Calculus 2nd edition single variable solutions

www.elsolucionario.net This page intentionally left blank www.elsolucionario.net Student’s Solutions Manual to accompany Jon Rogawski’s Single Variable CALCULUS SECOND EDITION BRIAN BRADIE Christopher Newport University ROGER LIPSETT W H FREEMAN AND COMPANY NEW YORK www.elsolucionario.net © 2012 by W H Freeman and Company ISBN-13: 978-1-4292-4290-5 ISBN-10: 1-4292-4290-6 All rights reserved Printed in the United States of America First Printing W H Freeman and Company, 41 Madison Avenue, New York, NY 10010 Houndmills, Basingstoke RG21 6XS, England www.whfreeman.com www.elsolucionario.net CONTENTS Chapter PRECALCULUS REVIEW 1.1 1.2 1.3 1.4 1.5 Real Numbers, Functions, and Graphs Linear and Quadratic Functions The Basic Classes of Functions Trigonometric Functions Technology: Calculators and Computers Chapter Review Exercises 13 16 23 27 Chapter LIMITS 31 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 31 37 46 49 57 61 66 73 76 82 Limits, Rates of Change, and Tangent Lines Limits: A Numerical and Graphical Approach Basic Limit Laws Limits and Continuity Evaluating Limits Algebraically Trigonometric Limits Limits at Infinity Intermediate Value Theorem The Formal Definition of a Limit Chapter Review Exercises Chapter DIFFERENTIATION 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 91 Definition of the Derivative The Derivative as a Function Product and Quotient Rules Rates of Change Higher Derivatives Trigonometric Functions The Chain Rule Implicit Differentiation Related Rates Chapter Review Exercises 91 101 112 119 126 132 138 147 157 165 5.4 5.5 5.6 Chapter APPLICATIONS OF THE INTEGRAL 6.1 6.2 6.3 6.4 6.5 Linear Approximation and Applications Extreme Values The Mean Value Theorem and Monotonicity The Shape of a Graph Graph Sketching and Asymptotes Applied Optimization Newton’s Method Antiderivatives Chapter Review Exercises 174 181 191 198 206 220 236 242 250 Chapter THE INTEGRAL 5.1 5.2 5.3 260 Approximating and Computing Area The Definite Integral The Fundamental Theorem of Calculus, Part I 260 274 284 290 296 300 307 317 Area Between Two Curves 317 Setting Up Integrals: Volume, Density, Average Value 328 Volumes of Revolution 336 The Method of Cylindrical Shells 346 Work and Energy 355 Chapter Review Exercises 362 Chapter EXPONENTIAL FUNCTIONS 370 Derivative of f (x) = bx and the Number e Inverse Functions Logarithms and Their Derivatives Exponential Growth and Decay Compound Interest and Present Value Models Involving y = k ( y − b) L’Hôpital’s Rule Inverse Trigonometric Functions Hyperbolic Functions Chapter Review Exercises 370 378 383 393 398 401 407 415 424 431 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 Chapter TECHNIQUES OF INTEGRATION 8.1 8.2 8.3 8.4 Chapter APPLICATIONS OF THE DERIVATIVE 174 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 The Fundamental Theorem of Calculus, Part II Net Change as the Integral of a Rate Substitution Method Chapter Review Exercises 8.5 8.6 8.7 8.8 Integration by Parts 446 Trigonometric Integrals 457 Trigonometric Substitution 467 Integrals Involving Hyperbolic and Inverse Hyperbolic Functions 481 The Method of Partial Fractions 485 Improper Integrals 503 Probability and Integration 520 Numerical Integration 525 Chapter Review Exercises 537 Chapter FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS 9.1 9.2 9.3 9.4 446 Arc