g f has a VA x = 1, f ¡1 has a HA y = iv f ¡1 (x) = ln x a E 12 000 A6 = E 17 919:50 A3:25 is the value after years months 8:64 years x=1 f An+1 -1 An+1¡=¡12¡000¡´¡(1.0835)n 12¡000 y=1 n x EXERCISE 4H.3 y=x ln(2x ¡ 1) b a A is y = ln x b y as its x-intercept is ln ³x + 1´ y = ln x y = ln(x + 2) 17:3 years b 50:7 b 25 years b 10 000 years 166 seconds 92:2 years c 115 years 8:05 sec 152 141 years c 166 years b 49 800 years 11:6 seconds a a a a REVIEW SET 4A c y = ln x has -2 -1 VA x = x y = ln(x ¡ 2) y = ln(x - 2) has VA x = y = ln(x + 2) has VA x = ¡2 y = ln(x2 ) = ln x, so she is correct This is because the y-values are twice as large for y = ln(x2 ) as they are for y = ln x a f ¡1 : x 7! ln(x ¡ 2) ¡ b i x < ¡5:30 ii x < ¡7:61 iii x < ¡9:91 iv x < ¡12:2 Conjecture HA is y = c as x ! ¡1, ex+3 ! and y ! ) HA is y = d VA of f ¡1 is x = 2, domain of f ¡1 is fx j x > 2g a b c ¡2 d 12 e f g 14 h ¡1 i a 12 b ¡ 13 c a + b + a x = 18 b x ¼ 82:7 c x ¼ 0:0316 a log P = log + x log b b log m = log n ¡ log p B5 x ¼ 2:32 ln 25 x = 10 a = REVIEW SET 4B a 32 b 23 c a + b a x = 1000 b x ¼ 4:70 c x ¼ 6:28 ¡ ¢ a log 144 b log2 16 c log4 80 fx j x R , x 6= 0g 0.627 10 -1 Range is fy j y R g y a T= y= x2 y c 12000 15:5 h An An¡=¡2000e0.57n d 2000 n g ¡1 (x) = 3x+2 ¡ a f 6:17 years, i.e., years 62 days 8:65 years, i.e., years 237 days -1.37 log5 ¡ i g is the reflection of f in the y-axis h is the reflection of g in the x-axis x magenta yellow 95 100 50 75 25 95 h 100 50 -1 75 25 95 100 50 75 25 95 100 p ii 8:4% = 0:7% = 0:007 r = + 0:007 = 1:007 12 after 74 months 50 b y g EXERCISE 4H.2 cyan a= REVIEW SET 4C ) approximately 2:8 weeks 75 y=-2 a y = ¡2x + log5 b M = 3(5¡2x ) c x = 10 x = 12 , y = 11 x = §3 4000 b n ¼ 2:82 25 x -1.37 6000 alongside x 8000 13:9 h a see graph -1.37 10000 6:9 h y=x y -2 3:90 h x=-2 x-intercept is 7, y-intercept is ¡1:37 2 - e2 x - x +1 x EXERCISE 4H.1 b p b K=n t a x ¼ 5:19 b x ¼ 4:29 c x ¼ ¡0:839 a 2A + 2B b A + 3B c 3A + 12 B d 4B ¡ 2A e 3A ¡ 2B a x ] ¡2, [, y R e y b VA is x = ¡2, For x ] ¡ 1, [ or ] 0:627, [ c -10 a t ln a x < ¡0:703 b x < 0:773 c x < 3:69 Domain is x ] 0, [ , f (x) for x ] 0, 1] y a b Domain is j a k ¼ 3:25 £ 2x b Q = P R c A = 400 a x ¼ 1:209 b x ¼ 1:822 a 2500 g d b 3290 years t - –––– c 42:3% Wt = 2500 × 3000 EXERCISE 4G.2 a b a b black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\868IB_HL-2_AN.CDR Friday, 12 December 2008 1:16:09 PM TROY IB_PD (869) ANSWERS a y b y = 3e ¡40 m 2400 m i iii y=e x ii a 585 m a y -1 -1\Qw_ x -5 (2,-9) t -40 b c ¡1 a 2x b + x c ¡ x a ln 24 b ln c ln d ln 125 a ln b ln c ln a x ¼ 5:99 b x ¼ 0:699 c x ¼ 6:80 d x ¼ 1:10 or 1:39 ¡1 12 i iii ii x-int’s are ¡1 and y-int is ¡5 b a, b y y (1' 0) x = ¡2 + log2 x Tw_ REVIEW SET 4D x x x a b y = (x - 2)2 - y y¡=¡2x+3 s(t) 869 y = |x| g ( x) = e y -x (2'-1) -5 x f ( x) = e x y x h( x) = e- x - y=-4 y = 2x When x = 0, y = 20 = X 2x > for all x as the graph is always above the y-axis X -3 Each function has domain fx j x R g Range of f is fy j y > 0g, Range of g is fy j y > 0g Range of h is fy j y > ¡4g x y y 10 y=e y = loge x x y = e3 x x x 3 a b ¡3 c ¡ 32 a ¼ e3:00 b ¼ e8:01 c ¼ e¡2:59 EXERCISE 5B.1 a ln 144 b ln( 32 ) c ln( 25 ) d ln e a, b c y i e1:2 a P = T Q1:5 b M = p N g ¡1 (x) a = ln ii b ³x + 5´ f ( x) = x + x=-5 g -1 2 x f ( x) = x f ( x) = x - -3 g ( x) = 2e x - a b y = f ( x) y=-5 y = f ( x) + y y y = f ( x) + c domain of g is fx j x R g, range is fy j y > ¡5g domain of g ¡1 is fx j x > ¡5g, range is fy j y R g a 13:9 weeks³ b 41:6 ´ weeks c 138 weeks y = f ( x) - x+4 , a=4 x , < x < ln b f ¡1 (x) = x e ¡1 a f (x) = ln d y = f ( x) + y y = f ( x) y = f ( x) + 1 x x yellow 95 100 50 75 25 95 100 50 y = f ( x) y = f ( x) - 75 25 95 50 75 25 95 100 50 75 25 100 magenta -2 x y = f ( x) - y x d 2x + 64x3 b 4x3 c x3 + 3x2 + 3x + 2x3 + 6x2 + 6x ¡ 4x b 2¡x + c 2x¡2 + d 2x+1 + 2 + 3x 2x + ¡ b c d x x x x¡1 cyan x c a 2x b x + c a y = f ( x) -2 EXERCISE 5A a d a If b > 0, the function is translated vertically upwards through b units If b < 0, the function is translated vertically downwards jbj units black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\869IB_HL-2_AN.CDR Friday, 12 December 2008 2:10:15 PM TROY y = f ( x) - -2 IB_PD (870) 870 ANSWERS a ³ y = x2 y ¡3 A translation of a ´ b y x y = f ( x - 3) -2 y = f ( x + 2) x x b i If a > 0, the graph is translated a units right ii If a < 0, the graph is translated jaj units left a b y y y = x3 a i (3, 2) ii (0, 11) iii (5, 6) b i (¡2, 4) ii (¡5, 25) iii (¡1 12 , 14 ) y y = |x| y = f ( x + 2) EXERCISE 5B.2 x -2 y = f ( x + 2) d y = ln x y y x = -2 y = f ( x) y = f ( x + 2) y = f ( x) y = f¡(x) = x y = f ( x - 1) c y = f ( x + 2) b y = f¡(x) = x y y = f ( x - 1) y a x -2 y = f ( x) x x x y = f ( x) x c y = f ( x - 1) y = f (x ¡a) is a horizontal translation of y = f (x) through a y = f¡(x) = x y = f ( x - 1) ³ ´ a y y x=1 y = 1x y = f ( x) y = f ( x) x y = f ( x) y = f ( x) y e y y = f ( x - 2) + y = x2 y = f¡(x) = e x d x f y = f ( x) y = f ( x) y y = f ( x) x -3 y = f ( x + 1) - y = f¡(x) = y = f ( x) x y = f¡(x) = ln x y = ex b y y = f ( x - 2) + a b y y y =3 y = f¡(x) = x3 y= x y = f ( x + 1) - c x f ( x) f ( x) x y= y=3 magenta 95 100 50 75 95 50 100 yellow f ( x) p affects the vertical stretching or compressing of the graph of y = f (x) by a factor of p If p > stretching occurs If < p < compression occurs 25 75 25 95 100 50 75 25 95 100 50 f ( x) x y = -4 75 f ( x) f ( x) y= y = f ( x + 1) - 25 y= y = f¡(x) = e x x y= y = f¡(x) = x y y= cyan x c y y = f ( x - 2) + y= y = -4 x x black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\870IB_HL-2_AN.CDR Tuesday, 22 January 2008 11:58:46 AM PETERDELL IB_PD (871) ANSWERS a c y = ( x - 1) b y 871 y y y = x2 y = 14 (2 x + 5) + y = x2 x x y = (2 x) c y = (2x + 3)2 V &- 52 , 1* Qw_ y = (2 x - 1) a i ( 32 , ¡15) ii ( 12 , 6) iii (¡1, 3) b i (4, 13 ) ii (¡6, 23 ) iii (¡14, 1) y 10 a -3 -Ew_ y = ( x + 3) a f (x) is translated horizontally unit left, then horizontally stretched by a factor of 2, then vertically stretched by a factor of 2, then translated units upwards x b y b i (0, ¡3) ii (2, 5) iii (¡4, ¡1) c i (0, ¡4) ii ( 32 , ¡2) iii ( 72 , ¡ 32 ) y y=x EXERCISE 5B.3 a y = 3x y y = e3 x y=e b x y = -e d y y=x x x y y = ln x x y = & 2x * c y y = ex x y = x2 x y y c b y = 3x x y = (3x)2 a y y = -3x y = x2 x c x x x y = -x2 y = - ln x y = ( x + 2) y = 2& 2x * y = x3 - e y = 2( x + 1) y f y x y = 2x y = ( 2x + 2) y = - x3 + -4 -2 x -1 x k affects the horizontal compressing of y = -2( x + 1) y = f (x) by a factor of k If k > it moves closer to the y-axis If < k < it moves further from the y-axis y = ¡f (x) is the reflection of y = f (x) in the x-axis a i f (¡x) = ¡2x + ii f (¡x) = x2 ¡ 2x + iii f (¡x) = j ¡x ¡ j b i ii x = -1 y y = x2 a y = 3(x - 2)2 + y y V(2' 1) x -Qw_ y = 2(x + 1) - x Qw_ V(-1'-3) x=2 y = 2( 2x - 3)2 + b -1 x y = x - 2x + y = x + 2x + y = 2( 2x - 3)2 iii y y y = x2 y = -2x + y = 2x + y = ( 2x - 3)2 x -3 cyan magenta 95 50 75 25 95 50 75 25 100 yellow 100 y = | -x - | x 95 100 50 75 25 95 100 50 75 25 V(6,¡0) y = (x - 3)2 V(3, 0) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\871IB_HL-2_AN.CDR Tuesday, 22 January 2008 12:06:27 PM PETERDELL y =|x -3| IB_PD (872) 872 ANSWERS y = f (¡x) is the reflection of y = f (x) in the y-axis a b a b a c y (3, 0) ii (2, 1) iii (¡3, ¡2) (7, 1) ii (¡5, 0) iii (¡3, 2) (¡2, ¡1) ii (0, 3) iii (1, 2) (¡5, ¡4) ii (0, 3) iii (¡2, 3) A rotation about the origin through 180o b (¡3, 7) (5, 1) i i i i -1 x y = h( x) y = h( x) + y = 12 h( x) y = h(- x) y = h( 2x ) -2 EXERCISE 5B.4 a b y y y=ƒ(x) EXERCISE 5C y=ƒ(x) ii y = iii y = 2x x x+3 +4 iv y = + v y = x 2(x + 3) b Domain is fx j x R , x 6= ¡3g Range is fy j y R , y 6= 4g a i y= x x -1 y=-ƒ(x) y=-ƒ(x) a i VA is x = 1, HA is y = c y ³ y=ƒ(x) ii Translate ¡1 ¡2 ´ , then vertically stretch, factor b i VA is x = ¡1, HA is y = x ³ ii Translate y=-ƒ(x) factor ¡3 ´ , reflect in the x-axis, then vertically stretch, c i VA is x = 2, HA is y = ¡2 a ³ b y=ƒ(-x) y ii Translate y=ƒ(x) y y=ƒ(-x) factor ´ , reflect in the x-axis, then vertically stretch, a i VA is x = ¡1, HA is y = ii x-intercept is ¡ 32 , y-intercept is y=ƒ(x) iii as x ! ¡1 (from left), y ! ¡1 x x c ¡2 as x ! ¡1 (from right), y ! as x ! ¡1, y ! (from below) as x ! 1, y ! (from above) y iv y x=-1 y=1 x x=-2 y=2 x=2 x -\Ew_ a A b B c D d C y ³ v Translate -1 y = f ( x) y = f ( x + 2) y = f (2x) y = f ( x) y = 12 f ( x) y = f ( 12 x) x iii as x ! (from left), y ! ¡1 as x ! (from right), y ! as x ! 1, y ! (from above) as x ! ¡1, y ! (from below) iv y y y = g ( x) y = g ( x) + y = - g ( x) y = g ( x + 1) y = g (- x) ´ b i VA is x = 2, HA is y = ii no x-intercept, y-intercept is ¡1 12 -2 ¡1 x -3 x -1\Qw_ cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 x=2 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\872IB_HL-2_AN.CDR Friday, 12 December 2008 2:23:44 PM TROY IB_PD (873) ANSWERS ³ ´ v Vertically stretch with factor 3, then translate a invariant points are (¡1, ¡1) and (1, ¡1) b invariant points are (0:586, 1), (2, ¡1) and (3:414, 1) c i VA is x = 3, HA is y = ¡2 ii x-intercept is 12 , y-intercept is ¡ 13 EXERCISE 5D.2 a iii as x ! (from left), y ! y b y as x ! (from right), y ! ¡1 as x ! ¡1, y ! ¡2 (from above) as x ! 1, y ! ¡2 (from below) iv y y = f (x) y = f ( x) -2 x=3 -2 x y = x(x + 2) x Qw_ b y f (x) f ( x) y=-2 x y = x(x + 2) y a -\Qe_ ³ ´ f ( x) x v Vertically stretch, factor ¡5, reflect in x-axis, then translate ¡2 d i VA is x = ¡ 12 , HA is y = 12 f ( x) iii as x ! ¡ 12 (from left), y ! x y=1 x (from below) as x ! ¡1, y ! 12 (from above) iv x f (x) as x ! ¡ 12 (from right), y ! ¡1 as x ! 1, y ! (1 12 ,-2) y c ii x-intercept is 15 , y-intercept is ¡1 f (x) (1 12 ,- 12 ) 12 873 x=4 y a b y f (x) y=2\Qw_ y f (x) f (x) x Qt_ -1 x=-\Qw_ v Vertically stretch, factor 74 , reflect in x-axis, then translate µ 1¶ ¡2 N N = 20 + f (x) a 70 weeds/ha b 30 weeds/ha c days d -2 x -2 c y f (x) f (x) 100 t+2 y=2 x 20 20 40 60 x=4 t 80 a e No, the number of weeds/ha will approach 20 (from above) f( x) f( x) EXERCISE 5D.1 a b y b y y f (x) y = ( x - 1)(x - 3) -2 y x -2 x -2 f (x) c y x f (x) f( x) y=- x2 y=2 y = -x2 y= ( x - 1)(x - 3) x cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 x=-4 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\873IB_HL-2_AN.CDR Friday, 12 December 2008 2:32:51 PM TROY x=4 IB_PD x (874) 874 ANSWERS 10 a F (x) = 4x ¡ b i (1, 3) ! (1, 6) ! ( 12 , 6) ! (1, 6) ! (1, 3) ii (0, 2) ! ( 12 , 1) and (¡1, 1) ! (0, ¡1) iii F ( 12 ) = and F (0) = ¡1 a (3, 0) b (5, 2) c (0, 7) d (2, 2) a i (0, 3) ii (1, 3) and (¡1, 3) iii (7, ¡4) and (¡7, ¡4) b i (0, 3) ii (1, 3) iii (10, ¡8) REVIEW SET 5A a b c 4x2 ¡ 4x d x2 + 2x e 3x2 ¡ 6x ¡ 2 a ¡15 b c ¡x2 + x + d ¡ 12 x ¡ 14 x2 e ¡x2 ¡ 3x + 11 g(x) = ¡x2 ¡ 6x ¡ REVIEW SET 5B x a ¡1 b y a x c y f ( x) = x y = f ( x + 2) y = f ( x + 2) y = f ( x + 2) - x -2 x x>a a, b g(x) = 3x3 ¡ 11x2 + 14x ¡ a -3 y -x a y x=-1 y=2 y y= -3 x=-c true ii false false iv true i iii i and ¡3 ii ¡3 x y=0 f (x ) x b y=¦(x) c x -c V(-1,-4) b c i 23 ii ¡2 iii i ¡1:1 ii 0:9 b c y=3x-2 -2 10 ¡ 3x x+2 d y y = f ( x) y = f (- x) y = - f ( x) y = f ( x + 2) y = f ( x) + V(¡1, ¡4) y x f ( x) = - x y = f (- x) y = - f ( x) y = f (2x) y = f ( x - 2) x a y=1 y y = f¡(x¡-¡2)¡+¡1 y c (2,-2) y = f¡(x) (4,-1) x b¡+¡c x=2 a a¡+¡c a -1 x y = f¡(x) x (2,-2) y x=4 f (x ) y = f¡(x) magenta yellow 95 100 50 25 95 100 50 75 25 95 100 50 75 (2,-2) x x=4 x¡=¡3\Qe_ 25 95 100 cyan y= x¡=¡1 -2 y = | f¡(x)¡| (2, 2) x 75 y¡=¡¦(x) 50 f¡(x) (2,-Qw_) (4, Qw_) 75 x=6 c 25 (0, 1), ( 11 , 1), all points on y = ¡1, x [2, 3] (4,¡2) (Qd_Q_¡,¡1) -1 y= y y¡=¡¦(x) ¦(x) d b c y¡=¡|¦(x)| b y2¡=¡¦(x-c) y1¡=¡¦(x) ¦(x) x=4 x b black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\874IB_HL-2_AN.CDR Friday, 12 December 2008 2:34:26 PM TROY IB_PD (875) ANSWERS a f (x) = 2x ¡ 2x ¡ b iii For y = log4 x, VA is x = 0, no HA x=1 y 2x - 2x - y= For y = log4 (x ¡ 1) ¡ 2, VA is x = 1, no HA y=x iv For y = log4 x, Tw_ domain is fx j x > 0g, range is fy R g y=1 For y = log4 (x ¡ 1) ¡ 2, domain is fx j x > 1g, range is fy R g x Tw_ 11 a, d y y= domain of f (x) is fx j x 6= 1g range of f (x) is fy j y 6= 1g c Yes, since it is a one-to-one function (passes both the vertical and horizontal line tests) 2x ¡ Also, the graph of d Yes, since f ¡1 (x) = f (x) = 2x ¡ f (x) is symmetrical about the line y = x 2x ¡ = 3x + x-int at x = 3 , ¡ 19 (x + y-int at y = 3 as x ! +1, y ! as x ! ¡1, y ! y¡=¡¦(|x|) ¡ 35 y¡=¡¦(x) (from the left), y ! +1 (from the right), y ! ¡1 EXERCISE 6A.1 a x = 0, ¡ 74 b x = 0, ¡ 13 c x = 0, a x= x e x= k x = 17 , ¡1 l x = ¡2, x a x = ¡1 § For y = ¡ 2, HA is y = ¡2, no VA iv For y = 2x , domain is f x j x R g, range is fy j y > 0g 2x¡1 d x=1§ b x= e x= § § p 19 p 37 20 p c x = ¡2 § f x= ¡ 12 p7 § p p EXERCISE 6B magenta yellow 95 25 95 100 50 75 25 a real distinct roots b a repeated root c real distinct roots d real distinct roots e no real roots f a repeated root a, c, d, f 95 100 50 p1 p p p h x = ¡ 49 § 97 i x = ¡ 74 § 497 p p p a x = ¡2 § 2 b x = ¡ 58 § 857 c x = 52 § 213 p p p d x = 12 § 12 e x = 12 § 25 f x = 34 § 417 y = log4 x 75 25 p g x= x p a x = § b x = ¡3 § c x = § p p p d x = ¡2 § e x = § f x = 12 § 12 y = log4 (x-1) - 28 15 d x = 1, p p7 EXERCISE 6A.3 p For y = 2x¡1 ¡ 2, domain is fx j x R g, range is fy j y > ¡2g 100 ¡ 32 p p iii For y = 2x , HA is y = 0, no VA cyan ¡ 32 a x = § b x = ¡3 § c x = § p p p d x = § e x = ¡3 § f x = § p p g x = ¡3 § 11 h x = § i no real solns y = -2 b i y = log4 (x ¡ 1) ¡ ii y x=1 , a x = ¡5 § b no real solns c x = § 2 p p p d x = § e x = ¡3 § f x = § p p g x = ¡1 § 10 h x = ¡ 12 § 12 i x = 13 § 95 f x = ¡ 23 , g x = ¡ 23 , h x = EXERCISE 6A.2p y y = 2x 11 d x = 0, b x = ¡ 12 , c x = ¡ 23 , d x = 13 , ¡2 a x = 2, b x = ¡3, c x = 0, e x = 12 , ¡1 f x = ¡2x x¡3 y = x-1 - 50 3 , i x = ¡ 14 , j x = ¡ 34 , 10 a i y = 2x¡1 ¡ ii 3 e x = 0, f x = 0, g x = 3, h x = 4, ¡2 i x = 3, j x = k x = ¡4, l x = ¡11, x = -\Te_ 75 x (from above) -\Et_ 25 y¡=¡|¦(x)| (from below) Ew_ 12 y = We_ (2,¡-1) - 12 y y= x y VA is x = ¡ 53 , 100 as x ! (1,¡1) f (x ) b (1, 1) and (2, ¡1) c x = 12 , e 50 as x ! ¡ 53 ¡ 53 3 y = f (x ) , HA is y = ) x= 12 75 y= 875 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\875IB_HL-2_AN.CDR Tuesday, 22 January 2008 12:24:38 PM PETERDELL IB_PD (876) 876 ANSWERS a ¢ = 16 ¡ 4m + - i m = ii m < iii m > + b ¢ = ¡ 8m i m= ii m < c ¢ = ¡ 4m i m= ii m < 9 iii m > + b ¢ = ¡ 4k2 i ¡1 < k < ii ¡1 k iii k = §1 iv k < ¡1 or k > c ¢ = k2 + 4k ¡ 12 i k < ¡6 or k > ii k ¡6 or k > iii k = ¡6 or iv ¡6 < k < d ¢ = k2 ¡ 4k ¡ 12 i k < ¡2 or k > ii k ¡2 or k > iii k = or ¡2 iv ¡2 < k < e ¢ = 9k2 ¡ 14k ¡ 39 - -1 m -3 a x = b x = c x = ¡4 d x = ¡2 e x = ¡3 f x = ¡2 + + -8 k + -1 k - f ¢ = ¡3k2 ¡ 4k i < k < ii 6k60 iii k = ¡ 43 or iv k < ¡ 43 or k > V(1' 3) V(-2' 1) d x y V(1'-3) @ = Qw_ (! - 3)W +2 Qs_E_ -5 V(3' 2) k y e f V(1' 4) 3\We_ y V(-2'-3) + x -\Re_ x k x=1 x=-2 -3\Wt_ @ = - Qe_ (! - 1)W + p a §3 b § c ¡5 and ¡2 d and ¡4 e and f ¡4 and h (touching)p p ¡2 g ¡1 p (touching) p i § j ¡2 § k § 11 l ¡4 § b i x = ¡2 a i x=1 ii (1, 4) ii (¡2, ¡5) p iii no x-intercept, iii x-int ¡2 § 5, a 3® = a6 , 2®2 = or a = ¡2, roots are ¡1 and ¡2 or k = 16, roots are ¡ 54 and a(8x2 ¡ 70x + 147) = 0, a 6= @ = - Aq_p_ (! + 2)W - a (2, ¡2) b (¡1, ¡4) c (0, 4) d (0, 1) e (¡2, ¡15) f (¡2, ¡5) g (¡ 32 , ¡ 11 ) h ( 52 , ¡ 19 ) i (1, ¡ 92 ) 2 3 y-intercept ¡1 y-intercept iv iv y b y = ¡(x ¡ 4)(x + 2) -2 x V(1,¡4) -2 -8 c d y = ¡3x(x + 4) c y = 2(x + 3)(x + 5) y y -2+~`5 x V(-2,-5) x¡=¡1 x i x= ( 54 , ¡ 98 ) ii ( 32 , ) x-intercepts 12 , 2, y-intercept iii x-intercepts 1, 2, y-intercept ¡2 i x= ii iii d iv iv 30 y -4 x= x -3 -5 x magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 V(54 ,- 89 ) y V( 32 , 14 ) 2 Qw_ cyan x -1 -2-~`5 y y y x¡=¡-2 EXERCISE 6D.1 a y = (x ¡ 4)(x + 2) x x=3 @ = -2(! - 1)W - a sum = 23 , product = 73 b sum = ¡11, product = ¡13 c sum = 65 , product = ¡ 14 5 p ¡8 + 60 < k < x x x=1 k EXERCISE 6C a¡2 a b a = 4, roots are 12 and k = 4, roots are ¡ 12 and 32 7x2 ¡ 48x + 64 = y c - ¡ 43 @ = 2(! + 2)W + y x=-2 k + i k < ¡ 13 or k > ii k ¡ 13 or k > 9 13 iii k = ¡ 13 or iv ¡ < k < 9 b y x=1 + + -2 + a @ = (! - 1)W + + + -6 -\Ql_E_ k = ¡ 35 roots are ¡1 and x x Or_ i k < ¡8 or k > ii k ¡8 or k > iii k = ¡8 or iv ¡8 < k < y 18 m Oi_ a ¢ = k2 + 8k ¡ 43 y = ¡ 14 (x + 2)2 f y -2 - iii m > e y = 2(x + 3)2 m black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\876IB_HL-2_AN.CDR Thursday, 24 January 2008 2:17:28 PM PETERDELL x -2 x= IB_PD x (877) ANSWERS i x= ii ( 23 , ) iii x-intercepts 13 , 1, y-intercept ¡1 e iv f y -1 x= g i ii iii ii ( 14 , ) iii x-intercepts ¡ 12 , 1, y-intercept i y = x¡ V&Qr_\' Oi_* -\Qw_ a x i ii iv 1)2 y = 2(x + (¡1, 3) iii b +3 y = 2(x ¡ 2)2 ¡ (2, ¡5) iii i ii iv y y @ = 2!W -¡8!¡+¡3 @ = 2!W ¡+¡4!¡+¡5 h i ii iii x = ¡3 (¡3, 1) x-int ¡2, ¡4, y-intercept ¡8 x V(2'-5) c y V(-3,¡1) V(3,¡9) x i y = 2(x ¡ 32 )2 ¡ ii ( 32 , iv -2 x ¡ 72 ) y -8 d y = 3(x ¡ 1)2 + (1, 2) iii i ii iv iii 1 x y iv x=4 (4, 5) p x-int § 5, y-intercept V(1' 2) V(4,¡5) x 4-2~`5 e i ii iv EXERCISE 6D.2 b y = (x + 2)2 ¡ y = ¡(x ¡ 2)2 + (2, 6) iii y i y = ¡2(x + 54 )2 + ii (¡ 54 , 49 ) f x y V&-\Tr_\\' Rk_O_* y = -2xX - 5x + 49 iii iv V(2' 6) y y x V&Ew_\'-\Uw_* 4+2~`5 x=4 a y = (x ¡ 1)2 + y = xX - x + y y = xX - x + x=-3 x=3 x V(-1' 3) -4 i ii iii x !=\Qr_ i y ¡ 14 V&Tw_\'-5\Qr_* y iv y ¢ 2 x=3 (3, 9) x-intercepts 0, 6, y-intercept iv x Qe_ x= iv V( 23 , 13 ) ¡ i 877 x -2 x V(1' 2) y = - xX + x + x ¡ c y = (x ¡ 2)2 ¡ a y = (x ¡ 2)2 + b y = (x + 3)2 ¡ V(-2'-6) d y = x+ y ¢ 2 ¡ ¡ c y = ¡(x ¡ 2)2 + d y = x + ¡ y e y = ¡2 x + V&-\Ew_\'-\Or_* f y = x¡ ¡ y -2 a c e a c V&Ew_\'-\Qr_* g y = (x ¡ 3)2 ¡ h y = (x + 4)2 ¡ 18 y y -2 magenta yellow 95 100 50 75 25 95 100 50 75 V(-4'-18) 25 95 100 50 75 25 V(3'-4) ¡ 47 y = 2(x ¡ 1)(x ¡ 2) b y = 2(x ¡ 2)2 y = (x ¡ 1)(x ¡ 3) d y = ¡(x ¡ 3)(x + 1) y = ¡3(x ¡ 1)2 f y = ¡2(x + 2)(x ¡ 3) y = 32 (x ¡ 2)(x ¡ 4) b y = ¡ 12 (x + 4)(x ¡ 2) y = ¡ 43 (x + 3)2 a y = 3x2 ¡ 18x + 15 b y = ¡4x2 + 6x + c y = ¡x2 + 6x ¡ d y = 4x2 + 16x + 16 e y = 32 x2 ¡ 6x + 92 f y = ¡ 13 x2 + 23 x + x x f y =3 x¡ EXERCISE 6E x V&-\Tw_\'-\Ef_E_* cyan 27 17 ¢ always > fas k2 > for all kg a = which is > and ¢ = k2 ¡ 16 ) positive definite when k2 < 16 i.e., ¡4 < k < x 95 y ¢ 2 100 33 50 ¡ 75 ¡ 25 e y = x+ + ¡ a cuts x-axis twice b touches x-axis c cuts x-axis twice d cuts x-axis twice e cuts x-axis twice f touches x-axis a a = which is > and ¢ = ¡15 which is < b a = ¡1 which is < and ¢ = ¡8 which is < c a = which is > and ¢ = ¡40 which is < d a = ¡2 which is < and ¢ = ¡23 which is < a = which is > and ¢ = k2 + 12 which is x V(2'-4) ¢ 2 ¡ EXERCISE 6D.3 x ¡ ¢ ¢ 2 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\877IB_HL-2_AN.CDR Tuesday, 22 January 2008 12:44:12 PM PETERDELL IB_PD (878) 878 ANSWERS a y = ¡(x ¡ 2)2 + b y = 2(x ¡ 2)2 ¡ c y = ¡2(x ¡ 3)2 + d y = 23 (x ¡ 4)2 ¡ e y = ¡2(x ¡ 2)2 + f y = 2(x ¡ 12 )2 ¡ 32 a x = 15 or ¡4 b x = ¡ 53 or c x = or a x = or b x = or c x = 12 or p EXERCISE 6F a (1, 7) and (2, 8) b (4, 5) and (¡3, ¡9) c (3, 0) (touching) d graphs not meet a (0:59, 5:59) and (3:41, 8:41) b (3, ¡4) touching c graphs not meet d (¡2:56, ¡18:81) and (1:56, 1:81) a (2, 4), (¡1, 1) b (1, 0), (¡2, ¡3) c (1, 4) d (1, 4), (¡4, ¡1) c = ¡9 y y¡=¡2xX-3x-7 m = or ¡8 10 ¡1 or 11 x a c < ¡9 b E.g., c = ¡10 REVIEW SET 6B REVIEW SET 6C solutions x p 13 a x = ¡ 52 § a x= p 37 § a max b x= 17 cm y-int = a ¡1, when x = b max 8, when x = ¡1 c max 13 , when x = 13 d ¡1 18 , when x = ¡ 14 REVIEW SET 6D f max 18 , when x = p ¡11§ 145 p § 473 b x= a x ¼ 0:586 or 3:414 b x ¼ ¡0:186 or 2:686 a two distinct real rational roots b a repeated root a m = 98 b m < 98 c m > 98 12:92 cm EXERCISE 6H a y a < ¡9, ¡1 < a < b x=2 y (-4' 6) @=(!-2)X-4 40 refrigerators, $4000 500 m by 250 m c 100 m by 112:5 m a 41 23 m by 41 23 m b 50 m by 31 14 m x x b 18 units a y = ¡ 34 x b cm by cm 125 10 40 11 157 -2 @=-\Qw_\(!+4)X+6 (2'-4) a b + a b + ¢ ¢ ¢ + an bn 12 m = 1 2 22 a1 + a2 + ¢ ¢ ¢ + a2n a b c 13 f (x) = x4 ¡ 2(a2 + b2 )x2 + (a2 ¡ b2 )2 , ¡4a2 b2 x = ¡1 (¡1, ¡3) y-intercept ¡1, p x-ints ¡1 § 12 x=-4 y d -1+\Qw_\~`6 -1-\Qw_\~`6 x -1 REVIEW SET 6A a ¡2, b x = ¡ 12 (¡ 12 , 92 ) c d e a no real solutions b two real distinct (4, 4) and (¡3, 18) k < ¡3 18 b 15 m by 30 m m = ¡5 or 19 k = 3, roots are ¡ 13 and b supports are 21 m, 34 m, 45 m, 54 m, 61 m, 66 m, 69 m least value = max = when x = and ¡5 or ¡7 and 5 or 14 18 and 20 or ¡18 and ¡20 15 and 17 or ¡15 and ¡17 15 sides 3:48 cm b cm by cm by cm 11:2 cm square 10 no 12 221 13 2:03 m 52:1 km h¡1 15 554 km h¡1 16 61:8 km h¡1 17 32 No, tunnel is only 3:79 m wide 4:8 m above ground level a y = ¡ 100 x2 + 70 V( 43 , 12 13 ) x= a = ¡2 which is < ) e 15 when x = 16 , @=-!X+2! EXERCISE 6G 14 18 19 V(1,¡1) y -10 y¡=¡x-10 p x = ¡ 72 § 265 a x = 72 § 237 b no real roots a c > ¡6 b e.g., c = ¡2, (¡1, ¡5) and (3, 7) 4x2 + 3x ¡ = @=2!X+4!-1 a x = b (2, ¡4) c ¡2 @=\Qw_\(!-2)X-4 y y d (-1'-3) b graph cuts x- a graph cuts xaxis twice axis twice &-\Qw_\' Ow_\* a neither b positive definite a y = 3(x ¡ 3)(x + 3) b y = ¡6(x ¡ 2)2 + 25 x=2 x -2 65 or 56 13:48 cm by 13:48 cm touch at (¡2, 9) 10 a(6x2 ¡ 10x ¡ 25) = 0, a 6= x -2 @=-2(!+2)(!-1) x=-\Qw_ y d b (¡ 32 , ¡ 15 ) c ¡3 ³ -3 @=2!X+6!-3 yellow 95 100 50 75 25 95 50 75 25 95 100 50 75 25 100 magenta ´ 600 ¡ 8x c 37 12 m by 33 13 m d 1250 m2 a(64x2 ¡ 135x ¡ 27) = 0; a 6= k<1 b A=x &-\Ew_\'-\Qs_T_* cyan ¡ 20 y = ¡4x2 + 4x + 24 a y = 3x2 ¡ 24x + 48 b y = 25 x2 + 16 x + 37 5 2 a = when x = ¡ b max = 18 when x = ¡ 54 x 95 15 100 ¡ 50 c y = ¡ 27 (x ¡ 1)(x ¡ 7) d y 75 a y =2 x+ ¢ 20 (x ¡ 2)2 = 29 (x + 3)2 a y = ¡6(x + 3)(x ¡ 1) b y = (2'-4) 25 ¡ REVIEW SET 6E black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\878IB_HL-2_AN.CDR Friday, 12 December 2008 2:36:24 PM TROY IB_PD (879) ANSWERS EXERCISE 7A p EXERCISE 7B.5 p c (z1 z2 z3 z4 ::::zn )¤ = z1¤ z2¤ z3¤ :::: zn¤ d (z n )¤ = (z ¤ )n a 3i b 8i c 12 i d i e i p p a (x + 3)(x ¡ 3) b (x + 3i)(x ¡ 3i) c (x + 7)(x ¡ 7) p p d (x+i 7)(x¡i 7) e (2x+1)(2x¡1) f (2x+i)(2x¡i) p p p p g ( 2x + 3)( 2x ¡ 3) h ( 2x + 3i)( 2x ¡ 3i) i x(x + 1)(x ¡ 1) j x(x + i)(x ¡ i) k (x + 1)(x ¡ 1)(x + i)(x ¡ i) l (x + 2)(x ¡ 2)(x + 2i)(x ¡ 2i) EXERCISE 7C.1 a 3x2 + 6x + b 5x2 + 7x + c ¡7x2 ¡ 8x ¡ d 4x4 + 13x3 + 28x2 + 27x + 18 p p x = §5 b x = §5i c x = § d x = §i e x = § 32 a f i l x = § 32 i g x = 0, x = §2 h x = 0, x = §2i p p x = 0, x = § j x = 0, x = §i k x = §1, x = §i x = §3, x = §3i a x = § 2i b x = ¡3 § 4i c x = ¡7 §pi p d x = 32 § 12 i e x = § i f x = 14 § i p p p a x = §i or §1 b x = § or §i c x = §3i or §2 p p d x = §i or §i e x = §1 f x = §i EXERCISE 7B.1 Re(z) z + 2i 5¡i Im(z) ¡1 0 z ¡3 + 4i ¡7 ¡ 2i ¡11i p i Re(z) ¡3 ¡7 0 Im(z) ¡2 ¡11 p b ¡ 15 + 25 i c a ¡ 25 + 15 i b ¡ 13 + i 13 + 15 i d 25 + x3 + x2 ¡ 4x + b x3 ¡ x2 ¡ 2x + 3x3 + 2x2 ¡ 11x + 19 d 2x3 ¡ x2 ¡ x + x5 ¡ x4 ¡ x3 + 8x2 ¡ 11x + 10 x4 ¡ 2x3 + 5x2 ¡ 4x + a c e f 2x3 ¡ 3x2 + 4x + b x4 + x3 ¡ 7x2 + 7x ¡ x3 + 6x2 + 12x + d 4x4 ¡ 4x3 + 13x2 ¡ 6x + 16x4 ¡ 32x3 + 24x2 ¡ 8x + 18x4 ¡ 87x3 + 56x2 + 20x ¡ 16 a c e f h j 6x3 ¡ 11x2 + 18x ¡ b 8x3 + 18x2 ¡ x + 10 ¡2x3 + 7x2 + 13x + 10 d 2x3 ¡ 7x2 + 4x + 2x4 ¡ 2x3 ¡ 9x2 + 11x ¡ 15x4 + x3 ¡ x2 + 7x ¡ g x4 ¡ 2x3 + 7x2 ¡ 6x + 4x4 + 4x3 ¡ 15x2 ¡ 8x + 16 i 8x3 + 60x2 + 150x + 125 x6 + 2x5 + x4 ¡ 4x3 ¡ 4x2 + a quotient is x, remainder is ¡3 b quotient is x ¡ 4, remainder is ¡3 c quotient is 2x2 + 10x + 16, remainder is 35 14 c 2x ¡ ¡ x+3 x¡2 11 d x2 + x ¡ e x2 + 4x + + 3x ¡ a x+2+ x¡2 f x2 + 3x + + c ¡ 25 + 35 i a ¡2 b ¡4 c d EXERCISE 7C.3 EXERCISE 7B.2 a b c d z = ¡ 4i z = 65 ¡ 72i m = ¡ 11 , n= 11 a 4, ¡ 32 c2 + d2 i £ ¤ £ ¤ + [2ab] i c z = a3 ¡ 3ab2 + 3a2 b ¡ b3 i magenta yellow 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a a = or (b = 0, a 6= ¡1) b a2 ¡ b2 = and neither a nor b is cyan b x¡ 2x x2 + p p p p b x = ¡ 12 , §i c z = 0, § i p p p p d x = 0, § e z = 0, §i f z = §i 2, § p p a (2x + 3)(x ¡ 5) b (z ¡ + i 7)(z ¡ ¡ i 7) p p c x(x + + 5)(x + ¡ 5) d z(3z ¡ 2)(2z + 1) p p e (z + 1)(z ¡ 1)(z + 5)(z ¡ 5) p p f (z + i)(z ¡ i)(z + 2)(z ¡ 2) 95 h bc ¡ ad i 124 x+4 b ¡3 § i c § d 0, §2 e 0, §i 100 a2 ¡ b + + x+2 a x = 1, ¡ 25 50 ¤ d2 c 3x ¡ + f §1, §i 75 a £ c2 x+1 x¡2 2x x2 + x + EXERCISE 7D.1 a a = ¡6, b = 10 b a = ¡2, b = ¡1 c a = ¡2, b = or a = 0, b = h ac + bd i b 2x + ¡ quotient is x2 + 2x + 3, remainder is quotient is x2 ¡ 3x + 5, remainder is 15 ¡ 10x p + 4x + 29), a 6= d ¡ 2x + 3) = 0, a 6= a(x2 ¡ 4x + 1) = 0, a 6= f a(3x2 + 2x) = 0, a 6= a(x2 + 2) = 0, a 6= h a(x2 + 12x + 37) = 0, a 6= z¤ a 2x + c x2 + x + + a(x2 EXERCISE 7B.4 19 3x ¡ 5x + d 2x + + x2 ¡ x + (x ¡ 1)2 4x + 15 ¡ 10x e x2 ¡ 2x + ¡ f x2 ¡ 3x + + (x + 1)2 (x ¡ 1)(x + 2) p z=i a(x2 ¡ 6x + 10) = a 6= a(x2 ¡ 2x + 10) = 0, a 6= a(x2 + quotient is x + 1, remainder is ¡x ¡ quotient is 3, remainder is ¡x + quotient is 3x, remainder is ¡2x ¡ quotient is 0, remainder is x ¡ a 1¡ EXERCISE 7B.3 a b c e g d x2 + 3x ¡ e 2x2 ¡ 8x + 31 ¡ i 25 a x = 0, y = b x = 3, y = ¡2 or x = 4, y = ¡ 32 c x = 2, y = ¡5 or x = ¡ 53 , y = d x = ¡1, y = b x+1¡ f x3 ¡ 2x2 + 52 x ¡ a x = 0, y = ¡2 b x = ¡2 c x = 3, y = 2 d x = ¡ 13 , y = ¡ 13 10 x¡4 a x+1+ ¡ i b 10 ¡ 4i c ¡1 + 2i d ¡ 3i ¡ 7i f 12 + i g + 4i h 21 ¡ 20i ¡3 + 7i b 2i c ¡2 + 2i d ¡1 + i ¡5 ¡ 12i f ¡5 + i g ¡6 ¡ 4i h ¡1 ¡ 5i = 1, i1 = i, i2 = ¡1, i3 = ¡i, i4 = 1, i5 = i, i6 = ¡1, i7 = ¡i, i8 = 1, i9 = i, i¡1 = ¡i, i¡2 = ¡1, i¡3 = i, i¡4 = 1, i¡5 = ¡i, i4n+3 = ¡i (1 + i)4 = ¡4, (1 + i)101 = ¡250 (1 + i) a = 3, b = ¡5 i 10 a c e f EXERCISE 7C.2 a e a e i0 a ¡ 10 ¡ 879 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\879IB_HL-2_AN.CDR Thursday, 24 January 2008 11:55:01 AM PETERDELL IB_PD (880) 880 ANSWERS a b c d P (z) = a(z ¡ 4)(z ¡ 3) a 6= P (z) = a(z + 2)(z + 1) a 6= P (z) = a(z ¡ 3)(z + 2z + 2) a 6= P (z) = a(z + 1)(z + 4z + 2) a 6= a b c d P (z) = a(z ¡ 1)(z ¡ 2) a 6= P (z) = a(z ¡ 2)(z + 1)(z + 3) a 6= P (z) = a(z ¡ 3)(z ¡ 2z + 2) a 6= P (z) = a(z ¡ 4z ¡ 1)(z + 4z + 13) a 6= EXERCISE 7E.2 a b c d e a a b c d EXERCISE 7D.2 a a = 2, b = 5, c = b a = 3, b = 4, c = a a = 2, b = ¡2 or a = ¡2, b = bp a = 3, b = ¡1 a = ¡2, b = 2, x = § i or ¡1 § + 1)(x ¡ 4)3 f P (x) = x2 (x + 2)(x ¡ 3) C b F c A d E e B f D P (x) = (x + 4)(2x ¡ 1)(x ¡ 2)2 P (x) = 14 (3x ¡ 2)2 (x + 3)2 P (x) = 2(x ¡ 2)(2x ¡ 1)(x + 2)(2x + 1) P (x) = (x ¡ 1)2 ¡8 ¢ x2 + 83 x ¡ , ¡1 § 2i d a c d e f g h a (repeated) and ¡ 14 a P (x) = (x ¡ 2)Q(x) + 7, P (x) divided by x ¡ leaves a remainder of b P (¡3) = ¡8, P (x) divided by x + leaves a remainder of ¡8 c P (5) = 11, P (x) = (x ¡ 5)Q(x) + 11 (x ¡ 1)(x ¡ + i)(x ¡ ¡ i) b (x + 3)(x + 2i)(x ¡ 2i) p p (2x ¡ 1)(x ¡ ¡ 3)(x ¡ + 3) (x ¡ 2)(x ¡ + 2i)(x ¡ ¡ 2i) (x ¡ 1)(2x ¡ 3)(2x + 1) p p (x + 2)(3x ¡ 2)(x ¡ i 3)(x + i 3) p p (x + 1)(2x ¡ 1)(x ¡ ¡ i 3)(x ¡ + i 3) (2x + 5)(x + 2i)(x ¡ 2i) ¡3:273, ¡0:860, 2:133 b ¡2:518, ¡1:178, 2:696 EXERCISE 7F P (x) = a(2x + 1)(x2 ¡ 2x + 10) a 6= p(x) = 4x3 ¡ 20x2 + 36x ¡ 20 p = ¡3, q = 52 other zeros are + 3i, ¡4 p a = ¡13, b = 34 other zeros are ¡ i, ¡2 § p p a = 3, P (z) = (z + 3)(z + i 3)(z ¡ i 3) p p k = 2, P (x) = (x + i 5)(x ¡ i 5)(3x + 2) a k = ¡8, P (x) = (x + 2)(2x + 1)(x ¡ 2) p p b k = 2, P (x) = x(x ¡ 3)(x + 2)(x ¡ 2) a = 7, b = ¡14 a a = 700, the time at which the barrier has a b a a = b a = a = ¡5, b = a = ¡3, n = a ¡3 b 3z ¡ EXERCISE 7D.4 returned to its original position a If k = 1, zeros are 3, ¡1 § i If k = ¡4, zeros are §3, b m = ¡ 10 85 85 b k = 36 000 , f (t) = 36 000 t(t ¡ 700)2 000 000 c 120 mm, at 233 milliseconds a i P (a) = 0, x ¡ a is a factor ii (x ¡ a)(x2 + ax + a2 ) March b i P (¡a) = 0, x + a is a factor ii (x + a)(x2 ¡ ax + a2 ) b a=2 V(t) 400 200 a cuts the x-axis at ® b touches the x-axis at ® c cuts the x-axis at ® with a change in shape 100 P (x) = 2(x + 1)(x ¡ 2)(x ¡ 3) P (x) = ¡2(x + 3)(2x + 1)(2x ¡ 1) P (x) = 14 (x+4)2 (x¡3) d P (x) = f a b d P (x) = ¡2(x + 3)(x + 2)(2x + 1) P (x) = (x ¡ 3)(x ¡ 1)(x + 2) P (x) = x(x + 2)(2x ¡ 1) c P (x) = (x ¡ 1)2 (x + 2) P (x) = (3x + 2)2 (x ¡ 4) + 4)(x ¡ V(t) = 100 2 a b c e 3)2 V (t) = ¡t3 + 30t2 ¡ 131t + 250 300 EXERCISE 7E.1 P (x) = p §i 10 p EXERCISE 7D.3 (x , 2 a x = ¡2, §i b x = ¡2, ¡ 12 , c x = (treble root) d x = ¡2, 32 , p e x = ¡3, 2, § f x = ¡ 12 , 3, § i a P (x) = (x + 3)2 (x ¡ 3) or P (x) = (x ¡ 1)2 (x + 5) b If m = ¡2, zeros are ¡1 (repeated) and 23 zeros are (x a ¡1, § b 1, § i c e § 12 , 3, ¡2 f 2, § 3i a a = ¡3, b = zeros are ¡ 12 , 2, §2i p b a = 1, b = ¡15 zeros are ¡3, 12 , § 14 , 243 P (x) = EXERCISE 7E.3p p ¡1 § i a a = ¡1, zeros are p § i 11 b a = 6, zeros are ¡ 23 , , If m = P (x) = 2(x + 1)2 (x ¡ 1)2 P (x) = (x + 3)(x + 1)2 (3x ¡ 2) P (x) = ¡2(x + 2)(x + 1)(x ¡ 2)2 P (x) = ¡ 13 (x + 3)(x + 1)(2x ¡ 3)(x ¡ 3) t V (t) = ¡t3 + 30t2 ¡ 131t + 250 9:938 m or 1:112 m (x+5)(x+2)(x¡5) 10 REVIEW SET 7A a a = 4, b = b a = 3, b = ¡4 c a = 3, b = ¡7 or a = 14, b = ¡ 32 a 12 + 5i b ¡1 + i c 18 + 26i Re(z) = a F b C c A d E e D f B p , Im(z) = ¡ 34 ³ z= ¡ 15 i ´ (a + 1)2 ¡ b2 2(a + 1)b +i (a + 1)2 + b2 (a + 1)2 + b2 w is purely imaginary if b = §(a + 1), a 6= ¡1 w= a P (x) = 5(2x ¡ 1)(x + 3)(x ¡ 2) b P (x) = ¡2(x+2)2 (x¡1) c P (x) = (x¡2)(2x2 ¡3x+2) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a 12x4 ¡ 9x3 + 8x2 ¡ 26x + 15 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\880IB_HL-2_AN.CDR Thursday, 24 January 2008 11:55:32 AM PETERDELL IB_PD (881) ANSWERS b 4x4 ¡ 4x3 + 13x2 ¡ 6x + c WXY, WXZ, WYX, WYZ, WZX, WZY, XWY, XWZ, 19x + 30 a ¡ 2x + ¡ b x¡5+ x+2 (x + 2)(x + 3) “If a polynomial P (x) is divided by x ¡ k until a constant remainder R is obtained then R = P (k):” p p 10 a = 7, b = or a = 4, b = § 11 1, ¡ 12 , § i x2 12 (z + 2)2 (z ¡ + i)(z ¡ ¡ i) 13 14 15 16 17 P (z) = z ¡ 6z + 14z ¡ 10z ¡ k = 3, b = 27, x = 3, ¡3; k = ¡1, b = ¡5, x = ¡1, p p k ] ¡1, 10 ¡ 10 [ or k ] 10 + 10, [ k = ¡2, n = 36 Another hint: Show that: (®¯)3 + (®¯)2 ¡ = 15 10 EXERCISE 8E x = 2, y = ¡2 ¡ 2i a = 0, b = ¡1 a = 0, b = or a = 3, b = 18 y P (x) = (x + 2)2 (x ¡ 1)(4x ¡ 3) a b 28 c 56 d 28 e k = or ABCD, ABCE, ABCF, ABDE, ABDF, ABEF, ACDE, ACDF, ACEF, x -2 10 12 13 14 -1 8x3 ADEF, BCDE, BCDF, BCEF, BDEF, CDEF, C46 = 15 Er_ 17 = 12 376 C = 126, C C = 70 C11 22x2 3x ¡ 11 P + ¡ 16x ¡ 11), a 6= p(x) = p ¡ (2z + 1)(z + i 5)(z ¡ i 5) P (z) = a(z ¡ 4z + 5)(z + 2zp + 10), a 6= k = ¡4, zeros are § 2i, ¡1 § 15 §2i, ¡1 § i 16 7 C313 = 286, C11 C212 = 66 C512 = 792 a C22 C310 = 120 b C12 C410 = 420 C33 C01 C611 = 462 REVIEW SET 7C x = ¡1, y = 2 z = ¡ 2i or + i a x = 0, y = b x = 5, y = ¡7 c x = 0, y = 0, or p x = 1, y = 0, or x = ¡ 12 , y = z = ¡ 3737 + 169 4416 i 169 , 10 a C11 C39 = 84 b C02 C48 = 70 c C02 C11 C37 = 35 11 a C516 = 4368 b C310 C26 = 1800 c C510 C06 = 252 or x = ¡ 12 , y = d C310 C26 + C410 C16 + C510 C06 = 3312 p ¡ 23 e C516 ¡ C510 C06 ¡ C010 C56 = 4110 267 214 12 a C26 C13 C27 = 945 b C26 C310 = 1800 a = ¡21, other zeros are + i, 12 a P (x) = a(2x ¡ 1)(x2 + 2), a 6= b P (x) = a(x2 ¡ 2x + 2)(x2 + 6x + 10), a 6= c C516 ¡ C09 C57 = 4347 13 C220 ¡ 20 = 170 14 a b 15 C = 126 10 k = 0, 4, ¡ 343 ; P (x) = (x + 2)2 (2x ¡ 1) when k = i C212 = 66 ii C111 = 11 i C312 = 220 ii C211 = 55 p 11 z = ¡ 12 , 2, §i 12 a a is real, k > b a is real, k p 13 a = 7, b = ¡20 14 k = § 2 15 quotient is x + 3x ¡ 9, remainder is 5x + 17, a = 4, b = ¡18 16 a the different committees of to be selected from men and women in all possible ways b Crm+n EXERCISE 8A 18 C210 £C27 = 945 19 C210 C29 + C29 C28 + C210 C28 + C210 C19 C18 EXERCISE 8F a x3 + 3x2 + 3x + b 27x3 ¡ 27x2 + 9x ¡ c 8x3 + 60x2 + 150x + 125 d 8x3 + 12x + x6 + a 13 b 20 c 19 d 32 EXERCISE 8C x4 c 11! 6! d 13! 10!3! e 3! 6! f p yellow 95 100 50 75 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 a W, X, Y, Z b WX, WY, WZ, XW, XY, XZ, YW, YX, YZ, ZW, ZX, ZY + x4 p p a + b 56 + 24 c 232 ¡ 164 a 64 + 192x + 240x2 + 160x3 + 60x4 + 12x5 + x6 b 65:944 160 601 201 EXERCISE 8D magenta x2 a 15 20 15 b i x6 + 12x5 + 60x4 + 160x3 + 240x2 + 192x + 64 ii 64x6 ¡ 192x5 + 240x4 ¡ 160x3 + 60x2 ¡ 12x + iii x6 + 6x4 + 15x2 + 20 + x152 + x64 + x16 £ 4! b 10 £ 10! c 57 £ 6! d 131 £ 10! e 81 £ 7! 62 £ 6! g 10 £ 11! h 32 £ 8! 11! b 9! c 8! d e 34 f n + g (n ¡ 1)! (n + 1)! cyan x3 24x2 d 16x4 ¡ 32x2 + 24 ¡ 4!16! 20! 25 a f a h 10! 8! b 8x3 a ¡ + ¡ 32x + 16 b 16x4 + 96x3 + 216x2 + 216x + 81 c x4 + 4x2 + + x42 + x14 1, 1, 2, 6, 24, 120, 720, 5040, 40 320, 362 880, 628 800 a b 30 c 17 d 30 e 100 f 21 a n b (n + 2)(n + 1) c (n + 1)n 7! 4! C412 C48 C44 = 5775 3! +C110 C29 C18 + C110 C19 C28 = 12 528 EXERCISE 8B a C612 = 462 b 17 a 24 a b c 24 42 1680 a 125 b 60 17 576 000 a b c 81 XYW, XYZ, XZW, XZY, YWX, YWZ, YXW, YXZ, YZX, YZW, ZWX, ZWY, ZXW, ZXY, ZYW, ZYX a AB, AC, AD, AE, BA, BC, BD, BE, CA, CB, CD, CE, DA, DB, DC, DE, EA, EB, EC, ED b ABC, ABD, ABE, ACB, ACD, ACE, ADB, ADC, ADE, AEB, AEC, AED, BAC, BAD, BAE, BCA, BCD, BCE, BDA, BDC, BDE, BEA, BEC, BED, CAB, CAD, CAE, CBA, CBD, CBE, CDA, CDB, CDE, CEA, CEB, CED, DAB, DAC, DAE, DBA, DBC, DBE, DCA, DCB, DCE, DEA, DEB, DEC, EAB, EAC, EAD, EBA, EBC, EBD, ECA, ECB, ECD, EDA, EDB, EDC at a time: 20 at a time: 60 a 120 b 336 c 5040 a 12 b 24 c 36 720 a 24 b 24 c 48 a 343 b 210 c 120 720, 72 a 648 b 64 c 72 d 136 a 120 b 48 c 72 10 a 628 800 b 241 920 a 48 b 24 c 15 12 a 360 b 336 c 288 a 15 120 b 720 a i 628 800 ii 28 800 b i 151 200 ii 33 600 a 40 320 b 5760 c 8640 11 13 14 15 REVIEW SET 7B a(x4 881 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\881IB_HL-2_AN.CDR Tuesday, 22 January 2008 12:51:12 PM PETERDELL IB_PD (882) 882 ANSWERS 2x5 + 11x4 + 24x3 + 26x2 + 14x + a 270 b 4320 EXERCISE 9B EXERCISE 8G a a 111 + ¡11¢ (2x) + :::: + a T6 = a a 25 ¡ b ¡9¢ 23 (¡3)3 3 ¡9¢ = 28 b 36 Conjecture: (An )2 ¡ 3(Bn )2 = for n Z + x EXERCISE 10A ¡ 13 (2x2 ) ¢ h 28 (¡1)4 ¡ 35 ¡ ¢ r n 11 r=0 r = 3n ¼c 2¼ c b i 0:641c a a 36o b g 18o h a 114:59o a j 2:39c 10 12 90 135 180 225 ¼ 3¼ ¼ 5¼ Degrees 270 315 360 Radians 3¼ 7¼ 2¼ Degrees 30 60 90 120 150 180 Radians ¼ ¼ ¼ 2¼ 5¼ ¼ Degrees 210 240 270 300 330 360 Radians 7¼ 4¼ 3¼ 5¼ 11¼ 2¼ i 6:53 cm ii 29:4 cm2 b i 10:5 cm ii 25:9 cm2 3:14 m b 9:30 m2 5:91 cm b 18:9 cm a 39:3o b 34:4o 0:75c , 24 cm2 b 1:68c , 21 cm2 c 2:32c , 126:8 cm2 10 cm, 25 cm2 65 cm2 a 11:7 cm b 11:7 c 37:7 cm d 185o a ® = 18:43 b µ = 143:1 c 387:3 m2 b h 24 11 227 m2 a ® = 5:739 b µ = 168:5 c Á = 191:5 d 71:62 cm a b y 1 x C(cos 199o , sin 199o ) 70 C(cos(¡35o ), sin(¡35o )) ii A(¡0:545, 0:839) B(¡0:326, ¡0:946) C(0:819, ¡0:574) yellow 95 100 50 0o 90o 180o 270o 360o 450o µ (radians) ¼ ¼ 3¼ 2¼ 5¼ sine cosine tangent 1 undef ¡1 0 1 undef µ (degrees) ¡1 undef a ¼ 0:6820 b ¼ 0:8572 c ¼ ¡0:7986 d ¼ 0:9135 e ¼ 0:9063 f ¼ ¡0:6691 75 25 95 100 50 75 25 95 100 50 75 25 (n + 1)! ¡ for all n Z + g for all n Z + f (n + 1)! n for all n Z + i , n Z+ h n+1 6n + Proposition: The number of triangles for n points within the original triangle is given by Tn = 2n + 1, n Z + x ii A(0:899, 0:438) B(¡0:829, 0:559) C(¡0:946, ¡0:326) b i A(cos 123o , sin 123o ) B(cos 251o , sin 251o ) a 4n ¡ b all n Z + , n > c 10 for all n Z + d n(n + 1), n Z + e (n + 1)! ¡ for all n Z + magenta a i A(cos 26o , sin 26o ) B(cos 146o , sin 146o ) p6 EXERCISE 9A cyan -1 x -2 = 84 a = §4 £ 34 £ (¡2)2 = 4860 11 q = § y 95 ¡6 c y 100 e ¼ 50 8 6:92c 45 75 £ 26 £ (¡3)6 d o 5¼ c 23¼ c 18 a a a a 25 10 3:83c g 15 20 15 + + + + + x x x x x x p 362 + 209 64:964 808 It does not have one ¡6¢ c 3¼ c m 3¼ c c n 4¼ f ¼c 108o c 135o d 10o e 20o f 140o 27o i 150o j 22:5o b 87:66o c 49:68o d 182:14o e 301:78o b 1+ 5:55c ¼ c 20 e EXERCISE 10C.1 a 43 758 teams b 11 550 teams c 41 283 teams a £ £ £ = 4536 numbers b 952 numbers (a + b)6 = a6 + 6a5 b + 15a4 b2 + 20a3 b3 + 15a2 b4 + 6ab5 + b6 a x6 ¡ 18x5 + 135x4 ¡ 540x3 + 1215x2 ¡ 1458x + 729 ¡6¢ k ¼ c 10 7¼ c l EXERCISE 10B REVIEW SET 8B ¡6¢ d 12 (¡1)100 = 1 a 262 £ 104 = 760 000 b £ 26 £ 104 = 300 000 c 26 £ 25 £ 10 £ £ £ = 276 000 a 45 b 120 a n(n ¡ 1) b n + a 3003 b 980 c 2982 28 a x3 ¡ 6x2 y + 12xy2 ¡ 8y3 b 81x4 + 216x3 + 216x2 + 96x + 16 a 252 b 246 a 24 b 20 000 10 60 11 k = ¡ 14 , n = 16 12 4320 13 a 900 b 180 ¡12¢ ¼c 4¼ c Radians b REVIEW SET 8A b c Degrees = 91 854 + 10x + 35x2 + 40x3 ¡ 30x4 b 84x3 c n = and k = ¡2 a = n P ¼c 3¼ c a ¡x in row n of Pascal’s triangle is 2n d After the first part let x = ¡9¢ for n Z + A4 = 97, B4 = 56 c The sum of the numbers 16 32 n Z+ n 2n + c For n = 1, 2, 3, 4, (An )2 ¡ 3(Bn )2 = + :::: ¡ ¢3 ¡21¢ ¡12¢ c b sum 1 1 3 1 1 10 10 a (x2 )6 ¢ (¡3)3 a ¡8¢ ¢20 ¡9¢ 3 628 799 628 800 = n2 for b Conjecture: un Conjecture: un = + :::: ¡x (2x) ¡ d T9 = ¡6¢ ¡ 18 + ¡ x3 11 210 10 a A1 = 2, B1 = 1; A2 = 7, B2 = 4; A3 = 26, B3 = 15; x b T4 = ¢ ¡x b ¢19 x ¡20¢ ¡x + ¡ ¡ ¢ 13 2 ¡ ¢15 + (2x)10 + (2x)11 10 (3x) ¢ (2x) ¡ x3 x 35 25 ¡ 19 (2x) 19 ¡ ¢ ¡11¢ ¡15¢ + x 14 x (2x)10 55 ¡15¢ (3x) ¡20¢ ¡17¢ ¡10¢ 14 ¡15¢ ¡15¢ c T10 = (3x) :::: + ¡ ¢ 14 ¡20¢ (2x)2 + :::: + ¡15¢ b (3x)15 + c (2x)20 + ¡11¢ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\882IB_HL-2_AN.CDR Tuesday, 22 January 2008 12:52:58 PM PETERDELL IB_PD (883) ANSWERS p 3 §5 p b cos µ = § c cos µ = §1 d cos µ = a cos µ = § a sin µ = a Quadrant p § 47 b sin µ = Degree measure c sin µ = d sin µ = §1 Radian measure cos µ sin µ tan µ ¼ +ve +ve +ve <µ<¼ ¡ve +ve ¡ve 3¼ ¡ve ¡ve +ve +ve ¡ve ¡ve < µ < 90 90 < µ < 180 0<µ< 180 < µ < 270 ¼<µ< 270 < µ < 360 3¼ ¼ 10 11 < µ < 2¼ p b cos µ = ¡ p b ¡2 c 10 a ¡ 2p p2 , 13 11 a sin x = p5 p 21 p p3 13 cos x = p ¡ 37 d 14 13 ¼ b 15o c 84o a b ¡2 tan µ c cos µ d sin µ e cos2 ® f sin2 ® g a sin µ b ¡2 sin µ c d ¡ cos µ e cos µ f sin µ Á ¡ µ = ¡(µ ¡ Á) a tan µ b ¡ tan µ c d tan µ e tan µ f tan µ 5¼ 2:175c c perimeter = 34:1 cm, area = 66:5 cm2 r = 8:79 cm, area = 81:0 cm2 67:4o or 112:6o a 10 600 m2 b 1:06 a ¡0:743 b ¡0:743 c 0:743 d ¡0:743 10 p b c 2 a y x a sin ¯ cos ¯ a i b c ¡ p1 p1 ¡1 ¡1 b c ¡ 12 p ¡ p tan ¯ ¡ p1 p 2 p 3 d p p p ¡ p e ¡ 14 p f g a e µ= h µ= 7¼ 11¼ 19¼ 23¼ , , , 6 ¼ 2¼ , c µ=¼ 3 c d EXERCISE 11B.1 a x = 28:4 b x = 13:4 c x = 3:79 a a = 21:25 cm b b = 76:9 cm c c = 5:09 cm 3¼ 7¼ , 2 µ = ¼2 EXERCISE 11B.2 ]C = 62:1o or ]C = 117:9o a ]A = 49:5o b ]B = 72:05o or 107:95o c ]C = 44:3o sin 27o ]ABC = 66o , BD = 4:55 cm sin 85o 6= No, x = 17:7, y = 33:1 11:4 9:8 a 88:7o or 91:3o b 91:3o c cosine rule as it avoids the ambiguous case p Area ¼ 25:1 cm2 x = + 11 2 0, ¼, 2¼ j µ= ¼ 2¼ 4¼ 5¼ , , , 3 10 a 36:2 cm2 b 62:8 cm2 11 a i and ii cm2 b i 21:3 cm2 ii 30:7 cm2 EXERCISE 11C 10 12 REVIEW SET 10A 21:1 km2 a 118 cm2 b 44:9 cm2 M(cos 73o , sin 73o ) ¼ (0:292, 0:956) yellow 95 100 50 75 a x = 34:1 b x = 18:9 a x = 41:5 b x = 15:4 25 95 50 75 25 95 100 50 75 100 magenta 17:7 m 207 m 23:9o 77:5 m 9:38o 69:1 m a 38:0 m b 94:0 m 55:1o AC ¼ 11:7 km, BC ¼ 8:49 km a 74:9 km2 b 7490 hectares 11 9:12 km ¼ 85 mm 13 10:1 km 14 29:2 m 15 37:6 km REVIEW SET 11A N(cos 190o , sin 190o ) ¼ (¡0:985, ¡0:174) P(cos 307o , sin 307o ) ¼ (0:602, ¡0:799) 25 p ¡3 + 73 c x = p53 p a x ¼ 10:8 b x ¼ 9:21 a x = b Area = 24 cm2 a 28:9 cm2 b 384 km2 c 28:3 cm2 x = 19:0 18:9 cm2 137 cm2 374 cm2 7:49 cm 11:9 m a 48:6o or 131:4o b 42:1o or 137:9o 14 is not covered + k¼ a x = + 22 b x = EXERCISE 10D p o 120 , 300o b ¡ p 3¼ 5¼ , f µ = ¼2 , 3¼ g µ= ¼ 3¼ 5¼ 7¼ 5¼ 11¼ , , , i µ = , 4 4 cyan ¼ 2¼ cos( 8¼ )= a 28:8 cm b 3:38 km c 14:2 m ]A = 52:0o , ]B = 59:3o , ]C = 68:7o 112o a 40:3o b 107o a cos µ = 0:65 b x = 3:81 2 h j k ¡1 l ¡ ¼ 11¼ 13¼ 23¼ , , , b µ = ¼3 , 5¼ b µ= a p sin( 8¼ )= p , ¡ 12 p , ¡2 EXERCISE 11A ¡ p1 a 30o , 150o b 60o , 120o c 45o , 315o d 120o , 240o e 135o , 225o f 240o , 300o a ¼4 , 5¼ b 3¼ , 7¼ c ¼3 , 4¼ d 0, ¼, 2¼ e ¼6 , 7¼ 4 2¼ 5¼ f 3, a sin( 2¼ )= a 12 b 12 c ¡ 12 a b sin µ 10 x = 47:5, AC = 14:3 cm or x = 132:5, AC = 28:1 cm 11 36:8 cm2 ¡ 12 b a 150o , 210o b 45o , 135o ¾ c e p ¡ 23 0, ¡1 a µ = ¼ + k2¼ b µ = d ¡ 12 p ¡ 23 a 0, ¡1 b 95 ¡ p1 ¡ p1 e 100 tan µ ¡ p1 d 50 cos µ c p p2 13 cos( 2¼ )= 75 p p sin µ b b a (0:766, ¡0:643) b (¡0:956, 0:292) 25 a 12 a ¡ p313 § REVIEW SET 10C EXERCISE 10C.3 d 3¼ a b c ¡2:478c d ¡0:4416c o o a 72 b 225 c 140o d 330o a 171:89o b 83:65o c 24:92o d ¡302:01o 11 a EXERCISE 10C.2 5¼ b 1:239c d sin µ = ¡ 12 13 d sin x = ¡ 12 , cos x = 13 14¼ 15 b c 174o 0:358 b ¡0:035 c 0:259 d ¡0:731 1, b ¡1, 0:961 b ¡0:961 c ¡0:961 d ¡0:961 p b ¡ 12 p1 15 2¼ a b sin x = 45 , cos x = ¡ 35 , cos x = ¡ p3 14 c sin x = ¡ c cos µ = a a a a a a 133o REVIEW SET 10B b i and ii and iii iv a sin µ = µ ¼ 102:8o 883 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\883IB_HL-2_AN.CDR Monday, 15 December 2008 12:36:50 PM TROY IB_PD (884) ANSWERS AC = 12:55 cm, ]A = 48:6o , ]C = 57:4o 113 cm2 7:32 m 204 m 530 m, bearing 077:2o 179 km, bearing 352o If the unknown is an angle, use the cosine rule to avoid the ambiguous case 10 a x = or b Kady can draw cm triangles: b The data is periodic i y = 32 (approx.) ii ¼ 64 cm iii ¼ 200 cm iv ¼ 32 cm c A curve can be fitted to the data a periodic b periodic c periodic d not periodic e periodic f periodic B 6m B' given could give two triangles: A 44° 8m a C b ¼ 2:23 m3 b p c y d 10 12 Data exhibits periodic behaviour b a y c y = -\Ew_\sin x x p 2p y = sin 3x p -0.5 2p 3p x 3p x 3p x -1 b y x y = sin &w_* x p -0.5 2p -1 c Not enough information to say data is periodic It may in fact be quadratic y = sin (-2x) y 0.5 y p -0.5 2p -1 10 12 x a Not enough information to say data is periodic 10¼ b B=3 c B= y d B= ¼ e B= ¼ 50 y = 2sin x + sin 2x 360 magenta yellow 50 75 25 95 100 50 720 x -2 75 25 95 100 50 75 25 c 6¼ d 95 100 5 a distance travelled 100 150 200 250 300 350 400 cyan ¼ b a B= height above 60 ground (cm) 50 40 30 20 10 50 ¼ 95 50 y x 0.5 75 2p p y y 25 y = \Ew_\sin x x 2p 0.5 Not enough information to say data is periodic It may in fact be quadratic y -1 -2 -1 a x x 10 y = -3sin x -1 -2 a x p EXERCISE 12A d y -1 -2 -3 a QS = 45 ¡ 36 cos Á b i Rb SQ = 52:5o or 127:5o ii 24:8 units c 23:2 units2 2p p -1 -2 -3 bG ¼ 74:4o 3:52 km 42 km a 2:18 pm b 157o ED a d2 = x2 ¡ (10 cos 20o )x + 25 b x = cos 20o b ii b = 103:8, d = 76:2 iii a = 95:4, c = 84:6 a max value 16 when x = b i y = 12 ¡ xp ii y2 = x2 + 64 ¡ 16x cos µ d max area = units2 when x = y = fi.e., isos ¢g y = 3sin x cm REVIEW SET 11B y cm 60° cm 11 a The information EXERCISE 12B.1 100 884 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\884IB_HL-2_AN.CDR Tuesday, 22 January 2008 1:49:02 PM PETERDELL IB_PD (885) ANSWERS b e y y = sin x + sin 2x + sin 3x 720 360 -2 y x 360 540 a 720 -4 0.5 p b 8¼ c ¼ a 4p 2p µ j translation ¼ 2p x y 4p 2p 2 a T ¼ 6:5 sin ¼6 (t ¡ 4:5) + 20:5 a T ¼ 4:5 sin ¼6 (t ¡ 10:5) + 11:5 T ¼ 13:1 sin(0:345)(t + 6:87) ¡ 5:43 a H ¼ sin(0:507)(t ¡ 3:1) b H x 9.3 15.5 -7 ¼ H = 10 sin 50 (t ¡ 25) + 12 y = sin x - p EXERCISE 12D 2p a y = cos x + x y -p y y = sin (x - 2) p 2p b y = cos x ¡ x y ¡ y = sin (x + 2) p c y = cos x ¡ 2p x ¼ ¢ -p p x p x y -p -1 -2 d ´ y -1 -2 c ¡2 ¡3 t -1 -2 -3 1 ¼ d 25 50 ¼ right EXERCISE 12C 4p EXERCISE 12B.2 b b 20 c ¶ 3.1 2 x i vert.¡ stretch factor 2, followed by a horiz.¡ compression, factor x y -0.5 -1 a 2p g reflection in the x-axis h translation -0.5 -1 0.5 p y = sin &x - \y_\* + ³ y -0.5 -1 x a vert translation ¡1 b horiz translation c vert.¡ stretch, factor d horiz.¡ compression, factor e vert.¡ stretch, factor 12 f horiz.¡ stretch factor -2 b 2¼ 2p y -1 @ = &_Asin _* x 180 0.5 Qw_ a p f p y = sin &x + \r_\* -1 -2 x -4 c y -2 y p y ¡ y = sin x + d y = cos x + ¼ ¢ y p 2p x -p p magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 -2 x -1 -1 cyan 885 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\885IB_HL-2_AN.CDR Friday, 12 December 2008 2:44:03 PM TROY IB_PD x (886) 886 ANSWERS e y= y cos x EXERCISE 12E.2 a i y = tan(x ¡ ¼2 ) We_ -p x p -2 f y= 2 y cos x y x p 2p 3p p 2p 3p p 2p 3p Ew_ -p -2 x p -4 ii y = ¡ tan x g y = ¡ cos x y -p ¡ ¢ ¼ h y = cos x ¡ -1 +1 x p y x y -2 -p ¡ i y = cos x + ¼ ¢ -4 x p iii y = tan 2x -2 ¡1 y 2 -p x x p -2 -2 j y = cos 2x y -4 ³ ´ -p EXERCISE 12F.1 y a x = 0:3, 2:8, 6:6, 9:1, 12:9 b x = 5:9, 9:8, 12:2 a x = 1:2, 5:1, 7:4 b x = 4:4, 8:2, 10:7 a x = 0:4, 1:2, 3:5, 4:3, 6:7, 7:5, 9:8, 10:6, 13:0, 13:8 b x = 1:7, 3:0, 4:9, 6:1, 8:0, 9:3, 11:1, 12:4, 14:3, 15:6 a i 1:6 ii ¡1:1 b i 1:557 ii ¡1:119 c i x = 1:1, 4:2, 7:4 ii x = 2:2, 5:3 x p -1 l y = cos 2x EXERCISE 12F.2 -p a x = 1:08, 4:35 b x = 0:666, 2:48 c x = 0:171, 4:92 d x = 1:31, 2:03, 2:85 x = ¡0:951, 0:234, 5:98 x p EXERCISE 12F.3 -3 c C: horizontal translation, D: vertical translation + c y = ¡5 cos a x = 2¼ , 4¼ , 8¼ , 10¼ , 14¼ b x = ¡ 11¼ , ¡ 7¼ , ¼6 , 5¼ 3 3 6 5¼ 7¼ 17¼ 5¼ ¼ c x = , , d x = ¡ , ¡¼, , ¼ e x = ¡ 13¼ , ¡ 3¼ , ¡ ¼6 , ¼2 , 11¼ , 5¼ f x = 0, 3¼ , 2¼ 6 2 x EXERCISE 12E.1 cyan magenta T X= yellow 95 N 100 50 O 45° 75 25 95 100 50 75 25 95 100 50 75 25 a b 0:27 c 0:36 triangle TON is isosceles, ON = TN d 0:47 e 0:70 f g 1:19 h 1:43 ¼ a x = ¼2 + k¼ ¼ b x = 12 + k ¼4 + k¼ 95 d ¡¼ ¢ 100 a y = cos 2x b y = cos ¡x¢ 50 2¼ , period 75 A: amplitude, B: ¼ 13¼ 25¼ , , b x = ¡ ¼3 , 5¼ 7¼ 5¼ 3¼ x = ¡ , ¡ , ¡ , ¡ ¼2 , ¼2 , 3¼ , 5¼ , 7¼ 2 x = ¼3 , 5¼ , 4¼ , 11¼ , 7¼ , 17¼ , 10¼ , 23¼ 6 6 a x= c 6¼ c 100 25 a 2¼ b reflection in x-axis y -p c horizontal stretch, factor k = a ¼ b ¼2 c ¼n -1 k y = cos a translation through x p ¡x¢ y black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\886IB_HL-2_AN.CDR Friday, 12 December 2008 2:44:36 PM TROY c x= ¼ ¡ ¼3 ¾ + k¼ IB_PD (887) ANSWERS a x = 0, a y ¼ , b x= ¼ ¼ 5¼ 9¼ , , 4 a y=sin¡x y b x= a x= c x= ¼ 5¼ or 3p ?? p c x= 3¼ or 7¼ b 4 ¼ 2¼ 7¼ 5¼ , , , 6 ¼ or -p 2p -5 5¼ b y -p -5 minute c minutes b x = 5¼ , 7¼ c x = 0:245, 3:387 a x = ¼3 , 5¼ 4 d no solutions e no solutions f x = 0:232, 1:803, 3:373, 4:944 40 H(t) (3, 20) EXERCISE 12I 10 a c e a f l t 400 b i 577 ii 400 650 It is the maximum population 150, after years e t ¼ 0:26 years H(t) = cos( ¼t ) + b t ¼ 1:46 sec a cos µ b ¡ sin µ c sin ® d ¡ cos ® e ¡ sin A on the 5th and 11th days 98:6 cents L¡1 on the 1st and 15th day f ¡ sin µ g EXERCISE 12H a csc x = 53 , sec x = 54 , cot x = 43 b csc x = ¡ p35 , sec x = 32 , cot x = ¡ p25 p b c d b ¡ p13 c ¡ p23 a c cot ¯ = sin x yellow A= 95 cos2 A = cos = ¡ S+D ¢ ¡ S+D ¢ 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 d p ¡ 23 sin 2A ¡ 12 cos 2A 22 sin A cos A = e sin cos µ sin µ ¡ cos µ sin µ + cos µ d cos(® ¡ ¯) tan µ h tan(3A) ¡ 21 21 b i 12 cos 2µ ¡ 12 cos 4µ ii 12 cos 5® ¡ 12 cos 7® iii cos 4¯ ¡ cos 6¯ iv cos 3µ ¡ cos 5µ 1 v cos 6A ¡ cos 10A vi 10 cos 4M ¡ 10 cos 10M 23 d cos p b i tan µ tan A + tan B + tan C ¡ tan A tan B tan C ¡ tan A tan B ¡ tan A tan C ¡ tan B tan C sin2 magenta sin µ + cos µ p 20 b i 12 cos 5µ + 12 cos 3µ ii 12 cos 8® + 12 cos 6® iii cos 4¯ + cos 2¯ iv cos 8x + cos 6x v 32 cos 5P + 32 cos 3P vi 18 cos 6x + 18 cos 2x f sin µ = ¡ 35 , cos µ = ¡ 45 , tan µ = 34 , cyan p + tan µ ¡ tan µ h 19 b i 12 sin 4µ + 12 sin 2µ ii 12 sin 7® + 12 sin 5® iii sin 6¯ + sin 4¯ iv sin 5µ + sin 3µ v sin 7® ¡ sin ® vi 16 sin 8A ¡ 16 sin 2A p csc µ = ¡ 53 , sec µ = ¡ 54 cos x a b c d cos x e cos x f sin2 x sin µ + ¡ p1 = e sin ¯ = ¡ p15 , cos ¯ = ¡ p25 , csc ¯ = ¡ 5, p , 2 + tan µ ¡ tan µ a cos µ b sin 3A c sin(B ¡ A) e ¡ cos(µ + Á) f sin(® ¡ ¯) g a cos 2® b ¡ sin 3Á c cos ¯ p p 7 a + b ¡2 ¡ 17 p 10 a ¡1 b tan(2A) 11 12 13 tan A = §1 14 tan ® = 25 15 62 16 tan(A + B + C) d undefined p p sin x = ¡ 47 , tan x = ¡ 37 , csc x = ¡ p4 , sec x = 43 , cot x = ¡ p3 p cos x = ¡ 35 , tan x = p2 , csc x = ¡ 32 , p sec x = ¡ p3 , cot x = 25 p p sin x = 521 , cos x = 25 , tan x = 221 , csc x = p5 , cot x = p2 21 21 p sin x = 12 , cos x = ¡ 23 , tan x = ¡ p1 , p sec x = ¡ p2 , cot x = ¡ 3 sec ¯ = ¡ (1 + sin µ)(1 ¡ sin µ) b tan ®(3 tan ® ¡ 2) (sec ¯ + csc ¯)(sec ¯ ¡ csc ¯) d (2 cot x ¡ 1)(cot x ¡ 1) (2 sin x + cos x)(sin x + cos x) b ¡ tan2 ¯ c sin2 µ d ¡ sin2 ® e sin2 ® g 13 h cos2 µ i j sin µ k cos µ cos ¯ + sin ¯ m sec µ + EXERCISE 12J i true ii true b 116:8 cents L¡1 a x 2p 2:51 t 5:49 20 a p -2p 30 a c d a a c d y = cot¡x i 7500 ii 10 300 b 10 500, when t = weeks i at t = 13 wks and 23 wks ii at t = 13 wks 20 m b at t = x 2p EXERCISE 12G a c d a d p -2p ¼ 5¼ 3¼ 13¼ 17¼ 7¼ , , , 12 , 12 , 12 12 x= y = sec¡x y=cos¡x x p ? 