Advanced engineering mathematics 9th edition applied solution

Solution Manuals Of ADVANCED ENGINEERING MATHEMATICS By ERWIN KREYSZIG 9TH EDITION This is Downloaded From www.mechanical.tk Visit www.mechanical.tk For More Solution Manuals Hand Books And Much Much More imfm.qxd 9/15/05 12:06 PM Page i INSTRUCTOR’S MANUAL FOR ADVANCED ENGINEERING MATHEMATICS imfm.qxd 9/15/05 12:06 PM Page ii imfm.qxd 9/15/05 12:06 PM Page iii INSTRUCTOR’S MANUAL FOR ADVANCED ENGINEERING MATHEMATICS NINTH EDITION ERWIN KREYSZIG Professor of Mathematics Ohio State University Columbus, Ohio JOHN WILEY & SONS, INC imfm.qxd 9/15/05 12:06 PM Page iv Vice President and Publisher: Laurie Rosatone Editorial Assistant: Daniel Grace Associate Production Director: Lucille Buonocore Senior Production Editor: Ken Santor Media Editor: Stefanie Liebman Cover Designer: Madelyn Lesure Cover Photo: © John Sohm/Chromosohm/Photo Researchers This book was set in Times Roman by GGS Information Services and printed and bound by Hamilton Printing The cover was printed by Hamilton Printing This book is printed on acid free paper Copyright © 2006 by John Wiley & Sons, Inc All rights reserved 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, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (508) 750-8400, fax (508) 750-4470 Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (201) 748-6011, fax (201) 748-6008, E-Mail: PERMREQ@WILEY.COM ISBN-13: 978-0-471-72647-0 ISBN-10: 0471-72647-8 Printed in the United States of America 10 imfm.qxd 9/15/05 12:06 PM Page v PREFACE General Character and Purpose of the Instructor’s Manual This Manual contains: (I) Detailed solutions of the even-numbered problems (II) General comments on the purpose of each section and its classroom use, with mathematical and didactic information on teaching practice and pedagogical aspects Some of the comments refer to whole chapters (and are indicated accordingly) Changes in Problem Sets The major changes in this edition of the text are listed and explained in the Preface of the book They include global improvements produced by updating and streamlining chapters as well as many local improvements aimed at simplification of the whole text Speedy orientation is helped by chapter summaries at the end of each chapter, as in the last edition, and by the subdivision of sections into subsections with unnumbered headings Resulting effects of these changes on the problem sets are as follows The problems have been changed The large total number of more than 4000 problems has been retained, increasing their overall usefulness by the following: • Placing more emphasis on modeling and conceptual thinking and less emphasis on technicalities, to parallel recent and ongoing developments in calculus • Balancing by extending problem sets that seemed too short and contracting others that were too long, adjusting the length to the relative importance of the material in a section, so that important issues are reflected sufficiently well not only in the text but also in the problems Thus, the danger of overemphasizing minor techniques and ideas is avoided as much as possible • Simplification by omitting a small number of very difficult problems that appeared in the previous edition, retaining the wide spectrum ranging from simple routine problems to more sophisticated engineering applications, and taking into account the “algorithmic thinking” that is developing along with computers • Amalgamation of text, examples, and problems by including the large number of more than 600 worked-out examples in the text and by providing problems closely related to those examples • Addition of TEAM PROJECTS, CAS PROJECTS, and WRITING PROJECTS, whose role is explained in the Preface of the book • Addition of CAS EXPERIMENTS, that is, the use of the computer in “experimental mathematics” for experimentation, discovery, and research, which often produces unexpected results for