Length and Surface Area Fluid Pressure and Force Center of Mass Taylor Polynomials Chapter Review Exercises www.elsolucionario.net 555 555 564 569 577 593 iii iv CALCULUS CON T EN TS Chapter 10 INTRODUCTION TO DIFFERENTIAL EQUATIONS 601 10.1 10.2 10.3 10.4 Solving Differential Equations Graphical and Numerical Methods The Logistic Equation First-Order Linear Equations Chapter Review Exercises Chapter 11 INFINITE SERIES 11.1 11.2 11.3 11.4 Sequences Summing an Infinite Series Convergence of Series with Positive Terms Absolute and Conditional Convergence 601 614 621 626 637 646 646 658 669 683 11.5 The Ratio and Root Tests 11.6 Power Series 11.7 Taylor Series Chapter Review Exercises 690 697 710 727 Chapter 12 PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS 742 12.1 12.2 12.3 12.4 12.5 Parametric Equations Arc Length and Speed Polar Coordinates Area and Arc Length in Polar Coordinates Conic Sections Chapter Review Exercises www.elsolucionario.net 742 759 766 780 789 801 PRECALCULUS REVIEW 1.1 Real Numbers, Functions, and Graphs Preliminary Questions Give an example of numbers a and b such that a < b and |a| > |b| solution Take a = −3 and b = Then a < b but |a| = > = |b| Which numbers satisfy |a| = a? Which satisfy |a| = −a? What about |−a| = a? solution The numbers a ≥ satisfy |a| = a and | − a| = a The numbers a ≤ satisfy |a| = −a Give an example of numbers a and b such that |a + b| < |a| + |b| solution Take a = −3 and b = Then |a + b| = | − + 1| = | − 2| = 2, but |a| + |b| = | − 3| + |1| = + = Thus, |a + b| < |a| + |b| What are the coordinates of the point lying at the intersection of the lines x = and y = −4? solution The point (9, −4) lies at the intersection of the lines x = and y = −4 In which quadrant the following points lie? (a) (1, 4) (b) (−3, 2) (c) (4, −3) (d) (−4, −1) solution (a) Because both the x- and y-coordinates of the point (1, 4) are positive, the point (1, 4) lies in the first quadrant (b) Because the x-coordinate of the point (−3, 2) is negative but the y-coordinate is positive, the point (−3, 2) lies in the second quadrant (c) Because the x-coordinate of the point (4, −3) is positive but the y-coordinate is negative, the point (4, −3) lies in the fourth quadrant (d) Because both the x- and y-coordinates of the point (−4, −1) are negative, the point (−4, −1) lies in the third quadrant What is the radius of the circle with equation (x − 9)2 + (y − 9)2 = 9? solution The circle with equation (x − 9)2 + (y − 9)2 = has radius The equation f (x) = has a solution if (choose one): (a) belongs to the domain of f (b) belongs to the range of f solution The correct response is (b): the equation f (x) = has a solution if belongs to the range of f What kind of symmetry does the graph have if f (−x) = −f (x)? solution If f (−x) = −f (x), then the graph of f is symmetric with respect to the origin Exercises Use a calculator to find a rational number r such that |r − π | < 10−4 solution r must satisfy π − 10−4 < r < π + 10−4 , or 9.869504 < r < 9.869705 r = 9.8696 = 12337 1250 would be one such number In Exercises 3–8, expressare thetrue interval of an involving absolute value Which of (a)–(f) for a in =terms −3 and b =inequality 2? (a) a2]< b [−2, solution (d) 3a 1 (f) < a b (0, 4) (−4, 4) solution The midpoint of the interval is c = (0 + 4)/2 = 2, and the radius is r = (4 − 0)/2 = 2; therefore, (0, 4) can be expressed as |x − 2| < [1, 