887 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\887IB_HL-2_AN.CDR Friday, 12 December 2008 2:45:28 PM TROY sin ¡ S¡D ¢ ¡ S¡D ¢ = 2 cos 2A + cos S + cos D cos D ¡ cos S IB_PD (888) 888 ANSWERS sin 3x cos 2x b cos 5A cos 3A c ¡2 sin 2® sin ® cos 4µ sin µ e ¡2 sin 4® sin 3® f sin 5® cos 2® ¡ ¢ ¡h¢ sin 3B sin B h cos x + h sin 2 ¡ ¢ ¡ ¢ ¡2 sin x + h sin h 2 24 a d g i 7 a b ¡ 25 a ¡ 79 b 19 p p a cos ® = ¡3 , sin 2® = p p b sin ¯ = ¡ 21 , sin 2¯ = ¡425 21 p a 13 b tan A = ¡ 73 p p tan ¼8 = ¡ a 9+52 b ¡ ¢ a b 10 b cos A = 13 a Domain fx j x 6= b Domain fx j x 6= ¼ ¼ tan A = p d cos 2¯ e ¡ cos 2Á i ¡ cos 2® ¡ cos 6¯ n cos 10® r ¡2 cos 6P a a a a T ¼ 7:05 sin x= a x= k Z g, Range fy j y is in R g 2¼ 4¼ 8¼ , , 9 c b 3¼ + k2¼ ¼ ¼ 5¼ , , b x ¡ 2¼ , ¡ ¼3 , ¼3 , 2¼ 3 = ¡ 5¼ , ¡ ¼6 , 2p y = cos x - y = cos x x p -0.5 2p y = cos &! - p\r_\* -1 c y y = 3cos 2x 11¼ p y = cos x -2 d y x 2p p y = 2cos &! - \e_\* + y = cos x p 2p -2 y = 4sin x a 28 milligrams per m3 b 8:00 am Monday a y = ¡4 cos 2x b y = cos ¼4 x + x a x ¼ 1:12, 5:17, 7:40 b x ¼ 0:184, 4:616 cyan magenta yellow 95 p ¡1 ¡ p 2 p b 2¡ 100 25 95 100 50 75 25 95 100 50 75 25 95 100 75 a 50 -4 75 2p p 50 x p y -2 25 2¼ + k¼ 2 5¼ , ¾ 0.5 or ¼, y b 7¼ ¼ , y = cos x REVIEW SET 12A ¼ 5¼ -3 as ¡1 sin x c b ¡ sin x sin 20x sin 2nx i sin 8x ii c sin x sin x sin(24 x) sin(26 x) i ii 24 26 sin(2n+1 x) sin x cos x cos 2x:::::: cos 2n x = 2n+1 sin 32µ sin µ b x= -1 x= 2¼ b x = ¼3 , ¼2 , 3¼ , 5¼ 7¼ 3¼ 11¼ x= x= , , no solutions f x = ¼6 , 5¼ , 7¼ , 11¼ 6 7¼ 3¼ 11¼ x = 0, ¼6 , ¼2 , 5¼ , ¼, , , , 2¼ h x = ¼4 6 b x = 0, -2 0, ¼, 7¼ , 11¼ , 6 ¼ 5¼ , ¼, d 3 ¡ sinn x a ¡ sin x b 4¼ 5¼ 10¼ 11¼ 16¼ 17¼ , , , , , 9 ¡7¼ ¡5¼ ¼ 3¼ , , 4, 4 EXERCISE 12M b b x= y a a x ¼ 0:896, 2:246 b x ¼ 3:33, 6:10 c x ¼ 0:730, 2:412, 3:871, 5:553 7¼ 11¼ 19¼ 23¼ , , , 6 REVIEW SET 12B EXERCISE 12L a x= c x= ¡ 10:5) + 24:75 10 a 5000 b 3000, 7000 c 0:5 < t < 2:5 and 6:5 < t p a cos x ¡ sin x = 29 cos(x + 1:19) b x = 0:761, ¼ d x = 0:761 (the solution x = ¼ has been lost) a c e g ¼ (t 2p x ¼ 0:392, 2:750, 6:675 b x ¼ 5:42 x ¼ 3:25, 4:69 b x ¼ 1:445, 5:89, 7:73 a x= 16 k = 2, b = ¼6 18 b µ = ¡ 8¼ , ¡ 4¼ , ¡ 2¼ , 9 19 a sin 3µ = ¡4 sin3 µ + sin µ b µ = 0, ¼4 , 3¼ , ¼, 5¼ , 7¼ , 2¼, 9¼ , 11¼ , 3¼ 4 4 21 x p -1 + k¼ , k Z g, Range fy j y > or y ¡1g c Domain fx j x 6= ¼ + k¼, k Z g, Range fy j y is in R g k¼ , p y = sin &! - \e_\* + y 6¼ sin 2® c 12 sin 2® ¡ cos 2M h cos 2® sin 6® l cos 8µ m p cos 4A q cos ® sin 2® b cos 2N g sin 4A k ¡ cos 6D 12 a cos A = x -1 24 25 10 a f j o 2p p EXERCISE 12K y = sin 3x y black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\888IB_HL-2_AN.CDR Friday, 12 December 2008 2:47:14 PM TROY IB_PD x (889) ANSWERS 10 a x = ¼2 , 3¼ , 5¼ , 7¼ b x = ¡¼, ¡ ¼3 , ¼, 5¼ 2 3¼ 7¼ a x = 0, , 2¼, , 4¼ b x = ¼6 , 2¼ , 7¼ , 5¼ a cos µ b ¡ sin µ c cos2 µ d ¡ cos µ e sin 2µ a sin2 ® ¡ sin ® + b ¡ sin 2® sin ® + cos ® a ¡ cos µ b 120 169 sin 119 169 b ¡x¢ = ³ c ³ a REVIEW SET 12C a a ¡ cos ® c p ¡ , 4 cos ® = p p7 2 ¼ ¼ ii x = + k¼ + k ¼2 c x ¼ 0:612 + k¼ p2 , 13 sin µ = ¡ p3 13 cos µ = a x p p b ¡387 c = ¼3 , 2¼ , 4¼ or 5¼ 3 p1 sin ® = c x= + k¼ a f a d g a p ¡ 18 d ¼ 5¼ 4¼ 11¼ , , , b x= 16 1000 1500 1000 800 2300 1200 1200 C D 46 35 58 23 69 86 A 58 17 b µ ¶ 40 50 65 35 40 35 40 55 44 40 35 @ 25 X= 18 a P = ( 27 35 39 ) ¶ c µ ¡14 Q= R ¶ 18 25 fr 13 st 36 19 mi ³ ¡1 ´ R ¶ 41 44 fr 24 42 st 67 43 mi 95 ¶ magenta = $291 N= @2A a i b i a b ¡1 ´ d ³ ¡3 ¡6 ¡12 ¡3 ³ C= ³ 12:5 9:5 78 669:5 65 589 µ1 ´ ´ ¶ ³ N= 2375 5156 2502 3612 income from day income from day ´ c $144 258:50 ³ ´ ³ ´ c 48 70 b 19 P= 52 76 22 d My costs at store A are E48, my friend’s costs at store B e store A are E76 a ³ does not exist ii ( 28 29 ) µ2 ¶ (8) ii 12 µ ¡2 ¶ (3 3) b 1 a ´ 100 50 75 25 95 100 50 75 25 2 µ a ³ 12 24 ´ b ³ ´ c A + (B + C) = EXERCISE 13B.2 c µ L B+A= C C A 0:07 B 0:90 C @ 0:41 A ¡0:28 ¡0:05 yellow R= EXERCISE 13B.6 ´ µ 16 18 15 13 21 16 10 22 24 a ´ ³ 48 12 1:79 b subtract c cost price B 28:75 C from sel@ 1:33 A ling price 2:25 3:51 b µ L a (A + B) + C = cyan 14 ¡9 14 ¡12 ¡14 ¡3 5 1 Number of columns in A does not equal number of rows in B a m = n b £ c B has columns, A has rows ³ B B @ 4 4 EXERCISE 13B.5 a x = ¡2, y = ¡2 b x = 0, y = a A+B= ¡ 12 229 101 R ¶ 19 fr 17 29 st 31 24 mi 2 b total points = ( 10 ) @ A = 56 points ¶ µ 187 ¶ 1:72 ii B 27:85 C @ 0:92 A 2:53 3:56 à VHS à games 70 231 95 b 0 ¡11 ¡3 100 137 49 d 50 92 52 µ 211 ¶Ã DVD X= 4 ´ 75 Saturday µ ¶ 102 L µ 23 a ¡6 ¡1 ³ 95 a i c ´ 12 ¡6 14 ¡5 ¡4 µFriday ¶ 85 100 a ³ ¶ b µ ¡14 ¡8 10 ¡2 ¡5 18 51 64 A 43 13 25 ¡1 ´ D µ4¶ 40 30 A 35 50 50 µ 20 a b ³ C 34 26 43 17 O e 2A + 2B h 4A ¡ B i 3B ¡ B c X = 2C ¡ 4B f X=A¡B X = 2C h X = 12 B ¡ A i X = 14 (A ¡ C) µ1 ¶ ³ ´ b ³ ´ c ¡ 14 ¡1 ¡6 b total cost = ( 27 35 39 ) 75 3 ´ B 34 34 34 17 µ4¶ 25 a ³ A 26 @ 43 34 a P = ( 10 ) EXERCISE 13B.1 a (11) b (22) c (16) ( w x y z ) groceries 0:15 0:95 1250 1000 A 1300 1200 @ 1500 B ´ 14 16 EXERCISE 13B.4 @ 2:35 A ³ ´ 1 4 3A b O c ¡C d ¡A ¡ B g ¡2A + C X=A¡B b X=C X = 12 A e X = 13 B X= a 1£4 b 2£1 c 2£2 d 3£3 a (2 ) b c total cost of 1:95 d 46 46 46 23 ¶ 10 60 m EXERCISE 13A µ ³ EXERCISE 13B.3 a ¡ sin µ b cos µ 10 1:5 m REVIEW SET 12D a ¡ ¾ ¼ 2¼ iii x = A ´ b 75 à DVD 136 à DVD b 27 à VHS 43 à VHS 102 à games 129 à games c total weekly average hirings 12F a a i x ¼ 1:33 + k¼ ii x ¼ 5:30 + k4¼ iii x ¼ 2:83 + k¼ b i x= ´ 11 14 11 35 @ 58 46 12 p sin 2® = 889 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_an\889IB_HL-2_AN.CDR Thursday, 11 March 2010 4:39:28 PM PETER ¶ b µ 10 ¡7 ¡4 ¡10 ¶ IB_PD (890) ANSWERS b ( 369 d 44 75 385 ) 420 $224 660 a ¶ 150 140 40 40 80 65 c¡ µ e 115 @ 136 A 46 106 a ¡3 b c ¡12 Hint: Let A = a jAj = ad ¡ bc jBj = wz ¡ xy 657 730 670 657 369 730 420 670 385 µ15 ³ µ 15 £ EXERCISE 13B.7 ³ AB = ¡1 ¡1 ´ ³ BA = AO = OA = O b I = ³ a a 0 A2 ´ ³ b ³ ´ 97 ¡59 118 38 a c ³ ´ , For example, A = 0 ´ 2 ³ b 0 ´ ³ = 3I, , gives A2 = 10 0 10 ´ ¡2 ¡ 23 ³ = 10I, 0:2 0:4 ¡0:1 0:3 ¡3 ¡1 ³ b i ¡4 ´ ³ h ´ ¡3 ¡1 ´³ ´ ¡3 ´ 59 , 13 d x= ´ ¡1 ´ x y ³ = 11 ¡5 ´ y = ¡ 25 13 , 34 ii y= µ X= ³ ´ 55 34 13 ¡ 27 ¡ 87 ¶ ¡1 , k 6= ¡3 k 2k³+ ´ k , k 6= 3k ³ ´³ ´ x y ´ k ¡2 ¡1 k + , k 6= ¡2 or ³ ´ , jAj = 10 11 = ´³ ´ k ¡1 ³ b Yes, x = 2:5, y = ¡1 ii x y ii k 6= ¡ 12 , x = ³ = 11 ´ , jAj = ¡2 ¡ 4k + 11k y= + 4k + 2k iii k = ¡ 12 , no solutions EXERCISE 13C.3 0 0 µ X= ³ b , b 6= ³ ³ ¡4 0 b x = ¡ 37 , y = ¡ 75 23 23 (k + 2)(k ¡ 1) ii ´ a 4 ¶ 0 A¡1 ´ ³ , µ = ¡1 0 ¡1 ¡1 2 ´ ³ , ¶ 1 ´ ³ , , (A¡1 )¡1 = ¡1 ¡1 ³ ¡1 ´ ´ b (A¡1 )¡1 (A¡1 ) = (A¡1 )(A¡1 )¡1 = I c (A¡1 )¡1 = A µ iv ´ 3 ¡ 13 µ ¶ 6 3 µ ii ¶ 2 ¶ v µ 6 3 iii ¶ µ vi ¶ 3 3 µ ¶ c (AB)¡1 = B¡1 A¡1 and (BA)¡1 = A¡1 B¡1 d (AB)(B¡1 A¡1 ) = (B¡1 A¡1 )(AB) = I magenta yellow 95 100 50 75 25 95 AB and B¡1 A¡1 are inverses 100 50 75 22 k = ii ´ 25 95 100 50 75 25 ³ = a i k = ¡3 ii a ¡2 b ¡1 c d a 26 b c ¡1 d a2 + a cyan a b X= a i EXERCISE 13C.1 ´³ ´ ³ b i ³ 10 g ³ c does not d exist e x = ¡40, y = ¡24 f x = ¶ 1¡a ´ ´ = y = ¡ 37 13 a a = 3, b = ¡4 b a = 1, b = p = ¡2, q = a A3 = 5A ¡ 2I b A4 = ¡12A + 5I ³ ¡1 b i 2 , jABj = (ad ¡ bc)(wz ¡ xy) 1 ¡1 ³ ´ x y c i k = ¡2 or A2 = b ´³ ´ 17 , 13 ´ , ³ a ³ ¡1 c x= ´ µ b ´ they are not square matrices y= ¶ c false as A(A ¡ I) = Oà does not imply that! A = O or A ¡ I = O a b d ³0 0´ ³1 0´ a ¡ a2 0 ³ a i 0 0 a b c d b A and B are not inverses since 32 , does not exist b when A is a square matrix AB = ´ a x= A2 + A b B2 + 2B c A3 ¡ 2A2 + A d A3 + A2 ¡ 2A AC + AD + BC + BD f A2 + AB + BA + B2 A2 ¡ AB + BA ¡ B2 h A2 + 2A + I i 9I ¡ 6B + B2 A3 = 3A ¡ 2I A4 = 4A ¡ 3I B3 = 3B ¡ 2I B4 = 6I ¡ 5B B5 = 11B ¡ 10I C3 = 13C ¡ 12I C5 = 121C ¡ 120I i I + 2A ii 2I ¡ 2A iii 10A + 6I A2 + A + 2I c i ¡3A ii ¡2A iii A a 0 ¡2 ¡4 ¡1 EXERCISE 13C.2 EXERCISE 13B.8 a e g a b c a b ³ ³ b ³ f ¡ 15 a AB = AB 6= BA 0 ´ ¡4 does e not exist 12 13 11 14 16 4 ³ 14 a ¶ c (( 125 195 225 ) ¡ ( 85 120 130 )) ³ a i ¡2 ii ¡8 iii ¡2 iv ¡9 v ¶ 12 13 11 14 16 4 µ 15 12 13 11 14 16 ¶ ¡ ( 85 120 130 ) £ 4 = $7125 µ15 12 13 11 14 16 ¶ b ( 125 195 225 ) £ 4 µ20 20 20 20 20 20 20 ¶ ¡ ( 85 120 130 ) £ 15 15 15 15 15 15 15 5 5 5 = ¡$9030, i.e., a loss of $9030 ( 125 195 225 ) £ ´ b AB = aw + by ax + bz cw + dy cx + dz ¢ 50 (3 2) d 75 a µ 125 ¶ 95 132 176 198 44 154 88 110 176 44 88 88 132 100 µ 22 c 25 890 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_an\890IB_HL-2_AN.CDR Tuesday, 13 January 2009 12:54:14 PM TROY IB_PD (891) ANSWERS ³1 ´ k 10 a a = 50 000, b = 100 000, c = 240 000 b yes c 2007, ¼ $284 000, 2009, ¼ $377 000 are inverses EXERCISE 13F.1 a x = 2, y = ¡3 b x = ¡1, y = c x = ¡2, y = ¡4 a intersecting b parallel c intersecting d coincident e intersecting f parallel a The second equation is the same as the first when divided A2 = 2A ¡ I, A¡1 = 2I ¡ A a A¡1 = 4I ¡ A b A¡1 = 5I + A c A¡1 = 32 A ¡ 2I 10 If A¡1 exists, i.e., jAj 6= throughout by The lines are coincident EXERCISE 13D.1 a 41 b ¡8 c d e ¡6 f ¡12 a x = or b When x = or 5, does not have an inverse a abc b c 3abc ¡ a3 ¡ b3 ¡ c3 k 6= ¡3 or ¡9 p ¡1 § 33 or b k = or b ¡34, a 16, ¡ 21 ¡ 17 16 16 B¡ 17 ¡ 29 B 16 16 B 5 @ 4 15 16 11 16 a B21 @ 2 3 5 11 16 15 16 ¡1 ¡ 34 ¡ 34 ¡ 16 a µ B¡ 15 B 34 B¡ 29 B 34 B 39 @ 34 = 2I, ¡ 74 ¡ 34 ¡ 34 ¡ 14 2 @¡1 a µ1 c µ1 1 b C A ¶ ¡1 ¡1 1 ¡1 ¡3 ¶µx¶ µ2¶ y z = ¡1 1 ¡1 ¡3 b c = ¶µ ¶ ¡1 a 15 2 2 ¡ 32 ¡1 ¡ 17 ¡ 61 17 58 17 ¡ 17 3¡t , t2R ii when y = s, x = ¡ 2s, s R a The system is inconsistent and so has no solutions The lines are parallel b The lines are coincident Infinitely many solutions ¡ 2t , t2R exist of the form x = t, y = b If k 6= ¡4, the system is inconsistent and so has no solutions If k = 4, the system has infinitely many solutions of the form x = t, y = 3t ¡ 2, t R 29 34 149 34 ¡ 157 34 17 ¡ 12 C ¡ 23 34 C C C A ¡ 83 34 C 87 34 ¡ 17 ³ ³ EXERCISE 13F.2 ¶ 4:5 7:5 ¡0:5 0:5 0:5 ¡3 ¡5 µ µ3¶ x = + 2t, y = t, z = 0, t R x = 4, y = ¡2, z = c x = 4, y = ¡3, z = no solution, system is inconsistent x = 2, y = ¡1, z = no solution, system is inconsistent ¡ 2t ¡ 5t , y = t, z = , t2R c x= 3 µ ¶ a ¡5 ¡5 0 k¡8 13 b If k 6= 8, no solutions, if k = 8, infinitely many 1:596 ¡0:996 ¡0:169 ¡3:224 1:925 0:629 ¡1:086 ¡0:396 µ2 µ7¶ ¶µx¶ ¡1 ¡1 y z = a b d a b ¶ 2t + 5 ¡ 9t , y= , solutions of the form x = 5 z = t (t is real) c The last row does not enable us to solve for z ¡2 magenta yellow 0 x= ¡3 ¡2 k ¡ 13 ¡6 ¡k + 13 ¶ 5t + ¡ 4t , y= , z = t (t is real) 3 c If k 6= 13, x = 73 , y = 13 , z = ¡1 µ 0 b x= ¡7 ¡5 a+1 a¡1 ¡ 2a a+1 ¶ ¡5t ¡ 11 19 ¡ 6t , y= , z = t (t is real) 7 50 c x = 17 a + 2, y = 27 a ¡ 2, z = 25 µ b If k = 13, infinitely many solutions of the form 95 100 50 75 25 95 100 50 75 25 95 100 50 b a 6= ¡4 6 , y= m+2 m+2 b If m = 2, there are infinitely many solutions of the form x = t, y = ¡ t (t is real) If m = ¡2, there are no solutions a 75 ´ k = 16 x = t, y = 3t ¡ 8, t R when k 6= 16 a x= y represents the cost per baseball in dollars, z represents the cost per basketball in dollars b 12 basketballs a 2x + 3y + 8z = 352 b x = 42, y = 28, z = 23 c E1 201 000 x + 5y + 4z = 274 x + 2y + 11z = 351 $11:80 per kg a 5p + 5q + 6r = 405 b p = 24, 15p + 20q + 6r = 1050 q = 27, 15p + 20q + 36r = 1800 r = 25 25 ¡21 c d is inconsistent and ) no solutions exist a 2a + ´ b A unique solution for m 6= or ¡2 AB = I, a = 2, b = ¡1, c = 3 MN = 4I, u = ¡1, v = 3, w = a x = 2:3, y = 1:3, z = ¡4:5 b x = ¡ 13 , y = ¡ 95 , z = 21 21 c x = 2, y = 4, z = ¡1 a x = 2, y = ¡1, z = b x = 4, y = ¡2, z = c x = 4, y = ¡3, z = d x = 4, y = 6, z = ¡7 e x = 3, y = 11, z = ¡7 f x ¼ 0:33, y ¼ 7:65, z ¼ 4:16 a x represents the cost per football in dollars, cyan k ¡ 16 d When a = ¡4, last row is 0 j ¡21 So, the system A b a µ¡5:5 b ¡1 0 a jAj = oranges 50 cents, apples 80 cents, pears 70 cents, cabbages $2:00, lettuces $1:50 b c ¡3 ¡5 0:050 ¡0:011 ¡0:066 0:000 0:014 0:028 ¡0:030 0:039 0:030 EXERCISE 13E ¡ 12 17 ¡ 11 B¡ @ C C C A µ2 0¶ a o 6:3 1CB a C B 6:7 C 1A@ p A = @ 7:7 A c 9:8 l 10:9 EXERCISE 13D.2 0 10 1 1 solutions for x and y c i when x = t, y = for all values of k except a k= b It gives no more information than the first Gives the same 95 = 100 A¡1 ´ A¡1 (kA) = I ¡1 A kA and k a X = ABZ b Z = B¡1 A¡1 X k 75 ³1 (kA) 891 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\891IB_HL-2_AN.CDR Thursday, 24 January 2008 12:00:01 PM PETERDELL IB_PD (892) 892 ANSWERS µ1 a b c a ¶ ¡1 1+m ¡7(m + 1) m ¡2(m + 1) ¡ m2 0 (m + 1)(m + 5) if m = ¡5, no solution if m = ¡1, infinitely many solutions if m 6= ¡5 or ¡1, unique solution µ1 k ¡(2 + 3k) ¡ k2 ¡k 0 (3k + 25)(k ¡ 1) 6(k ¡ 1) ¶ If a = ¡ 27 , ¡7t , x= c P (x) = ¡ 29 x2 + y= 172 x c no solution ¶ µ b ! c (11 12) d BA does not exist 2 ¡2 ¡1 ¡2 ¶ c @ ¡ 32 ¡4 A 11 x = 3t, y = ¡7t, z = 2t, t is real REVIEW SET 13B z = t, t R 71 ¡ a + 3b + c = ¡10 ) a = ¡ t, b = ¡4 ¡ 3t, c = 10t (t is real) b There are three unknowns and only two pieces of information c x2 + y2 + 4x + 2y ¡ 20 = 10 When k 6= 27, there are no solutions When k = 27, there are infinite solutions of the form x = 2¡t, y = 2t + 3, z = t (t is real) x = 5, y = 4, z = 3t , Ã1 unique solution if k 6= 34 , no solution if k = x = 1, y = ¡2, z = ¡1 a ¡2a + 4b + c = ¡20 EXERCISE 13F.3 4t 5t x = , y = , z = t, t R 7 If a 6= ¡ 27 , x = y = z = b d k 6= or ¡ 25 a x = ¡ 2t, y = t, z = 3t + 1, t R b x = 18 ¡ 5t, y = t, z = 7t ¡ 22, t R ¶ 4 + 2t ¡ 11t , y= , z = t (t is real) 5 c k = ¡ 25 µ a b k = 1, infinitely many solutions of the form x= µ4 a ³ ´ 0 a ( 10 ) b thousand $ d µ4 0 ¶ c ( 15 18 21 ) µ5¶ e CA does not exist d Max profit = $20 448 when producing 2966 ´ 2n b Conjecture is: Mn = a A2 = ³ A5 = ´ ³ , A3 = 242 243 ³ ´ ´ ³ a P2 = ³ ¡2 ¡1 ³ b Pn = a ³ e µ i ¡2 ´ ³ ´ ¡1 ³ c ´ ´ , P4 = ¡2 ¡8 ³ g ´ ³ k cyan ³ ¡4 ¡3 ´ ³ d ¡1 ¡4 ´ ³ µ ´ c 3 12 ¡ 11 ¶ ´ magenta ¡1 ¡2 ¶ ´ A¡1 = 53 A ¡ 2I a a = ¡3, b = 18, c = 48 ) s(t) = ¡3t2 + 18t + 48 b 48 m c seconds A5 = 779A + 290I, A6 = 4185A + 1558I a d = 80 b a = 2, b = 8, c = 10 ¡ 12 yellow 50 x = 1, y = ¡1, z = A3 = 27A + 10I, A4 = 145A + 54I, ¶ 75 ´ 2:9 ¡0:3 ¡0:3 2:1 ´ When t = there are no solutions When t = there are infinite solutions of the form x = + s, y = ¡ s, z = s (s is real) 6t ¡ ¡8 3(t ¡ 5) , y= , z= When t 6= or 3, x = t¡3 t¡3 t¡3 ´ 95 µ1 ¡1 ¡3 11 a i jBj 6= ii AB = BA b k R , but k 6= 3, ¡2, ftechnologyg Does not have unique solutions when t = or ´ 100 c X= ³ d ¡4 ¡2 REVIEW SET 13D ³ 95 100 50 13 not possible ³ ¡6 ¡8 l f X= ¡2 ¡10 ´ 2 ¡5 h 25 75 25 95 µ 14 ¶ e X= 100 50 ¶ 75 25 d X= ¡ 12 µ solutions exist If k = 1, no solutions exist X= ¡3 ¡10 y= c Unique solution for k 6= ¡3 or If k = ¡3, infinitely many a Y = B ¡ A b Y = 12 (D ¡ C) c Y = A¡1 B d Y = CB¡1 e Y = A¡1 (C ¡ B) f Y = B¡1 A µ does not exist REVIEW SET 13C , n Z+ ¡3 ¡2 12 , b x = ¡1, or ¡4 fusing technologyg ³ ´ ³ a b 10 ¡12 ´ b x= ¶ 20t + 14 ¡13t ¡ , y= , z = t, t R 9 14 b when m 6= a a = 0, b = 5, c = 1, d = ¡4 b a = 2, b = ¡1, c = 3, d = a x = 0, y = ¡ 12 ¡4 x= for all n Z + ¡11 ³ j ´ 80 , 81 : ³ ¡5 b 2A ¡ I $56:30 AB = I, BA = I, A¡1 = B x = 2, y = 1, z = ¶ ´ ¡3 ³ f ¶ , P3 = b ¡5 ¡4 ¡2 ´ n+1 n ¡n ¡ n REVIEW SET 13A ³ ´ 3n ¡ 3n ¡ 23 d Yes as A¡1 = ³ 26 , A4 = 27 b Conjecture is: An = µ M4 = , µ a ´ X= ³ ¡2 1 ´ a x = 6, y = ¡2, z = b x = 32 , y = ¡ 76 , z = ¡ 76 95 ³ ³ 100 M3 = , ´ 50 ³ 75 ´ 25 a M2 = ³ EXERCISE 13G black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\892IB_HL-2_AN.CDR Thursday, 24 January 2008 12:00:59 PM PETERDELL IB_PD (893) ANSWERS If k = §2, there are no solutions c ¡2 ¡ 2k k+4 , y= k2 ¡ k ¡4 If k = and m 6= 20 there are no solutions If k = and m = 20 there are infinite solutions of the 10 + 3s ¡ t , y = s, z = t, s, t are real form x = µ ¶ 10 + 3s ¡ m¡20 m ¡ 20 k¡2 , y = s, x = If k 6= 2, z = k¡2 d If k 6= §2, x = b k 6= or Scale: cm º 30 km¡h-1 32° 25 km EXERCISE 14A.2 a p, q, s, t b p, q, r, t c p and r, q and t d q, t e p and q, p and t a true b true c false d false e true f false 11t + 13 ¡ t , y= , z = t, t is real 9 d When k = 23 , the system is inconsistent and has no solutions EXERCISE 14B.1 a b b Opera E32 a 3x + 2y + 5z = 267 ¡9 6 ¡3 ´ b³ ¡10 A3 A4 = ¡I, A8 = A ¡ I µ0 e not possible ¡6 22 ¡12 11 A5 = ¡A, c E200 Play E18 Concert E27 2x + 3y + z = 145 x + 5y + 4z = 230 x = 2, y = 1, z = d p+q q REVIEW SET 13E ³ 8° Scale: cm º 10 km c If k = 2, x = a 150 km¡h-1 ´ c µ ¡2 10 ¡7 ¡6 ¡1 ¶ ¶ c d p+ q p+ q q p A6 = ¡A + I, = I, A7 q p p+q p q p = A, e f a A6n+3 = (A6 )n A3 = ¡I, A6n+5 = ¡A + I b A¡1 = ¡A + I a a = 1, b = ¡1 b x = ¡5, y = 4, z = q q p p+q p+q p EXERCISE 14A.1 a b c ¡ ! ¡! ¡ ! ¡! a AC b BD c AD d AD a i 135° ii 35 m q+p 70° 25 m¡s-1 30 N q p q p p+q d b yes 50 m¡s-1 10° EXERCISE 14B.2 a a b b 100 m¡s-1 -q p-q 45° p-q 75 m¡s-1 p a b c d -q N 30 N N 146° p-q p-q p 45° scale: cm º 10 m¡s-1 a 40 b m¡s-1 -r yellow 95 100 50 75 25 95 100 50 75 25 95 50 75 25 95 100 50 75 25 100 magenta -q q p p+q-r -q p scale: cm º 10 N cyan p -q black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\893IB_HL-2_AN.CDR Friday, 12 December 2008 2:55:33 PM TROY p -r p-q-r IB_PD 893 (894) 894 ANSWERS c EXERCISE 14C.1 a -p b -q ³ ´ ¡ ! ¡ ! ¡! ¡! AB b AB c d AD e f AD t = r + s b r = ¡s ¡ t c r = ¡p ¡ q ¡ s r=q¡p+s e p=t+s+r¡q p = ¡u + t + s ¡ r ¡ q i r + s ii ¡t ¡ s iii r + s + t i p + q ii q + r iii p + q + r ³ a ³ f a 24:6 km b east of south a 82:5 m b 23:3o west of north c 48:4 seconds ³ a EXERCISE 14B.4 c -r 2s f ³ Qw_\r a -\Ew_\s e -s ³ s a r r r f s ³ 2r+3s a h g s s s r+3s b c p q p = 2q p q ´ ³ ³ , CA = X ³ e ¡3 ³ p ¡! 34 units f DA = ´ ¡5 ¡5 µ ´ ´ ³ f 11 e ´ ´ ¡5 ´ ¶ ´ p , DA = 10 units P(0,¡-1,¡2) Z Y -1 X d OP = p p OP = p units Y ii P(-1,¡-2,¡3) Z Y 26 units 14 units ii a i p c i 21 units ii Z iii (¡ 12 , 12 , (1, ¡ 12 , OP = p -2 -1 Y X 14 units p b i 14 units ii (1, ¡ 12 , 32 ) p 0) d i 14 units ii (1, 12 , ¡ 32 ) 2) yellow 95 100 50 75 25 95 100 50 75 25 95 50 75 100 magenta a isosceles b right angled c right angled d straight line p (0, 3, 5), r = units a (0, y, 0) b (0, 2, 0) and (0, ¡4, 0) b a parallelogram 25 ´ ¡5 ¡3 ´ ´ b P(3,¡1,¡4) M ¡1 Z N 17 ¶ ³ d ³ ¡5 ¡5 OP = units P 95 ´ ¡5 P(0,¡0,¡-3) i X 100 ³ ³ e ³ c ³ f e ³ ´ ´ ´ ´ ¡6 Y c 50 ³ d d Z q 75 ³ c ´ X 25 ´ p = -3q p p µ h ´ ´ 14 ¡1 ¡1 ´ b ³ c ³ d ¡4 ´ ¡3 ´ ´ ¡1 b a q ¡5 ³ ¡1 ¡8 ¡1 ´ ¡6 EXERCISE 14D p = Qe_ q cyan ³ c 11 g ´ ³ ¡ ! e a ³ b e CA = ¡5 q d ¡1 ³ ³ e p p p p p 13 units b 17 units c units d 10 units e 29 units p p p p a 10 units b 10 units c 10 units d 10 units p p p p e 10 units f units g units h units p p i units j units ³ ´ ³ ´ p p ¡ ! ¡ ! ¡1 a AB = , AB = 37 units b BA = ¡6 , BA = 37 units ³ ´ ³ ´ p p ¡ ! ¡ ! ¡4 c BC = ¡1 , BC = 17 units d DC = , DC = 58 units Qw_\&r+3s* p ¡7 ´ ³ c ´ ¡2 ¡1 ¡1 h ´ a s a -3 EXERCISE 14C.4 s r Qw_\r+2s ³ ³ ´ a b ³ ´ d c ¡6 ¡4 ¡3 ´ ´ ¡3 ¡15 ³ s r Qw_\r ³ b EXERCISE 14C.3 f 2r-s ³ ´ ¡5 ´ ¡2 g ´ ¡9 ³ ´ ³ c b ¡5 ¡3 ¡3 ³ d ´ ¡2 ´ ¡6 b EXERCISE 14C.2 9.93o b ³ a EXERCISE 14B.3 a -1 -5 r-q-p h¡1 d r a a d f a b c black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\894IB_HL-2_AN.CDR Friday, 12 December 2008 2:56:24 PM TROY IB_PD (895) ANSWERS EXERCISE 14E.1 a Z T(3,¡-1,¡4) ¡1 ¡3 ¡! ¶ ¡4 ¡1 µ µ ¶ ¶ ¡4 µ µ ¡5 ¶ ¡! b MN = µ a f a d a c ¶ ¡5 ¡1 ¶ ¡ ! , DC = µ ¡5 ¡1 µ ³ ´ µ c y= ³ ´ ¡ ! ¡! , CM = a M(1, 4) b CA = µ5¶ µ ¡ ! AB = ¡2 ¡ @ b x= , AB = A c p ³ ´ a ! cyan ¡1 magenta b PR = QS ¶ µ b ¡1 ¶ µ c p 11 units ¡5 ¶ @ 12 A d p 26 units p units ³ ¡1 ´ b ¡ p217 ³ ¡1 ¡4 ´ c p6 18 µ ¡1 ¶ d ¡ 53 µ ¡1 ¶ ¡2 ¡2 a a a a b 22 c 29 d b c 14 d ¡1 b 94:1o a t = b t = ¡8 c yellow ¶ 95 ¡5 p 14 a k ´ ¡2 , k 6= b k ³ ´ ³ ´ ¡2 , k 6= c k ³ ´ ³ ´ , k 6= d k , k 6= e k , k 6= µ e 66 e 52 f g h 14 e f b c t = or d t = ¡ 32 ¡ ! ¡ ! 100 ¶ 50 d ¡4 10 75 µ 25 ¶ 50 c ¡9 p ³ µ ¡6 bC is a right angle 10 AB ² AC = 0, ) BA p p 11 b AB = 14 units, BC = 14 units, ABCD is a rhombus c 0, the diagonals of a rhombus are perpendicular 12 a 101:3o or 78:7o b 116:6o or 63:4o 13 a b ¡9 c 63:4o or 116:6o d 71:6o or 108:4o ¡5 95 ¶ 75 b ¡3 25 µ ¡3 µ ¡2 ¶ a t = ¡ 32 b t = ¡ 67 c t = ¡1§ d impossible Show a ² b = b ² c = a ² c = b t = ¡ 56 µ ¡3 ¶ c ¡2 95 ¶ 100 50 25 ¡2 75 µ 14 a µ ¡3 ¶ b ¡1 ¡ ! , QS = EXERCISE 14I ¡1 x= 29 units 100 µ ¡1 ¶ à A a k = §1 b k = §1 c k = d k = e k = § 23 p a units b units c units d ¼ 6:12 units a p15 (i + 2j) b p113 (2i ¡ 3k) c p133 (¡2i ¡5j ¡2k) C(5, 1, ¡8), D(8, ¡1, ¡13), E(11, ¡3, ¡18) a parallelogram b parallelogram c not parallelogram a D(9, ¡1) b R(3, 1, 6) c X(2, ¡1, 0) ¡ ! ¡ ! ¡ ! a BD = 12 a b AB = b ¡ a c BA = ¡b + a ¡! ¡! ¡! 