open-ended problems, deeper insights, and relations among practical problems These changes in the problem sets will help students in solving problems as well as in gaining a better understanding of practical aspects in the text It will also enable instructors to explain ideas and methods in terms of examples supplementing and illustrating theoretical discussions—or even replacing some of them if so desired imfm.qxd 9/15/05 12:06 PM Page vi vi Instructor’s Manual “Show the details of your work.” This request repeatedly stated in the book applies to all the problem sets Of course, it is intended to prevent the student from simply producing answers by a CAS instead of trying to understand the underlying mathematics Orientation on Computers Comments on computer use are included in the Preface of the book Software systems are listed in the book at the beginning of Chap 19 on numeric analysis and at the beginning of Chap 24 on probability theory ERWIN KREYSZIG im01.qxd 9/21/05 10:17 AM Page Part A ORDINARY DIFFERENTIAL EQUATIONS (ODEs) CHAPTER First-Order ODEs Major Changes There is more material on modeling in the text as well as in the problem set Some additions on population dynamics appear in Sec 1.5 Electric circuits are shifted to Chap 2, where second-order ODEs will be available This avoids repetitions that are unnecessary and practically irrelevant Team Projects, CAS Projects, and CAS Experiments are included in most problem sets SECTION 1.1 Basic Concepts Modeling, page Purpose To give the students a first impression what an ODE is and what we mean by solving it Background Material For the whole chapter we need integration formulas and techniques, which the student should review General Comments This section should be covered relatively rapidly to get quickly to the actual solution methods in the next sections Equations (1)–(3) are just examples, not for solution, but the student will see that solutions of (1) and (2) can be found by calculus, and a solution y ϭ e x of (3) by inspection Problem Set 1.1 will help the student with the tasks of Solving yЈ ϭ ƒ(x) by calculus Finding particular solutions from given general solutions Setting up an ODE for a given function as solution Gaining a first experience in modeling, by doing one or two problems Gaining a first impression of the importance of ODEs without wasting time on matters that can be done much faster, once systematic methods are available Comment on “General Solution” and “Singular Solution” Usage of the term “general solution” is not uniform in the literature Some books use the term to mean a solution that includes all solutions, that is, both the particular and the singular ones We not adopt this definition for two reasons First, it is frequently quite difficult to prove that a formula includes all solutions; hence, this definition of a general solution is rather useless in practice Second, linear differential equations (satisfying rather general conditions on the coefficients) have no singular solutions (as mentioned in the text), so that for these equations a general solution as defined does include all solutions For the latter reason, some books use the term “general solution” for linear equations only; but this seems very unfortunate im01.qxd 9/21/05 10:17 AM Page 2 Instructor’s Manual SOLUTIONS TO PROBLEM SET 1.1, page 10 12 14 16 y ϭ Ϫe؊3x/3 ϩ c y ϭ (sinh 4x)/4 ϩ c Second order First order y ϭ ce0.5x, y(2) ϭ ce ϭ 2, c ϭ 2/e, y ϭ (2/e)e0.5x ϭ 0.736e0.5x y ϭ ce x ϩ x ϩ 1, y(0) ϭ c ϩ ϭ 3, c ϭ 2, y ϭ 2e x ϩ x ϩ y ϭ c sec x, y(0) ϭ c/cos ϭ c ϭ _12␲, y ϭ _12␲ sec x Substitution of y ϭ cx Ϫ c into the ODE gives yЈ2 Ϫ xyЈ ϩ y ϭ c Ϫ xc ϩ (cx Ϫ c 2) ϭ Similarly, y ϭ _14x 2, yЈ ϭ _12x, _1 x Ϫ x(_1 x) ϩ _1 x ϭ 4 thus 18 In Prob 17 the constants of integration were set to zero Here, by two integrations, y Љ ϭ g, v ϭ yЈ ϭ gt ϩ c1, y ϭ _21gt ϩ c1t ϩ c2, y(0) ϭ c2 ϭ y0, and, furthermore, v(0) ϭ c1 ϭ v0, hence y ϭ _12gt ϩ v0 t ϩ y0, as claimed Times of fall are 4.5 and 6.4 sec, from t ϭ ͙100/4 ෆ9ෆ and ͙200/4 ෆ9ෆ 20 yЈ ϭ ky Solution y ϭ y0 ekx, where y0 is the pressure at sea level x ϭ Now y(18000) ϭ y0 ek⅐18000 ϭ _12y0 (given) From this, ek⅐18000 ϭ _1 , y(36000) ϭ y ek⅐2⅐18000 ϭ y (ek⅐18000)2 ϭ y (_1 )2 ϭ _1 y 0 22 For year and annual, daily, and continuous compounding we obtain the values ya(1) ϭ 1060.00, yd(1) ϭ 1000(1 ϩ 0.06/365)365 ϭ 1061.83, yc(1) ϭ 1000e0.06 ϭ 1061.84, respectively Similarly for years, ya(5) ϭ 1000 ⅐ 1.065 ϭ 1338.23, yd(5) ϭ 1000(1 ϩ 0.06/365)365⅐5 ϭ 1349.83, yc(5) ϭ 1000e0.06⅐5 ϭ 1349.86 We see that the difference between daily compounding and continuous compounding is very small The ODE for continuous compounding is ycЈ ϭ ryc SECTION 1.2 Geometric Meaning of y ؅ ‫ ؍‬ƒ(x, y) Direction Fields, page Purpose To give the student a feel for the nature of ODEs and the general behavior of fields of solutions This amounts to a conceptual clarification before entering into formal manipulations of solution methods, the latter being restricted to relatively small—albeit important—classes of ODEs This approach is becoming increasingly important, especially because of the graphical power of computer software It is the analog of conceptual studies of the derivative and integral in calculus as opposed to formal techniques of differentiation and integration Comment on Isoclines These could be omitted because students sometimes confuse them with solutions In the computer approach to direction fields they no longer play a role im01.qxd 9/21/05 10:17 AM Page Instructor’s Manual Comment on Order of Sections This section could equally well be presented later in Chap 1, perhaps after one or two formal methods of solution have been studied SOLUTIONS TO PROBLEM SET 1.2, page 11 Semi-ellipse x 2/4 ϩ y 2/9 ϭ 13/9, y Ͼ To graph it, choose the y-interval large enough, at least Ϲ y Ϲ 4 Logistic equation (Verhulst equation; Sec 1.5) Constant solutions y ϭ and y ϭ _12 For these, yЈ ϭ Increasing solutions for Ͻ y(0) Ͻ _12, decreasing for y(0) Ͼ _12 The solution (not of interest for doing the problem) is obtained by using dy/dx ϭ 1/(dx/dy) and solving x ϩ c ϭ Ϫ2/(tan _12 y ϩ 1); dx/dy ϭ 1/(1 ϩ sin y) by integration, thus y ϭ Ϫ2 arctan ((x ϩ ϩ c)/(x ϩ c)) Linear ODE The solution involves the error function 12 By integration, y ϭ c Ϫ 1/x 16 The solution (not needed for doing the problem) of yЈ ϭ 1/y can be obtained by separating variables and using the initial condition; y 2/2 ϭ t ϩ c, y ϭ ͙2t ෆ Ϫ1 18 The solution of this initial value problem involving the linear ODE yЈ ϩ y ϭ t is y ϭ 4e؊t ϩ t Ϫ 2t ϩ 20 CAS Project (a) Verify by substitution that the general solution is y ϭ ϩ ce؊x Limit y ϭ (y(x) ϭ for all x), increasing for y(0) Ͻ 1, decreasing for y(0) Ͼ (b) Verify by substitution that the general solution is x ϩ y ϭ c More “squareshaped,” isoclines y ϭ kx Without the minus on the right you get “hyperbola-like” curves y Ϫ x ϭ const as solutions (verify!) The direction fields should turn out in perfect shape (c) The computer may be better if the isoclines are complicated; but the computer may give you nonsense even in simpler cases, for instance when y(x) becomes imaginary Much will depend on the choice of x- and y-intervals, a method of trial and error Isoclines may be preferable if the explicit form of the ODE contains roots on the right SECTION 1.3 Separable ODEs Modeling, page 12 Purpose To familiarize the student with the first “big” method of solving ODEs, the separation of variables, and an extension of it, the reduction to separable form by a transformation of the ODE, namely, by introducing a new unknown function The section includes standard applications that lead to separable ODEs, namely, the ODE giving tan x as solution the ODE of the exponential function, having various applications, such as in radiocarbon dating a mixing problem for a single tank Newton’s law of cooling