5] [−4, 0] solution The midpoint of the interval is c = (1 + 5)/2 = 3, and the radius is r = (5 − 1)/2 = 2; therefore, the interval [1, 5] can be expressed as |x − 3| ≤ (−2, 8) June 7, 2011 www.elsolucionario.net LTSV SSM Second Pass CHAPTER PRECALCULUS REVIEW In Exercises 9–12, write the inequality in the form a < x < b |x| < solution −8 < x < 11 |2x + 1| < |x − 12| < solution −5 < 2x + < so −6 < 2x < and −3 < x < In Exercises |3x −13–18, 4| < 2express the set of numbers x satisfying the given condition as an interval 13 |x| < solution (−4, 4) 15 |x − 4| < |x| ≤ solution The expression |x − 4| < is equivalent to −2 < x − < Therefore, < x < 6, which represents the interval (2, 6) 17 |4x − 1| ≤ |x + 7| < solution The expression |4x − 1| ≤ is equivalent to −8 ≤ 4x − ≤ or −7 ≤ 4x ≤ Therefore, − 74 ≤ x ≤ 94 , which represents the interval [− 74 , 94 ] In Exercises |3x +19–22, 5| < 1describe the set as a union of finite or infinite intervals 19 {x : |x − 4| > 2} solution x − > or x − < −2 ⇒ x > or x < ⇒ (−∞, 2) ∪ (6, ∞) 21 {x : |x − 1| > 2} {x : |2x + 4| > 3} √ √ solution x − > or x − < −2 ⇒ x > or x < −1 (this will never happen) ⇒ x > or x < − ⇒ √ √ (−∞, − 3) ∪ ( 3, ∞) 23 Match (a)–(f) with (i)–(vi) {x : |x + 2x| > 2} (a) a > (b) |a − 5| < (c) a − (i) (ii) (iii) (iv) (v) (vi) (f) ≤ a ≤ a lies to the right of a lies between and The distance from a to is less than 13 The distance from a to is at most a is less than units from 13 a lies either to the left of −5 or to the right of solution (a) On the number line, numbers greater than appear to the right; hence, a > is equivalent to the numbers to the right of 3: (i) (b) |a − 5| measures the distance from a to 5; hence, |a − 5| < 13 is satisfied by those numbers less than 13 of a unit from 5: (iii) (c) |a − 13 | measures the distance from a to 13 ; hence, |a − 13 | < is satisfied by those numbers less than units from : (v) (d) The inequality |a| > is equivalent to a > or a < −5; that is, either a lies to the right of or to the left of −5: (vi) (e) The interval described by the inequality |a − 4| < has a center at and a radius of 3; that is, the interval consists of those numbers between and 7: (ii) (f) The interval described by the inequality < x < has a center at and a radius of 2; that is, the interval consists of those numbers less than units from 3: (iv) 25 Describe {x : x + 2x < 3} as an interval Hint: Plot y = x + 2x − x < as an interval Describe x : solution The inequality x + 1x + 2x < is equivalent to x + 2x − < In the figure below, we see that the graph of y = x + 2x − falls below the x-axis for −3 < x < Thus, the set {x : x + 2x < 3} corresponds to the interval −3 < x < www.elsolucionario.net June 7, 2011 LTSV SSM Second Pass S E C T I O N 1.1 Real Numbers, Functions, and Graphs y y = x2 + 2x − −4 −3 −2 10 x −2 27 Show that if a > b, then b−1 > a −1 , provided that a and b have the same sign What happens if a > and b < 0? Describe the set of real numbers satisfying |x − 3| = |x − 2| + as a half-infinite interval solution Case 1a: If a and b are both positive, then a > b ⇒ > ab ⇒ b1 > a1 Case 1b: If a and b are both negative, then a > b ⇒ < ab (since a is negative) ⇒ b1 > a1 (again, since b is negative) Case 2: If a > and b < 0, then a1 > and b1 < so b1 < a1 (See Exercise 2f for an example of this) < 12|x , then 29 Show that xif satisfy |a − 5|both < 12 |xand Which − |b 3| − < 8| and − 5||(a