1 d OD = b + a e AD = b ¡ a f DA = 12 a ¡ b 13 a 5 ¶ 10 11 12 ¡6 ¡5 ¶ ¡ ! , CB = 11 p3 11 Y(¡7, 8, ¡6) p § 411 d y= ³ ´ @ c µ ¡1 ¶ p units 50 b y= ¶ ¡ 12 a ¡7 : b ¡1 : a a = 7, b = ¡1 b a = ¡ 72 , b = ¡ 21 25 ¶ EXERCISE 14H a B(¡1, 10) b B(¡2, ¡9) c B(7, 4) a x= p= 20 ) e X(3, 13 , 53 ) f a + b b p = 75 a ¡ 25 b n m a + m+n b m+n ¡ ! a PR = ¶ e x = 15 (4s ¡ t) f x = 3(4m ¡ n) µ p= 1a opposite direction A, B and C are collinear and AB = BC d A, B and C are collinear and AC = BC a x = 12 q b x = 2n c x = ¡ 13 p d x = 12 (r ¡ q) ¡1 T( 73 , ¡3, EXERCISE 14F.1 a y= p : b : c : d : e ¡2 : ¡2 : Q(¡ 13 , 1, 43 ) b R(¡5, ¡7, 13) c S(2, 54 , 34 ) ¡ 23 a S = (¡2, 8, ¡3) b ABCD is a parallelogram µ p ¶ r = 3, s = ¡9 a = ¡6, b = ¡4 ¡ ! ¡ ! a 23 b ¡ 43 a AB k CD, AB = CD ¡ ! ¡ ! b RS k KL, RS = 12 KL @¡1 A @¡2 A b = 6, c = ¡6 b a = 4, b = 2, c = b = 2, c = b a = 1, b = b = ¡1, c = s = 4, t = ¡7 b r = ¡4, s = 0, t = µ ¡2 EXERCISE 14G EXERCISE 14E.2 ¡ ! µ h EXERCISE 14F.2 p 42 units c MN = ¶ a AB = ¶ 16 a r = 2, s = ¡5 b r = 4, s = ¡1 b AB = ¡1 ¡2 p p ¡! ¡ ! a OA = , OA = 30 u b AC = ¡1 , AC = 30 u ¡5 µ ¶ p ¡ ! c CB = ¡1 , CB = 35 units p p a 13 units b 14 units c units a a = 5, a a = 13 , c a = 1, a r = 2, ¡4 15 a 11 units b 14 units e f ¡ p1 11 11 p p p B C @ A c 38 units d units ¡3 11 B p1 C p 26 units p BA = 26 units ¶ µ ¡4 ¶ ¡1 ¡ ! , AB = 2 ¡ ! , BA = ¡ ! , OB = µ µ A g p 11 ¶ a NM = ¡ 72 95 p 26 units ¡1 100 ¡! OA = p f @ 75 µ OT = ¶ ¡1 X µ ¡ ! c µ Y -1 a AB = b ¡ ! OT = 895 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\895IB_HL-2_AN.CDR Tuesday, 13 January 2009 12:54:55 PM TROY IB_PD (896) 896 ANSWERS bC ¼ 62:5o , the exterior angle 117:5o AB 15 16 a 54:7o b 60o c 35:3o 17 a 30:3o b 54:2o 18 a M( 32 , 52 , 32 ) b 51:5o EXERCISE 14J.3 µ0¶ µ0¶ 21 a = , b= , c= 0 a ² b = a ² c, but b 6= c 23 a Hint: Square both sides will A ¡ ! ¡ ! gram Find AB and OC, etc p 133 b units2 a Yes b No 10 k = REVIEW SET 14A a O 60 m¡s-1 B b 8° Scale: cm º 10 m¡s-1 with a vector which is meaningless µ 11 a ¶ a a b c ¡i ¡ j ¡ k d i ¡ 6j + 2k µ ¡11 ¶ µ ¡1 ¡1 a ¶ µ0¶ b 45 m N µ ¡1 c ¶ µ d ¡1 a b y y y-2x -x -x ¡ ! ¡ ! ¡ ! ¡! a PQ b PR 4:845 km, 208o a AC b AD ¶ a AB = CD, [AB] k [CD] b C is midpoint AB a p+r=q b l+m=k¡j+n a r + q b ¡p + r + q c r + 12 q d ¡ 12 p + 12 r ³ ´ ¶ 22 b k c (¡ i + j ¡ 2k)n ¡15 d (5i + j + 4k)n n, k R , n, k 6= ³ a 11 a a £ b b c 2(b £ a) d µ x x+y a £ (b + c) = (a £ b) + (a £ c) µ ¡4 ¶ Scale: cm º 10 m 60° ¡2 , a ² (a £ b) = = b ² (a £ b) £ b is a vector perpendicular to both a and b i£i=0 j£j=0 k£k=0 i £ j = k j £ i = ¡k j £ k = i k £ j = ¡i i £ k = ¡j k £ i = j a £ a = a £ b = ¡b £ a µ1¶ b 17 c 17 a£b= a b µ2¶ b c units3 23 10 24 ¡7 25 a ² b is a scalar and so a ² b ² c is a scalar ‘dotted’ EXERCISE 14J.1 p c S = 12 fja £ bj + ja £ cj + jb £ cj + j(b ¡ a) £ (c ¡ a)jg C a b Consider the parallelo- units2 p units2 19 a t = or ¡3 b r = ¡2, s = 5, t = ¡4 20 a 74:5o b 72:45o µ1¶ p 101 69 units2 p a D(¡4, 1, 3) b 307 units2 p p p a units b ( 42 + + + 6) units2 a (3, 1, 0), (1, 3, 3), (4, 2, 3), (4, 3, 3) b ¼ 79:0o p k = § 33 a b ¡5 ´ ³ c ¡4 ´ 12 a k ¶ µ2¶ b ¶ or p ¡ 610 µ ¡5 ¡7 ³ ¶ 10 a p µ 14 a x = a i £ k = ¡j, k £ i = j a a ² b = ¡1 a £ b = b cos µ = ¡ p1 28 ¡! ¡1 µ ¡! ¡ ! ¶ µ b OA £ OB = ¡ ! OB = ¡3 p p27 28 d sin µ = µ ¡1 ¶ ¶ p p27 28 magenta µ ¡3 ¶ + µ ¶ ¡3 11 b 12 µ b m = 5, n = ¡ 12 q µ 95 100 50 75 ¡3 ¡26 p ³ ¡4 ´ ³ ´ 10 units d ¡10 11 p 109 units ´ 16 r = 4, s = 162 units c ¶ c ¡1 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\896IB_HL-2_AN.CDR Friday, 12 December 2008 2:59:18 PM TROY p p 61 units µ 74 units p ¡8 ¶ : t = § 80:3o b 10 a ¡1 b yellow p µ ¡6 ¶ 40:7o a 25 95 100 50 75 ³ c b x= ¡ ! a p ¡! ¡ ! j OA £ OB j= 83 25 95 100 50 75 25 p ¶ ¡1 ´ 13 units c a PQ = ¡! ¡ ! c Area ¢OAB = j OA jj OB j sin µ p ¡! ¡ ! = j OA £ OB j= 283 units2 ¡ ! ¡ ! a OC is parallel to AB b a £ b = b £ c cyan p ¡1 ¡13 REVIEW SET 14B c sin µ = ³ b 17 a q + r b r + q, DB = AC, [DB] k [AC] µ1¶ 2 ´ 12 a 17 units b 13 a p + q b EXERCISE 14J.2 a OA = ¡4 ¡2 95 ¡1 ¡5 ¡7 100 , k 6= 0, µ 50 14 a p 10 75 µ ¶ 25 13 k ¡5 ¡7 µ p 46 units c (¡1, 12 , ) ¶ c 60o IB_PD (897) ANSWERS b ¼ 124o , b b ¼ 45:0o L ¼ 11:3o , M 11 K 12 63:95o 13 c = 50 14 a a ² b is a scalar, so a ² b ² c is a scalar dotted with a z z+w ³ ´ z or ¡3 k = ¡ ! ¡ ! a AC = ¡p + r, BC = ¡q 1+r µ ¡12 ¡7 ¶ ¡ 23 µ ¡2 ¡4 ¶ B t = 56 a ¡ 12 b + 23 c 10 11 If µ is acute, u ² v = a 5¡i p 14 units2 units b 4+i z1+1 z1 2i c + 52 i z1 p1 i p2 ¡ p2 ¡ p2 + C A Qw_\z1 ¡3 : (¡ 16 , 17 23 , 3) 3 units z2 z2+\Qw_\z1 p k p p 199, If µ is obtuse, u ² v = ¡ 199 d ¡ 12 i z1 + 13 ¼ 26:4o z1 REVIEW SET 14E p p AB = AC = 53 units and BC = 46 units ) ¢ is isosceles r = 3, s = ¡ 52 , t = 14 (0, 0, 1) and (0, 0, 9) µ ¡11 ¶ a x= µ2¶ b x= ¡10 a r = ¡2, s = a k = § 12 z1+2i b ¼ 41:81o c R( 23 , 43 , 53 ) a = or ¡ 36 5 a (1, 1, 2) and (3, ¡1, 0) b @ 12 t = -z -2 q p 14 units a 7i ¡ 3j + 10k b ¡ ! + -z z §7 a 11 b c b @¡3 A REVIEW SET 14D a PQ = p a i p + q ii w-3z -1 z-w b b ¼ 58:7o K ¼ 64:4o , b L ¼ 56:9o , M 5 k , k 6= 72:35o or 107:65o -z b ¡2 + 3i -w a ¡13 b ¡36 t = z 2z-w REVIEW SET 14C 10 a w d ¡7i -w w vector, which is meaningless b b £ c must be done first otherwise we have a scalar crossed with a vector which is meaningless 15 a k = § p7 b k = § p1 33 c ¡1 + 5i a 4+i 897 15 b I µ p ¡1 ¶ d a b ¼ 62:2o µ5¶ R b µ ¡! ¡3 ¡4 ¡! ¡! So, LM ² KM = ) ¶ p units2 ¡! , KM = 3i-z c k=0 µ ¡2 ¶ g ¡2 ¡1 z z R f I 2-z -z h z R R z z* -4 z*+i i I z-4 z+2 z+2 qR q z* I I b = 90o M I R e I -z c -2z z 12 a t = ¡4 b LM = I z 3z z b § p414 (3i ¡ 2j + k) b ¡ p514 10 ¼ 16:1o 11 a 1 a a = ¡2, b = a 10 b ¼ 61:6o z z-4 R EXERCISE 15A.1 EXERCISE 15A.2 I zn a I zx -4 b z R zz z* magenta yellow 95 100 50 75 25 95 100 50 75 25 95 -5 100 50 75 25 95 100 50 75 25 zv R z* -4 z -2 zc cyan R zb I black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\897IB_HL-2_AN.CDR Friday, 12 December 2008 3:00:38 PM TROY IB_PD R (898) 898 ANSWERS z¤ = z a = 11, b = ¡ EXERCISE 15B.6 a ¡1:41 + 1:01i b 1:27 ¡ 3:06i c ¡2:55 ¡ 1:25i a cis (¡0:927) b 13 cis (¡1:97) c 17:7 cis (2:29) p a cis ¼4 b 19 cis (¡2:50) EXERCISE 15B.1 p a b 13 c 17 d e p p p a b c d e f p p h p1 i j k 5 l 5 p g ² jz ¤ j = jzj ² zz ¤ = jzj2 ² jzwj = jzj jwj ² jz n j n = jzj ¯ ¯ ¯z¯ ¯ ¯ £ jwj = jzj w ² p ¯ ¯ ¯z¯= ¯w¯ a a(x2 + 2x + 4) = 0, a 6= b a(x2 ¡ 2x + 2) = 0, a 6= EXERCISE 15C jzj jwj a 32 b ¡1 c ¡64i d e 220 = 048 576 a b · a a2 + b2 ¡ (a ¡ 1)2 + b2 ¸ f i p ¡ ¡ ¢ p if k > 0, ¡k cis ¡ 3¼ if k < 0, f ¡5 ¯ ¯ c ¯ z1 ¯ = p + 32 i 12 a cis ¼ b r cis µ + ¡ ¢ ¼ a cis (¡µ) b cis µ ¡ a cos b cos ¡¼¢ 12 d clock rotn of 12 ¼ z z+ 1, ¡ 12 § i about ¡1, ::::: then z ¤ = r cis (¡µ) ¡ arg z = ¡µ tan µ ¡ tan3 µ ¡ tan2 µ p , cos ¡¼¢ ¡ 5¼ ¢ 12 , tan , cos ¡ 4¼ ¢ ¡ ! ´3 = z + 3z + § p 2 a z= b z= p i p 3 c (z ¡ 1)¤ = sin( ) cis ¡ ¡ ¼ ³ arg z2 ¡ z1 z3 ¡ z2 ´ = p d a=¡ 2¼ a z= magenta cis p cis p cis yellow 25 95 100 50 75 25 ¡µ ¼ 95 iv p ¡¼¢ ¡ 3¼ ¢ 12 , 12 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\898IB_HL-2_AN.CDR Friday, 12 December 2008 3:03:31 PM TROY p i 2 ¡ p i ¡ ¢ I cis 3¼ ¡ ¡5¼ ¢ R ¡ ¢ cis ¡¼ -1 I ¡1 + i ¡¼¢ p cis 12 -~`2 ~`2 = ¡1 + i, ¡ ¡7¼ ¢ + R cis -1 -1 cis ¢ 100 50 ¼ 75 25 95 100 50 75 25 p ¡ 32 i, 3i, ¡ 3 ¡ 32 i -1 ¼ a ei¼ = ¡1, ei = i c i µ2 ii ¼2 + µ iii 2µ ¡ cyan ¡ 7¼ ¢ ¡ ¡5¼ ¢ , cis 95 ¯ ¯ ¯ z2 ¡ z1 ¯ = c b ¯ ¯ z ¡z ¡ ¡¼ ¢ cis ¼ , 100 ¡ ¡ 2µ ¡ 7¼ ¢ ¡ 7¼ ¢ , cis 50 ¢ ¡Á ¼¢ cis + ¢ ¡ Á Á + ¡ 3¼ ¢ 75 b z¡1= 12 , tan -1 cis arg(z ¡ 1) = = + 2µ ¡ ¼2 ¡ 11¼ ¢ ¢ Á z2 I a j¡zj = 3, arg (¡z) = µ ¡ ¼ b jz¤ j = 3, arg (z ¤ ) = ¡µ c jizj = 3, arg (iz) = µ + ¼2 p d j(1 + i)zj = 2, arg((1 + i)z) = µ + ¼4 ¡ arg ¼ p p ¡ i, 2i, ¡ ¡ i a z = §2, §2i p p p p b z = § i 2, ¡ § i a jz ¡ 1j = , ¡ ¡i ¢ + z z EXERCISE 15B.5 sin Á , Á sin( ) ¯ ¡ ¢ p p p p 2+ ¼ , sin 12 = 6¡ 4 p p p p = ¡ 2¡ , sin 11¼ = 6¡ 12 = ¡ 11¼ ¢ ¢ c True cis 3µ ¡ n¼ ¢ ¯ ¯ ¯= d ¯ ¡i z2 EXERCISE 15D.1 p i ¼ p 3¡i) (¡ 64 14 Hint: When n = 1, 2i sin µ = z ¡ z1 : Now cube both sides a jzj = 2, arg(z) = µ b cis (¡µ) c cis (µ + ¼) d cis (¼ ¡ µ) ¡ ´ p i + ¡ ! ¡ ! BC is a 90o rotation of BA about B ¡! b OD ´ z1 + z3 ¡ z2 a cos 4µ = cos4 µ ¡ cos2 µ + b sin 4µ = cos3 µ sin µ ¡ cos µ sin3 µ ³ a cis 3µ b cis 2µ c cis 3µ d + 12 i p p e + i f g ¡2i h ¡4 i 4i + ¡1 p 11 a AB ´ z2 ¡ z1 , BC ´ z3 ¡ z2 Hint: Notice that p 2 ¼ n zn ¡1¢ ¡ ! ¼ 13 c a ¡1 b ¡1 c , 10 b tan 3µ = a b EXERCISE 15B.4 b ¡ ¼2 < Á ¡¼¢ ii x = tan p ³ ¡ ¼¢ p ¡ ¼¢ cos ¡ 12 +i sin ¡ 12 f ¯ c i x= a 2i b 2+4 2i c 3+2i d 1¡i e ¡ 524 288 a ¯z ¯ = 8, arg(z ) = 3µ b ¯iz ¯ = 4, arg(iz2 ) = ¢ ¼ a cis b cis ¼2 c cis ¼ d cis ¡ ¡ ¢ ¡ ¢ p p e cis ¼4 f 2 cis ¡ ¼4 g cis 5¼ h cis not possible if k = p p p ¯ ¯ EXERCISE 15B.3 ¼ (2:180 + 0:498i) 14 + i = cis = cis a n = 4k, k any integer b n = + 4k, k Z p ¼ µ p a i units ii (1, 4) b i 5 units ii (¡ 32 , 2) a i w + z ii w ¡ z a reflection in the R-axis b anti-clockwise rotation of ¼ about c reflection in the I-axis d clockwise rotation of ¼2 about z = + 6i p p e a jzj cis ¡2b + i b 10 11 (a ¡ 1)2 + b2 EXERCISE 15B.2 k cis (1¡i) 64 d h ¡¼¢ cis a 128 ¡ 128i b 1024 + 1024 3i c b jz1 z2 z3 ::::zn j = jz1 j jz2 j ::::: jzn j and that jz n j = jzjn p c d e p p p e + i f 16 + 16 3i R ¡ ¢ p cis ¡7¼ 12 IB_PD (899) ANSWERS p p ¡¼¢ p cis b z= ¡ 11¼ ¢ cis ¡ ¢ p cis 11¼ 12 , 12 ¡ ¡5¼ ¢ cis a Yes a I =1+i ¡ 2¼ ¢ cis -~`2 cis cis cis ¡ p ¼ -1 p ¡ 12 i e z= p p p p f p , 24 ¡ 13¼ ¢ p cis p cis p cis p ¡¼¢ cis I 24 , 24 ¡ ¡11¼ ¢ p , 24 ¡ ¡23¼ ¢ R ¡ ¡23¼ ¢ cis ¡µ¢ p ¡ ¡3¼ ¢ cis 20 ¡¼¢ cis p , cis , 20 p cis , 20 20 p1 ¡ p1 i, p1 + p1 i, ¡ p1 p i z + = (z ¡ a 16 cis ¡ ¼2 w cis 18 p 11 R ¡ ¢ p cis ¡17¼ 18 or ¡ ¢ p cis ¡5¼ 18 ´ ¢ b i cis ¡ 7¼ ¢ w cis ¡ ¡3¼ ¢ cis -1 ¡ 7¼ ¢ magenta p ¡a 2b where w = cis ¡ 2¼ ¢ d 31 w R w4 p ¡ i , 2i , ¡ ¡ i p ¡ p i p a + 2i b 2 c 175 a = 0, b = ¡1 a x = 0, y > ¡ ¢ c jiz ¤ j arg (iz ¤ ) = 4, 20:3 cis ¡ ¡7¼ ¢ 20 ¼ = 15 b i ¡ 7¼ ¢ , 20:3 cis ¡ ¢ 2¼ 95 100 50 75 25 2¼ black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\899IB_HL-2_AN.CDR Thursday, 24 January 2008 12:09:57 PM PETERDELL -4 cos µ = ) R iii c i w 95 = ¡µ ¡µ ii -2 100 z ¡ ¡3¼ ¢ I v yellow ¡1¢ 13 a a(z ¡ cos 2¼ z + 1) = 0, a 6= b a(z2 + z ¡ 1) = 0, a 6= q 50 ¯ ¯ ¡ ¢ d ¼ ¡2:034 b 3x2 +3y ¡20x+12 p 32 § i32 , ¡3 a n = b ¡n =¢ ¡2 c n ¡= ¡1 ¢ 0:3 ¡ 17¼ ¢ ¼ 10 z = 20:3 cis 20 , 20:3 cis 9¼ , cis 20 , 20 ¡ ¡¼ ¢ +2i sin 75 25 95 100 50 75 (n = 0, 1, 2) and w = cis 25 wn = a ¯z ¯ = 64, arg z = 3µ b ¯ z1 ¯ = 14 , arg ¡¼¢ R ii cos ¼ =¼ µ³ ´ ¶ z and w4 I ¯ ¯ I p p 2z + 1)(z + 2z + 1) 95 w3 arg((1 ¡ i)z) = ® ¡ REVIEW SET 15B C~`4 -1 cis 1¡ 100 50 75 25 w2 , 5¼ b ¡ 11¼ 12 ¡ 3¼ ¢12 p c ¡k cis b = p1 ¢ ¢ ¡ 7¼ ¢ I p a i z = wn ¡ (n = 0, 1, 2) and w = cis 2¼ ii z = 2wn + (n = 0, 1, 2) and w = cis cyan z12 z22 w2 EXERCISE 15D.2 iii z = ¡ ³ ¡ ¼3 ¼ , a , Im 2b = 10 a z = 1, w, b a jzj = and arg z = ¡ 2¼ b z = cis (¡2¼) = c Simplifies to (z + z ¤ ) ¡ where z + z ¤ is always real ¡ b cis ¡ 20 ¡ 3¼ ¢ p i, ¡ ¢ cis ¡ ¼3 , n = 3k, k is an integer ¡ p1 + ¡ ¼2 w3 cis ¡ ¢ p cis ¡3¼ 20 ¡ ¡11¼ ¢ cis ¡ µ³2 ´ ¶ z R 20 p ¡ ¡19¼ ¢ cis 24 ¡ ¡19¼ ¢ cis cis ¡ nµ ¢ a ¯ z12 ¯ = 1, arg 1+i ¡ ¡11¼ ¢ cos ¯ 2¯ ¯z ¯ ¡ ¡11¼ ¢ cis =1+i ¡ 13¼ ¢ a cis I 20 ¡ 8¼ ¢ -1 Real part is 16 Imaginary part is 16 a 2x + 4y = ¡1 b y = ¡x jzj = 4 a Re ¡ 13¼ ¢ cis b (1 ¡ i)z = cis ® ¡ 24 24 REVIEW SET 15A p ¡¼¢ p cis 24 ¡ ¢ p z = cis ¡5¼ , 18 ¡ 7¼ ¢ p cis 18 , ¡ ¢ p cis ¡17¼ 18 z= ¡ 13¼ ¢ cis ¡ 6¼ ¢ cos nµ ¡ cos µ ¡ cos [(n + 1)µ] + ¡ cos µ 2n cosn p R cis = -1 cis =w z = + 2i or ¡ i a y = x b y = p13 x + 1, x > c 7x2 + 16y2 = 112 R d z= 5 ¡ w5 ¡ 2¼ ¢ EXERCISE 15E p + 12 i ¡ 12 i ¡ , I cis ¡ 4¼ ¢ b Hint: The LHS is a geometric series I + 12 i or , ¡ 8¼ ¢ c p cis ¡ 6¼ ¢ ~`2 ¡ ¢ p cis ¡5¼ 12 c z= , ¡ 4¼ ¢ R 12 b z = cis 0, 1+i 899 ii ) µ= arg v = 2¼ arg w = ¡ 2¼ ¼(4m ¡ 7) m= IB_PD ¼ =0 (900) 900 ANSWERS REVIEW SET 15C b c d a a reflection in R-axis b anti-clockwise rotation of ¼ about c anti-clockwise rotation of ¼2 about ®2 ®3 ®4 +2 +2 +2 ®+2 , , , where ® = cis ® ¡ ®2 ¡ ®3 ¡ ®4 ¡ p ¡1 + i = cis ( 2¼ ) , m = 3k , k is an integer p p 10 ¡ 2i, ¡2 ¡ 2i, 4i b z= ¯ ¯ ¡ ¡ 2¼ ¢ , arg (1 ¡ z) = 12 jzj 5, ¡ ¼4 < arg z EXERCISE 16B.1 ¡ µ ¼ ¯ ¯ ¯ z2 ¡ z1 ¯ = 1, 15 b ¯ ¯ z ¡z ³ y=-x z2 ¡ z1 z3 ¡ z2 arg ´ w2 b 0, w ¡ 1, w3 w4 EXERCISE 16A.1 ³ ´ ³ x y a i = ³ ´ b i x y = x y = ³ ³ ´ x y d i ³ = ³ +t ¡6 +t ³ ³ ´ +t ¡1 11 ´ ³ +t ¡2 x1 y1 ³ = ¡5 ³ ´ x2 y2 ´ a ii x = ¡6 + 3t ´ ´ ³ ´ y = 11 + t, t R ¡120 ¡40 ´ ³ +t 15 ´ x y a ³ = y z µ0¶ = µx¶ +t µ ¡2 ¶ 1 ¡2 cyan a ¡120 ¡40 x B 10 C(22, 25) 20 A(-4, 6) D(1, 16) 10 ´ ³ c 80 60 ´ A(2, 4), B(8, 0), C(4, 6) BC = BA p = 52 units ) isosceles ¢ A(¡4, 6), B(17, 15), C(22, 25), D(1, 16) B(17, 15) x yellow 50 25 95 100 50 75 10 25 100 magenta b y , t2R 95 ³ +t line ¶ 50 ´ c t2R 75 25 200 100 C µ1¶ 95 100 50 75 a b c a µ ´ A y = +t , t2R z x = ¡ t, y = + 2t, z = ¡1 + 6t, t R x = 2t, y = ¡ t, z = ¡1 + 3t, t R x = 3, y = 2, z = ¡1 + t, t R x = ¡ 2t, y = + t, z = + t, t R 25 c ¡4 ¡3 b µ2¶ +t ³ line line 95 b ¶ ¡1 = 0, ) direction vectors are ? ´ y 100 µx¶ ¡7 ´ EXERCISE 16B.2 75 µ ¡60 80 ³ ´ + (t ¡ a) ³ ´ b (7,¡-13) = ³ f at 2:30 pm ³ ´ ³ ´ ¡1 ¡3 a A(3, ¡4) and B(4, 3) b For A , for B ¡2 c 97:1o d at t = 1:5 hours (5,¡-8) y z ² (3,¡-3) µx¶ p ´ 20p5 d 10 d ¯ 60 ¯ = 100 km e at 1:45 pm and dmin ¼ 31:6 km ii x = ¡1 ¡ 2t Points are: (¡1, 4), (1, 3), (5, 1), (¡3, 5), (¡9, 8) b k = ¡5 When t = ¡2, x = 0, y = ) no a (1, 2) b y (1,¡2) p x c 29 cm s¡1 a 52 m s¡1 x2 (t) = 15 ¡ 4(t ¡ a), y2 (t) = ¡ 3(t ¡ a) ¯³ ´¯ ¯ 80 ¯ y = 7t, t R a When t = 1, x = 3, y = ¡2 ) yes EXERCISE 16A.2 ³ c ¡2 ³ = 1:37:42 pm ³ y = + 2t, t R x = ¡1 + 2¸, y = ¡ ¸, ¸ R p iii d Torpedo is fired at 1:35:28 pm and the explosion occurs at ii x = ¡ 8t ´ ´ ) y = ¡4 + 4t, t R ³ ´ ¡6 ¡4 iii m s¡1 ´ ) x1 (t) = ¡5 + 3t, y1 (t) = ¡ t p b speed = 10 km min¡1 c a minutes later, (t ¡ a) have elapsed a ii x = + t ¡8 ´ ´ d 10:12 am e ) ´ 12 3:5 iii 13 m s¡1 For B it is ¡2 p p ¡1 c For A, speed is km h For B, speed is km h¡1 ³ ´ ³ ´ ³ ´ c i ¡4 ´ ³ ³ b ³ ¡ 1, ¡ 1, ¡1 w + w2 + w3 + w + , , , w ¡ w2 ¡ w3 ¡ w ¡ c ´ ¡4 = 0, ) perpendicular 28:6o b For A it is ¡ 2¼ ¢ w + 12 , w2 + 12 , w3 + 12 , w4 + 120 ¡90 ´ ´ a A is at (4, 5), B is at (1, ¡8) 2¼ = 17 z = 1, w, w2 , w3 , w4 where w = cis , ³ a not included 16 a = ¡3, b = ¡8, c = 30 a R ´ ³ ² ¡2 10 12 ³ c i (¡2, ¡7) ii , ³ b i (0, ¡6) ii ® arg(1 + z) = ³ a i (¡4, 3) ii ¡ ¢ ¡ ¢ 14 + z = cos ®2 cis ®2 , ¡®¢ j1 + zj = cos 0) b (0, 4, 1) c (4, 0, 9) 75:5o 75:7o I ¼ , EXERCISE 16A.3 ¢ ¡µ¢ (¡ 12 , (0, 7, 3) and ( 20 , ¡ 19 , ¡ 11 ) a (1, 2, 3) b ( 73 , 23 , 83 ) 3 11 a ¯(2z)¡1 ¯ = 12 , arg (2z)¡1 = ¡µ b j1 ¡ zj = sin x = 3t, y = 1, z = ¡ 4t, t R x = 1, y = ¡ 3t, z = 5, t R x = 5t, y = ¡ 2t, z = ¡1 + 4t, t R black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\900IB_HL-2_AN.CDR Friday, 12 December 2008 3:04:52 PM TROY 20 IB_PD (901) ANSWERS a A is at (2, 3), B(8, 6), C(5, 0) b AB = BC = a P is at (10, 4), Q(3, ¡1), R(20, ¡10) ³ a Either no solutions or an infinite number of solutions b i a1 = ka2 , b1 = kb2 , c1 = kc2 for some k ii a1 = ka2 , b1 = kb2 , c1 = kc2 , d1 = kd2 for some k c i Planes meet in a line x = ¡2 + 3t, y = t, z = 5, t R ii Planes meet in a line x = ¡2t, y = t, z = +3t, t R iii Planes are coincident ´ ¡ ! ¡ ! 10 ¡14 , PQ ² PR = c ]QPR = 90o d 74 units2 a A is at (2, 5), B(18, 9), C(14, 25), D(¡2, 21) ³ ¡ ! b AC = p 12 20 ´ ¡ ! and DB = ³ 20 ¡12 ´ ) p 544 units ii 544 units iii c Diagonals are perpendicular and equal in length, and as their midpoints are the same, i.e., (8, 15), ABCD is a square i If k 6= ¡2, planes meet in a line with infinitely many solutions b If k = 16, planes are coincident, with infinitely many solutions If k 6= 16, planes are parallel with no solutions EXERCISE 16C a Meet at a point (1, ¡2, 4) They intersect at (1, 2, 3), angle ¼ 10:9o Lines are skew, angle ¼ 62:7o They are parallel, ) angle 0o They are skew, angle ¼ 11:4o They intersect at (¡4, 7, ¡7), angle ¼ 40:2o They are parallel, ) angle = 0o k 6= 16, the lines are parallel and so there are no solutions If k = 16, the lines are coincident We ) have infinitely many solutions of the form x = t, y = 3t ¡ 8, t R : If a 6= ¡4, we have a unique solution The lines meet at a b c d e f If ¡ a+88 , 4a+16 ¡21 2a+8 + 5t 9¡t , y= , z = t, t R 3 c Meet in a line x = 3t ¡ 3, y = t, z = 5t ¡ 11, t R d No solutions as planes are parallel and intersected by 3rd plane e Two planes are coincident and the other cuts obliquely at the line x = 52 + 12 t, y = ¡ 32 + 32 t, z = t, t R b Meet in a line x = f Meet at the point (3, ¡2, 0) If k = the planes meet in a line x = ¡10t, y = ¡1 ¡ 7t, z = t, t R If k 6= 5, the line of intersection of any two planes is parallel to the third ) no solutions ¢ If a = ¡4, the lines are parallel and so not A unique solution exists if m 6= ¡1 or m 6= ¡5 intersect (no solns) No case for infinite number of solns p3 p p510 units 3 units units p a p units b 2 units 26 p111 1498 a units b d p p 2769 13 ¼ 2:87 units c p 6180 10 units e units f t+7 t¡5 ,y= , z = t, 3 t R If m = ¡5, the system is inconsistent ) no solutions If m = ¡1, planes meet in a line x = They meet at the point ¼ 4:05 units REVIEW SET 16A ¼ 7:86 units ³ ´ a ¶ µ b ¡1 ¶ µ0¶ c µ1¶ d 0 p3 a b a b c d or (¡1, ¡2, 10 p 19 p , p5 ³ units yellow ´ ³ is parallel to ³ ¡2 ´ ³ ² 10 10 ´ ´ ³ ² ¡5 ´ =0 ¡5 ´ =0 µ ¶ µ ¶ ¡4 y = +t , t R b ¡4x + 5z = 24 z ¡1 c (¡5, 2, 9) or (11, 2, ¡11) a ¼ 15:8o b ¼ 65:9o µx¶ µ ¶ µ ¡3 ¶ y = ¡1 + t a , t R b P( 67 , 87 , ¡ 37 ) z ¡2 (6, ¡1, ¡10) b ¼ 28:6o c 14 units 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 ¡2 a a ¼ b the line and plane are parallel ) c ¼ 11:3o d ¼ 30:7o a ¼ 83:7o b ¼ 84:8o c ¼ 86:2o d ¼ 73:2o e ¼ 62:3o a 3x + 2y ¡ z = ¡1 b (0, 1, 3) 0o magenta ³ ´ X23, x1 = + t, y1 = ¡ 3t, t > Y18, x2 = 13 ¡ t, y2 = ¡ 2a + at, t > interception occurred at 2:22:30 pm µ = 192:7o , ¼ 4:54 units per minute µx¶ p26 138 13:1o m = 10 REVIEW SET 16B units EXERCISE 16E cyan , t2R c K(7, 17), L(22, 11), M(33, ¡5), N(3, 7) d 261 units2 ¡ 13 ), d = jd2 ¡ d1 j units b b x = ¡6 + 4t, y = ¡ 3t, t R and NK is perpendicular to MN as ) units 18 p a2 + b2 + c2 19 2x ¡ y + 2z = ¡1 and 2x ¡ y + 2z = 11 17 a ³ ´ ´ ¢ ³ 50 p2 units b N( , p units b units 14 a N(3:4, 1:2, 1), d = 16 a ¡3 ¡68 64 , 29 29 i ¡6i + 10j ii ¡5i ¡ 15j iii (¡6 ¡ 5t)i + (10 ¡ 15t)j t = 0:48 h c shortest dist ¼ 8:85 km, so, will miss reef 75 b B(3, 6, ³ +t , b KL is perpendicular to NK as 25 13 a k = ³ ´ 29 a KL is parallel to MN as units ¡ 11 ) ´ x y = +t p 10(3i ¡ j) (1, ¡3, 0) 10 X axis at (2, 0, 0) 11 a y¡3z = ¡7 b x¡z = ¡2 c 3x¡y = 12 y¡2z = ¡ 32 ¡6 ¡ 94 a (¡4, 3) b (28, 27) c 10 m s¡1 d a y=0 b z=4 a ii ¡2x + 6y + z = 18 b ii ¡5x + 3y + 12z = 12 c ii ¡y + z = fmany vector forms existg a x = + t, y = ¡2 ¡ 3t, z = 4t, t R b x = + t, y = ¡ t, z = ¡1 ¡ 2t, t R x = ¡ t, y = ¡1 + 3t, z = ¡ 3t, t R (1, 2, 0) x = + t, y = ¡2 + 2t, z = ¡ 5t, t R a (0, ¡4, 9) b (1, ¡2, 4) c (¡5, ¡14, 34) p a (¡1, ¡1, 4); units b (0, 1, ¡3); 11 units c (¡ 17 , ¡ 26 , ¡ 17 ); 7 ³ = ³ ´ a 2x ¡ y + 3z = b 3x + 4y + z = 19 c x ¡ y ¡ 2z = ¡1 d x + 3y + z = 10 ¡1 x y a EXERCISE 16D µ x = ¡ 2s + 3t, y = s, z = t, s, t R a If k = ¡2, planes are coincident with infinitely many solutions 95 ´ ¡ ! ¡7 ¡5 , PR = EXERCISE 16F 100 ³ ¡ ! b PQ = p 45 units 901 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\901IB_HL-2_AN.CDR Friday, 12 December 2008 3:05:26 PM TROY IB_PD (902) 902 ANSWERS a 17 units b ( 83 , ¡ ! ¡3 µ a PQ = ¶ , ) 3 p ¡ ! ¡ ! j PQ j = 26 units, QR = c t(p + 10) = q + has infinitely many solutions for t when µ ¡4 ¶ p + 10 = and q + = 0, ) p = ¡10, q = ¡2 ¡1 A a B a b x = + t, y = 4t, z = ¡ 3t, t R µx¶ µ2¶ µ ¶ µ ¡4 ¶ y = +¸ c + ¹ ¡1 , ¸, ¹ R z ¡3 p a units b (1, 2, 4) c 116 units 10 a 5x + y + 4z = b x = 5t, y = t, z = 4t, t R c ( 14 , 14 , 7) O y meet at the µ ¡1 ¶ = t, t R If k 6= ¡2 the planes ¡1) 10 a t = p 2§ 16 p 14 = 2¡ ³ 2+ 10 p 14 ¡1 ¶ µ ¡2 ¶ p1 74 b ¡1+ p , ¡1¡ ! p2 , à p8 74 j+ 3+ p ¡ p p ¡ p3 74 c k = ¡7 or 11 ¡ ! 10 a PQ = ¡ ! ¡! p 3¡ p , and i+ ¼ 26:4o ´ p4 ! ¡2 ¡4 j¡ p3 74 k 15 b ¼ 41:8o ¡ ! 12 a A(2, ¡1, 0) c r = ¡3 +u p 14 units2 f normal is 28 p8 74 c positively skewed d The modal travelling time was between 10 and 20 minutes a column graph b histogram ¶ ¶ µ µx¶ y z ¡4 ¶ µ = µ¡2¶ ¶ µ +¸ a ¶ ¡1 µ5¶ a ( 15 , 17 , ) 5 ¡ ! and AC = a c a c b (¡1, 3, 1) c 6x ¡ 8y ¡ 5z = ¡35 cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 0 50 40 30 20 10 b c d Seedling height 300 325 350 375 400 425 450 length (cm) 20 58:33% i ii 1218 512 mm i 5:61 ii iii b i 16:3 ii 17 iii 18 i 24:8 ii 24:9 iii 23:5 A : 6:46 B : 6:85 b A : B : The data sets are the same except for the last value, and the last value of A is less than the last value of B, so the mean of A is less than the mean of B d The middle value of the data sets is the same, so the median is the same b k = ¡1 10 EXERCISE 17B.1 ¡ ! ¡ ! ¡ ! ¡ ! and AB² AC = 0, ) AB ? AC b 14x ¡ 34y ¡ z = ¡11, ¼ 2:42 units p c x = ¡ 2t, y = ¡ t, z = ¡1 + 6t, t R d units ¡ ! a AB = ¡1 15 no of matches d 3x¡y+2z = ¡2 ¡2 frequency frequency 10 units3 REVIEW SET 16D Leaf 368888 00000222444455556666788889 0 5 78 122234578 025556 j means 12 minutes 74 µ g b Stem k or ¡ p1 i ¡ µ height (cm) 170 175 180 185 190 195 200 205 numerical data and ´ 11 a OM = 12 (OB + OC) d : e ¡ ! ¡ ! So, BD = 3BY, etc c The modal class is 185-190 cm, as this occurred the most frequently d slightly positively skewed a Continuous numerical, but has been rounded to become discrete , t2R ¡4 +t p2 , p p ¡ p a ¶ 10 4x + 2y + z = 3, ¼ 64:1o à ¡1 120 130 140 150 160 170 µ units 100 ³ µ 47 48 49 50 51 52 53 54 55 y z ¡ ! and BY = b t = ¡ 13 95 a ¶ Heights of basketball players b D(¡1, ¡1, 2) c ( 16 , 56 , 23 ) units a intersecting b cos µ = µx¶ b ¶ frequency ¡1 50 p31 110 3 ¡3 a Heights can take any value from 170 cm to 205 cm, e.g., 181:37 cm b 15 10 2 a They not meet, the line is parallel to the plane b units2 11 b 2x + 3y + 6z = 147 c 14 units d (5, 7, 3) and (9, 13, 15) frequency a 14x + 29y ¡ 4z = 32 b ¼ 55:9o c r = a n= p 2 EXERCISE 17A REVIEW SET 16C µ µ ¡ ! p4 114 e C c BD = b m=1 c x¡y¡z =0 d t=2 1 a+b a X(7, 3, ¡1), D(7, 1, ¡2) b Y(5, 3, ¡2) 75 12 a a (4, 1, ¡3) and (1, ¡5, 0) 7:82o 25 x= = ¡ 11 + t, z , point ( 43 , ¡ 14 a1 a2 b b x = 7, y = + 13 t, z = ¡4 + 13 t, t R c M (7, 34 , ¡3 14 ) 11 If k = ¡2, the planes meet in the line , OABC is a rhombus So, its diagonals bisect its angles b black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\902IB_HL-2_AN.CDR Tuesday, 13 January 2009 12:55:25 PM TROY IB_PD (903) ANSWERS c median ¼ 35 cm d actual median = 34:5, i.e., a good approx a mean: $29 300, median: $23 500, mode: $23 000 b The mode is the lowest value, so does not take the higher values into account 160 c No, since the data is positively skewed, the median is not in the 11 120 100 80 60 40 median 20 score EXERCISE 17B.2 10 number of phone calls 1011 mode, median (2) mean (2.96) have won its case by arguing that a larger sample would have found an average of 50 matches per box a b c a b c 90 100 b 42 c 20 d 13 e 29 f 39 g 16 124 cm ii Q1 = 116 cm, Q3 = 130 cm 124 cm ii 130 cm tall 29 cm ii 14 cm d over 14 cm ii iii iv v 10 ii iii iv v Length (x cm) 24 x < 27 27 x < 30 30 x < 33 33 x < 36 36 x < 39 39 x < 42 42 x < 45 Frequency 10 9 10 iii range = iv IQR = b i = 0, Q1 = 4, median = 7, Q3 = 8, max = ii C frequency 18 27 29 30 iii range = iv IQR = c i = 117, Q1 = 127, med = 132, Q3 = 145:5, max = 151 ii 115 120 125 130 135 140 145 150 155 a cumulative frequency Statistic value Q1 median Q3 max value 25 20 15 10 b i Year 9: 11, Year 12 10 14 16 17:5 Year 12: 11:5 ii Year 9: 5, Year 12: c i true ii true length (cm) cyan magenta yellow 95 50 75 25 95 100 50 45 75 42 25 39 95 36 100 50 33 75 30 100 27 25 24 95 Year 7:5 10 12 a median = 6, Q1 = 5, Q3 = b c median 100 30 50 80 iii range = 34 iv IQR = 18:5 35 75 70 a i 35 ii 78 iii 13 iv 53 v 26 b i 65 ii 27 a was 98, was 25 b greater than or equal to 70 c at least 85 marks d between 55 and 85 e 73 f 30 g ¼ 67 a i = 3, Q1 = 5, median = 6, Q3 = 8, max = 10 ii EXERCISE 17C 25 60 EXERCISE 17D.2 31:7 a 70 b ¼ 411 000 litres, i.e., ¼ 411 kL c ¼ 5870 L 3 a 125 people b ¼ 119 marks c 25 d 137 marks 50 median = 2:45, Q1 = 1:45, Q3 = 3:8 range = 5:2, IQR = 2:35 i greater than 2:45 ii less than 3:8 iii The minimum waiting time was minutes and the maximum waiting time was 5:2 minutes The waiting times were spread over 