Torricelli’s law of outflow im25.qxd 9/21/05 2:06 PM Page 396 396 Instructor’s Manual 24 By Theorem 1, the load Z is normal with mean 40N and variance 4N, where N is the number of bags Now gives the condition 2000 Ϫ 40N P(Z Ϲ 2000) ϭ ⌽( ᎏᎏ ) ϭ 0.95 2͙N ෆ 2000 Ϫ 40N ᎏᎏ м 1.645 2͙N ෆ by Table A8 The answer is N ϭ 49 (since N must be an integer) SECTION 25.4 Testing of Hypotheses Decisions, page 1058 Purpose Our third big task is testing of hypotheses This section contains the basic ideas and the corresponding mathematical formalism Applications to further tasks of testing follow in Secs 25.5–25.7 Main Content, Important Concepts Hypothesis (null hypothesis) Alternative (alternative hypothesis), one- and two-sided Type I error (probability ␣ ϭ significance level) Type II error (probability ␤; Ϫ ␤ ϭ power of a test) Test for ␮ with known ␴ (Example 2) Test for ␮ with unknown ␴ (Example 3) Test for ␴ (Example 4) Comparison of means (Example 5) Comparison of variances (Example 6) Comment on Content Special testing procedures based on the present ideas have been developed for controlling the quality of production processes (Sec 25.5), for assessing the quality of produced goods (Sec 25.6), for determining whether some function F(x) is the unknown distribution function of some population (Sec 25.7), and for situations in which the distribution of a population need not be known in order to perform a test (Sec 25.8) SOLUTIONS TO PROBLEM SET 25.4, page 1067 If the hypothesis p ϭ 0.5 is true, X ϭ Number of heads in 4040 trials is approximately normal with ␮ ϭ 2020, ␴ ϭ 1010 (Sec 24.8) Hence P(X Ϲ c) ϭ ⌽([c Ϫ 2020]/͙1010 ෆ) ϭ 0.95, c ϭ 2072 Ͼ 2048, not reject the hypothesis Left-sided test, ␴ 2/n ϭ 4/10 ϭ 0.4 From Table A8 in App we obtain c Ϫ 30.0 P(X ෆ Ϲ c)␮ϭ30.0 ϭ ⌽( ᎏ ) ϭ 0.05 ͙0.4 ෆ Hence c ϭ 30.0 Ϫ 1.645͙0.4 ෆ ϭ 28.96 and we reject the hypothesis im25.qxd 9/21/05 2:06 PM Page 397 Instructor’s Manual 397 (b) Right-sided test We get xෆ Ͻ c ϭ 31.04 and not reject the hypothesis We obtain ␩(28.5) ϭ P(X ෆ Ͼ 28.96)␮ϭ28.5 ϭ Ϫ P(X ෆ Ϲ 28.96)␮ϭ28.5 28.96 Ϫ 28.50 ϭ Ϫ ⌽( ᎏᎏ ) ͙0.4 ෆ ϭ Ϫ ⌽(0.73) ϭ Ϫ 0.7673 ϭ 0.2327 The test is two-sided The variance is unknown, so we have to proceed as in the case of Example From the sample we compute t ϭ ͙10 ෆ (0.811 Ϫ 0.800)/͙0.000 ෆ54 0ෆ ϭ 4.73 Now from Table A9 in App with n Ϫ ϭ degrees of freedom and the condition P(Ϫc Ϲ T Ϲ c) ϭ 0.95 we get c ϭ 2.26 Since t Ͼ c, we reject the hypothesis 10 Hypothesis ␮ ϭ 30 000, alternative ␮ Ͼ 30 000 Using the given data and Table A9, we obtain t ϭ ͙50 ෆ (32 000 Ϫ 30 000)/4000 ϭ 3.54 Ͼ c ϭ 1.68 Hence we reject the hypothesis and assert that the manufacturer’s claim is justified 12 Hypothesis H0: Not better Alternative H1: Better Under H0 the random variable X ϭ Number of cases cured in 200 cases is approximately normal with mean ␮ ϭ np ϭ 200 • 0.7 ϭ 140 and variance ␴ ϭ npq ϭ 42 From Table A8 and ␣ ϭ 5% we get (c Ϫ 140)/͙42 ෆ ϭ 1.645, c ϭ 140 ϩ 1.645͙42 ෆ ϭ 150.7 Since the observed value 148 is not greater than c, we not reject the hypothesis This indicates that the results obtained so far not establish the superiority 14 We test the hypothesis ␴02 ϭ 25 against the alternative that ␴ Ͻ 25 As in Example 4, we now get Y ϭ (n Ϫ 1)S 2/␴02 ϭ 27S2/25 ϭ 1.08S From Table A10 with 27 degrees of freedom and the condition P(Y Ͻ c) ϭ ␣ ϭ 5% we get c ϭ 16.2 As in Example we compute the corresponding c* ϭ 16.2/1.08 ϭ 15.00 Now the observed value of S is s ϭ 3.5 ϭ 12.25 Since this is less than c* and the test is left-sided, we reject the hypothesis and assert that it will be less expensive to replace all batteries simultaneously 16 We test the hypothesis ␴x2 ϭ ␴y2 against the alternative ␴x2 Ͼ ␴y2 We proceed as in Example By computation, v0 ϭ sx2/sy2 ϭ 350/61.9 ϭ 5.65 im25.qxd 9/21/05 2:06 PM Page 398 398 Instructor’s Manual For ␣ ϭ 5% and (5, 6) degrees of freedom Table A11 in App gives 4.39 Since 5.65 is greater, we reject the