perihelion Show that the vertices of to thethe hyperbola = have x-coordinates and + e cos θ e+1 e−1 distance from the sun) solution We define an xy-coordinate system so that the orbit is an ellipse in standard position, as shown in the figure y A'(−a, 0) Sun F1(c, 0) A(a, 0) x The aphelion is the length of A F1 , that is a + c By the given data, we have c ⇒ c = 0.25a a a − c = 2.7 ⇒ c = a − 2.7 0.25 = e = Equating the two expressions for c we get 0.25a = a − 2.7 0.75a = 2.7 ⇒ a = 2.7 = 3.6, c = 3.6 − 2.7 = 0.9 0.75 The aphelion is thus A F0 = a + c = 3.6 + 0.9 = 4.5 billion miles www.elsolucionario.net Kepler’s Third Law states that the ratio T /a 3/2 is equal to a constant C for all planetary orbits around the sun, where T is the period (time for a complete orbit) and a is the semimajor axis (a) Compute units of days and kilometers, given that the semimajor axis of the earth’s orbit is 150 × 106 km June C 15,in2011 (b) Compute the period of Saturn’s orbit, given that its semimajor axis is approximately 1.43 × 109 km LTSV SSM Second Pass S E C T I O N 12.5 Conic Sections 797 Further Insights and Challenges 65 Verify Theorem solution Let F1 = (c, 0) and F2 = (−c, 0) and let P (x, y) be an arbitrary point on the hyperbola Then for some constant a, P F1 − P F2 = ±2a y F2 = (−c, 0) F1 = (c, 0) x P = (x, y) Using the distance formula we write this as (x − c)2 + y − (x + c)2 + y = ±2a Moving the second term to the right and squaring both sides gives (x − c)2 + y = (x + c)2 + y ± 2a (x − c)2 + y = (x + c)2 + y ± 4a (x + c)2 + y + 4a (x − c)2 − (x + c)2 − 4a = ±4a (x + c)2 + y xc + a = ±a (x + c)2 + y We square and simplify to obtain x c2 + 2xca + a = a (x + c)2 + y = a x + 2a xc + a c2 + a y c2 − a x − a y = a c2 − a x2 y2 − =1 a2 c2 − a For b = c2 − a (or c = a + b2 ) we get x2 y2 x y − =1⇒ − = a b a b 67 Verify that if e > 1, then Eq (11) defines a hyperbola of eccentricity e, with its focus at the origin and directrix at −2 x = d Verify Theorem in the case < e < Hint: Repeat the proof of Theorem 5, but set c = d/(e − 1) solution The points P = (r, θ) on the hyperbola satisfy P F = eP D, e > Referring to the figure we see that P F = r, P D = d − r cos θ Hence r = e(d − r cos θ) r = ed − er cos θ r(1 + e cos θ) = ed ⇒ r = ed + e cos θ www.elsolucionario.net June 15, 2011 LTSV SSM Second Pass (1) 798 C H A P T E R 12 PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS y D rcos q d − r cos q D r P q x F x=d Remark: Equality (1) holds also for θ > π2 For example, in the following figure, we have P D = d + r cos (π − θ) = d − r cos θ y D r cos (p − q ) d P r q x x=d Reflective Property of the Ellipse In Exercises 68–70, we prove that the focal radii at a point on an ellipse make equal angles with the tangent line L Let P = (x0 , y0 ) be a point on the ellipse in Figure 25 with foci F1 = (−c, 0) and F2 = (c, 0), and eccentricity e = c/a L R1 = ( 1, 1) y P = (x0, y0) R2 = ( 2, 2) F1 = (−c, 0) F2 = (c, 0) FIGURE 25 The ellipse x x y + = a b line 69 Points R1 and R2 in Figure 25 are defined so that F1 R1 and F2 R2 are perpendicular y x to the tangent Show that of the tangent line at P is Ax + By = 1, where A = 02 and B = 02 (a) Show, with A the andequation B as in Exercise 68, that a b α1 + c α −c A = = β1 β2 B (b) Use (a) and the distance formula to show that F1 R1 β = F2 R2 β2 (c) Use (a) and the equation of the tangent line in Exercise 68 to show that β1 = B(1 + Ac) , A2 + B β2 = B(1 − Ac) A2 + B solution (a) Since R1 = (α1 , β1 ) and R2 = (α2 , β2 ) lie on the tangent line at P , that is on the line Ax + By = 1, we have Aα1 + Bβ1 = and Aα2 + Bβ2 = A Similarly, the slope of The slope of the line R1 F1 is α1β+c and it is perpendicular to the tangent line having slope − B β2 the line R2 F2 is α2 −c and it is also perpendicular to the tangent line Hence, A α1 + c = β1 B and A α2 − c = β2 B www.elsolucionario.net June 15, 2011 LTSV SSM Second Pass S E C T I O N 12.5 Conic Sections 799 (b) Using the distance formula, we have R1 F1 = (α1 + c)2 + β12 Thus, α1 + c +1 β1 R1 F1 = β12 (1) A Substituting in (1) gives By part (a), α1β+c = B R1 F1 = β12 A2 +1 B2 (2) Likewise, R2 F2 = (α2 − c)2 + β2 = β2 α2 − c +1 β2 (3) A , substituting in (3) gives but since α2β−c = B 2 R2 F2 = β22 A2 +1 B2 (4) Dividing, we find that β12 = β22 R2 F2 R1 F1 so R1 F1 R2 F2 β = 1, β2 as desired (c) In part (a) we showed that ⎧ ⎪ ⎨ Aα1 + Bβ1 = β1 B ⎪ = ⎩ α1 + c A Eliminating α1 and solving for β1 gives β1 = B(1 + Ac) A2 + B (5) Similarly, we have ⎧ ⎪ ⎨ Aα2 + Bβ2 = β2 B ⎪ = ⎩ α2 − c A Eliminating α2 and solving for β2 yields β2 = B (1 − Ac) A2 + B (6) Here is another proof of the Reflective Property 71 (a) Prove that P F1 = a + x0 e and P F2 = a − x0 e Hint: Show that P F1 − P F2 = 4x0 c Then use the defining (a) Figure 25Psuggests L is the unique line that intersects the ellipse only in the point P Assuming this, prove that property F1 + P Fthat = 2a and the relation e = c/a QF1 + QF2 > P F1F+RP F2 for all points Q on the tangent line other than P F2 R (b) Verify that 1of 1Least = Distance (Example in Section 4.7) to prove that θ = θ (b) Use the Principle PF PF (c) Show that sin θ1 = sin θ2 Conclude that θ1 = θ2 solution (a) Consider a point Q = P on the line L (see figure) Since L intersects the ellipse in only one point, the remainder of the line lies outside the ellipse, so that QR does not have zero length, and F2 QR is a triangle Thus QF1 + QF2 = QR + RF1 + QF2 = RF1 + (QR + QF2 ) > RF1 + RF2 since the sum of lengths of two sides of a triangle exceeds the length of the third side But since point R lies on the ellipse, RF2 + RF2 = P F1 + P F2 , and we are done www.elsolucionario.net June 15, 2011 LTSV SSM Second Pass 800 C H A P T E R 12 PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS y Q R P x F2 F1 (b) Consider a beam of light traveling from F1 to F2 by reflection off of the line L By the principle of least distance, the light takes the shortest path, which by part (a) is the path through P By Example in Section 4.6, this shortest path has the property that the angle of incidence (θ1 ) is equal to the angle of reflection (θ2 ) 73 Show thatthat y =the x 2length /4c is QR the equation parabola with directrix y =P−c, Show in Figureof26a is independent of the point focus (0, c), and the vertex at the origin, as stated in Theorem solution The points P = (x, y) on the parabola are equidistant from F = (0, c) and the line y = −c y P(x, y) F(0, c) x y = −c That is, by the distance formula, we have PF = PD x + (y − c)2 = |y + c| Squaring and simplifying yields x + (y − c)2 = (y + c)2 x + y − 2yc + c2 = y + 2yc + c2 x − 2yc = 2yc x = 4yc ⇒ y = x2 4c Thus, we showed that the points that are equidistant from the focus F = (0, c) and the directrix y = −c satisfy the equation y = x4c 75 Derive (13) in andstandard (14) in position: the