5:2 minutes EXERCISE 17B.3 40 i ii Q1 = 4, Q3 = iii iv i 17:5 ii Q1 = 15, Q3 = 19 iii 14 iv i 24:9 ii Q1 = 23:5, Q3 = 26:1 iii 7:7 iv 2:6 a a i b i c i a i b i a i 2:61 ii iii b This school has more children per family than the average Australian family c positive d The mean is larger than the median and the mode a i 69:1 ii 67 iii 73 b i 5:86 ii 5:8 iii 6:7 a i 5:63 ii iii b i 6:81 ii iii c the mean d yes a mean = $163 770, median = $147 200 (differ by $16 570) b i mean selling price ii median selling price a ¼ 70:9 g b ¼ 210 g c 139 g 10:1 cm 10 a mean for A ¼ 50:8, mean for B ¼ 49:9 b No, as to the nearest match, A is 51 and B is 50 11 17:25 goals per game 12 and 12 13 a i $31 500 ii $28 000 iii $33 300 b The mean a a b i 40 ii 40 30 a 2270 h b ¼ 69% c 62 or 63 EXERCISE 17D.1 Phone calls in a day 15 12 3 a i 49 ii 49 iii 49:03 b no c The sample of only 30 is not large enough The company could b 20 a ¼ 61 b ¼ 87 students c ¼ 76 students d 24 (or 25) students e 76 marks : : a 26 years b 36% c i 0:527 ii 0:030 frequency value 11 as an outlier d The mean takes into account the larger numbers of phone calls e the mean cumulative frequency 140 centre a mean: 3:19, median: 0, mode: b The data is very positively skewed so the median is not in the centre c The mode is the lowest value so does not take the higher values into account d yes, 21 and 42 e no a 44 b 44 c 40:2 d increase mean to 40:3 116 3144 km $185 604 x = 15 10 a = 37 12 14:77 13 and a b c 1:43 a i 2:96 ii iii b c positively skewed with data 903 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\903IB_HL-2_AN.CDR Thursday, 11 March 2010 11:32:51 AM PETER 10 11 12 13 IB_PD (904) 904 ANSWERS e a Minx = 33, Q1 = 35, Q2 = 36, Q3 = 37, Maxx = 40 b i ii c ‘old’ ‘new’ 30 31 32 33 34 36 35 37 39 38 40 41 a b ¼ 28:3% c cm d IQR ¼ 2:4 cm e 10 cm, which means that 90% of the seedlings have a height 60 80 100 120 140 160 180 200 lifespan (hours) of 10 cm or less a 27 b 29 c 31:3 d IQR ¼ 4:3 e 28 10 s f For the ‘old type’ of globe, the data is bunched to the right of the median, hence the distribution is negatively skewed For the ‘new type’ of globe, the data is bunched to the left of the median, hence the distribution is positively skewed g The manufacturer’s claim, that the ‘new type’ of globe has a 20% longer life than the ‘old type’ seems to be backed up by the 25% higher mean life and 19:5% higher median life EXERCISE 17E a x ¼ 4:87, Minx = 1, Q1 = 3, Q2 = 5, Q3 = 7, Maxx = b 10 EXERCISE 17F.1 frequency a Sample A b c score d x ¼ 5:24, Minx = 2, Q1 = 4, Q2 = 5, Q3 = 6:5, Maxx = frequency a discrete c set b c d a 10 Shane 5 wickets per innings Brett 10 0 10 wickets per innings d There are no outliers for Shane Brett has outliers of and which must not be removed h EXERCISE 17F.3 a b a b a b a b a b Shane Brett 10 i Generally, Shane takes more wickets than Brett and is a more consistent bowler a continuous c For the ‘New type’ globes, 191 hours could be considered an Mean Median Range IQR Old type 107 110:5 56 19 a 16% b 84% c 97:4% d 0:15% times a b 32 c 136 a 458 babies b 444 babies REVIEW SET 17A The mean and median are ¼ 25% and ¼ 19% higher for the ‘new type’ of globe compared with the ‘old type’ New type 134 132 84 18:5 a Diameter of bacteria colonies yellow 95 100 50 75 25 95 50 75 25 95 100 50 75 25 100 magenta b 489 3557 115688 012345566779 leaf unit: 0:1 cm 01279 c The distribution is slightly negatively skewed The range is higher for the ‘new type’ of globe (but has been affected by the 191 hours) The IQR for each type of globe is almost the same cyan x ¼ 1:72 children, s ¼ 1:67 children ¹ ¼ 1:72 children, ¾ ¼ 2:83 children2 x = 14:5 years, s ¼ 1:75 years ¹ ¼ 14:5 years, ¾ ¼ 3:18 years2 x = 37:3 toothpicks, s ¼ 1:45 toothpicks ¹ ¼ 37:3 toothpicks, ¾ ¼ 2:16 toothpicks2 x = 47:8 cm, s ¼ 2:66 cm ¹ ¼ 47:8 cm, ¾ ¼ 7:31 cm2 x = $390:30, s ¼ $15:87 ¹ ¼ $390:30, ¾ ¼ 253:18 dollars2 EXERCISE 17G outlier However, it could be a genuine piece of data, so we will include it in the analysis d Bullets: range = 11, x = 5:7 We suspect the Rockets, they have two zeros Rockets: s = 3:9 à greater variability Bullets: s = 3:29 standard deviation We suspect variability in standard deviation since the factors may change every day b i sample mean ii sample standard deviation c less variability a x = 69, s = 6:05 b x = 79, s = 6:05 c The distribution has simply shifted by 10 kg The mean increases by 10 kg and the standard deviation remains the same a x = 1:01 kg; s = 0:17 b x = 2:02 kg; s = 0:34 c Doubling the values doubles the mean and the standard deviation p = 6, q = a = 8, b = b x = 8:7 a 0:809 b 0:150 c the extreme value greatly increases the standard deviation a sn ¼ 6:77 kg b ¹ ¼ 93:8 kg, ¾ ¼ 46:4 kg2 a x ¼ 77:5 g, sn ¼ 7:44 g b ¹ = 77:5 g, ¾2 = 58:9 g2 a 32:4 b 9:86 min2 Brett’s distribution is positively skewed f Shane has a higher mean (¼ 2:89 wickets) compared with Brett (¼ 2:67 wickets) Shane has a higher median (3 wickets) compared with Brett (2:5 wickets) Shane’s modal number of wickets is (14 times) compared with Brett, who has a bi-modal distribution of and (7 times each) g Shane’s range is wickets, compared with Brett’s range of wickets Shane’s IQR is wickets, compared with Brett’s IQR of wickets Brett’s wicket taking shows greater spread or variability s 4.97 12:6 EXERCISE 17F.2 e Shane’s distribution is reasonably symmetrical x 25 30:5 b Andrew a Rockets: range = 11, x = 5:7; 15 15 Andrew Brad set 10 frequency x s 95 B 1:06 100 50 a A 75 25 c black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\904IB_HL-2_AN.CDR Thursday, 11 March 2010 11:33:15 AM PETER i ii 3:15 cm 4:5 cm IB_PD (905) ANSWERS a highest = 97:5 m, lowest = 64:6 m REVIEW SET 17B b use groups 60 - < 65, 65 - < 70 , 70 - < 75 , etc c A frequency distribution table for distances thrown by Thabiso distance (m) tally freq (f ) 60 - < 65 j 65 - < 70 jjj © 70 - < 75 © jjjj 75 - < 80 jj © 80 - < 85 © jjjj jjj © 85 - < 90 © jjjj j 90 - < 95 jjj 95 - < 100 jj Total 30 d 30 20 0 10 20 30 11 12.5 60 ,65) ,70) ,75) ,80) ,85) ,90) ,95) 100) 105) 65 70 75 80 85 90 5, 0, [ [6 [ [ [ [ [ [9 [10 distance (m) ¼ 81:1 m ii 83:1 m < e i a = and b = or a = and b = a 15 16.5 18 EXERCISE 18A a 0:78 b 0:22 a 0:487 b 0:051 c 0:731 a 43 days b i ¼ 0:047 ii ¼ 0:186 iii 0:465 a ¼ 0:089 b ¼ 0:126 Cumulative frequency 50 EXERCISE 18B 40 a fA, B, C, Dg b fBB, BG, GB, GGg c fABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, 30 20 BADC, BCAD, BCDA, BDAC, BDCA, CABD, CADB, CBAD, CBDA, CDAB, CDBA, DABC, DACB, DBAC, DBCA, DCAB, DCBAg d fGGG, GGB, GBG, BGG, GBB, BGB, BBG, BBBg median 10 Score 0 9.95 19.95 29.95 39.95 a 49.95 b die b ¼ 25:9 c ¼ 12:0 d x ¼ 26:0, s ¼ 8:31 Girls pos skewed 36:3 sec 7:7 sec Boys approx symm 34:9 sec 4:9 sec coin T H b The girls’ distribution is positively skewed and boys’ distribution is approximately symmetrical The median swim times for boys is 1:4 seconds lower than for girls but the range of the girls’ swim times is 2:8 seconds higher than for boys The analysis supports the conjecture that boys generally swim faster than girls with less spread of times c b A B A B Min 11 11:2 i Range 2:6 Q1 11:6 12 ii IQR 1:2 Median 12 12:6 Q3 12:6 13:2 Max 13 13:8 c i We know the members of squad A generally ran faster because their median time is lower ii We know the times in squad B are more varied because their range and IQR is higher yellow 50 75 25 95 100 50 75 25 95 50 75 25 95 100 50 75 25 100 magenta die black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\905IB_HL-2_AN.CDR Thursday, 11 March 2010 4:41:27 PM PETER spinner A b H T spinner spinner X Y Z X Y Z X Y Z a i 101:5 ii 98 iii 105:5 b 7:5 c x = 100:2, s ¼ 7:59 a x ¼ 33:6 L, s ¼ 7:63 L b ¹ ¼ 33:6 L, ¾ ¼ 58:7 L2 a 12 % b 95% c 68% 10 a 58:5 sec b sec a 88 students b m ¼ 24 T c spinner d 5-cent 10-cent H H T a die 1 spinner D C B A a cyan die 95 shape centre (median) spread (range) 100 a 50 b ¹ = $103:50, ¾ ¼ 378:15 $2 a x = $103:50, s ¼ $19:40 a 68% b 95% c 81:5% d 13:5% a mean is 18:8, standard deviation is 2:6 b 13:6 to 24:0 a 2:5% b 84% c 81:5% 10 a i x ¼ 3:88 cm ii s2 ¼ 0:0571 cm2 b ¹ ¼ 3:88 cm, ¾ ¼ 0:0591 cm2 11 a a = 4, b = b c 2:5 12 a 120 students b 65 marks c 54 and 75 d 21 marks e 73% of them f 81 marks 11 40 x ¼ 49:6, s ¼ 1:60 Does not justify claim Need a larger sample ¼ 414 customers range = 19, lower quartile = 119, upper quartile = 130, s ¼ 6:38 60 margin (points) 10 f Frequency 40 Frequency histogram displaying the distance Thabiso throws a baseball 905 d B C D coin spinner A B H C A B T C draw draw P P B W P B B W P W B W IB_PD (906) 906 ANSWERS EXERCISE 18C.1 a b c a b i a 14 b 19 a b a 15 ii d e 15 f b c c 124 1461 36 d 18 e 237 1461 d f 12 g h fAKN, ANK, KAN, KNA, NAK, NKAg a b c b i ii iii coin T T H H 1 b c d b 10 c h 36 18 18 b i c j 13 18 11 36 e a 16 d +4 ¡2 + 3 ¡3 + 5 18 25 36 f iv g 36 49 b c 216 343 a 0:0096 b 0:8096 a b a 0:56 b 0:06 c 0:14 d 0:24 a b i ¢ = ¡ ¢4 ¡ ¢5 +5 ¡ ¢3 ¡ ¢2 4 = ¡ ¢4 ¡ ¢1 4 12 125 c iii + ¡ ¢4 ¡ ¢ ¡ ¢ ¡ ¢ + 10 + 53 512 ii B c b ¡ ¢ ¡ ¢4 135 512 + q5 ¡ ¢¡ ¢ ¡ ¢ ¡ ¢ 16 81 = +5 ¡ ¢2 ¡ ¢3 16 5pq4 = 32 2 2 1 +6 23 +4 23 3 2 8 ii 23 = iii 27 ¡ ¢3 ¡ ¢ A 15 16 125 b ¡ ¢4 = + 10p2 q ¡ ¢3 ¡ ¢2 ¡ ¢5 4 47 128 iii a A = f1, 2, 3, 6g, B = f2, 4, 6, 8, 10g b i n(A) = ii A [ B =f1, 2, 3, 4, 6, 8, 10g iii A \ B =f2, 6g a b ii EXERCISE 18D.1 a ¢ ¡ ¢4 + 10p3 q EXERCISE 18H.1 10 i iii a ii 10( 12 )2 ( 12 )3 = = 5p4 q a ¼ 0:154 b ¼ 0:973 a ¼ 0:0305 b ¼ 0:265 ¼ 0:000 864 ¼ 0:0341 times spinner 10 cent H T a 32 b i 10 a cent q)5 +10 EXERCISE 18C.2 ¼ 0:9602 c ¡ + a iv £ 99 ¼ 0:0002 100 98 97 £ 99 ¼ 0:0398 100 a b i 5( 12 )4 ( 12 ) = + BACD, BADC, BCAD, BCDA, BDAC, BDCA, CABD, CADB, CBAD, CBDA, CDAB, CDBA, DABC, DACB, DBAC, DBCA, DCAB, DCBAg 19 45 p5 a a fABCD, ABDC, ACBD, ACDB, ADBC, ADCB, £ c to start with a (p + a fGGG, GGB, GBG, BGG, GBB, BGB, BBG, BBBg b i 18 ii 18 iii 18 iv 38 v 12 vi 78 98 100 97 99 a (p + q)4 = p4 + 4p3 q + 6p2 q + 4pq3 + q b 4( 12 )3 ( 12 ) = d b EXERCISE 18G fremember leap yearsg 33 A B A B A B d 27 125 EXERCISE 18D.2 a a c a 14 55 100 b b 100 £ a 55 £ 99 100 99 98 £ 15 30 b 15 c A ¼ 0:0006 e 97 100 ¼ 0:000 006 d £ 96 99 £ 95 98 f ¼ 0:912 b A EXERCISE 18E a 1st spin 2nd spin B R Y B R Y B R Y Qw_ B Qr_ Qr_ Qw_ Qw_ Qr_ R Qr_ Qr_ Qr_ Qw_ Y Qr_ Qr_ b c 16 d e 17 40 11 30 a 19 25 a b 10 21 b a b rain Qt_ no rain 19 30 50 Qw_ win b Qw_ lose Aw_p_ win Qw_Op_ lose c 15 c 15 d b 15 c 15 d A A B A'ÇB A B A È B' B A' ÇB' b B A B 15 magenta yellow 95 100 50 75 25 95 C 100 50 75 25 95 15 100 50 a d A c 19 B a 10 75 25 95 100 50 75 25 d A' These are all possibilities, so their probabilities must sum to cyan 25 A b 10 a c b EXERCISE 18F 20 49 13 25 b a P(win) = 0:032 B a 29 b 17 c 26 d a 65 b c d 52 a 19 b 12 c 45 d 58 e 13 f 20 40 40 a Rt_ a B black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\906IB_HL-2_AN.CDR Friday, 12 December 2008 3:30:05 PM TROY C IB_PD (907) ANSWERS c d A B A 907 REVIEW SET 18A B ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BCAD, BCDA, BDAC, BDCA, CABD, CADB, CBAD, CBDA, CDAB, CDBA, DABC, DACB, DBAC, DBCA, DCAB, DCBA C e C A f B A a 2 a B b b 8 c 25 a 24 25 b c 11 12 P(N wins) 44 125 = N (0.4) N (0.4) = 0:352 N (0.4) R (0.6) R (0.6) C C N (0.4) N (0.4) R (0.6) EXERCISE 18H.2 a LHS set and the second with the RHS set a A = f7, 14, 21, 28, 35, , 98g a B = f5, 10, 15, 20, 25, , 95g i n(A) = 14 ii n(B) = 19 iii iv 31 b+c a+b+c+d a i b a+b+c+d ii a+b+c a+b+c+d iii iv a b 13 c 15 15 10( ) ( ) ¼ 0:205 b 5( 45 )4 ( 15 ) £ 499 £ 498 ¼ 0:000 000 193 500 b 1¡ a+b+c a+b+c+d 496 500 495 499 £ £ 494 498 b P(A or B) = P(A) + P(B) ¡ P(A and B) M P 22 18 25 b i ii a a 13 20 5 a a b b b b 20 W' (0.05) 10 5 c c 20 d 11 50 c d b REVIEW SET 18B 14 e 25 e 14 23 a 0:46 b f BBBB, BBBG, BBGB, BGBB, GBBB, BBGG, BGBG, BGGB, GGBB, GBBG, GBGB, BGGG, GBGG, GGBG, GGGB, GGGG a 70 163 37 40 b 10 25 c = 5 33 a b 19 66 c 11 d 16 33 a 0:45 b 0:75 c 0:65 10 a 0:0484 b 0:3926 11 23 12 15 13 P(A[B [C) = P(A) + P(B) + P(C) ¡ P(A \ B) a Two events are independent if the occurrence of each event does EXERCISE 18J ¡ P(B \ C) ¡ P(A \ C) + P(A \ B \ C) P(R \ S) = 0:2 and P(R) £ P(S) = 0:2 ) not influence the occurrence of the other For A and B independent, P(A) £ P(B) = P(A and B) b Two events, A and B, are disjoint if they have no common outcomes P(A or B) = P(A) + P(B) a 0:09 W (0.36) R (0.25) b 0:52 W' (0.64) are independent events 12 10 c No, as P(A \ B) 6= P(A) £ P(B) W' (0.64) ¡ 0:9 £ 0:8 £ 0:7 = 0:496 using a Venn diagram and P(A \ B) ¡ 12 ¢4 ¡ ¢ a i 13 13 20 a i iii 13 ii 10 13 25 a b ¼ 0:544 0:9 a b No, as P(C \ D) 6= P(C) P(D) 10 31 70 ¡3 63 125 +4 + ¡ ¢ ¡ ¢3 ¡ ¢3 ¡ ¢ ¡ ¢4 5 +6 b i ¡ ¢2 ¡ ¢2 216 625 ii 328 625 yellow 50 75 25 10 a ¼ 0:0205 b ¼ 0:205 95 ¡ ¢4 a 20(0:6)3 (0:4)3 ¼ 0:276 b 6(0:6)(0:4)5 + (0:4)6 ¼ 0:0410 100 50 75 25 95 100 50 75 25 95 100 50 75 = +15(0:6)2 (0:4)4 + 6(0:6)(0:4)5 + (0:4)6 X wins X wins Y wins Y wins Y wins a 0:0435 b ¼ 0:598 a ¼ 0:773 b ¼ 0:556 99 10 ¼ 0:424 0:0137 15 148 13 83 10 a 19 b 19 10 a 0:95 b ¼ 0:306 c 0:6 11 a 0:104 b ¼ 0:267 c ¼ 0:0168 25 = (0:6)6 + 6(0:6)5 (0:4) + 15(0:6)4 (0:4)2 + 20(0:6)3 (0:4)3 X wins X wins X wins X wins Y wins Y wins Y wins EXERCISE 18L + 21 31 ¢ (0:6 + 0:4)6 1 ¼ 0:655 12 ¼ 0:318 ¼ 0:530 a 10 b a ¼ 0:0288 b ¼ 0:635 c ¼ 0:966 15 a ¼ 0:0962 b ¼ 0:0962 magenta b +4 EXERCISE 18K cyan W (0.36) R' (0.75) a 0:35 b 0:85 c 0:15 d 0:15 e 0:5 91 14 a 216 b 26 15 Hint: Show P(A0\ B ) = P(A0) P(B ) 95 b 100 30 a 10 a ¼ 0:0238 b ¼ 0:976 10 a 11 a ¼ 0:259 b ¼ 0:703 15 23 23 d 0:9975 W (0.95) W' (0.05) 11 20 14 25 ¼ 0:023 86 W' (0.05) 22 study both + ( 45 )5 ¼ 0:737 W (0.95) W (0.95) EXERCISE 18I a R (0.6) R (0.6) For each of these draw two diagrams, shade the first with the black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\907IB_HL-2_AN.CDR Thursday, 11 March 2010 11:33:42 AM PETER IB_PD (908) 908 ANSWERS REVIEW SET 18C EXERCISE 19D.1 88 a P(M\C) = 0:85, P(M )P(C) ¼ 0:801, so not independent b 14 a ¼ 0:692 b ¼ 0:558 ¼ 0:127 49 15 8 a 10 a 1:54 £ 10¡6 b b 1:28 £ 13 c i 10¡8 11 a V(t) 100 13 ii 80 60 40 REVIEW SET 18D 20 time (h) a The occurrence of each event does not affect the occurrence of the other P (1) = , 32 10 , 32 P (2) = P (3) = 10 , 32 P (4) = 0.2 , 32 b c d e a a 0:93 b 0:8 c 0:2 d 0:65 n = 11 a 0:07831 b 0:07663 B [probability p(1 ¡ q)(1 + q)] is more likely than 10¡000 A [probability p(1 ¡ q)(2 ¡ p)] a b 14 c 15 58 10 ( 13 + 23 )5 , ¼ 0:313 148 243 11 50 12 a x!1 =0 x a b ¡ 23 20 17 e f g h 20 n 10 25 50 100 500 a i as x ! 0¡ , f (x) ! as x ! 1, f (x) ! as x ! 0+ , f (x) ! 0+ as x ! ¡1, f (x) ! 0+ i vertical asymptote x = as x ! 0+ , y ! ¡1 j horizontal asymptote y = 12 as x ! 1, f (x) ! as x ! ¡1, f (x) ! 12¡ k oblique asymptote y = ¡x as x ! 1, y ! as x ! ¡1, y ! (¡x)+ l horizontal asymptotes y = 0, y = 1000 as x ! 1, y ! 1000¡ as x ! ¡1, y ! 0+ EXERCISE 19C a b c d e AU 2:485 2:393 2:363 2:348 2:336 80 n 10 50 100 500 1000 10 000 AL 0:160 00 0:202 50 0:240 10 0:245 03 0:249 00 0:249 50 0:249 95 converges to 100 AU 0:600 00 0:550 00 0:510 00 0:505 00 0:501 00 0:500 50 0:500 05 iii n 10 50 100 500 1000 10 000 AL 0:549 74 0:610 51 0:656 10 0:661 46 0:665 65 0:666 16 0:666 62 AU 0:749 74 0:710 51 0:676 10 0:671 46 0:667 65 0:667 16 0:666 72 iv n 10 50 100 500 1000 10 000 AL 0:618 67 0:687 40 0:738 51 0:744 41 0:748 93 0:749 47 0:749 95 AU 0:818 67 0:787 40 0:758 51 0:754 41 0:750 93 0:750 47 0:750 05 black 50 Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\908IB_HL-2_AN.CDR Thursday, 24 January 2008 2:43:33 PM PETERDELL AU 0:360 00 0:302 50 0:260 10 0:255 03 0:251 00 0:250 50 0:250 05 AL 0:400 00 0:450 00 0:490 00 0:495 00 0:499 00 0:499 50 0:499 95 75 25 95 50 75 25 95 100 50 75 25 95 100 50 75 25 100 yellow AL 2:185 2:273 2:303 2:318 2:330 n 10 50 100 500 1000 10 000 f 2¼ magenta 60 ii b ¼r2 c Area of circle = ¼r2 c ¡ sin x cyan 40 EXERCISE 19D.2 as x ! ¡3¡ , f (x) ! as x ! 1, f (x) ! 3¡ as x ! ¡3+ , f (x) ! ¡1 as x ! ¡1, f (x) ! 3+ vertical asymptote x = 0, oblique asymptote y = 2x as x ! 0¡ , y ! ¡1 as x ! 1, y ! (2x)+ as x ! 0+ , y ! as x ! ¡1, y ! (2x)¡ horizontal asymptote W = 5000 d horizontal asymptote y = as t ! 1, W ! 5000¡ as x ! 1, y ! 1¡ as x ! ¡1, y ! 1¡ horizontal asymptote y = as x ! 1, f (x) ! 0+ as x ! ¡1, f (x) ! 0¡ vertical asymptotes x = ¡2, x = 1, horizontal asymptote y = as x ! 1, f (x) ! 0+ as x ! ¡2¡ , f (x) ! ¡1 as x ! ¡2+ , f (x) ! as x ! ¡1, f (x) ! 0¡ as x ! 1¡ , f (x) ! as x ! 1+ , f (x) ! ¡1 oblique asymptote y = x as x ! 1, y ! x¡ as x ! ¡1, y ! x+ vertical asymptote x = 0, horizontal asymptote y = n b Hint: P (n) 10 000 c $375 000 < profit < $875 000 d $531 250 < profit < $781 250 e Use upper sum with 100 intervals c ¡1 d e f a vertical asymptote x = ¡3, horizontal asymptote y = c P(n) 2000 g h ¡2 i ¡3 p 32 q r 76 EXERCISE 19B b 4000 a b c 11 d 16 e f j k ¡8 l 12 m 12 n o lim 0.8 6000 EXERCISE 19A 0.6 8000 13 50 b 0.4 maximum speed is 100 km h¡1 , minimum speed is 68:4 km h¡1 Hint: V (t) > 68:4 km h¡1 4:84 km < distance from Adelaide < 20:64 km 9:28 km < distance from Adelaide < 17:18 km 95 P (5) = , 32 32 100 b P (0) = IB_PD (909) ANSWERS b i a ii n 10 50 100 200 1000 10 000 iii iv y a+1 c area = Rational bounds for ¼ 2:9045 < ¼ < 3:3045 3:0983 < ¼ < 3:1783 3:1204 < ¼ < 3:1604 3:1312 < ¼ < 3:1512 3:1396 < ¼ < 3:1436 3:1414 < ¼ < 3:1418 2 n 50 100 500 y= x n 10 50 100 500 a 0.2 0.4 0.6 0.8 Z c AU 3:3349 3:1615 3:1516 3:1436 dx ¼ 3:1416 + x2 r(t) 25 20 15 x AL 0:5497 0:6105 0:6561 0:6615 0:6656 c AU 0:7497 0:7105 0:6761 0:6715 0:6676 10 R1p t (h) x dx ¼ 0:67 b Hint: 10 mg r(t) 25 mg c 35:9 mg < amount < 50:5 mg d Use more intervals of shorter length cos x , h and x are in radians c cos x a y ¦(x) 0.8 0.6 0.4 0.2 y = + x3 x 0.5 b AL 2:9349 3:1215 3:1316 3:1396 a b x 0.2 0.4 0.6 0.8 EXERCISE 19D.3 y x b 0.8 0.6 0.4 0.2 upper rectangles b n = 10 000 a y lower rectangles n 50 100 500 AL 3:2016 3:2214 3:2373 1.5 c AU 3:2816 3:2614 3:2453 R2p a c y y = e- x d e ¡1 f ¡ 12 a vertical asymptote x = 2, oblique asymptote y = x + as x ! 2¡ , f (x) ! ¡1 as x ! 1, f (x) ! (x + 3)+ as x ! 2+ , f (x) ! as x ! ¡1, f (x) ! (x + 3)¡ b horizontal asymptote y = ¡3 as x ! 1, y ! as x ! ¡1, y ! ¡3+ c no asymptotes d vertical asymptote x = 0, horizontal asymptote y = as x ! 0+ , f (x) ! ¡1 as x ! 1, f (x) ! 0+ e vertical asymptote x = 32 f vertical asymptote x = x REVIEW SET 19 a ¡4 b b maximum is 0:5, minimum is 0:167 c 0:95 < area < 1:28 d Use more intervals of shorter length + x3 dx ¼ 3:24 a 18 b 4:5 c 2¼ a b a as as + x ! 32 , x ! 0¡ , 0.5 1.5 b AU ¼ 0:977 units2 , AL ¼ 0:761 units2 c AU ¼ 0:8733 units2 , AL ¼ 0:8560 units2 10 a 2¼ b R2 12 a A = 17 , B = 25 b (4 ¡ x2 ) dx ¼ 4 13 a y ! ¡1 x 21 y y ! ¡1 y¡=¡sinX¡x c ¼ y y= x + x2 p p EXERCISE 20A cyan magenta 95 100 50 75 95 50 75 100 yellow 25 a b c 3x2 d 4x3 a b 2x ¡ c 3x2 ¡ 4x x 25 95 100 50 75 25 95 100 50 75 25 909 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\909IB_HL-2_AN.CDR Thursday, 24 January 2008 2:46:46 PM PETERDELL IB_PD (910) 910 ANSWERS a ¡1 (x + 2)2 b a p x+2 b ¡ c ¡ p 2x x p c x3 d ¡ x4 a 8(4x ¡ 5) b 2(5 ¡ 2x)¡2 c 12 (3x ¡ x2 )¡ £ (3 ¡ 2x) d ¡12(1 ¡ 3x)3 e ¡18(5 ¡ x)2 2x + Function x Derivative Function x¡2 Derivative ¡2x¡3 x2 2x1 x¡3 ¡3x¡4 x3 3x2 x2 x4 4x3 x¡ ¡ 12 x¡ x¡1 ¡x¡2 xn nxn¡1 dy = 3x2 , dx dx dy £ = b dx dy e f ¡ 12 f m ¡ p 2x x j ¡ a n 8x ¡ c a 6x2 ¡ 14x b 2¼x c ¡ 5x3 d 100 e 10 f 12¼x2 p x a b c 2x ¡ 10 d ¡ 9x2 e + 2 4x f 6x2 ¡ 6x ¡ 16 a b ¡ 729 13 c ¡7 d e e f ¡11 , g(x) = 2x + x p 10 c f (x) = x, g(x) = x2 ¡3x d f (x) = , g(x) = 3x¡x2 x a f (x) = x3 , g(x) = 3x + 10 b f (x) = 1 a u¡2 , u = 2x ¡ b u , u = x2 ¡ 3x yellow 95 100 50 75 b 25 95 100 50 75 25 95 100 50 75 ¡ 3x d ¡ 28 27 y = ¡7x, contact at (¡3, 21) c 2u , u = ¡ x2 d u , u = x3 ¡ x2 e 4u¡3 , u = ¡ x f 10u¡1 , u = x2 ¡ 25 (1 ¡ 3x) + 32 x(1 ¡ 3x)¡ a y = ¡7x + 11 b 4y = x + c y = ¡2x ¡ d y = ¡2x + a 6y = ¡x + 57 b 7y = ¡x + 26 c 3y = x + 11 d x + 6y = 43 p a y = 21 and y = ¡6 b ( 12 , 2) c k = ¡5 d y = ¡3x + a a = ¡4, b = b a = 2, b = a 3y = x + b 9y = x + c 16y = x ¡ d y = ¡4 a y = 2x ¡ 74 b y = ¡27x ¡ 242 c 57y = ¡4x + 1042 d 2y = x + a = 4, b = a (¡4, ¡64) b (4, ¡31) c does not meet the curve again a y = (2a ¡ 1)x ¡ a2 + 9; y = 5x, contact at (3, 15) +3 x2 magenta 2x(2x + 1) ¡ 2x2 (2x + 1)2 p ¡1 (1 ¡ 2x) + x x b EXERCISE 20F a f (g(x)) = (2x + 7)2 b f (g(x)) = 2x2 + p p c f (g(x)) = ¡ 4x d f (g(x)) = ¡ x cyan dy is undefined at x = ¡1) dx ii x andpx = b i x = ¡2 § 11 ii x = ¡2 dy is zero when the tangent to the function is horizontal c dx (gradient 0), i.e., at its turning points or points of horizontal inflection dy is undefined at vertical asymptotes dx EXERCISE 20D.1 ¡1 x = or b i never (note: 3 dy dy = 4+ 2, is the gradient function of y = 4x ¡ dx x dx x from which the gradient at any point can be found dS dS = 4t + ms¡1 , is the instantaneous rate of change b dt dt in position at the time t, i.e., it is the velocity function dC dC = + 0:004x $ per toaster, is the instantaneous c dx dx rate of change in cost as the number of toasters changes EXERCISE 20D.2 11 d (x2 ¡ 3) ¡ 2x2 d (x2 ¡ 3)2 (1 ¡ 2x)2 2x(3x ¡ x ) ¡ (x2 ¡ 3)(3 ¡ 2x) (3x ¡ x2 )2 a b c ¡ 324 a f f (g(x)) = 13 3(2 ¡ x) + (1 + 3x) (2 ¡ x)2 f 1 a p +1 b c d 2¡ p p p x x x x x2 p ¡25 e ¡ p f 6x ¡ 32 x g h + 2p p x x 2x3 x 2x x x2 + c EXERCISE 20E.2 + x2 x p p y = 0, y = 27x + 54 c y = 0, y = ¡ 14x + 14 95 k 2x ¡ l 2x + x x o 3x2 + 12x + 12 x2 p ¡ x2 )3 + x(x ¡ x2 )2 (1 ¡ 2x) a ¡48 b 406 14 a 3x2 b 6x2 c 14x d 2x + e ¡4x f 2x + g 3x2 + 6x + h 20x3 ¡ 12x i ¡1 x (x 100 p ¡1 x (x ¡ 3)2 + x(x ¡ 3) 10x(3x2 ¡ 1)2 + 60x3 (3x2 ¡ 1) 50 e d 75 d ¡ 27 c 2x(3 ¡ x) ¡ 12 x2 (3 ¡ x)¡ EXERCISE 20C Hint: Substitute y = x3 a 2x(2x ¡ 1) + 2x2 b 4(2x + 1)3 + 24x(2x + 1)2 a b 10 c ¡ 25 a 12 b 108 dx = 13 y¡ dy dy =1 dy f EXERCISE 20E.1 a ¡4 b c ¡12 d 45 a ¡1 b 34 c ¡ 32 d ¡12 e ¡1 f ¡ 289 a 14 b c ¡ 27 d 14 e f (g(x)) = b ¡18 c ¡8 d ¡4 e ¡ 32 a EXERCISE 20B (2x3 3 a ¡ p13 ¡1 x 2 ¡ x2 )¡ £ (6x2 ¡ 2x) g ¡60(5x ¡ 4)¡3 2 h ¡4(3x ¡ x2 )¡2 £ (3 ¡ 2x) i 6(x2 ¡ )2 £ (2x + ) x x f 25 ¡2 (2x ¡ 1)2 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\910IB_HL-2_AN.CDR Monday, 15 March 2010 9:09:01 AM PETER IB_PD (911) ANSWERS 10 a b c y f (x) = x2 16x + a3 y = 24a REVIEW SET 20C A is ( 32 a, 0), a B is (0, 1 dy = 3x2 (1 ¡ x2 ) ¡ x4 (1 ¡ x2 )¡ dx 1 (2x ¡ 3)(x + 1) ¡ 12 (x2 ¡ 3x)(x + 1)¡ dy = b dx x+1 24 ) a2 Area = 18 units2 , a area ! as a ! d x x = ¡ 12 , 11 a Domain fx j x < 2g c 8x + 3y = ¡19 p 12 a = 14 , point of intersection ( 12 , 23 ) 13 b = 3 BC = EXERCISE 20G 12 ¡ 6x 20 d e 24¡48x f x4 (2x ¡ 1)3 a b 12x¡6 c 2x d2 y d2 y d3 y 30 = ¡6x, = ¡6 b = 2¡ , dx2 dx3 dx2 x d2 y d3 y = ¡ 94 x¡ , = 45 x¡ dx dx3 a c d e d3 y 120 = dx3 x dP = 4t ¡ 12 $1000 per year dt dP is the rate of change in profit with time c dt d i t years ii t > years e minimum profit is $100 000 when t = i dP f = Profit is increasing at $4000 per dt t=4 year after years i dP = 28 Profit is increasing at $28 000 per dt t=10 year after 10 years i dP = 88 Profit is increasing at $88 000 per dt t=25 year after 25 years d2 y d3 y =2+ , = dx (1 ¡ x) dx3 (1 ¡ x)4 p dy = a x = b x = 0, § a dx (1 ¡ x)2 REVIEW