hypothesis and assert that the variance of the first population is greater than that of the second 18 The test is two-sided As in Example we compute from (12) 12 Ϫ 15 t0 ϭ ͙9 ෆ ᎏ ϭ Ϫ3.18 ͙4ෆ ϩ4 Now Table A9 with ϩ Ϫ ϭ 16 degrees of freedom gives from P(Ϫc Ͻ T Ͻ c) ϭ 0.95 the value c ϭ 2.12 Since Ϫ3.18 Ͻ Ϫ2.12, we reject the hypothesis SECTION 25.5 Quality Control, page 1068 Purpose Quality control is a testing procedure performed every hour (or every half hour, etc.) in an ongoing process of production in order to see whether the process is running properly (“is under control,” is producing items satisfying the specifications) or not (“is out of control”), in which case the process is being halted in order to search for the trouble and remove it These tests may concern the mean, variance, range, etc Main Content Control chart for the mean Control chart for the variance Comment on Content Control charts have also been developed for the range, the number of defectives, the number of defects per unit, for attributes, etc (see the problem set) SOLUTIONS TO PROBLEM SET 25.5, page 1071 The assertion follows from (6c) in Sec 24.8 As the numeric values we obtain ␮ Ϯ 3␴/͙nෆ ϭ Ϯ 0.09/͙6ෆ ϭ Ϯ 0.037, 10 12 14 16 so that LCL ϭ 0.963 and UCL ϭ 1.037 LCL and UCL will correspond to a significance level ␣, which, in general, is not equal to 1% LCL ϭ n␮ Ϫ 2.58␴ ͙nෆ, UCL ϭ n␮0 ϩ 2.58␴ ͙nෆ, as follows from Theorem in Sec 24.6 LCL ϭ 5.00 Ϫ 2.58 • 1.55/͙4 ෆ ϭ 3, UCL ϭ Decrease by a factor ͙2ෆ ϭ 1.41 By a factor 2.58/1.96 ϭ 1.32 (see Table 25.1 in Sec 25.3) Hence the two operations have almost the same effect The sample range tends to increase with increasing n, whereas ␴ remains unchanged Trend of sample means to increase Abrupt change of sample means The random variable Z ϭ Number of defectives in a sample of size n has the variance npq Hence X ෆ ϭ Z/n has the variance ␴ ϭ npq/n2 ϭ 0.05 • 0.95/100 ϭ 0.000 475 This gives UCL ϭ 0.05 ϩ 3␴ ϭ 0.115 From the given values we see that the process is out of control im25.qxd 9/21/05 2:06 PM Page 399 Instructor’s Manual 399 SECTION 25.6 Acceptance Sampling, page 1073 Purpose This is a test for the quality of a produced lot designed to meet the interests of both the producer and the consumer of the lot, as expressed in the terms listed below Main Content, Important Concepts Sampling plan, acceptance number, fraction defective Operating characteristic curve (OC curve) Acceptable quality level (AQL) Rejectable quality level (RQL) Rectification Average outgoing quality limit (AOQL) Comments on Content Basically, acceptance sampling first leads to the hypergeometric distribution, which, however, can be approximated by the simpler Poisson distribution and simple formulas resulting from it, or in other cases by the binomial distribution, which can in turn be approximated by the normal distribution Typical cases are included in the problem set SOLUTIONS TO PROBLEM SET 25.6, page 1076 We expect a decrease of values because of the exponential function in (3), which involves n The probabilities are 0.9098 (down from 0.9825), 0.7358, 0.0404 P(A; ␪ ) ϭ e؊20␪(1 ϩ 20␪ ) from (3) From Fig 539 we find ␣ and ␤ For ␪ ϭ 1.5% we obtain P(A; 0.015) ϭ 96.3%, hence ␣ ϭ 3.7% Also ␤ ϭ P(A; 0.075) ϭ 55.8%, which is very poor AOQ(␪ ) ϭ ␪e؊30␪(1 ϩ 30␪), AOQЈ ϭ 0, ␪max ϭ 0.0539, AOQ(␪max) ϭ 0.0280 The approximation is ␪ 0(1 Ϫ ␪ )2 and is fairly accurate, as the following values show: ␪ Exact (2D) Approximate 0.0 0.2 0.4 0.6 0.8 1.0 1.00 0.63 0.35 0.15 0.03 0.00 1.00 0.64 0.36 0.16 0.04 0.00 10 From the definition of the hypergeometric distribution we now obtain P(A; ␪ ) ϭ ( 20 Ϫ 20␪ 20␪ )( ) Ͳ 20 ( ) (20 Ϫ 20␪ )(19 Ϫ 20␪ )(18 Ϫ 20␪ ) ϭ ᎏᎏᎏᎏ 6840 This gives P(A; 0.1) ϭ 0.72 (instead of 0.81 in Example 1) and P(A; 0.2) ϭ 0.49 (instead of 0.63), a decrease in both cases, as had to be expected 12 Ϫ 0.995 ϭ 0.05, 0.855 ϭ 0.44 im25.qxd 9/21/05 2:06 PM Page 400 400 Instructor’s Manual 