text as follows Write the coordinates of P with respect to the rotated axes Consider twoEqs ellipses in Figure 21 in polar form x = r cos α, y = r sin α Explain why P has polar coordinates (r, α + θ ) with respect to the y formulas standard x and y-axes and derive Eqs (13) and (14) usingx the2 addition for cosine and sine + =1 E1 : a1 b1 are (r, α), then the line from the origin to P solution If the polar coordinates of P with respect tox the2 rotated y axes E2 : x-axis (the + x -axis) = the x -axis makes an angle of θ with the has length r and makes an angle of α with the rotated Since a2 b x-axis, it follows that the line from the origin to P makes an angle of2α + θ with the x-axis, so that the polar coordinates of PWe with thesimilar standard are (r, α + θ) Writeexists (x , ya )factor for ther rectangular coordinates to to Eaxes scaling if there > such that for all (x,of y)Ponwith E1 ,respect the point sayrespect that Eto is under the rotated axes for the P with respect to the standard (rx, ry) liesand on (x, E2 y) Show thatrectangular E1 and E2coordinates are similarofunder scaling if and only if theyaxes haveThen the same eccentricity Show that any two circles are similar under scaling x = r cos(α + θ) = (r cos α) cos θ − (r sin α) sin θ = x cos θ − y sin θ y = r sin(α + θ) = r sin α cos θ + r cos α sin θ = (r cos α) sin θ + (r sin α) cos θ = x sin θ + y cos θ www.elsolucionario.net LTSV SSM Second Pass If we rewrite the general equation of degree (Eq 12) in terms of variables x and y that are related to x and y by 15, 2011 Eqs (13)June and (14), we obtain a new equation of degree in x and y of the same form but with different coefficients: Chapter Review Exercises 801 CHAPTER REVIEW EXERCISES Which of the following curves pass through the point (1, 4)? (b) c(t) = (t , t − 3) (a) c(t) = (t , t + 3) (c) c(t) = (t , − t) (d) c(t) = (t − 3, t ) solution To check whether it passes through the point (1, 4), we solve the equations c(t) = (1, 4) for the given curves (a) Comparing the second coordinate of the curve and the point yields: t +3=4 t =1 We substitute t = in the first coordinate, to obtain t = 12 = Hence the curve passes through (1, 4) (b) Comparing the second coordinate of the curve and the point yields: t −3=4 t =7 We substitute t = in the first coordinate to obtain t = 72 = 49 = Hence the curve does not pass through (1, 4) (c) Comparing the second coordinate of the curve and the point yields 3−t =4 t = −1 We substitute t = −1 in the first coordinate, to obtain t = (−1)2 = Hence the curve passes through (1, 4) (d) Comparing the first coordinate of the curve and the point yields t −3=1 t =4 We substitute t = in the second coordinate, to obtain: t = 42 = 16 = Hence the curve does not pass through (1, 4) Find parametric equations for the circle of radius with center (1, 1) Use the equations to find the points of intersection Find parametric equations for the line through P = (2, 5) perpendicular to the line y = 4x − of the circle with the x- and y-axes solution Using the standard technique for parametric equations of curves, we obtain c(t) = (1 + cos t, + sin t) We compare the x coordinate of c(t) to 0: + cos t = 2π t =± cos t = − Substituting in the y coordinate yields + sin ± 2π √ =1±2 √ =1± www.elsolucionario.net June 15, 2011 LTSV SSM Second Pass 802 C H A P T E R 12 PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS Hence, the