SET 20A 2x + x2 x = (¡2, ¡25) a = 52 , b = ¡ 32 a = 12 y = 4x + 2 a 6x ¡ 4x3 b + dM = 8t(t2 + 3)3 dt dA = dt b ³ 10 a ¡ p ¡ b x ¡ x x c (x2 x ¡1 t(t + 5) ´3 ³ ´ x2 1+ a 19 000 m3 per minute b 18 000 m3 per minute ¡ 2(t + 5) ds = 28:1 ¡ 9:8t represents the instantaneous dt velocity of the ball c t = 2:867 secs The ball has stopped and reached its maximum height d 41:49 m e i 28:1 m s¡1 ii 8:5 m s¡1 iii ¡20:9 m s¡1 s0 (t) > ball travelling upwards s0 (t) ball travelling downwards d2 s ds , i.e., the f 5:777 sec g is the rate of change of dt2 dt ¡1 instantaneous acceleration b 69:58 m s a 1:2 m b t3 ¡ 3x)¡ (2x ¡ 3) 11 a ¡2(5 ¡ 4x)¡ b ¡4(5 ¡ 4x)¡ c ¡24(5 ¡ 4x)¡ REVIEW SET 20B EXERCISE 21B dy dy = + 3x¡2 b = 4(3x2 + x)3 (6x + 1) dx dx dy = 2x(1 ¡ x2 )3 ¡ 6x(1 ¡ x2 )2 (x2 + 1) dx a c a i Q = 100 ii Q = 50 iii Q = b i decr unit per year ii decr p12 units per year c Hint: Consider the graph of y = 7, y = ¡25 5y = x ¡ 11 3(x + a f (x) = p 3)2 x ¡1 x (x ¡ + ¡5 dQ = p <0 dt t for all t > 3)3 x p Area = units2 k=2 a $118 000 b f a d e EXERCISE 21A d3 y = 6(2x ¡ 3)(5x2 ¡ 15x + 9) + 6(x2 ¡ 3x)(10x ¡ 15) dx3 y = 16x ¡ (Hint: normal is y = ¡3x + 8) d2 y d2 y a = 36x2 ¡ b = 6x + 34 x¡ dx2 x dx2 a = 9, b = ¡16 A = 9, B = 2, f 00 (¡2) = ¡18 a = 64 4y = 3x + a b y = ¡ 2x+ k k y = 4x c A(2k, 0) B(0, ) k d2 y d3 y ¡24 = 3, = dx2 x dx3 x d2 y = 6(x2 ¡ 3x)(5x2 ¡ 15x + 9), dx2 127 p 10 dQ dt dQ against t dt t b f (x) = 4x3 x2 + + x5 (x2 + 3)¡ units2 a 0:5 m b t = 4; 9:17 m, t = 8; 12:5 m, t = 12; 14:3 m c t = 0: 3:9 m year¡1 t = 5: 0:975 m year¡1 Hint: tangent is 4y = ¡57x ¡ 99 t = 10: 0:433 m year¡1 a = ¡1, b = a = and the tangent is y = 3x ¡ which meets the curve again at (¡4, ¡13) magenta yellow 50 25 95 100 50 75 25 95 100 50 75 a 25 95 100 50 75 25 A = ¡14, B = 21 cyan 97:5 dH = > 0, for all t > 0, the tree is dt (t + 5)2 dH ! as t increases always growing, and dt i E4500 b i decr of E210:22 per km h¡1 ii E4000 ii incr of E11:31 per km h¡1 d as 95 3267 152 b f 00 (x) = ¡ 14 x¡ 100 area = x3 75 a f 00 (x) = ¡ black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\911IB_HL-2_AN.CDR Friday, 12 December 2008 3:33:29 PM TROY IB_PD 911 (912) ANSWERS p dC = at v = 500 000 i.e., 79:4 km h¡1 dt c t = Stone is 367:5 m above the ground and moving skyward a The near part of the lake is km from the sea, the furthest part is km dy dy = 10 = 0:175, height of x ¡ x + 35 x = 12 ; dx dx hill is increasing as gradient is positive dy = ¡0:225, height of hill is decreasing as x = 12 ; dx gradient is negative ) top of the hill is between x = 12 and x = 12 c 2:55 km from the sea, 63:1 m deep dV ³ ´ dV t dt 80 = ¡1250 ¡ a t dt 80 d a b b c a This is where the rate of change is a minimum, however it is out of the bounds of the model (you cannot make < jeans!) a 7m b (h + 5) m c 5m d av velocity = (2t + h + 3) m s¡1 , = increasing for x > 0, never decreasing decreasing for x > 0, never increasing incr for x and x > 4, decr for x p2 increasing for ¡ decreasing for ¡ 32 ¡ k l m cm s¡2 c cm s¡2 = v (1) n t a velocity at t = b acceleration at t = b i b The object is initially cm to the right of the origin and is moving yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 b s(0) = m above the ground, v(0) = 98 m s¡1 skyward 50 t magenta x + x decreasing for x ¡1, x > p p a increasing for x >p and x ¡ decreasing for ¡ 6p x < ¡1, ¡1 < x 0, x < 1, < x b increasing for x > decreasing for x < 1, < x 25 - a(t): cyan x ii increasing for ¡1 x < 1, < x s t + -1 a v(t) = 98 ¡ 9:8t, a(t) = ¡9:8 + - c i d At t = 2, s(2) = cm to the left of the origin e f 06t62 + 10 decreasing for x ¡1, x > origin and is accelerating to the right at cm s¡2 t p ii increasing for ¡1 x < c The object is instantaneously stationary, cm to the left of the v(t): p -1 to the left at cm s¡1 It is accelerating at cm s¡2 to the right + , x> , p2 decreasing for x ¡1, x > t 3 ii increasing for ¡1 x + a(t): t s(t): p2 + - 95 - + v(t): p2 -1 a v(t) = 2t ¡ 4, a(t) = -1 6x6 x ¡ 32 + 25 p p increasing for x ¡ 3, x > + p p decreasing for ¡ x + increasing for x > 1, decreasing for x increasing for ¡1 x 1, x > decreasing for x ¡1, x p p increasing for ¡ 6px 1, x > + p2 decreasing for x ¡ 2, x + a i EXERCISE 21C.2 t decreasing for x ¡ + - + never b i never ii ¡2 < x x > d i all real x ii never ii x 1, x > x > ii x < 0, < x h decr for x ¡ 12 , x > 3, incr for ¡ 12 x i increasing for x > 0, decreasing for x p p j increasing for x > ¡ 32 + 25 and x ¡ 32 ¡ 25 s0 (1) d p cm s¡2 = v (t) i.e., the instantaneous accn at time t s(t): x > ii x ii 16x65 x < 4, increasing for x > 0, decreasing for x decreasing for all x increasing for x > ¡ 34 , decreasing for x ¡ 34 a ¡14 cm s¡1 b (¡8 ¡ 2h) cm s¡1 c ¡8 cm s¡1 = s0 (2) i.e., velocity = ¡8 cm s¡1 at t = d ¡4t = s0 (t) = v(t) ´ x b = v(0) c = a b c d e f g lim (2t + h + 3) = s0 (t) ! 2t + as h ! p 1+h+1 t 20 i i i i h!0 ³ 16 a c e f EXERCISE 21C.1 s¡1 - + EXERCISE 21D.1 d C 00 (x) = 0:0018x + 0:04, C 00 (x) = when x = ¡22:2 s¡1 g Show that a = of making one more pair of jeans if 220 pairs are currently being made c $56:58 This is the actual increase in cost to make an extra pair of jeans (221 rather than 220) cm s¡2 b c i t and t ii t d 28 m Hint: s0 (t) = v(t) and s00 (t) = a(t) = g a C (x) = 0:0009x2 + 0:04x + dollars per pair b C (220) = $56:36 per pair This estimates the additional cost t -1250 was first opened) d2 V 125 This shows that the rate of change of V is c = dt2 constantly increasing, i.e., the outflow is decreasing at a constant rate dP = 0, the population is not changing over time, a When dt i.e., it is stable b 4000 fish c 8000 fish a + - + b x(2) = 20, x(4) = 16 b at t = (when the tap s¡1 at 49 m s¡1 Its speed is decreasing t = 12 Stone is 470:4 m above the ground and moving groundward at 19:6 m s¡1 Its speed is increasing 490 m e 20 seconds v(t) = 12 ¡ 6t2 , a(t) = ¡12t s(0) = ¡1, v(0) = 12, a(0) = Particle started cm to the left of the origin and was travelling to the right at a constant speed of 12 cm s-1 p p p p t = 2, s( 2) = ¡ d i t > ii never v(t) = 3t2 ¡ 18t + 24 a(t) = 6t ¡ 18 100 c 75 912 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\912IB_HL-2_AN.CDR Friday, 12 March 2010 4:41:29 PM PETER IB_PD (913) ANSWERS EXERCISE 21D.2 d e g h i a A - local B - local max C - horiz inflection + + b -2 x c i x ¡2, x > ii ¡2 x d + -4 + For b we have intervals where the function is increasing (+) or decreasing (¡) For d we have intervals where the function is above (+) and below (¡) the x-axis e a V.A x = and x = ¡1, H.A y = no V.A.’s, H.A y = ¡1 f V.A x = ¡2, H.A y = V.A x = 2, parabolic asymptote y = 3x2 + 7x + 14 V.A x = ¡1, OA y = 2x ¡ V.A x = 12 , OA y = 32 x + 54 a i H.A ¡ 1y ¢= 1, V.A.s x = and x = ¡2 ii , is a local maximum 25 iii x-intercepts are and 1, y-intercept is iv y x local max &\Qw_\' wA_t_\* b ƒ(x)¡ ƒ(x) -~`2 ~`2 (0,-2) local y=1 horizontal inflection (0,¡1) x x -1 y= x x=-2 c local max (-1,¡4) ƒ(x) ƒ(x) d -~`2 -2 e (-1,-1) local f (1,-1) local -1 ƒ(x)¡ ƒ(x) (2,¡9) x x2 - y= x +1 local (0,-1) horizontal inflection c i H.A y = 1, V.A.s x = ¡4 and x = ¡1 ii (2, ¡ 19 ) is a local min., (¡2, ¡9) is a local max iii x-intercepts are and 1, y-intercept is iv y x (no stationary points) x g x=3 y=1 x (1,¡0) local x ~`2 ƒ(x) h ƒ(x)¡ y= (1,¡0) -3 -3 x horizontal inflection x (¡Qr_¡,- Qr_¡) local y=1 local (2,-\Qo_\) ƒ(x)¡ (0,¡1) local max -2 x local max (0,-8) (-1,-9) local x x local max (-2,-9) j x - 5x + x + 5x + (-2,-27) local ƒ(x)¡ i x -x x2 - x - b i H.A y = 1, no V.A ii (0, ¡1) is a local minimum iii x-intercepts are and ¡1, y-intercept is ¡1 iv y local max (0,¡0) 913 x=-4 x=-1 d i H.A y = 1, V.A x = ¡1 ii (2, ¡ 13 ) is a local minimum iii x-intercepts are and 1, y-intercept is iv (1,-9) local b , local if a > 0, local max if a < a = 2a a a = ¡12, b = ¡13 b (¡2, 3) local max (2, ¡29) local x=¡ P (x) = ¡9x3 ¡ 9x2 + 9x + a greatest value 63 (at x = 5) least value ¡18 (at x = 2) b greatest value = (at x = and x = 0) y= y=1 x=-1 yellow 95 100 50 75 25 95 (¡1, ¡2) is a local 100 50 75 25 95 100 50 75 25 95 100 50 iii x-intercept is 0, y-intercept is 75 a H.A y = 0, V.A x = and x = ¡2 b H.A y = 0, V.A x = ¡2 c H.A y = 0, no V.A 25 EXERCISE 21E magenta a local (2,-\Qe_) i H.A y = ii (1, 2) is a local max least value = ¡16 (at x = ¡2) Maximum hourly cost = $680:95 when 150 hinges are made per hour Minimum hourly cost = $529:80 when 104 hinges are made per hour cyan x - 6x + ( x + 1) black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\913IB_HL-2_AN.CDR Tuesday, 22 January 2008 4:34:45 PM PETERDELL h f (x) = ¡4(x + 1)(x ¡ 1) (x2 + 1)2 IB_PD i (914) 914 ANSWERS y iv b i VA x = ¡1, oblique asymptote y = x + local max (1,¡2) x2 + 2x + , no turning points (x + 1)2 iii x-intercepts and ¡3, y-intercept ii f (x) = y=0 x y= local (-1,¡-2) iv 4x x2 + b i H.A y = 0, V.A.s x·= and x = ¡1 f (x) ii no stationary points y ¸ ¡4(x2 + 5) = (x ¡ 5)2 (x + 1)2 -3 x y = x+ iii x-intercept is 0, y-intercept is iv y x=-1 y= x + 3x x +1 x = -1 x=5 c i VA x = 2, oblique asymptote y = ¡2x + ii f (x) = y=0 x y= ¡2(x ¡ 3)(x ¡ 1) , (x ¡ 2)2 iii no x-intercept, y-intercept is iv y 4x x - 4x - 2 c i H.A y = 0, V.A x = h f (x) = ii (¡1, ¡1) is a local minimum i ¡4(x + 1) (x ¡ 1)3 y = -2x + - iii x-intercept is 0, y-intercept is iv y f ( x) = x ii (4, 14 ) is a local maximum f (x) = iii x-intercept is 1, y-intercept is ¡ 34 iv x=-2 y = -2x + x-2 p p x2 (x2 ¡ 3) , local max at (¡ 3, ¡32 ), 2 (x + 1) (x ¡ 1) p p local at ( 3, 3 ), horizontal inflection at (0, 0) iii x-intercept is 0, y-int is x=-1 y y=x i ¡3(x ¡ 4) & 3, (x + 2)3 f ( x) = -3 * y= y x=1 x x2 (x2 + 3) , horizontal inflection at (0, 0) (x2 + 1)2 iii x and y intercepts of ii f (x) = 3x - ( x + 2) y iv (x + 1)(x + 3) , (x + 2)2 x3 x -1 e i No VA exists, OA is y = x y=x a VA x = ¡2, oblique asymptote y = x + ii f (x) = 3 * x &- , local max (4,¡Qr_¡) -\Er_ (3'-7) f (x) = iv h x d i VA x = 1, x = ¡1 OA y = x x=1 d i H.A y = 0, V.A x = ¡2 (1' 1) x = 4x ( x - 1) ii local (-1,-1) local at (1, 1), local max at (3, ¡7) y= x3 x2 + local at (¡1, 2), local max at (¡3, ¡2) x iii no x-intercept, y-intercept is 12 iv y x=-2 y= (-1' 2) x + 4x + x+2 y= x+2 EXERCISE 21F cyan magenta yellow 95 100 no inflection b horizontal inflection at (0, 2) non-horizontal inflection at (2, 3) horizontal inflection at (¡2, ¡3) horizontal inflection at (0, 2) ) f no inflection non-horizontal inflection at (¡ 43 , 310 27 50 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 (-3'-2) a c d e 75 x black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\914IB_HL-2_AN.CDR Friday, 12 December 2008 3:37:56 PM TROY IB_PD (915) ANSWERS a i local minimum at (0, 0) v ii no points of inflection iii decreasing for x 0, iv p2 iv concave down for ¡ ƒ(x) increasing for x > function is concave up for all x v (0,0) local real x (0,0) horizontal inflection iv concave down for x 0, concave up for x > c i f (x) 6= 0, no stationary points ii no points of inflection iii incr for x > 0, v (~` We_, Uo_ ) x h i ii iii iv x never concave up d i local max at (¡2, 29) local at (4, ¡79) ii non-horizontal inflection at (1, ¡25) iii increasing for v ƒ(x)¡ iv (~`2,-1) local v no stationary points no inflections increasing for x > 0, never decreasing concave down for x > 0, never concave up ƒ(x)¡ x Ql_Y_ a y ¦'(x) max NSPI non-horizontal inflection -4 nonhorizontal inflection local max (0,¡3) (-~`2,-1) local never decr local max (-2,¡29) x ƒ(x)¡ iv concave down for x > 0, x ¡2, x > decreasing for ¡2 x concave down for x 1, concave up for x > p32 (-~` We_, Uo_ ) ƒ(x)¡ b i horizontal inflection at (0, 0) ii horizontal inflection at (0, 0) iii increasing for all v , x> ƒ(x)¡ nonhorizontal inflection x p2 6x6 p concave up for x ¡ 915 SPI (1,-25) x -3 x ¦(x) -80 b (4,-79) local y max e i horiz inflection at (0, ¡2) local at (¡1, ¡3) ii horizontal inflection at (0, ¡2) ) non-horizontal inflection at (¡ 23 , ¡ 70 27 -2 iii increasing for x > ¡1, decreasing for x ¡1 iv concave down for v non-horizontal ƒ(x)¡ ¡ 23 x concave up for x ¡ 23 , x > inflection (0,-2) ¦ '(x) ¦ ''(x) horizontal inflection -2 local (-1,-3) f i local at (1, 0) ii no points of inflection iii increasing for x > 1, decreasing for x iv concave up v x b ¦ '(x) y max y for all x ¦(x) ¦ (x) y a x -1 x ¦'(x) &-\We_\'- Uw_Pu_\* -2 NSPI ¦ (x) ¦ ''(x) y¡=¡(x-1)4 x x c local (1, 0) y p p g i local minimum at (¡ 2, ¡1) and ( 2, ¡1), local maximum at (0, 3), p2 ii non-horizontal inflection at ( , ) x p2 yellow 95 100 50 75 25 95 100 50 75 25 95 50 75 25 95 100 50 75 25 100 magenta ¦ (x) NSPI non-horizontal inflection at (¡ , 79 ) p p iii increasing for ¡ x 0, x > p p decreasing for x ¡ 2, x cyan ¦ '(x) ¦ ''(x) max black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\915IB_HL-2_AN.CDR Tuesday, 22 January 2008 4:43:28 PM PETERDELL IB_PD (916) 916 ANSWERS ³2 + x´ EXERCISE 21G 50 fittings 250 items 10 blankets 25 km b c y (m) 17 a QR = m c Hint: All solutions < can x be discarded as x > d 416 cm 18 between A and N, 2:578 m from N p3 19 at grid reference (3:544, 8) 20 : 21 e 63:7% h¡1 Lmin = 28:28 m, x = 7:07 m d EXERCISE 21H a 14.14 m f x (m) 7.07 m area equation d 100 and substitute into the surface x2 a 5.62 cm x (cm) 10 a 4.22 cm 8.43 cm e 10.84 cm c x y c y3 d f a b c a f µ = 293:9o ¡25 y3 ¡1 y3 b V ¡1 dV = dq 3V ¡ q q (°) + + 17 m -1 12 2y x2 a a + s and t > i $535 ii $1385:79 i ¡$0:27 per km h¡1 ii $2:33 per km h¡1 51:3 km h¡1 local maximum at (¡2, 51), local minimum at (3, ¡74) non-horizontal inflection at ( 12 , ¡11:5) d local max (-2, 51) y non-horizontal inflection (\Qw_\,-11\Qw_\) x local (3,¡-74) , x > c base is 1:26 m square, height 0:630 m x2 ¡2y(x + y ) dy = b 4x ¡ 5y = 14 dx x(x + 6y ) y= y max 24 Largest D(x) = 24, which is not an acceptable solution as can be seen in the diagram NSPI y=¦(x) SPI x 24 m y=¦'(x) 12 m 16 a Hint: Use the cosine rule + b 3553 km2 4.605 c 5:36 pm t magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 cyan t 1\Qw_ b k=9 x2 + (24 ¡ x)2 12 f + Speed is increasing for t concave up for x > p - 2x + 2y x2 - a(t): t c concave down for x 12 , the time taken to get from B to C (Proof: Use sign diagram or second derivative test Be sure to check the end points.) 12 3:33 km 13 r = 31:7 cm, h = 31:7 cm 14 m from the 40 cp globe e 3x2 ¡ 2y 2x f d2 q 2q ¡ 3V ¡ = dV (1 ¡ V )2 b x ¡2, x > decreasing for ¡2 x 11 a For x < or x > 6, X is not on AC c x = 2:67 km This is the distance from A to X which minimises d[D(x)]2 = 4x ¡ 48 dx c Smallest D(x) = 17:0 ¡2x ¡ y x e 6y ¡ 8x2 9y d b increasing for b 2x 3y -5 36 500 V (cmC) 15 a D(x) = dy dx dy dx dy i 2xy + x2 dx e 4y3 e ¡ µ ¢2 360 h y+x dy dx a(0) = ¡18 cm s¡2 (reducing speed) c At t = 2, particle is cm to the left of the origin, is stationary and is accelerating towards the origin d t = 1, s = and t = 2, s = ¡1 b Hint: Show that 2¼r = AC c Hint: Use the result from b and Pythagoras’ theorem e dy dx d ¡y ¡2 b s(0) = cm to left of origin, v(0) = 12 cm s¡1 towards origin b cm £ cm a x 63:66 c x = 63:66 m, l = m (i.e., circular) µ 10 a Hint: Show that AC = 360 £ 2¼ £ 10 100 ¡ x y 5.42 cm r (cm) 36 dy dx x b ¡ 3y v(t): 15 ¡ µ ¢2 q g ¡2y ¡3 dy dx a v(t) = (6t2 ¡ 18t + 12) cm s¡1 a(t) = (12t ¡ 18) cm s¡2 r = 5:42 cm d V = 13 ¼ c 3y2 REVIEW SET 21A a recall that Vcylinder = ¼r2 h and that L = 1000 cm3 b recall that SAcylinder = 2¼r2 + 2¼rh c d A = 554 cm2 , A (cmX) 1500 dy dx dy dx a b ¡ 19 f 450 ¡1 y 2 a ¡ SAmin = 213:4 cm2 , x = 4:22 cm e y (cmX) b ¡3 j y2 + 2xy a 2x cm b V = 200 = 2x £ x £ h c Hint: Show h = dy dx black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\916IB_HL-2_AN.CDR Tuesday, 22 January 2008 4:50:22 PM PETERDELL IB_PD (917) ANSWERS REVIEW SET 21B d 917 y d[A(x)]2 = 5000x ¡ 4x3 b dx Area is a maximum when x ¼ 35:4, A = 1250 m2 120 ¡360 a v(t) = 15 + cm s¡1 , a(t) = cm s¡2 (t ¡ 1)3 (t ¡ 1)4 local max (-0.828,¡0.343) local (4.828,¡11.66) x -2 b At t = 3, particle is 30 cm to the right of the origin, moving to the right at 30 cm s¡1 and decelerating at 22:5 cm s¡2 c 06t<1 cm from each end 2y2 ¡ 2x dy = a k = ¡1 b c 11x ¡ 2y = 24 dx 3y2 ¡ 4xy 1 a v(t) = ¡ p , a(t) = p t 4t t + - v(t): t + a(t): Ae_y_ e p < 0:343 or p > 11:66 a Hint: Use Pythagoras to find h as a function of x and then substitute into the equation for the volume of a cylinder b radius = 4:08 cm, height = 5:77 cm km b Hint: Show that x (length of pipe)2 = (LQ + 1)2 + (8 + x)2 then simplify a LQ = c 11:2 km (when x = km) t k When x = b x(0) = 0, v(0) is undefined, a(0) is undefined c Particle is 24 cm to the right of the origin and is travelling to the ³ 1¡ p1 ´ max right at 2:83 cm s¡1 Its speed is increasing d e a b c e Changes direction at t = , 36 0:083 cm to the left of the origin Particle’s speed is decreasing for t 36 y-int at y = ¡1 x-int at x = 1, x = ¡1 x2 + > for all real x (i.e., denominator is never 0) local minimum at (0, ¡1) x y=¦(x) y non-horizontal inflection at 1 &- , - 2* local (0,¡-1) EXERCISE 22A a 4e4x b ex c ¡2e¡2x d non-horizontal inflection at 1 & , - 2* x g 2e + 3e¡x h k 20e2x l 40e¡2x p ¡0:02e¡0:02x 2e¡x ex +2+ (e¡x + 1)2 h b local maximum at ¡2 , 32 27 ¢ , local minimum at (2, 0), non-horizontal inflection at c ¡4 y axis intercept at (0,¡0) , 16 27 ¢ non-horizontal inflection &\Re_\' Qw_Yu_\* local max &\We_\, Ew_Wu_\* a VA at x = OA is y = x + b local max at (¡0:828, 0:343) a c a a a d 28.0 m 56.0 m local at (4:828, 11:66) c x-intercepts at x = and x = ¡2, y-intercept at y = e2x p 2x e + 10 ln N = ln 50 + 2t b ln P = ln 8:69 ¡ 0:0541t ln S = ln a ¡ kt b 12 c ¡1 d ¡ 12 e f g 15 eln b eln 10 c yellow ³ 95 100 50 75 25 95 100 50 75 25 95 50 75 25 100 magenta eln a d h ex ln a x = ln b no real solutions c no real solutions x = ln e x = f x = ln or ln g x = h x = ln i x = ln ¡ x e¡x EXERCISE 22B b A = 200x ¡ 2x2 ¡ 12 ¼x2 c 95 c 1¡x ex a local maximum at (1, e¡1 ) b local max at (¡2, 4e¡2 ), local at (0, 0) c local minimum at (1, e) d local maximum at (¡1, e) x local (2,¡0) 100 ¡1 x e¡x g d ¢¡ 6e3x ¡ 2e¡x + xe¡x e¡x ¡ e ¡ ¡ e¡x f p (1 ¡ e3x )3 ¡ 2e¡x dn y n b =k y dxn d2 y dy and Hint: Find and substitute into the equation dx dx2 ¡(54e¡2x + 3x2 e3y + 8xy ) dy = dx 3x2 (xe3y + 4y2 ) a y-intercept at y = 0, x-intercept at x = and x = 50 xex ¡ ex x2 d REVIEW SET 21C 75 xex ¡ 12 ex p x x ¡e¡x (1 ¡ e¡x )2 a 4ex (ex + 2)3 b y=¦(x) 25 x e ¡e¡ f 2e¡x ex ¡ e¡x ¡1 i ¡2xe¡x j e x £ 2 x x m 2e2x+1 n 14 e o ¡4xe1¡2x e 2xe3x + 3x2 e3x f x cyan x e2 2 a ex + xex b 3x2 e¡x ¡ x3 e¡x c y y=|¦'(x)| y=¦'(x) NSPI x y black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\917IB_HL-2_AN.CDR Tuesday, 22 January 2008 4:53:29 PM PETERDELL p 3+ ´ ³ or ln p 3¡ ´ IB_PD (918) 918 ANSWERS e3 + (¼ 10:54) b no, ) there is no y-int 4¡ c gradient = d x > 12 e f 00 (x) = < for (2x ¡ 1)2 all x > , so f (x) is concave down a 2x ln b 5x ln c 2x + x2x ln 3x2 x3 a x= ¡ ln ln ¡ ¡ x ln e f 6x x2 3x a (ln 3, 3) b (ln 2, 5) c (0, 2) and (ln 5, ¡2) a f (x): x-int at x = ln 3,¡ y-int ¢ at y = ¡2 g(x): x-int at x = ln 53 , y-int at y = ¡2 d x2x 2x f f (x) x = Qw_ d Hint: Show: x ! 0, f (x) ! ¡1, and as x ! 1, f (x) ! Hint: Show that f (x) > for all x > y = f (x) y y=3 (ln 5, 2) y = g(x) ln Te_ x e y =¡ x+ y = -3 a P = ( 12 ln 3, 0) Q = (0, ¡2) x for all x -2 c f (x) is concave down &\Qw_\\ln¡3, 0* non-horizontal inflection below the x-axis and concave up above the x-axis f (x) (e, 1) normal has equation x f (x) = ¡ex + + e concave down for x · ¡2, concave up for x ¸ ¡2 y inflection (-2'-\qD_y_) ¼ 63:43o a k = 50 ln (¼ 0:0139) b i 20 grams ii 14:3 grams iii 1:95 grams c days and minutes (216 hours) d i ¡0:0693 g h¡1 ii ¡2:64 £ 10¡7 g h¡1 (0' 0) x b 2x + 1 ¡ 2x x ¡ x2 c f ¡ ln x 2x2 j e¡x ¡ e¡x ln x k x g ex ln x + ex x ¡ 95 ¢ d ¡ x ln x x h ln(2x) p +p x x l g ¡ ¡1 ¡ 2x ¡2 2x + b x h x ln x e 2x ln x + x p 2x ln x ln x ¡ p x(ln x)2 i c 1+ 2x d ln (¼ 0:231) 0:728 units of alcohol produced per hour f (x) does not have any x or y-intercepts as x ! 1, f (x) ! 1, as x ! ¡1, f (x) ! (below) local minimum at (1, e) e ey = ¡2x ¡ f (x) 11 a A = b k = c 12 a b c d ¡1 x[ln x]2 i local (1, e) 1 ¡ x 2(2 ¡ x) 1 ¡ + f g x+3 x¡1 x 3¡x 3x ¡ 2x 1 2x + + ¡ h i x x +1 x2 + 2x x¡5 1 dy = 2x ln a i ii iii log3 x + b x ln x ln 10 ln dx x magenta vertical asymptote x=0 t t 13 a v(t) = 100 ¡ 40e¡ cm s¡1 , a(t) = 8e¡ cm s¡2 b s(0) = 200 cm on positive side of origin yellow 95 100 50 75 25 95 v(0) = 60 cm s¡1 100 50 75 25 95 100 50 75 25 95 100 50 75 25 e cyan ln 2t a k = ¡1 ln 15 15 d i decreasing by 11:7o C/min ii decreasing by 3:42o C/min iii decreasing by 0:998o C per minute 10 a 43:9 cm b 10:4 years c i growing by 5:45 cm per year ii growing by 1:88 cm per year 4x3 + 1 [ln(2x + 1)]2 a ln b c d e x x +x x¡2 2x + 1 ¡ ln(4x) x2 dW = ¡ 50 ln £ 20e¡ 50 dt (¼ 0:123) b 100o C e Hint: You should find (-1'-\Qr_) EXERCISE 22C a 3y = ¡x + ln ¡ , , B is (0, ¡ 2e) y = ¡ c x f x-intercept at x = , y-intercept at y = as x ! 1, y ! , as x ! ¡1, y ! 0¡ local minimum at (¡1, ¡ 14 ) y¡=¡4x¡-¡2x a ¡2 ¢ e 2 x+ ¡1 e2 e y = ea x + ea (1 ¡ a) so y = ex is the tangent to y = ex from the origin a x>0 b f (x) > for all x > 0, so f (x) is always increasing Its gradient is always positive f 00 (x) < for all x > 0, so f (x) is concave down for all x > A is y d b f (x) = ex + 3e¡x > a b c d e a4 b ¡ 2ae2a ¡ a da = db 4abe2a ln b ¡ 3a3 b2 + b EXERCISE 22D ln (0, -2) x e 3+1 as x ! 1, f (x) ! as x ! ¡1, f (x) ! ¡3 (above) g(x): as x ! 1, g(x) ! (below) as x ! ¡1, g(x) ! ¡1 c intersect at (0, ¡2) and (ln 5, 2) b f (x): black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\918IB_HL-2_AN.CDR Friday, 12 March 2010 4:42:57 PM PETER a(0) = cm s¡2 IB_PD (919) ANSWERS c as t ! 1, v(t) ! 100 cm s¡1 (below) e after 3:47 sec d y=100 v(t) (cm s-1) the velocity is decreasing the fastest y 25000 20000 15000 10000 5000 t (sec) (5, 5ln + 1) A(t) d Point of inflection is at (1, 3:38) This is when e t>1 a 60 14 a at 4:41 months old b c 06t6 ¼13¡890 x minimum (e-1, 0.632) 919 10 b ¼ 13 900 ants c ¼ 24 000 ants d Yes, 25 000 ants e after 3:67 months t (years) 2C bees b 37:8% increase c Yes, C bees d 3047 bees 0:865C e B (t) = 1:73t and so B (t) > for all t > e (1 + 0:5e¡1:73t )2 ) B(t) is incr over time a ³ p1 2¼ 15 a There is a local maximum at 0, ´ f f (x) is incr for all x and decr for all x > ³ b Inflections at ¡1, p1 2e¼ ´ ³ and p1 2e¼ 1, ´ y 5000 4000 3000 2000 1000 c as x ! 1, f (x) ! (positive) as x ! ¡1, f (x) ! (positive) d local max &0, non-horizontal inflection &-1, - x2 y= e 2p f (x) * 2p &1, ³ 16 20 kettles 17 C = p1 , (¡ ) 1.5 2.5 a 3x2 ex +2 e 2 y = x+ xex ¡ 2ex x3 b e ¡ e 19 a Hint: They must have the same y-coordinate at x = b 1 and the same gradient c a = 2e d y = e¡ x ¡ 12 c ey (2y + 1) ¡ xey (2y + 1) y = ex + y=9 p (ln 5, 8) 20 after 13:8 weeks 21 a = 2e , b = ¡ 18 23 a i ¡e¡x (x+1) ii e¡x (x) iii ¡e¡x (x¡1) iv e¡x (x¡2) b f (n) (x) = (¡1)n e¡x (x ¡ n + 2) 24 a eax (ax+1), aeax (ax+2), a2 eax (ax+3), a3 eax (ax+4) b f (n) (x) = an¡1 eax (ax + n) (0, 4) y=3 y = - 5e -x a y-intercept at y = ¡1, no x-intercept b f (x) is defined for all x 6= c f (x) < for x < and < x and f (x) > for x > EXERCISE 22E a 200 E f 00 (x) > for x > 1, f 00 (x) < for x < So, the gradient of the curve is negative for all defined values of x and positive for all x > The curve is concave down for x < and concave up for x > x d tangent is y = e2 y y= e 150 100 50 t i 177:1 units ii 74:7 units When t ¼ 0:667 hours = 40 i 0:1728 h ¼ 10 ii 91 t = 43 or 80 This is when the effectiveness is decreasing the fastest t cyan b s(0) = 80 m, magenta yellow 95 100 50 25 95 100 50 d v(0) = ¡48 m s¡1 , a(0) = 0:8 m s¡2 c as t ! 