14 For ␪ ϭ 0.05 we should get P(A; ␪ ) ϭ 0.98 (Figure 539 illustrates this, for different values.) Since n ϭ 100, we get np ϭ and the variance npq ϭ • 0.95 ϭ 4.75 Using the normal approximation of the binomial distribution, we thus obtain, with c to be determined, c 100 c Ϫ ϩ 0.5 Ϫ Ϫ 0.5 )0.05 x0.95100؊x Ϸ ⌽( ᎏᎏ ) Ϫ ⌽( ᎏᎏ ) ϭ 0.9800 x ͙4.75 ෆ ͙4.75 ෆ ͚( xϭ0 The second ⌽-term equals ⌽(Ϫ5.5/͙4.75 ෆ) ϭ ⌽(Ϫ2.52) ϭ 0.0059 (Table A7) From this, ⌽((c Ϫ 4.5)/͙4.75 ෆ) ϭ 0.9859, so that Table A8 gives ෆ ϭ 9.27 c ϭ 4.50 ϩ 2.19͙4.75 Hence choose or 10 as c 16 By the binomial distribution, P(A; ␪ ) Ϸ n ͚ ( x) ␪ (1 Ϫ ␪)n؊x x xϭ0 ϭ (1 Ϫ ␪)n؊1[1 ϩ (n Ϫ 1)␪], which in those special cases gives (1 Ϫ ␪ )(1 ϩ ␪ ) ϭ Ϫ ␪ (n ϭ 2) (1 Ϫ ␪ ) (1 ϩ 2␪ ) (n ϭ 3) (1 Ϫ ␪ ) (1 ϩ 3␪ ) (n ϭ 4) 18 We have P(A; ␪ ) Ϸ (1 Ϫ ␪ )5, hence AOQ(␪ ) Ϸ ␪ (1 Ϫ ␪ )5 Now [␪ (1 Ϫ ␪ )5]Ј ϭ gives ␪ ϭ 1/6, and from this, AOQL ϭ AOQ(_16) ϭ 0.067 SECTION 25.7 Goodness of Fit ␹ 2-Test, page 1076 Purpose The ␹ 2-test is a test for a whole unknown distribution function, as opposed to the previous tests for unknown parameters in known types of distributions Main Content Chi-square (␹ 2) test Test of normality Comments on Content The present method includes many practical problems, some of which are illustrated in the problem set Recall that the chi-square distribution also occurred in connection with confidence intervals and in our basic section on testing (Sec 25.4) im25.qxd 9/21/05 2:06 PM Page 401 Instructor’s Manual 401 SOLUTIONS TO PROBLEM SET 25.7, page 1079 The hypothesis that the coin is fair is accepted, in contrast with Prob 1, because we now obtain (7 Ϫ 5)2 (3 Ϫ 5)2 ␹02 ϭ ᎏ ϩ ᎏ ϭ 1.6 Ͻ 3.84 5 From the sample and from Table A10 with n Ϫ ϭ degrees of freedom we obtain ␹02 ϭ (92 ϩ 82 ϩ 112 ϩ 42 ϩ 102 ϩ 22)/30 ϭ 12.87 Ͼ c ϭ 11.07 Accordingly, we reject the hypothesis that the die is fair K ϭ classes (dull, sharp) Expected values 10 dull, 390 sharp; degree of freedom; hence 49 49 ␹2ϭ_ ϩ_ ϭ 5.03 Ͼ 3.84 10 390 Reject the claim Here, two things are interesting First, 16 dull knives (an excess of 60% over the expected value!) would not have been sufficient to reject the claim at the 5% level Second, 49/10 contributes much more to ␹02 than 49/390 does; in other applications the situation will often be qualitatively similar The maximum likelihood estimates for the two parameters are xෆ ϭ 59.87, sˆ ϭ 1.504 K Ϫ Ϫ ϭ degrees of freedom From Table A10 we get the critical value 9.21 Ͼ ␹02 ϭ 6.22 Accept the hypothesis that the population from which the sample was taken is normally distributed ␹02 is obtained as follows x 58.5 59.5 60.5 61.5 ϱ x Ϫ xෆ ᎏ s x Ϫ xෆ ⌽( ᎏ ) s Expected Observed Terms in (1) Ϫ0.91 Ϫ0.25 0.42 1.08 0.181 0.402 0.662 0.860 1.000 14.31 17.51 20.50 15.68 11.00 14 17 27 13 0.01 0.01 2.06 3.78 0.36 ␹02 ϭ 6.22 Slightly different results due to rounding are possible 10 Let 50 ϩ b be that number Then 2b 2/50 Ͼ c, b Ͼ 5͙cෆ, 50 ϩ b ϭ 60, 63, 64 12 ␮ˆ ϭ xෆ ϭ 23/50 ϭ 0.46, degree of freedom, since we estimated ␮ We thus obtain (6 Ϫ 3.34)2 (33 Ϫ 31.56)2 (11 Ϫ 14.52)2 ␹02 ϭ ᎏᎏ ϩ ᎏᎏ ϩ ᎏᎏ ϭ 3.04 Ͻ 3.84 31.56 14.52 3.34 Hence we accept the hypothesis 14 Expected 2480/3 ϭ 827 cars per lane; accordingly, _ ␹02 ϭ 827 (832 ϩ 232 ϩ 107 2) ϭ 22.81 Ͼ 5.99 (2 degrees of freedom, ␣ ϭ 5%) Hence we reject the hypothesis (Note that even ␣ ϭ 1% or 0.5% would lead to the same conclusion.) 