1, v(t) ! ¡40 m s¡1 (below) e t = 6:93 seconds 95 t a v(t) = ¡8e¡ 10 ¡ 40 m s¡1 , a(t) = 45 e¡ 10 m s¡2 ft > 0g 100 50 75 25 95 100 50 a 60 cm b i 4:244 years ii 201:2 years c i 16 cm per year ii 1:95 cm per year t 75 x=1 4 25 x 75 b Local max in c is at t = 23 Point of inflection in e is at t = v a y = e2 Show that f (t) = Ae¡bt (1 ¡ bt) Show that f 00 (t) = Abe¡bt (bt ¡ 2) 75 i ii 25 a x -1 b c d e REVIEW SET 22A * 2ep 18 267 torches e x a as t ! 1, e¡bt ! 0, etc x ´ 3047 0.5 non-horizontal inflection * 2ep C¡=¡4571 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\919IB_HL-2_AN.CDR Friday, 12 March 2010 4:44:34 PM PETER v(t) m s-1 t v(t)=-40 -48 IB_PD (920) 920 ANSWERS A( 12 , e¡1 ) 100 or 101 shirts, $938:63 profit Tangent is y = ln 3, so never cuts x-axis 10 a 2x (1 + x ln 2) · (x2 + 2)(x ¡ 3) ¡ x3 b x cos x ¡ sin x l tan x + x sec2 x x2 p sin x a 2x cos(x2 ) b ¡ p sin( x) c ¡ p x cos x j cos x ¡ x sin x k 3x2 2x + + x2 + x¡3 ¡ x3 ¸ d sin x cos x e ¡3 sin x cos2 x f ¡ sin x sin(2x) + cos x cos(2x) g sin x sin(cos x) REVIEW SET 22B + c f 00 (x) = ex , d k ¡ y 10 a f (x) = a x = ln ex +3 ex ¡2¢ x b f (x) = a a b a -1 c y (3¼ ,1) max ¼ ( , 1) max (0, 0) 25 11 a x = t + - y = cos 2x -1 ¼ ( , -1) (3¼ ,-1) (2¼, 0) x ¼ 3¼ , 2 x x -1 y¡=¡sec¡x local max (p,-1) ¼ 2¼ y local max 3~`3 ) (¼6 , x= 5¼ (5¼6 , EXERCISE 23A -3~`3 (2¼, 2) 2¼ ¼ x (3¼ , 0) stationary inflection -2 ) local cos(2x) b cos x ¡ sin x c ¡3 sin(3x) ¡ cos x cos(x + 1) e sin(3 ¡ 2x) f sec2 (5x) cos( x2 ) + sin x h 3¼ sec2 (¼x) i cos x + sin(2x) 13 a x(0) = ¡1 cm v(0) = cm s¡1 a(0) = cm s¡2 p b At t = ¼4 seconds, the particle is ( ¡ 1) cm left of the magenta yellow 95 100 50 75 95 100 50 75 25 95 100 50 75 25 100 cyan 25 p origin, moving right at cm s¡1 , with increasing speed c At t = 0, x(0) = ¡1 cm, at t = ¼, x(¼) = cm, at t = 2¼, x(2¼) = ¡1 cm d for t ¼2 and ¼ t 3¼ 2x ¡ sin x b sec2 x ¡ cos x c ex cos x ¡ ex sin x ¡e¡x sin x + e¡x cos x e cot x f 2e2x tan x + e2x sec2 x ¡ ¢ cos(3x) h ¡ 12 sin x2 i sec2 (2x) 95 3¼ x 12 y = sin(2x) + cos x [0, 2¼] x 75 [0, 2¼] local (2p, 1) x= (1,¡1) 50 2¼ y = sin x (¼, 0) local (0, 1) f (x) is increasing for all x > and is concave downwards for all x > y normal is x + 2y = 25 (2¼, 1) max (2¼, 1) is a local minimum 0 ¼ c f (x) has a period 2¼ y ¼ d x = w_ 12 197 or 198 clocks per day P is (ln a, 1) 10 a x > b Sign diag of f (x) Sign diag of f 00 (x) ¼ b (0, 1) is a local minimum, (¼, ¡1) is a local maximum a d g a d g (¼, 1) max x (3¼ ,-1) local or b x = e2 15 d x= max 2¼ ¼ 10 10 cm min¡1 , a(t) = cm min¡2 t t 10 10 b s(e) = 25e ¡ 10, v(e) = 25 ¡ , a(e) = e e c As t ! 1, v(t) ! 25 cm min¡1 from below d e t = 10 v(t) 13 p b y (0, 1) ¼ (2 , 1) local max y = sin x y y = -x 3x ¡ (x + 2) x(x + 2) ¡ ¼ rising b rising at 2:731 m per hour ¡34 000¼ units per second b V (t) = i m s¡1 ii m s¡1 iii ¼ 1:11 m s¡1 a v(t) = 25 ¡ c ¡12 cos2 ( x2 ) tan4 ( x2 ) a y = x b y = x c 2x ¡ y = y = e x-x -1 l b The answers of a are cycled over and over -2 cos(2x) sin3 (2x) sin(2x) cos2 (2x) j d2 y d3 y d4 y dy = cos x, = ¡ sin x, = ¡ cos x, = sin x dx dx2 dx3 dx4 a thus f (x) is concave up for all x cos x sin2 x h ¡12 sin(4x) cos2 (4x) i ¡ 3x2 ¡ dy dy = = ¡ b dx x ¡ 3x dx x+3 x ex+y (y2 + 1) dy = c d xx +1 (2 ln x + 1) dx 2y ¡ ex+y (y + 1) (0, ln ¡ 1) a x = ln b x = ln or ln a local minimum at (0, 1) b As x ! 1, f (x) ! 1, as x ! ¡1, f (x) ! ¡x (above) a black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\920IB_HL-2_AN.CDR Thursday, 11 March 2010 4:40:00 PM PETER IB_PD x (921) ANSWERS EXERCISE 23B b c a b a sec x + x sec x tan x b ex (cot x ¡ csc2 x) £ ¡ ¢ â đô c sec(2x) tan(2x) d ¡e¡x cot x2 + 12 csc2 x2 e x csc x[2 ¡ x cot x] f £ £ p csc x ¡ ¤ x ¤ cot x g tan x h csc(x2 ) ¡ 2x2 cot(x2 ) cos(x) cos(2x) ¡ sin(x) sin(2x) ¡2e¡2x tan x + e¡2x sec2 x f (x) = cos x + sin(2x), f 00 (x) = ¡3 sin x + 16 cos(2x) 1 ¡1 x cos(4x) ¡ 4x sin(4x), 1 f 00 (x) = ¡ 14 x¡ cos(4x) ¡ 4x¡ sin(4x) ¡ 16x cos(4x) x(0) = cm, x0 (0) = cm s¡1 , x00 (0) = cm s¡2 f (x) = a b t = ¼4 sec and 3¼ sec c cm a for x ¼2 and 3¼ x 2¼ b increasing for 3¼ x 2¼, decreasing for x c y p ¡ x csc2 x ¡ 12 x¡ cot x cos x sin x + 2x ´¡ i p x 2x x sin2 x ¢ p ¡ p p a 2x ¡ y = ¼4 ¡ b 2x + y = 2¼ + 3 p p p p p a x ¡ 3y = ¼6 ¡ b 6x + y = ¼ + Hint: Let µ = arctan(5) ) tan µ = 5, etc v ¼ p ¡3 p 16 ¡ x2 c ¡ 9x2 d a p 20 10 ¼ 21:1 m per minute 3:60 m s¡1 a f (x) = for all x, so f (x) is a constant ( ¼2 ) b ¼4 + k¼, k Z c ¼4 p a i km ii 10 km a V (r) = 89 ¼r3 m3 b x2 x3 x6 1 ii iii iv ¡ v ¡ vi x 3x xn+1 : b the antiderivative of xn is n+1 a i a is decreasing at 7:5 units per second increasing at cm per minute a 4¼ m2 per second b 8¼ m2 per second increasing at 6¼ m2 per minute decreasing at 0:16 m3 per minute 20 cm per minute p 256 cm per minute decreasing at a 0:2 m s¡1 b 90 m s¡1 p radians per second 100 rad sec¡1 b decr 100 250 13 at rad sec¡1 vi 3e 100 5x e iii 2e x iv 100e0:01x v b the antiderivative of ekx is p vii x ¼x e ¼ kx e k d (x + x2 ) = 3x2 + 2x dx ) antiderivative of 6x2 + 4x = 2x3 + 2x2 d 3x+1 (e b ) = 3e3x+1 dx ) antiderivative of e3x+1 = 13 e3x+1 p d p (x x) = 32 x c dx p p x = 23 x x ) antiderivative of d (2x + 1)4 = 8(2x + 1)3 d dx ) antiderivative of (2x + 1)3 = 18 (2x + 1)4 rad sec¡1 b 100¼ rad sec¡1 17 b p 120 m min¡1 magenta yellow 25 95 100 50 75 25 95 100 50 75 sin(5x) x 25 95 100 50 75 25 cyan x ii x3 a REVIEW SET 23A a cos(5x) ln(x) + 2x e 2 a i m s¡1 12 increasing at 0:128 radians per second p 13 0:12 radians per minute 14 37 m s¡1 p 15 a 23 ¼ cm s¡1 b cm s¡1 200 p ¼ 13 dr = ¡ 375¼ ¼ ¡0:006 79 m min¡1 dt EXERCISE 24A EXERCISE 23E 16 a m above the floor p dy x sin x = cos x ¡ p dx cos x ¡1 dy dy = ex [cot(2x) ¡ csc2 (2x)] c = p b dx dx ¡ x2 p p p a 3x ¡ y = 33¼ ¡ b 4x + y = + ¼3 ¼ 109:5o c µ = 30o o hour p 34 53 sec when µ = 36:9 c m 9:866 m 1:340 m from A e AP + PB is a minimum when µ = Á 11 a decr at p a and tan(® + µ) = x x ³3´ ³2´ p µ = arctan ¡ arctan c x= x x The p maximum angle of view (µ) occurs when Sonia’s eye is m from the wall 10 decreasing at cos µ dy =¡ dx sin µ REVIEW SET 23B EXERCISE 23D y2 x2 + =1 b 9 b a tan ® = v(2) = cm s¡1 c units2 , when µ = ¼4 , 3¼ , 5¼ , 7¼ 4 a a2 + b2 ¡ 2ab cos µ = c2 + d2 ¡ 2cd cos Á f ¡1 x dy dy ex = arcsin x + p = ex arccos x ¡ p b dx dx ¡ x2 ¡ x2 dy e¡x = ¡e¡x arctan x + c dx + x2 dy = ¡p c , x ]¡a, a[ dx a2 ¡ x2 d = ¼ cm s¡1 i.e., for 2n t 2n + 1, n f0, 1, 2, 3, g ¡1 p 25 ¡ x2 a b 95 2x + x4 e b ¡3¢ 100 + 4x2 2p b t 1, t 3, t 5, etc EXERCISE 23C.2 a x 3p ?? p a v(0) = cm s¡1 , v( 12 ) = ¡¼ cm s¡1 , v(1) = cm s¡1 , b x = ¡ 13 50 p p ? 75 a x= ¼ f ( x) = cos x EXERCISE 23C.1 a b ¡ ¼2 c ¼4 d ¡ ¼4 e ¼6 f 5¼ g ¼3 3¼ ¼ h i ¡ j ¼ ¡0:874 k ¼ 1:238 l ¼ ¡1:551 921 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\921IB_HL-2_AN.CDR Friday, 12 December 2008 4:12:15 PM TROY IB_PD (922) ANSWERS EXERCISE 24B 24 23 units2 a b c d e 3:48 units2 f units2 g 3:96 units2 a 4:06 units2 b 2:41 units2 c 2:58 units2 ii ¡ 16 units2 c = R 7x6 ; dx = R = 3x2 + 2x; R = 2e2x+1 ; = R 3p x; p 2x+1 e (2x + 1)3 dx = x dx = Z dy =¡ p ; dx 2x x h (3x2 + 2x) dx = x3 + x2 + c R = 8(2x + 1)3 ; f +c e2x+1 dx = Z p x x ¡2 dy = p ; dx ¡ 4x (2x d d (ln jAxj) = (ln jAj + ln jxj) = dx dx x Recall that: a f (x) = ¡e¡2x + b f (x) = x2 + ln j1 ¡ xj + ¡ ln Both are correct + 1)4 + c +c c f (x) = 23 x ¡ 18 e¡4x + 18 e¡4 ¡ Z p dx = ¡ p + c x x x Z p a x3 x2 x5 ¡ ¡ + 2x + c b 23 x + ex + c 5 c 3ex ¡ ln jxj + c d 25 x ¡ ln jxj + c x i 5ex + a x + ln jxj + c h x 12 1 x 2 x2 + x ¡ ln jxj + c ¡1 h 2x + 8x 20 ¡ x ¡ x + 3 g 43 x +c i +c x x x +x+c 12 x2 a f (x) = x2 ¡ x + b f (x) = x3 + x2 ¡ p p c f (x) = ex + x ¡ ¡ e d f (x) = 12 x2 ¡ x + d (2x + 5)4 + c b (4x 32 ¡ 3)8 + c e a x2 3 + (ln x)4 x2 +c x2 + tan x + c yellow sin x + c b +c cos x p x3 ¡ sin x + b f (x) = sin x + cos x ¡ 2 3 EXERCISE 24F.2 a ¡ 13 cos(3x) + c b 95 100 50 75 25 f (x) = c f (x) = 23 x ¡ tan x ¡ 23 ¼ d sin 95 100 50 75 25 95 100 50 75 25 a ex sin x + ex cos x, ex sin x + c b ¡e¡x sin x + e¡x cos x, e¡x sin x + c c cos x ¡ x sin x, sin x ¡ x cos x + c d 11 3 µ2 + cos µ + c c 25 t + tan t + c d 2et + cos t + c e sin t ¡ ln jtj + c f 3µ ¡ ln jµj + tan µ + c +c a y = 13 (2x ¡ 7) + b (¡8, ¡19) ¯ ¯ ¯ ¯ ¡ ln ¯x3 ¡ 3x¯+c ¡ tan x + c p f ¡ cos x ¡ sin x + + c g 27 x3 x + 10 cos x + c p h 19 x3 ¡ 16 x2 + sin x + c i tan x + cos x + 43 x x + c ¡2 +c 3(2x ¡ 1)3 p + c f ¡4 ¡ 5x + c magenta +c ex p g ¡ 35 (1 ¡ x)5 + c h ¡2 ¡ 4x + c cyan p x ¯ ¯ ¯ ¯ ¡2 ln ¯5x ¡ x2 ¯+c f d tan x ¡ cos x + c e +c c 2(3 ¡ 2x) ¡ 4) (x2 +x)5 +c + c d 2e a ¡3 cos x ¡ 2x + c b 2x2 ¡ sin x + c c a f (x) = (3x 1 x3 +1 e EXERCISE 24F.1 b f (x) = 4x + 4x ¡ 4x + EXERCISE 24D a 1 +c i +c h ¡ 8(1 ¡ x2 )4 2(x2 + 4x ¡ 3) ¯ ¯ e f (x) = ¡ ln ¯x3 ¡ x¯ + c f x 3 g f ¡ a f (x) = 14 x4 ¡ 52 x2 + 3x + c b f (x) = 43 x ¡ c f (x) = 3ex ¡ ln jxj + c a f (x) = 13 x3 + 12 x2 + x + c f (x) = 13 x3 ¡ 16 x+5 (x3 +2x+1)5 +c +c 27(3x3 ¡ 1)3 (2+x4 )4 +c e + x) + c c f (x) = ¡ 13 (1 ¡ x2 ) + c d f (x) = ¡ 12 e1¡x + c +c c y = ¡ +c x d y = ¡ + c e y = 2e ¡ 5x + c f y = x + x3 + c x p a y = x ¡ 2x2 + 43 x3 + c b y = 23 x ¡ x + c c y = x + ln jxj + + c x a y = 6x + c b y = (x3 a f (x) = ¡ 19 (3 ¡ x3 )3 + c b f (x) = ln jln xj + c + p + 1)5 + c b x2 + + c c ¯ ¯ ¯ ¯ d ln ¯x3 ¡ x¯+c e ¡ 2x + c + ln j2x + 1j + c a ln ¯x2 + 1¯ + c b ¡ 12 ln ¯2 ¡ x2 ¯ + c c ln ¯x2 ¡ 3x¯ + c 2x2 + x + c x (x3 x3 ln j2x ¡ 1j ¡ e ¡ex¡x + c f e1¡ x + c ¡ 2x + c c 2ex + x 2 a e1¡2x + c b ex + c c ln jxj + 13 x3 ¡ ex + c + 32 x2 ¡ 2x + c b dx = 4x2 ¡ +c ¡ ln jxj + c d ¡2x¡ ¡ 8x + c e f ¡ 15 x5 + 34 x 2x ¡ dx = ln jx + 2j ¡ ln jx ¡ 2j + c x2 ¡ d 25 g x + ln jxj + c f 3 EXERCISE 24E a e ¡2x 5x¡2 e + c c ¡ 13 e7¡3x + c 5 ln j2x ¡ 1j + c e ¡ ln j1 ¡ 3xj + c ¡e¡x ¡ ln j2x + 1j + c g 12 e2x + 2x ¡ 12 e¡2x + c ¡ 12 e¡2x ¡ 4e¡x + 4x + c i 12 x2 + ln j1 ¡ xj + c c y = ¡ 12 e¡2x + ln j2x ¡ 1j + c +c EXERCISE 24C.2 ¡1 + 35 x5 + x3 + x + c a y = x ¡ 2ex + 12 e2x + c b y = x ¡ x2 + ln jx + 2j + c p dx = ¡ 12 ¡ 4x + c ¡ 4x x +c ln(5 ¡ 3x + x2 ) + c (5 ¡ 3x + x2 is > 0) 10 ln 11 x ln x ¡ x + c x a 2ex + 52 e2x + c b iii ¡6 d ¡¼ x ¡ 12 x4 + 13 x3 + c + c d x ¡ 23 x3 + 15 x5 + c e ¡ 83 (5 ¡ x) + c f d x6 x c b EXERCISE 24C.1 dy dx dy dx dy dx dy dx dy dx (2x ¡ 1)3 + ¡ 12 (1 ¡ 3x)4 ¡x¢ 95 c i ¡ 13 a p units2 +c e 100 34 units2 50 75 922 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\922IB_HL-2_AN.CDR Friday, 12 December 2008 4:13:04 PM TROY ¡ 32 sin(4x) + c c e¡x +c cos(2x) + tan(2x) + c IB_PD (923) ANSWERS ¡ + c g ¡ cos 2x + ¡ x + c i ¡2 tan sin(2x) ¡ 12 cos(2x) + sin(8x) + cos x + c 16 j l a x c e 32 ¡x¢ ¢ + x + x + 32 ¡ sin(4x) + c d x + 12 x i l cos(x2 ) cos(3x) + sin(4x) + c sin(2x) + c +c j (cos x)¡2 11 +c ¡ 16 x+c k cosec3 (2x) +c sin5 x ¡ ¡x¢ x a ¡2 ln b 13 2 2(ln x) , 12 (ln x)2 + c 23 16 x tan x + ln jcos xj + c x (2x + 3)n+1 + c if n 6= ¡1, 2(n + 1) sec(3x) + c p sec2 x + c i ¡2 cot x + c b c e ¡ (¼ 1:718) d 12 12 a a b b 1:557 c 20 13 (ln 2)2 f p 24 R1 b a a R R4 p x dx = 4:67 x7 dx = R1 8 f b c a 6:5 b ¡9 c d ¡2:5 a u2 c p a a 12 49 b f (x) = 3x3 54 + a A = 2, + 6x ¡ Z b a ¼ + 4x ¡ dx = ¡ 2x + B = ¡5 p a 3¡ b 13 (e9 ¡e) tan x, p 10 a = ln 11 e¡¼ R p y ¡ u2 y = x - 3x y = x -3 c (0,-3) d b ln y (0, 0) tan x dx = ln(sec x)+c x (3, 0) (1,-2) units2 ln (¼ ¡0:0236) y = ex -1 (ln¡2, 1) y=2 y = -1 magenta yellow 50 75 25 95 enclosed area = ln ¡ (¼ 0:0794) units2 100 50 75 25 ¡ 2x + ln jxj + c 95 100 50 75 25 95 ¡ ln jxj + 3x + c b cyan ¡ 27 ¢ x y = - 2e -x x 100 50 75 25 a f ln u2 iii 13 units2 REVIEW SET 24B ¡2e¡x ¼ 2 (¼ 4:70) u2 e b i, ii ln jcos(2x)j + c c esin x + c R p is x2 ¡ + c x dy = p ; dx x2 ¡ 4 5x2 ¡ ln jcos xj + c, for n = p sin8 x + c b for n 6= 2, ( ¡ 1) u2 a 10 23 units2 a x+c b ¡ 32 ln j1 ¡ 2xj+c c ¡ 12 e1¡x +c d ¡ 13 e4¡3x +c ¡ 12 ln p =2 x+ a 12 u2 b + e¡2 (¼ 1:14) u2 c 27 u2 d 14 u2 a ¡5 b REVIEW SET 24A f (x) d 12 u2 e 2e ¡ ¡4 g(x) dx + 3e2 + 12e ¡ 13 a 18 units2 b ln (¼ 1:39) units2 c ln (¼ 1:10) u2 d a ¡4 b 6:25 c 2:25 R9 b e d 20 56 units2 e 13 units2 f 12 23 units2 f (x) dx b ¡ , EXERCISE 25A a 13 units2 b 63 34 units2 c e ¡ (¼ 1:718) units2 c 5¼ + 12ex + b n p (¡ x) dx = ¡4:67 d + cos1¡ x + c, n ¡1 13 ln d ln e ln f ln j2x + 3j + c p is never x p x > for all x, ) f (x) > for all x as a = or §3 12 ¡ 4e 11 (¡x7 )dx = ¡ 18 ¼ ¼ 12 10 a = ¼ b a 2¼ b ¡4 c a + d 0:0337 e c EXERCISE 24G.2 R4 ¼ e ln( 27 ) (¼ ¡0:6264) 3n+1 , n 6= ¡1 (¼ 0:2402) g h ln (¼ 3:892) i 2n + a 2¡ b a e3x a a e 23 f ln (¼ 1:099) g 1:524 h i e ¡ (¼ 1:718) p c 3¡1 d 6e2x if n = ¡1, EXERCISE 24G.1 a +c cos3 x ln jsin(3x)j + c c ¡ cot x + c +c h + x)3 + c a y = 15 x5 ¡ 23 x3 + x + c b y = 400x + 40e¡ + c sin7 x + c sec x + c e ¡ csc x + c f ¡2 csc (3x2 REVIEW SET 24C sin4 (2x) + c c ln jsin x ¡ cos xj + c (3x2 + x)2 (6x + 1) dx = 12 a A = 4, B = ¡2, C = ¡2 ¡ ¢ b ln jxj ¡ ln jx + 1j ¡ ln jx ¡ 1j + c c ln 16 25 ln jsin(2x) ¡ 3j + c sin3 x + sin x + c ¡ecos x + c b etan x + c ln jsin xj + c b sin(2x) + c (x + 1)3 + ln jx ¡ 2j + c µ ¶ ³ jx + 2j ´ (x ¡ 1) a ln + c b ln +c (x ¡ 1)2 jx + 2j tan4 10 A = 1, B = 2, C = 1, D = 4, +c c e ¡(2 + sin x)¡1 + c f 9x ¡ sin x + b 12 f (x) = 13 x3 ¡ 32 x2 + 2x + 16 12 sin(x2 ) + c p p 23 ( ¡ 2) a f (x) = 14 x4 + 13 x3 ¡ 10 x + b 3x + 26y = 84 ln sin(6x) + c + sin x + a 15 d (3x2 + x)3 = 3(3x2 + x)2 (6x + 1) dx sin(2x) + c a ¡ cos x + 23 cos3 x ¡ 15 cos5 x + c b a d a d g a 2x ¡ sin x + c b b ¡2(cos x) + c c ¡ ln jcos xj + c g ln j1 ¡ cos xj + c h ¡ 12 ¡ 13 x sin(8x) + c f sin(2x) + c 9x + 3e2x¡1 + 14 e4x¡2 + c +c R sin(2x) + c b sin5 x + c (sin x) + d ¡ 23 4 ¢ ¡ 2x + c c k ¼ ¢ x sin(4x) + a ¡¼ 95 h sin ¡ tan ¡¼ 100 2x e f 923 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\923IB_HL-2_AN.CDR Thursday, 11 March 2010 4:40:47 PM PETER e units2 IB_PD (924) 924 ANSWERS EXERCISE 25C y E4250 a P (x) = 15x ¡ 0:015x2 ¡ 650 dollars b maximum profit is $3100, when 500 plates are made c 46 x 954 plates (you can’t produce part of a plate) 14 400 calories 76:3o C x 1 + 120 a y = ¡ 120 (1 ¡ x)4 ¡ b 2:5 cm (at x = m) y2 = 4x x y = 2x ( ¼4 , a 1) b ln a p enclosed area = u2 b y 9¼ units2 a y= units2 (¼ 7:07 units2 ) x a + ¼ b ¡2 c ¼ b 13 a 14 a ¡2 Area = 32 34 21 12 units2 R b R ¡1 ¡2 R3 (x3 ¡ 7x ¡ 6) dx + R ¯ ¯ ¯x3 ¡ 7x ¡ 6¯ dx ¡1 R a A= 11 a C1 is y = sin 2x, C2 is y = sin x b A( ¼3 , ii A = p ) (7x + ¡ c 12 u2 b x3 ) dx R a R5 f (x) dx + R [g(x) ¡ f (x)] dx R d a jg(x) ¡ f (x)j dx y2 = x -1 x (2,-1) y = x-3 -3 f (x) dx a (2, ¡1) and (5, 2) b 4:5 units2 p k = 16 Hint: Show that the areas represented by the integrals can be arranged to form a £ e unit rectangle a v(0) = 25 m s¡1 , v(3) = m s¡1 b as t ! 1, v(t) ! c 25 EXERCISE 25B.1 v(t) m/s v(t ) = i travelling forwards ii travelling backwards (i.e., opposite direction) 100 (t + ) 16 km km from starting point (on positive side) t(sec) velocity (km h-1) d seconds e a(t) = ¡200 , t>0 f (t + 2)3 21 12 units2 a = ln 3, b = ln 10 ¡ ¢ ¼ m units2 (p,¡1) p EXERCISE 25B.3 p x -1 a 40 m s¡1 b 47:77 m s¡1 c 1:386 seconds d as t ! 1, v(t) ! 50 f v(t) (m s-1) e a(t) = 5e¡0:5t and as 50 x = p4 v(t ) = 50 - 10e -0.5t 40 y¡=¡3 a 12 cm b cm a 16 cm b 12 cm left a 41 units b 34 units b units ex > for all x, a(t) > for all t 134:5 m kL4 k= 11 ¡ 12 a b ¼ ¡0:7292 b 0:2009 units2 13 a sec x, ln jtan x + sec xj + c b i y ii 0:965 units2 10 12 14 16 18 20 t (mins) EXERCISE 25B.2 g c b (5, 2) y 16 a 2:88 units2 b 4:97 units2 17 k ¼p1:7377 18 b ¼ 1:3104 19 k ¼ 2:3489 20 a = 40 30 20 10 metres b 3:33 cm [f (x) ¡ g(x)] dx c b units2 c 101 34 units2 15 a C1 is y = cos2 x, C2 is y = cos(2x) b A(0, 1) B( ¼4 , 0) C( ¼2 , 0) D( 3¼ , 0) E(¼, 1) 9:75 km d jf (x) ¡ g(x)j dx or 269 cm R [f (x) ¡ g(x)]dx + R units2 f (x) dx ¡ b c b a + d f (x) dx = ¡ (area between x = and x = 5) 110 m a ¢ 0:08 x REVIEW SET 25A 10 a 40 12 units2 b units2 c units2 12 a i A = ¡ dC dV = 12 x2 + and = ¼r2 dV dx 3:82020 units e 0:974 km £ 2:05 km -3 R 30 0:005 x 12 1:05o x3 ¡ Extra hint: 3 c 2:375 cm d x + y =9 -3 ¡ 0:01 x = 3p REVIEW SET 25B 29:6 cm 4:5 units2 ¼ units2 100 + 100 a 105 milliamps b as t ! 1, I ! 100 I(t) = 900 m a Show that v(t) = 100 ¡ 80e¡ 20 t m s¡1 and as t ! 1, v(t) ! 100 m s¡1 b 370:4 m t(sec) no, R3 t f (x) dx = ¡ (area from x = to x = 3) a local maximum at (1, 12 ), local minimum at (¡1, ¡ 12 ) b as x ! 1, f (x) ! (above) as cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 x ! ¡1, f (x) ! (below) black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_an\924IB_HL-2_AN.CDR Tuesday, 13 January 2009 3:28:34 PM TROY IB_PD (925) ANSWERS c f ( x) = x 1+ x 2 d ƒ(x) ln (¼ 0:805) units2 R4 c R6 yA dx = 2¼, yB dx = ¡ ¼2 m = 1, c = y a (1, Qw_ ) local max 3¼ d 925 x k = 13 x local (-1,- Qw_ ) y a p y¡2 b f (y) = p c 334¡ y=x +2 3 40 12 units2 y Hint: ¼ 4:01 units2 y = sin x 10 a ¼ Qw_\¼ x y x y¡=¡sin¡x y¡=¡sin2¡x 31:2 units2 y x = - p2 10 a x p x= y = - sec x ¡ 1¡ b ³ 11 -4 ¼2 ´ ¼ p ¢ units2 ¡ u2 12 m = 13 a a ¼ 0:8767 b ¼ 0:1357 u2 ¼ 14 b 8:66 years c 12:21% p.a x EXERCISE 26A 127¼ g ¼2 a 36¼ units3 b 8¼ units3 c b x = ¡ ¼2 and x = ¼2 c x-intercepts are x = ¡ 5¼ , ¡ 3¼ , ¡ ¼4 , 4 e ¼ 3¼ 5¼ , , 4 t (seconds) m d ¼ 2:59 units2 483¼ d at t = 2, 23 m from its starting point, it changes direction It changes direction again at t = 4, 13 m from its starting point, and at t = 5, it is 23 m from its starting point c d units3 h 146¼ units3 ¼ (e c ¼2 ¼ a m units3 1024¼ units3 256¼ e units3 ¼ (e c ¡ 1) units3 e units3 b units3 ¼2 ¡ f ¼ 2¡ 10 a p units3 ¢ ¼ c 2¼ units3 d ¼ units3 units3 y y = sin x + cos x a y ¡ 1) units3 48¼ units3 p a a = ¡3 b A has x-coordinate 1 B Y Z p ? X p ? b ¼ x p ¼(1) < ¼ < i.e., b partition as R¼ R0¼ y Z sin x dx < ¼(1) sin x dx < ¼ + ¢ x p ? units3 y b 2¼2 units3 y = sin(2 x) x p ? EXERCISE 26B Y A 4 c units2 B ¡¼ 11 a From the graph, area ¢OBX < area under the curve < area OXYZ ) 255¼ units3 40¼ units3 units3 b 30:2 a 8¼ units3 b b The particle moves in the positive direction initially, then 13 units3 a a sphere of radius r 23 f units3 units3 a 63¼ units3 b ¼ 198 cm3 a a cone of base radius r and height h ¡ ¢ b y = ¡ hr x + r c V = 13 ¼r2 h + 250¼ 3 a 186¼ units3 b REVIEW SET 25C + units3 a 18:6 y-intercept is y = d (¼ ¡ 2) units2 12 ln units2 13 ¼ 2:35 m a v(t): 992¼ a A is at (¡1, 3), B(1, 3) b C 136¼ 15 units3 a A is at (2, e) b ¼(e2 + 1) units3 95 100 50 75 95 100 yellow 25 a A is at (1, 1) b x 50 25 95 magenta p 3p ? 100 50 p ? 75 95 100 50 75 25 cyan 25 p ? 75 X black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\925IB_HL-2_AN.CDR Friday, 12 December 2008 4:19:36 PM TROY 11¼ units3 162¼ units3 IB_PD (926) 926 ANSWERS 9¼ a A is at (5, 1) b units3 y EXERCISE ¯ 27D y = + - x2 ¯ a ln ¯ex ¡ e¡x ¯ + c b b see diagram alongside: 7x ln (3x+5)6 18 +c c d ln j2 ¡ cos xj+c e x tan x+ln jcos xj+c f y = - - x2 x V = 36¼ units3 , which is independent of r REVIEW SET 26 a ¼ 3¼ 32 ¡ 31¼ units3 units p units3 a b a arcsin x + c b arcsin g arctan (2x) + c e p 2 ³ arctan x p 2 a ´ 93¼ units3 128¼ units3 c EXERCISE 27A d d 18¼ units3 ¡x¢ arcsin(2x) + c f +c h arctan +c c arctan ¡ 2x ¢ 128¼ 96¼ 5 units3 units3 ¡x¢ arcsin x -1 k c = p l = f (x) for all x n o For y to have meaning ¡ x2 > which has solution x ] ¡ 1, 1[ Area = ¼ p units2 b (x ¡ 3) (x + 1) ¡(3¡x2 ) 5 a p 28 ¡ p 44 15 a x ¡ arctan b ¡x¢ ln(x2 p 48 6¡54 + c b 12 i arcsin arctan ¡ 2x ¢ ¡2 x +c h p 1054 35 p arcsin x ¡ 12 x [ln x]2 ) + c ¢ ln x + c j 16 ln(1 + ³ ln p ¡ x2 + c x+c +c ¯ ¯ ¯ ¯ ¡ +c sin6 x sin8 x ¡ + ¢ ¯ ¯ ¡ ¢ 15 a y = sin10 x 10 ¯ ¯ ¡ ¢ +c +c 10 e x2 b r = ln x d y = ¡ ln(¡ c y = Aee y = e¡x y= p x2 + c) ¡ 2(x + 1)e¡x ¡ 11 y = 2x2 + 12 y = tan x2 + p ¼ ¢ ¡2 x2 + 2x ¡ b Horizontal asymptote y = 0, p vertical asymptotes x = ¡1 § p c y = dx line passing through (0, 0), gradient d “diameter meets the tangent to the circle at right angles” 16 b x2 + y2 = c circle centre (0, 0), radius = t 17 v = 4(1:5) m s¡1 , v ¼ 6:64 m s¡1 18 0:8% 19 t = sec g m s¡1 b t = 14 ln( 53 ) ¼ 0:128 sec 21 a V0 is the original volume of water, V is the volume of water that has evaporated ) V0 ¡ V is the volume remaining b ¼ 17:7% r2 dh = 22 hours 23 a b ¼ 14:4 hours dt ¼h2 ¡ 2¼rh 20 a v ! EXERCISE 27E.2 y = ex (x2 + c) y = ex (x + c) or y = Ae¡x ¡ 12 ex yellow 25 x(x2 + 5y2 )2 = k y = Axe2x 95 24 ¼ 12:5 minutes 25 12 midnight 100 50 75 25 95 100 50 75 25 95 100 50 75 + a ¡x2 e¡x ¡ 2xe¡x ¡ 2e¡x + c b 12 ex (sin x + cos x) + c c ¡ 12 e¡x (cos x + sin x) + c d ¡x2 cos x + 2x sin x + cos x + c a u2 eu ¡ 2ueu + 2eu + c b x(ln x)2 ¡ 2x ln x + 2x + c p p p a ¡u cos u + sin u + c b ¡ 2x cos 2x + sin 2x + c p p p 23 3x sin 3x + 23 cos 3x + c 25 arcsin 13 f (x) = 2e¡ x 14 y = x2 ¡ 9, a = §3 a xex ¡ ex + c b ¡x cos x + sin x + c c 13 x3 ln x ¡ 19 x3 + c d ¡ 13 x cos 3x + 19 sin 3x + c e 12 x sin 2x + 14 cos 2x + c f x tan x + ln jcos xj + c g x ln x ¡ x + c h x(ln x)2 ¡ 2x ln x + 2x + c i x arctan x ¡ 12 ln(x2 + 1) + c p (ln x) +2 ln x+2 ¡ + c i 23 (x ¡ 3) + x ¡ + c x ¡ 15 (sin x sin 4x + cos x cos 4x) + c ln x2 ¡ 2x + + 52 arctan x¡1 +c 1 sin x ¡ sin x + c m ln x + + arctan x2 p arcsin x2 + ¡ x2 + c 12 ¡ x ¡ ln j2 ¡ xj ¡ 2¡x + (2¡x) +c e2x+2 10 y = (x + 1)2 EXERCISE 27C +c g +r ´ x2 +16 magenta arctan2 x ¡p ¢ x + c c x ln 2x ¡ x + c ¯ ¯ ¯ ¯ e ln ¯ p x2 ¯ + c x +1 ¡ x ¢ xp9¡x2 + c b arcsin (sin x ¡ cos x) + c z = e3r ¡ ¢ p ¡1 p k 16 ¡ x2 + c l arcsin x2 ¡ 14 x(2 ¡ x2 ) ¡ x2 + c 16x cyan ¡x e +c a y = Ax3 b y = Ae cos3 x2 ) + c jxj arctan x+1 p ¡x¢ y = Ae2x a p = c ¡ arctan p c + 9) + c d ln(1 + ¡ ¢ p e x2 ¡ ¡ arccos x2 + c f cos x ¡ g a y = 10e4x b M = 20e¡3t c y = 16 t ¡ 2n d P = e ¡ e y = (x ¡ 3) + 15 (3¡x2 ) +c d 15 (t2 +2) ¡ 23 (t2 +2) +c ¡p ¢ p e x ¡ ¡ arctan x ¡ + c c 12 a y = Ae5x b M = Ae¡2t c y2 = 4x + c d P = 32 t + c e Q = Ae2t ¡ 32 f t = Q2 + 3Q + c + 2(x ¡ 3) + c ¡ 45 (x + 1) + 23 (x + 1) + c (x ¡ 3) + EXERCISE 27E.1 EXERCISE 27B a p h j +c 1¡x2 ii p f +c 1¡(¡x)2 d b i f (¡x) = p y m a +c ¡ 3x ¢ ln jsin 2xj+c (x ¡ 3) + (x ¡ 3) + c ln (x + 2) + +c o ¡ +c n ln jcos xj + 2x cos x+2 ³ ´ b ¼ 124 units3 2¼ units3 65¼ units3 128¼ units3 b units3 95 a ¼ ´ 100 ³ 50 ¼2 a 312¼ units3 b 402¼ units3 c (x + 3) ¡ 34 (x + 3)4 + c h x3 + 32 x2 + 3x + ln jxj + c g 5 i ¡x2 e¡x ¡2xe¡x ¡2e¡x +c j 25 (1 ¡ x) ¡ 23 (1 ¡ x) +c p p ¡ ¢ 2 x k x 41¡x ¡ x 1¡x + arcsin + c l 32 arccos x2 + c 8 75 -2 c 24¼2 ¼ 237 units3 +c black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\926IB_HL-2_AN.CDR Friday, 12 March 2010 4:46:08 PM PETER IB_PD (927) ANSWERS a REVIEW SET 27A 16 (4 ¡ x) ln(x2 + 1) + ¡ 32 (4 ¡ x) + roll x arctan x ¡ c a 12 e¡x (sin x ¡ cos x) + c b ex (x2 ¡ 2x + 2) + c (9 c p ¡ x2 ) ¡ 9 ¡ x2 + c (6, 1) (5, 1) (4, 1) (3, 1) (2, 1) (1, 1) 1 y = 20 ¡ 4ex ¼ 0:005 27 sec f (x) = 3e x y = x2 + Ax p arctan( 2x ) + c c 80 5¡ 3 p 2 a cos x + x sin x + c b x ¡ + arccos( x ) + c a arcsin( x3 ) + c b 124 15 P (9) = probability Fe_y_ Se_y_ 1 a continuous b discrete c continuous d continuous