16 We test the hypothesis H0 that the number of defective rivets is the same for all three machines Since (7 ϩ ϩ 12)/600 ϭ 0.045 im25.qxd 9/21/05 2:06 PM Page 402 402 Instructor’s Manual the expected number of defective rivets in 200 should be Hence ␹02 ϭ [(7 Ϫ 9)2 ϩ (8 Ϫ 9)2 ϩ (12 Ϫ 9)2]/9 ϭ 1.56 Now from Table A10 with degrees of freedom we get c ϭ 5.99 and conclude that the difference is not significant 2 2 18 xෆ ϭ 107, ␹02 ϭ _ 107 (13 ϩ 12 ϩ ϩ ϩ ) ϭ 3.252 Ͻ 9.49 (4 degrees of freedom), so that we accept the hypothesis of equal time-efficiency 20 Team Project n ϭ • 77 ϭ 231 (a) aj ϭ 231/20 ϭ 11.55, K ϭ 20, ␹02 ϭ 24.32 Ͻ c ϭ 30.14 (␣ ϭ 5%, 19 degrees of freedom) Accept the hypothesis (b) ␹02 ϭ 13.10 Ͼ c ϭ 3.84 (␣ ϭ 5%, degree of freedom) Reject the hypothesis (c) ␹02 ϭ 14.7 Ͼ c ϭ 3.84 (␣ ϭ 5%, degree of freedom) Reject the hypothesis SECTION 25.8 Nonparametric Tests, page 1080 Purpose To introduce the student to the ideas of nonparametric tests in terms of two typical examples selected from a wide variety of tests in that field Main Content Median, a test for it Trend, a test for it Comment on Content Both tasks have not yet been considered in the previous sections Another approach to trend follows in the next section SOLUTIONS TO PROBLEM SET 25.8, page 1082 Under the hypothesis that no adjustment is needed, shorter and longer pipes are equally probable We drop the pipes of exact length from the sample Then the probability that under the hypothesis one gets or fewer longer pipes among 18 pipes is, since np ϭ and ␴ ϭ npq ϭ 4.5, 18 18 ͚ ( x ) ( ᎏ2 ) xϭ0 Ϫ Ϫ _12 Ϫ ϩ _12 Ϸ ⌽( ᎏᎏ ) Ϫ ⌽( ᎏᎏ ) ͙4.5 ෆ ͙4.5 ෆ ϭ ⌽(Ϫ2.6) Ϫ ⌽(Ϫ4.5) ϭ 0.0047 and we reject the hypothesis and assert that the process needs adjustment Let X ϭ Number of positive values among values If the hypothesis is true, a value greater than 20°C is as probable as a value less than 20°C, and thus has probability 1/2 Hence, under the hypothesis the probability of getting at most positive value is P ϭ (_1 )9 ϩ • (_1 )9 ϭ 2% 2 Hence we reject the hypothesis and assert that the setting is too low We drop from the sample Let X ϭ Number of positive values Under the hypothesis we get the probability 9 P(X ϭ 9) ϭ ( ) ( ᎏ ) ϭ 0.2% im25.qxd 9/21/05 2:06 PM Page 403 Instructor’s Manual 403 Accordingly, we reject the hypothesis that there is no difference between A and B and assert that the observed difference is significant Under the hypothesis the probability of obtaining at most negative differences (80 Ϫ 85, 90 Ϫ 95, 60 Ϫ 75) is 15 15 15 15 ( ᎏ ) [1 ϩ ( ) ϩ ( ) ϩ ( )] ϭ 1.76% We reject the hypothesis and assert that B is better 10 Hypothesis H0: q25 ϭ Alternative H1, say, q25 Ͼ If H0 is true, a negative value has probability p ϭ 0.25 Reject H0 if fewer than c negative values are observed, where c results from P(X Ϲ c)H0 ϭ ␣ 14 n ϭ values, with transpositions, namely, 101.1 before 100.4 and 100.8, so that from Table A12 we obtain P(T Ϲ 2) ϭ 0.117 and we not reject the hypothesis 16 We order by increasing x Then we have 10 transpositions: 418 Ͼ 301, 352, 395, 375, 388 395 Ͼ 375, 388 465 Ͼ 455 521 Ͼ 455, 490 Hypothesis no trend, alternative positive trend, P(T Ϲ 10) ϭ 1.4% by Table A12 in App Reject the hypothesis 18 n ϭ values, with transpositions, namely, 41.4 before 39.6 43.3 before 39.6, 43.0 45.6 before 44.5 Table A12 gives P(T Ϲ 4) ϭ 0.007 Reject the hypothesis that the amount of fertilizer has no effect and assert that the yield increases with increasing amounts of fertilizer SECTION 25.9 Regression Fitting Straight Lines Correlation, page 1083 Purpose This is a short introduction to regression analysis, restricted to linear regression, and to correlation analysis; the latter is presented without proofs Main Content, Important Concepts Distinction between regression and correlation Gauss’s least squares method Sample regression line, sample regression coefficient Population regression coefficient, a confidence interval for it im25.qxd 9/21/05 2:06 PM Page 404 404 Instructor’s Manual Sample covariance sxy Sample correlation coefficient r Population covariance ␴XY Population correlation coefficient ␳ Independence of X and Y implies ␳ ϭ (“uncorrelatedness”) Two-dimensional normal distribution If (X, Y) is normal, ␳ ϭ implies independence of X and Y Test for ␳ ϭ SOLUTIONS TO PROBLEM SET 25.9, page 1091 ͚x i ϭ 24, ͚yi ϭ 13, n ϭ 4, ͚xi yi ϭ 79, (n Ϫ 1)sx2 ϭ 20, by (9a) Hence we obtain from (11) ⅐ 79 Ϫ 24 ⅐ 13 k1 ϭ ᎏᎏ ϭ 0.05 ⅐ 20 This gives the answer y Ϫ 3.25 ϭ 0.05(x Ϫ 