e discrete f discrete g continuous h continuous a i height of water in the rain gauge ii x 200 mm iii continuous b i stopping distance ii x 50 m iii continuous c i number of switches until failure ii any integer ¸ iii discrete a x b YYYY YYYN YYNN NNNY NNNN P(x = 1) = P(x = 0) = a 28 15 28 10 28 56 15 56 30 56 10 56 d 15 26 36 36 36 36 36 36 D 10 11 12 P(D = d) 36 36 36 36 36 b 0.4 0.2 1 5 1 1 1 1 N c Probability 36 10 36 36 36 36 36 d EXERCISE 28C 102 days a yellow 95 95 100 50 75 25 95 50 75 25 100 100 magenta 100 has in each game b k = 0:23 c i P(x > 2) = 0:79 ii P(1 x 3) = 0:83 cyan b 25 30 times $1:50 15 days a i 0:55 ii 0:29 iii 0:16 b i 4125 ii 2175 iii 1200 a $3:50 b No a i 16 ii 13 iii 12 b i $1:33 ii $0:50 iii $3:50 c lose 50 cents d lose $50 a $2:75 b $3:75 P (xi ) > b P (5) < which is not possible a a The random variable represents the number of hits that Sally 50 75 P 25 95 Die P(x) 50 b D probability that Jason will hit one or more home runs in a game 75 Die 6 7 8 9 10 10 11 10 11 12 11 a i ¼ 0:819 ii ¼ 0:164 iii ¼ 0:0164 b 0:001 15 12 a Die in a game 25 P(D = d) c P (1) + P (2) + P (3) + P (4) + P (5) = 0:4512 and is the 10 11 12 sum X Die 1 a k = 0:2 b k = 17 a P (2) = 0:1088 b a = 0:5488, the probability that Jason does not hit a home run P(X = x) 10 a EXERCISE 28B d b x X Wi_ P(X = x) c P (3) = 0:0247k; P (4) = 0:0123k Ri_ P(x) b k = 2:6130 P(x > 2) = 0:2258 a P (0) = 0:6648 b P(x > 1) = 0:3352 YYNY YNYN NNYN YNYY YNNY NYNN NYYY NNYY YNNN NYNY NYYN (x = 4) (x = 3) (x = 2) (x = 1) (x = 0) c i x = ii x = 2, or a x = 0, 1, 2, b HHH HHT TTH TTT HTH THT THH HTT (x = 3) (x = 2) (x = 1) (x = 0) d a k = 12 b k = 12 25 a P (0) = 0:1975k; P (1) = 0:0988k; P (2) = 0:0494k; EXERCISE 28A , , 8 (6, 6) (5, 6) (4, 6) (3, 6) (2, 6) (1, 6) He_y_ dN = kN (k a constant) b ¼ 8:97 £ 106 bacteria dt a y = 1¡ b HA y = 1, p p x + 4x + VA x = ¡2 + 3, x = ¡2 ¡ 18 remains P(x = 2) = (6, 5) (5, 5) (4, 5) (3, 5) (2, 5) (1, 5) c a c P(x = 3) = 18 , (6, 3) (6, 4) (5, 3) (5, 4) (4, 3) (4, 4) (3, 3) (3, 4) (2, 3) (2, 4) (1, 3) (1, 4) roll 2 ; P (3) = 36 ; P (4) = 36 ; 36 P (6) = 36 ; P (7) = 36 ; P (8) = 36 ; P (10) = 36 ; P (11) = 36 ; P (12) = 36 ; 36 ; 36 P (5) = p c mH = ¡2, mL ¼ ¡0:00496 b 0:0248 metres (6, 2) (5, 2) (4, 2) (3, 2) (2, 2) (1, 2) b P (0) = 0; P (1) = 0; P (2) = REVIEW SET 27B ¡ 27 (4 ¡ x) + c 927 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\927IB_HL-2_AN.CDR Friday, 12 December 2008 4:21:13 PM TROY IB_PD (928) 928 ANSWERS EXERCISE 28D.1 xi P (xi ) a k = 0:03 b ¹ = 0:74 c ¾ = 0:9962 P (1) = 10 , P (2) = 10 , P (3) = 10 , ¹ = 2:5, ¾ = 0:6708 a P (0) = 0:216, P (1) = 0:432, P (2) = 0:288, P (3) = 0:064 xi P (xi ) 0:216 0:432 b ¹ = 1:2, ¾ = 0:8485 a xi P (xi ) 0:1 b ¹ = 3:0, 0:2 c M P (Mi ) 0:4 0:2 probability 0.2 0:1 36 36 36 36 36 11 36 In 1, median = 0, mode = In 3, median = 1, mode = In 7, median = 5, mode = xi P (xi ) In 2, median = 3, mode = In 5, median = 3, mode = a 3:4 b 1:64 c ¼ 1:28 a k = 0:3 b 6:4 c 0:84 a b c d e f g p a a = 0:15, b = 0:35 a a = ¡ 84 b c x px 16 16 16 b i 0:6 ii ¼ 0:611 15 15 15 a a = 0:25, b = 0:35 b 0:99 a = 14 , E(X) = 11 , Var(X) = 12 275 144 ¼ 1:91 pi a 1:5 b px = X f 52px one coin 100 times b 12 11:6 d The binomial distribution does not apply as the result of each x 0:7 0:2 91 92 (1:694)x e¡1:694 , x = 0, 1, 2, 3, x! 156 156 132 132 75 74 33 32 11 3 1 p p 3+ 33 b i ¼ 0:506 ii ¼ 0:818 yellow 95 100 50 75 95 100 50 75 25 95 50 75 100 magenta 25 a 0:271 b 0:271 c 0:677 a 0:901 b years a ¼ 0:150 b ¼ 0:967 c mode = flaw per metre a m ¼ 6:8730 b ¼ 0:177 c ¼ 0:417 25 X f 500px a m= a i ¹ = 3, ¾ = 1:2247 100 2:4 c ¾ ¼ 1:292 and m ¼ 1:302, so, ¾ is very close to m in value a ¼ 0:0498 b ¼ 0:577 c ¼ 0:185 d ¼ 0:440 EXERCISE 28E.2 cyan 6:5 The fit is excellent p ¼ 0:800 c ¼ 0:200 ¼ 0:840 c ¼ 0:160 d ¼ 0:996 ¼ 0:002 46 c ¼ 0:131 d ¼ 0:710 ¼ 0:807 b b b b 12 13:0 (7:13)x e¡7:13 , x = 0, 1, 2, 3, x! i ¼ 0:0204 ii ¼ 0:0752 iii ¼ 0:839 iv ¼ 0:974 b draw is dependent upon the results of previous draws e The binomial distribution does not apply, assuming that ten bolts are drawn without replacement We not have a repetition of independent trials 95 18 17:4 a 1:694, so px = blue marble with the same chances each time 50 x (1:5)x e¡1:5 for x = 0, 1, 2, 3, 4, 5, 6, x! c The binomial distribution applies as we can draw out a red or a 75 a i ¼ 7:13 ii px = b The binomial distribution applies, as this is equivalent to tossing 25 The fit is excellent two possible outcomes (H or T) and each toss is independent of every other toss q 3pq 3p2 q p3 EXERCISE 28F a The binomial distribution applies, as tossing a coin has one of ¹ = 1:2, ¾ = 1:07 ¹ = 0:65, ¾ = 0:752 EXERCISE 28E.1 ¼ 0:268 ¼ 0:476 ¼ 0:0280 ¼ 0:998 2 ¹ = 5, ¾ = 1:58 a xi 1 17 and respectively a E(aX + b) = E(aX) + E(b) = aE(X) + b b i 13 ii ¡5 iii 13 a i 13 ii 16 b i ¡7 ii 16 c i ii a 2E(X)+3 b 4E(X )+12E(X)+9 c 4E(X )¡4fE(X)g2 a a a a x iii This distribution is negatively skewed and is the exact reflection of b EXERCISE 28D.3 5 probability 0.4 0.3 0.2 0.1 b i ii 16 iii The distribution is positively skewed c i ¹ = 4:8, ¾ = 0:980 ii EXERCISE 28D.2 probability 0.4 0.3 0.2 0.1 16 2 0:262 0:393 0:246 0:082 0:015 0:002 0:000 ii or rolling a die iii The distribution is bell-shaped b i ¹ = 1:2, ¾ = 0:980 P (1) = P (2) = P (3) = ::::: = P (6) = a 0:3125 0.1 Tossing a coin P(head) = P(tail) = x px or 0:2344 0.3 0:064 b ¹ = 4:472, ¾ = 1:404 a or 0:0938 ii P (¹ ¡ ¾ < x < ¹ + ¾) ¼ 0:8 P (¹ ¡ 2¾ < x < ¹ + 2¾) ¼ i ii ¾ = 1:0954 $390 a 0:288 or 0:0156 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\928IB_HL-2_AN.CDR Friday, 12 December 2008 4:22:33 PM TROY IB_PD (929) ANSWERS µ 10 a y = e¡x 1+x+ x2 ¶ b c d e y b x c P (x) ) ¡ 12 x2 e¡x = which is < for all x P (x) decreases as x increases f (x) > for all x [ 0, 0:9 ] and area under curve = ¹ ¼ 0:590, median = 0:6, mode = 0:6 Var(X) ¼ 0:0299, ¾ ¼ 0:173 0:652 The task can be performed between 18 minutes and 42 minutes 65:2% of the time EXERCISE 29B.1 Z+ y 0.15 xi P (xi ) 0.1 b 0.05 20 a 84:1% b 2:3% c i 2:15% ii 95:4% d i 97:7% ii 2:3% a i 34:1% ii 47:7% b i 0:136 ii 0:159 iii 0:0228 iv 0:841 a ¼ 41 days b ¼ 254 days c ¼ 213 days a 0:0388 b 25 of them a p = 0:3 b 0:850 EXERCISE 29B.2 REVIEW SET 28B a 0:341 b 0:383 c 0:106 a 0:341 b 0:264 c 0:212 d 0:945 e 0:579 f 0:383 a a ¼ 21:4 b a ¼ 21:8 c a ¼ 2:82 b 0:975 a 0:849 b a k = 0:05 0:156 a 12 a k = 145 2:56 £ 10¡6 c 0:991 d 0:000 246 b ¹ = 1:7, ¾ = 0:954 ¹ = 6:43, ¾ = 2:52 480 b ¼ 17:0 b 408 , ¼ 1:19 c median = 3, mode = 145 EXERCISE 29C.1 ³ a E 13 ¼ 0:238 ¼ 0:736 as p = 0:1 e4 10 m = 1:2, ¼ 0:879 b Var ¦(x) x¡=¡2 c i ii iii iv 0:8 x and Rk k=2 2.5 cyan ¢ , etc magenta 90:4% b 4:78% 83 b 61:5% c 23 eels ¹ ¼ 23:6, ¾ ¼ 24:3 ¾ = 21:6 b 54:4% ¾ = 0:0305 b 0:736 3 a a = ¡ 10 b y y= -3 10 yellow 25 x(x - 3) ¾ ¼ 0:501 mL k ¼ 0:885 ¼ 0:207 100 50 95 0.9 75 25 95 100 50 0:378 a a 0:003 33 a ¹ = 52:4, a ¹ = 2:00, c i 1:2 ii 1:5 iii ¼ 1:24 iv 0:24 d 13 20 75 25 95 100 50 ¹ ¾ ¹ = 64, ¾ = a i 81:85% ii 84:1% b 0:8185 ¹ = 31:2 x 75 X¡ REVIEW SET 29A ¦(x) 0.6 25 , etc EXERCISE 29D ¾ ¢ a k ¼ 0:878 b k ¼ 0:202 c k ¼ ¡0:954 a k ¼ ¡0:295 b k ¼ 1:165 c k ¼ ¡1:088 b i k ¼ 0:303 ii k ¼ 1:037 a k ¼ 79:1 b k ¼ 31:3 a+b a+b , median = , mode is undefined b ¹= b¡a 2 b¡a (a ¡ b) , ¾= p c Var(X) = 12 12 a median ¼ 0:347 b mode = ¼ a a = 18 b ¹ ¼ 0:0852 c 0:0334 d 0:0501 ¡1 = Var ¹ ¾ EXERCISE 29C.3 (5 ¡ 12y) dy = a k= a= 10 a X¡ a 0:159 b 0:3085 c 0:335 a 0:348 b 0:324 c 0:685 a 0:585 b 0:805 c 0:528 b k = 13 fk = 12 does not satisfy k 12 g c If k = 12 the graph goes below the horizontal axis d ¹ = 54 , median ¼ 0:116 , 32 ¾ a a a a 95 12 ¾ ¡1 EXERCISE 29C.2 p a b = 30 b i mean ¼ 1:55 ii ¼ 0:483 a k ¼ 1:0524 b median ¼ 0:645 4 a k = ¡ 375 b c ¼ 3:46 d 13 e 19 a k6 =E 0:885 b 0:195 c 0:3015 d 0:947 e 0:431 0:201 b 0:524 c 0:809 d 0:249 e 0:249 0:383 b 0:950 a a = 1:645 b a = ¡1:282 Physics 0:463, Chemistry 0:431, Maths 0:990, German 0:521, Biology 0:820 b Maths, Biology, German, Physics, Chemistry 65:6% EXERCISE 29A b ´ ³X ¡ ¹´ a m=4 b a a = ¡ 32 X ¡¹ ¾ 100 60 with variations around the mean occurring symmetrically as a result of random variations in the production process 10 a 0:0516 b No 11 a i ¹ = 1:28, ¾ = 1:13 ii ¹ = 1:14, ¾ = 0:566 b 0:366 a k= 40 a, b The mean volume (or diameter) is likely to occur most often 50 x ¾=1 p = 0:18, a 0:302 b 0:298 c 0:561 p = 0:04, n = 120 a ¹ = 4:8 b ¾ = 2:15 a $4 b $75 a 14 b x = 16 c med = 14, mode = a= A C b ¹ = 2, 0:0625 0:25 0:375 0:25 0:0625 75 a B 0.2 REVIEW SET 28A a a= 929 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\929IB_HL-2_AN.CDR Friday, 12 December 2008 4:23:52 PM TROY IB_PD x (930) 930 ANSWERS a 1438 students b 71 marks c IQR ¼ 20:2 marks ¹ ¼ 80:0 cm, median and mode are also 80:0 cm a k = ln b ¼ 0:474 c ¹ = ¡ ln , ¾ = ¡ 2(ln 4)2 10 ¼ 0:0708 35 a A = 12 , B = ¡ 12 REVIEW SET 29B 39 a ¼ 38 b x = tan( 16 ), tan( 5¼ ), tan( 9¼ ), tan( 13¼ ) 16 16 16 a i 2:28% ii 84% b 0:840 a 0:260 b 29:3 weeks a a = 6:3 grams b b = 32:3 grams a b c d a R2 ax2 (2 ¡ x) dx = gives a = 0:32968 b ¼ 0:796 Z dx = ¼ + x2 a ¼ b F (x) = c ¹= ¼ a a= b mode = ¯ ¼ p P n=0 ¡1¡ 2a 41 a P = Q = ¯ µ+Á = 1r z 43 c z4 = r4 cis 4µ, cis (¡µ), iz ¤ = r cis ( ¼2 ¡ µ) p p 45 x = + 46 x = 2, y = 18 or x = 64, y = 47 m = ¡ 19 or 49 x = ¡3 or ¼ ¡7:64 n P 50 a b ¼ xn n! 42 b ¯cos( µ¡Á )¯, which is 6= ln 2, Var(X) = an+1 (a sin nµ ¡ sin(n + 1)µ) + a sin µ a2 ¡ 2a cos µ + 40 b ex = 0:6 f (x), i.e., k = an+1 (a cos nµ ¡ cos(n + 1)µ) ¡ a cos µ + a2 ¡ 2a cos µ + b mode occurs at the maximum of f (x) and so mode is median is 1:23 R 1:2 P(0:6 < x < 1:2) = x (2 ¡ x) dx ¼ 0:392 4 c p 36 a ¡ ¡ x2 +c b arctan x+ 12 ln(1+x2 )+c c arcsin x+c ³ ln ´2 p k=1 p = n+1¡1 p k+1 k+ 52 There are about 1574 trees; remain steady at 1000 · ¼ c median ¼ 1:08 d ¹ = 16 15 53 a $865:25 b R = r P 100m r )mn 100m r mn ¡ ) 100m (1 + ¼ 0:403 a ¹ ¼ 61:218, ¾ ¼ 22:559 b ¼ 0:756 10 a k = b median = 27 c ¹ ¼ 2:75, Var(X) ¼ 2:83 54 EXERCISE 30 55 a 11 a b 57 ¡ ( 121 )10 ¼ 0:430 58 n = 59 128 c a - 60 107 576 66 70 72 74 75 b p = 879 876 choices 56 10 61 63 ® ¼ 6:92o 65 63 ¼ 3:97 km 69 a k = b p 10 11 ¼ 0:953 units 71 a = x ) tan3 2(n ¡ 1) p 3¡ ¼ 0:382 ¼ ln units2 y = arccos(3 ¡ 73 x + tan x + c Either A or B must occur, or A and B are disjoint a ¼ 1:48 units b ¼ 3:82 units 76 a i A0 \ B ii B p ¼ 77 ¡23 ¡ 84 2i 78 µ = ¡ 11¼ , ¡ 7¼ , 12 , 5¼ 79 ¡1 + 2i 12 12 12 Re p (m + 2)(m + 1)(n + 2)(n + 1) ¡21¢ p 5+i p p p p ( 3) cis (arctan( 52 )); a = 3, µ = arctan( 52 ) p p z = cis ( 13 arctan( 52 ) + k 2¼ ), k = 0, 1, p ¼ Im cis (¡ ), b p cis ( 5¼ ), 12 p 13p 5p 12 cis ( 13¼ ) 12 12 ¸ (1 + 1¡x 80 y = e x+1 (x + 1)2 a (1 ¡ i)2 = ¡2i, (1 ¡ i)4n = (¡4)n b 256 c § i a z= ¡ ¢ cos( 2¼ ) + i sin( 2¼ ) , w= 3 c cos( 11¼ )= 12 u1 = 2, un = cot( µ2 ) a p p ¡ 2+ , 3n ¡ 3n + sin( 11¼ )= 12 3, n > ¡ 1 cos( ¼4 ) + i sin( ¼4 ) p p 6¡ y = e2x¡1 arctan x ¡ 12 x + 12 arctan x + c ln (or log3 2) 10 x > a x=3 b x= ln x 11 x = ¡ 3¼ or 24 25 13 a 14 x = 2¼ b or + 81 a x ] ¡ 1, 0[ [ ]2, 1[ b x x¡2 c 4x ¡ 3y = 12 ¡ ln 82 a 24 b 16 83 a or §1 b A¡1 = A 49 25 p 85 86 ¼ 1:73 m ¢ ¼ ¡ 24 12 a ¡e2 b e2 ¡ 5¼ 15 d = 16 (0, ¡1, ¡1) c ¡ 336 625 17 a y = 2x ¡ c exactly one d ] 32 , [ 144° 25 cm 625 527 d b y 15 cm 88 a cis (¡ ¼2 ) b z = cis (¡ ¼6 ), cis ( ¼2 ), cis (¡ 5¼ ) y¡=¡¦(x) (2,¡1) c A x e ¡8i Im radius =¡2 -3 30° 26 b ¼ab B p p l 32 b ¡ 2 34 a A = 1, B = 0, C = ¡1 b 12 ln ¡ 32 ln Re 30° C cyan magenta yellow 95 100 50 75 25 95 100 50 89 b ¢PQR is right angled at Q and is isosceles as QR = QP 75 25 95 100 50 75 25 95 100 50 75 25 31 a circle, centre (0, 0), radius black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\930IB_HL-2_AN.CDR Thursday, 11 March 2010 11:40:04 AM PETER IB_PD (931) ANSWERS 90 a a = (r + r1 ) cos µ, b = (r ¡ 1r ) sin µ b r = or z is real and non-zero 91 a 2p2 ¡ p4 b p ¼ 0:541 92 a A2 = ³ 4 ´ b A3 = ³ ¡n¢ k 94 a ¡ (1 ¡ p)n b 139 a = 43 , b = 141 w = ´ 12 pk (1 ¡ p)n¡k r c p=1 11 b 50 < k < 54 i.e., k ]50, 54[ (3,¡54) x b (0, 6) A translation of 102 b x = cos( 2¼ ), cos( 8¼ ), cos( 14¼ ) 9 104 a An = ³ c Sn = ³ S20 = ´ 2n ¡ 1 translation of 2n+1 ¡ ¡ n n 097 150 097 130 20 e (¡2, p 156 157 159 160 161 162 x x b yellow ¼ 7:82o a µ 1 ¡2 95 100 95 100 ¼ 50 75 50 a f (x) = ¡x2 + 6x ¡ 13 b f (x) = ¡(x ¡ 3)2 ¡ y = sin( ¼2 x) ¡ 158 x = 2, y = 4, z = ¡1 ¼ a a = 13, b = 12, c = 30 , d = 15 b ¼ 24:93 m a ( 15 , 17 , ) c (¡1, 3, 1) d 6x ¡ 8y ¡ 5z = ¡35 5 ¡2 p p2 ¡4 p sin µ, µ ¡ 25 d x= 95 50 75 25 95 100 50 75 25 100 magenta B 1 q ¶ c p 6= p = 0, q 6= p=q=0 t2R i ii iii p 30 166 a 5i ¡ 2j + k b (5i ¡ 2j + k) 167 52 units2 168 a D(7, 1, ¡2) X(7, 3, ¡1) Y(5, 3, ¡2) ¡ ! ¡ ! b Hint: Show BD = k BY 170 a t = 23 b t = ¡ 13 171 a x = b x = 0:2 or 0:3 172 a x < 15 or x > 27 b ¡6 < x < or x > 173 a ¼ 0:785 b ¼ 0:995 174 y = 12 x + 14 sin 2x + 175 y2 = 14 x2 ¡ 177 ¼ 50 000 guinea pigs 131 x = ¡ 11¼ , ¡ 3¼ , ¡ 7¼ , 12 12 cyan C d x = ¡2 ¡ 4t, y = t, z = + 2t, 163 b k = ¡1 c p = ¡2, q = 164 AB = I a = 2, b = ¡1, c = p 3n ¡ , n Z + 126 a k p 128 b r km c 1000 km h¡1 130 b ¼ , ¼ , 5¼ 12 12 ¼ 5¼ 4¼ 5¼ = , , , 133 x = ¡ 10 b x = ¡1 c x = ee x < b x < 12 c x < 138 A -1 p 125 a k = b un = 136 a a d 75 p 40 3 124 a un = cos µ tann¡2 µ b u1 = and un+1 = un2 cos µ, n Z + 135 a y -1 -1 132 x f (3, ¡2) A reflection in y = x -2 1 ´ 116 a ¼ 0:549 b ¼ 0:001 72 117 ¾ = + 118 a ¼ 375 b ¼ 0:0366 119 a 66 b ¼ 275 120 a x ln x ¡ x + c b k = e Rm c m is the solution of ln x dx = 12 123 b y -1 25 b mean = 800, sd = ¡1:5¢ c 105 f (n) = ¡ 3n + 106 ¼ 6:40 cm 107 a ¼ 0:34 b ¾ ¼ 108 a ¾ ¼ 3:599 86 b ¼ 0:781 109 17 110 ¼ 0:114 111 a ¼ 0:242 b ¼ 0:769 32 112 ¼ 0:842 113 a = 35 114 10 115 a k = ) ´ 3n2 followed by a vertical stretch 155 for all n Z + 2n+1 ¡ ¡2¢ d ( 12 , 3) A horizontal compression factor followed by a 3x ¡ 10 103 a y2 = x2 + 64 ¡ 16x cos µ, cos µ = 2x p c cm2 when ¢ is isosceles 2n 10 ¡2¢ of factor c (¡2, ¡5) ³ y= [a2 ¡ ¡ b2 ] + i[2ab] , purely imaginary if a2 ¡ b2 = (a + 1)2 + b2 and ab 6= 154 a (0, 4) A translation of (5,¡50) 15 , 149 f (x) = 13 (x + 3)(4x + 1)(2x ¡ 3)2 150 a x2 + 4x + b a = ¡4, real zero is 151 b h(x) = (x ¡ t)(x ¡ 2t)(x ¡ 3t) c (¡3, 27), (¡2, 16), (¡1, 5) 152 a i x2 ¡ 2x + (1 + k2 ) ii k = 0, §1, §2, §3 b pq = ¡1, ¡2, ¡5, ¡10 c x = 1, ¡2, § 2i 153 P (z) = (z2 + 2)(z2 ¡ 2z + 5) ii b cm or 2:2 cm c AB = cm is not possible 98 b 1, 1, 101 a y 140 x = 143 a 25¼ units2 b 25¼ sin( ® ) units2 2 x 144 9b = 2a2 145 a = ¡1, b = ¡7 146 a = 2, n = 147 a P (x) = (ax + b)(2x2 ¡ 3x + 1) c P (x) = (3x + 7)(2x2 ¡ 3x + 1) 148 f (x) = (x + 1)2 (2x ¡ 5), x = ¡1 or x > 52 pr (1 ¡ p)n¡r d no real solutions ¡ ¢ ¡ ¢ 95 a n n2 committees b n3 committees d i 96 a or a = ¡ 43 , b = ¡ 23 142 n ¡ ¢ P n r=k 931 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\931IB_HL-2_AN.CDR Thursday, 11 March 2010 11:40:14 AM PETER IB_PD (932) 932 ANSWERS 178 b y = 203 34 unit2 204 VAs x = ¡ ¼2 , x = ¡ ¼4 , x = 624 51 t + 311( 311 ) c i 624 people ii about 67 people iii ¼ 4:24 pm 179 a 7¼ p1 ei(¡ 12 ) 205 b 12 208 a 4 ¼ , 47 67 206 3¼ x= No HAs 31 17 or 3x 2x + b x¡2 x¡1 p p 3y ¡ 2x dy = b ( 7, 0), (¡ 7, 0) dx 2y ¡ 3x 211 x = a2 ¡ p ¡ 65 ¡ 2x + 29 ¡y2 ¡ yexy dy = dx xexy + 2xy ¡ cos y 215 3:76 units2 216 217 x < 62 371 212 y = ¡1 + 3earctan x 213 ¡x2 cos x + 2x sin x + cos x + c 214 y= -4 p (¡ 3¼ , ¡ 42 ) 209 a i 13 ii 11 b 23 21 21 210 a A(2, ¡3, 1) b B(¡1, 2, 3) c p = or y= f (x) y¡=¡¦(x) -8 ( ¼4 , 207 a 180 a 479 001 600 b i 43 545 600 ii 257 600 iii 159 667 200 iv 58 060 800 c i 5775 ii 1575 181 20:6 cm 182 a 8008 b 5320 c 2211 183 n = or 184 a ¡224 b 880 c ¡40 185 a when x = ¼2 b sin2 2x + cos 2x = needs to be solved c (0:999, 2:028) and (2:143, 2:028) p p 13¡2 186 x = or 187 a = e6 ¡ 188 189 a = 5, b = 190 y p ), or ¡5 < x < ¡1 p p + 2) ¡ x ¡ ¡ x + + ln( x + + 1) + c 218 (x 219 ¡ y cos(xy) dy = dx x cos(xy) + 2y 220 a P(0, 3, ¡8) b µ ¼ 18:8o c 5x ¡ 11y ¡ 7z = 23 221 a < ¡8 or a > 222 40¼ cm s¡1 223 a = 224 a x y -4 x = -2 -1 x=3 191 x = or 192 a = ¡2, b = 3; A11 = ³ p 193 x = (y = 0) 194 + 2i, ¡1 § 3i 195 ¼ (e ¡ 1) units3 196 Translate through 197 199 a c 200 a b c unit2 ¡ ¢ ¡10 ¡2 ¡1 Z b V =¼ ¡ x tan b local ¡ 5t , y = ¡ 2t, z = t, t R a = b a = 4; x = 12 a+8 a , y = ¡1 ¡ a, z = a 6= or 4; x = ¡ ¡ a a when a = 2; x = ¡7, y = ¡3, z = VA is x = ¡1, HA is y = 2(x ¡ 1) f (x) = ; local (1, 12 ) (x + 1)3 ¡4(x ¡ 2) f 00 (x) = ; inflection (2, 59 ) (x + 1)4 dx p at (¡ 12 , ¡ 2e ) p § 23 f 00 (x) = e1¡2x (16x3 ¡12x) local max at ( 12 , p e ) c x = or d as x ! 1, f (x) ! (above) as x ! ¡1, f (x) ! (below) e y e ) (12 , x y (- 12 ,- f k= (1' Qw_\) y¡=¡1 -1 x x¡=¡-1 b 1:39 units cyan = p 50 75 95 2x ¡ ¡1¢ d x [¡1, 1] but x ¼ = ¡0:571 or 0:476; range y R magenta yellow 95 100 50 95 100 50 75 25 ¡ (1 + x ¡ x2 )2 100 95 100 50 75 202 x [¡1, 0] [ [1, 2] factor 2, followed by a translation f (x) 25 ¡2 201 a v = t+2 75 -2 25 -3 -4 e ) p1 226 a k = ¡2, a = ¡1 c x = + t, y = ¡3 ¡ 2t, z = t, t R d (3, ¡3, 0) e ¼ 61:9o 228 a = 8, b = 25, c = 26; z > ¡2 229 a ¼3 b x ¼ ¡0:571, t ¼ 0:476 c A horizontal stretch factor 13 , followed by a vertical stretch NSPI (2' To_\) 25 ¢2 225 a f (x) = e1¡2x (1¡4x2 ) ¡ x2 unit2 d p then reflect in x-axis p 198 2x ¡ x2 , x ´ black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\932IB_HL-2_AN.CDR Thursday, 11 March 2010 11:43:22 AM PETER IB_PD (933) INDEX magenta yellow 95 100 50 75 95 100 50 25 95 100 50 75 25 95 100 50 75 25 cyan 75 126 273 286 688 709 250 422 430 56 66 41 354 628 590 627 153 562 492 223 226 801 505 451 430 605 591 485 461 473 399 485 325 223, 560 56 59 549 534 147, 158 184 178 422 383 absolute value function ambiguous case amplitude angular velocity antiderivative arc length Argand diagram argument arithmetic sequence arithmetic series asymptotic augmented matrix average acceleration average speed average velocity axis of symmetry Bayes’ theorem bimodal binomial coefficient binomial expansion binomial experiment box-and-whisker plot Cartesian equation Cartesian form chain rule chord class interval coincident lines coincident planes collinear points column graph column matrix combination common difference common ratio complement complementary events completing the square complex conjugates complex number complex plane component form 25 composite function compound interest concave downwards concave upwards constant term continuous coplanar lines coplanar points cosine rule critical value cubed factor cubic function cumulative frequency De Moivre’s theorem decreasing function definite integral dependent events derivative function determinant differential equation direction vector discrete discriminant displacement displacement function divisor domain dot product double angle formulae echelon form empty set equal matrices equal vectors Euler form Euler’s equation expectation experimental probability exponential equation exponential function factor theorem factorial rule first derivative five-number summary frequency histogram function INDEX black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_an\933IB_HL-2_AN.CDR Friday, 25 January 2008 11:39:05 AM PETERDELL 933 27 62 647 647 188 786 462 415 270 28 204 126 500 439 635 584, 730 539 593 343, 348 774 451 786 149 375, 746 627 191 23 402 315 355 550 327 374 431 97 791 528 87 126 199 217 616 505 487 19 IB_PD (934) INDEX magenta yellow 95 100 50 75 25 95 100 50 25 95 100 50 75 25 95 100 50 75 25 cyan matrix mean median midpoint mode modulus mutually exclusive natural exponential natural logarithm negative definite negative matrix negative vector negatively skewed non-stationary inflection normal normal curve normal vector nth roots nth roots of unity Null factor law number sequence oblique asymptote ogive one-to-one function optimum solution order of matrix overspecified system parabola parallel lines parallel planes parameter parametric equation Pascal’s triangle percentile period periodic function permutation point of inflection Poisson distribution polar form polynomial population mean population standard deviation population variance 292 56, 229 59 68 639 639 593 485 42, 642 639 46 292 44 339 21 422 178 661 635 538, 558 803 630 628 628 715 717 771 503 461 473 549 44 342, 350 571 193 126 688 639, 653 639, 653 126 683 503 579 46 19 general sine function general term geometric sequence geometric series global maximum global minimum gradient function histogram horizontal asymptote horizontal inflection horizontal line test horizontal translation identity function identity matrix image imaginary axis imaginary number implicit relation increasing function independent events independent trials initial conditions instantaneous acceleration instantaneous velocity integral integrating constant integration by parts interquartile range intersecting lines intersecting planes intersection inverse operation invertible matrix limit linear factor linear function linear speed local maximum local minimum logarithmic function logistic function lower quartile lower rectangles many-to-one function mapping 75 934 black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\934IB_HL-2_AN.CDR Friday, 12 March 2010 4:47:41 PM PETER 324 489, 795, 814 489, 796, 814 389 489, 796, 814 35, 425 550 110 110, 673 160 331 374 490 648 611 817 467 441 444 145 54 42, 642 500 46 652 325 363 166 461 473 515 260 227 508 286, 292 286 219, 560 648 807 430 144 794 794 794 IB_PD (935) INDEX magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 cyan standard deviation stationary point statistic surge function tangent tangent ratio translation two-dimensional grid underspecified system union unique solution unit circle unit vector universal set upper quartile upper rectangles variance vector vector equation vector product velocity vector Venn diagram vertical asymptote vertical line test vertical translation volume of revolution zero zero matrix zero vector 373, 388 160 490 286, 292 814 788 608 234 148 126 204 191 609 184 786 23, 503 574 244 422 188 184 41, 126 19 528 191 198 411 193 325 532, 549 372 402 411 616, 653 47 28, 636, 653 272 204 342, 350 462 758 325 204 position vector positive definite positively skewed principal axis probability density function probability distribution product rule proposition quadratic formula quadratic function quadruple factor quotient quotient rule radical conjugates random variable range rational function rational number real axis real polynomial real quadratic reciprocal function relation relative frequency remainder remainder theorem right hand rule root row matrix sample space scalar scalar product scalar triple product second derivative self-inverse function sign diagram sine rule single factor singular matrix skew lines solid of revolution square matrix squared factor black V:\BOOKS\IB_books\IB_HL-2ed\IB_HL-2ed_AN\935IB_HL-2_AN.CDR Friday, 12 March 2010 4:46:58 PM PETER 935 512 639 515 683 591 254 133 536 363 549 344, 348 253 400 549 503 579 795, 814 372 450 402 450 549 42, 642 19 292 758 193 331 377 IB_PD (936) cyan magenta yellow 95 100 50 75 25 95 100 50 75 25 95 100 50 75 25 95 100 50 75 NOTES 25 936 black Y:\HAESE\IB_HL-2ed\IB_HL-2ed_AN\936IB_HL-2_AN.CDR Friday, 25 January 2008 4:22:42 PM PETERDELL IB_PD (937)