6), hence y ϭ 2.95 ϩ 0.05x n ϭ 5, ͚x i ϭ 85, xෆ ϭ 17, ͚yi ϭ 75, yෆ ϭ 15, ͚xi yi ϭ 1184, hence (n Ϫ 1)sx2 ϭ 4sx2 ϭ 74 Hence (11) gives ⅐ 1184 Ϫ 85 ⅐ 75 k1 ϭ ᎏᎏ ϭ Ϫ1.230 ⅐ 74 The answer is y Ϫ 15 ϭ Ϫ1.230(x Ϫ 17), hence y ϭ 35.91 Ϫ 1.230x n ϭ 9, ͚xi ϭ 183, xෆ ϭ 20.33, ͚yi ϭ 440, yෆ ϭ 48.89, ͚xi yi ϭ 7701, (n Ϫ 1)sx2 ϭ 8sx2 ϭ 944 From (11) we thus obtain ⅐ 7701 Ϫ 183 ⅐ 440 k1 ϭ ᎏᎏᎏ ϭ Ϫ1.32 ⅐ 944 This gives the answer y Ϫ 48.89 ϭ Ϫ1.32(x Ϫ 20.33), hence y ϭ 75.72 Ϫ 1.32x Equation (5) is y Ϫ 1.875 ϭ 0.067(x Ϫ 25); thus y ϭ 0.2 ϩ 0.067x 10 Equation (5) is y Ϫ 10 ϭ 1.98(x Ϫ 5); thus y ϭ 0.1 ϩ 1.98x The spring modulus is 1/1.98 12 c ϭ Ϯ3.18 from Table A9 with n Ϫ ϭ degrees of freedom (corresponding to 2_12% and 97_12%, by the symmetry of the t-distribution) From the sample we compute 4sx2 ϭ 82 000, 4sxy ϭ 354 100, 4sy2 ϭ 530 080 im25.qxd 9/21/05 2:06 PM Page 405 Instructor’s Manual 405 From this and (7) we get 354 100 k1 ϭ ᎏ ϭ 4.31829 82 000 Also q0 ϭ 993 and K ϭ 0.2 The answer is CONF0.95{4.1 Ϲ ␬1 Ϲ 4.5} 14 Multiplying out the square, we get three terms, hence three sums, ͚ (x j Ϫ xෆ)2 ϭ ϭ ͚x ͚x ͚ ͚ ͚ Ϫ 2xෆ xj2 Ϫ ᎏ n j j ϩ nxෆ xi xj ϩ n( ᎏ n ͚ xj) and the last of these three terms cancels half of the second term, giving the result SOLUTIONS TO CHAP 25 REVIEW QUESTIONS AND PROBLEMS, page 1092 ෆ ϭ 0.392 and 22 From Table 25.1 in Sec 25.3 we obtain k ϭ 1.96 • 4/͙400 CONF0.95{52.6 Ϲ ␮ Ϲ 53.4} 24 xෆ ϭ 26.4, k ϭ 2.576 • 2.2/͙5ෆ ϭ 2.534 by Table 25.1, Sec 25.3 This gives the answer CONF0.99{23.86 Ϲ ␮ Ϲ 28.94} 26 n Ϫ ϭ degrees of freedom, F(c1) ϭ 0.025, c1 ϭ 0.22, F(c2) ϭ 0.975, c2 ϭ 9.35 from Table A10 in App 5; hence k1 ϭ 0.7/0.22 ϭ 3.182, k2 ϭ 0.7/9.35 ϭ 0.075 by Table 25.3 in Sec 25.3 The answer is CONF0.95{0.075 Ϲ ␴ Ϲ 3.182} 28 k ϭ 2.26 • 0.157/͙10 ෆ ϭ 0.112; CONF0.95{4.25 Ϲ ␮ Ϲ 4.49} 30 n ϭ 41 by trial and error, because F(c) ϭ 0.975 gives c ϭ 2.02 (Table A9 with 40 d.f.), so that L ϭ 2k ϭ 2sc/͙n ෆ ϭ 0.099 by Table 25.2 32 From Table A10 in App and Table 25.3 in Sec 25.3 we obtain c ϭ _1 (͙253 ෆ Ϫ 2.58)2 ϭ 88.8, c2 ϭ _12(͙253 ෆ ϩ 2.58)2 ϭ 170.9, so that k1 ϭ 244/c1 ϭ 2.76 and k2 ϭ 244/c2 ϭ 1.43 The answer is CONF0.99{1.42 Ϲ ␴ Ϲ 2.75} 34 The test is two-sided We have ␴ 2/n ϭ 0.025, as before Table A8 gives c Ϫ 15.0 P(X ෆ Ͻ c)15.0 ϭ ⌽( ᎏ ) ϭ 0.975, ͙0.025 ෆ c ϭ 15.31 and 15.0 Ϫ 0.31 ϭ 14.69 as the left endpoint of the acceptance region Now xෆ ϭ 14.5 Ͻ 14.70, and we reject the hypothesis 36 The test is right-sided From Table A9 with n Ϫ ϭ 14 degrees of freedom and P(T Ͼ c)␮ ϭ 0.01, thus P(T Ϲ c)␮ ϭ 0.99 im25.qxd 9/21/05 406 2:06 PM Page 406 Instructor’s Manual we get c ϭ 2.62 From the sample we compute 36.2 Ϫ 35.0 t ϭ ͙15 ෆ ᎏᎏ ϭ 4.90 Ͼ c ͙0.9 ෆ and reject the hypothesis 38 The hypothesis is ␮0 ϭ 1000, the alternative ␮ 1000 Because of the symmetry of the t-distribution, from Table A9 with 19 degrees of freedom we get for this two-sided test and the percentages 2.5% and 97.5% the critical values Ϯ2.09 From the sample we compute t ϭ ͙20 ෆ(991 Ϫ 1000)/8 ϭ Ϫ5.03 Ͻ c ϭ Ϫ2.09 and reject the hypothesis 40 2.58 • ͙0.0002 ෆ4ෆ/͙2ෆ ϭ 0.028, LCL ϭ 2.722, UCL ϭ 2.778 42 2.58͙0.0004 ෆ/͙2ෆ ϭ 0.036, UCL ϭ 3.536, LCL ϭ 3.464 44 P(T Ϲ 3) ϭ 6.8% (Table A12) No im25.qxd 9/21/05 2:06 PM Page 407 im25.qxd 9/21/05 2:06 PM Page 408 im25.qxd 9/21/05 2:06 PM Page 409 im25.qxd 9/21/05 2:06 PM Page 410 ... INSTRUCTOR’S MANUAL FOR ADVANCED ENGINEERING MATHEMATICS imfm.qxd 9/15/05 12:06 PM Page ii imfm.qxd 9/15/05 12:06 PM Page iii INSTRUCTOR’S MANUAL FOR ADVANCED ENGINEERING MATHEMATICS NINTH EDITION ERWIN... Comment on “General Solution? ?? and “Singular Solution? ?? Usage of the term “general solution? ?? is not uniform in the literature Some books use the term to mean a solution that includes all solutions, that... actual solution methods in the next sections Equations (1)–(3) are just examples, not for solution, but the student will see that solutions of (1) and (2) can be found by calculus, and a solution