intersection points with the y-axis are (0, ± √ 3) We compare the y coordinate of c(t) to 0: + sin t = π t =− sin t = − or π Substituting in the x coordinates yields + cos − + cos π π √ =1+2 √ =1+ = − cos Hence, the intersection points with the x-axis are (1 ± π =1−2 √ √ =1− √ 3, 0) Find a parametrization c(θ) of the unit circle such that c(0) = (−1, 0) Find a parametrization c(t) of the line y = − 2x such that c(0) = (2, 1) solution The unit circle has the parametrization c(t) = (cos t, sin t) This parametrization does not satisfy c(0) = (−1, 0) We replace the parameter t by a parameter θ so that t = θ + α, to obtain another parametrization for the circle: c∗ (θ) = (cos(θ + α), sin(θ + α)) (1) We need that c∗ (0) = (1, 0), that is, c∗ (0) = (cos α, sin α) = (−1, 0) Hence cos α = −1 sin α = ⇒ α=π Substituting in (1) we obtain the following parametrization: c∗ (θ) = (cos(θ + π), sin(θ + π)) Find a path c(t) that traces the line y = 2x + from (1, 3) to (3, 7) for ≤ t ≤ Find a path c(t) that traces the parabolic arc y = x from (0, 0) to (3, 9) for ≤ t ≤ solution Solution 1: By one of the examples in section 12.1, the line through P = (1, 3) with slope has the parametrization c(t) = (1 + t, + 2t) But this parametrization does not satisfy c(1) = (3, 7) We replace the parameter t by a parameter s so that t = αs + β We get c∗ (s) = (1 + αs + β, + 2(αs + β)) = (αs + β + 1, 2αs + 2β + 3) We need that c∗ (0) = (1, 3) and c∗ (1) = (3, 7) Hence, c∗ (0) = (1 + β, + 2β) = (1, 3) c∗ (1) = (α + β + 1, 2α + 2β + 3) = (3, 7) We obtain the equations 1+β =1 + 2β = α+β +1=3 ⇒ β = 0, α = 2α + 2β + = Substituting in (1) gives c∗ (s) = (2s + 1, 4s + 3) www.elsolucionario.net June 15, 2011 LTSV SSM Second Pass Chapter Review Exercises 803 Solution 2: The segment from (1, 3) to (3, 7) has the following vector parametrization: (1 − t) 1, + t 3, = − t + 3t, 3(1 − t) + 7t = + 2t, + 4t The parametrization is thus c(t) = (1 + 2t, + 4t) In Exercises 9–12, express the=parametric infor the0form (x) draw arrows specifying the direction of motion Sketch the graph c(t) (1 + cos t,curve sin 2t) ≤ t y≤ = 2πf and c(t) = (4t − 3, 10 − t) solution We use the given equation to express t in terms of x x = 4t − 4t = x + t= x+3 Substituting in the equation of y yields y = 10 − t = 10 − x+3 x 37 =− + 4 That is, y=− x 37 + 4 11 c(t)c(t) = =3 (t −3 +, t1, + t2 − t t 4) solution We use the given equation to express t in terms of x: x =3− t =3−x t t= 3−x Substituting in the equation of y yields y= 3−x = + + 3−x 2/(3 − x) (3 − x)3 In Exercises 13–16, x = tan t, ycalculate = sec t dy/dx at the point indicated 13 c(t) = (t + t, t − 1), t = solution The parametric equations are x = t + t and y = t − We use the theorem on the slope of the tangent dy line to find dx : dy dy 2t dt = = dx dx 3t + dt We now substitute t = to obtain 2·3 dy = = dx t=3 14 · 32 + 15 c(t) = (et − 1, sin t), t = 20 c(θ ) = (tan2 θ, cos θ), θ = π4 dy solution We use the theorem for the slope of the tangent line to find dx : dy dy (sin t) cos t dt = dx = t = t dx (e − 1) e dt We now substitute t = 20: dy cos 20 = 20 dx t=0 e www.elsolucionario.net c(t) = (ln t, 3t − t), P = (0, 2) June 15, 2011 LTSV SSM Second Pass 804 C H A P T E R 12 PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS 17 Find the point on the cycloid c(t) = (t − sin t, − cos t) where the tangent line has slope 12 solution Since x = t − sin t and y = − cos t, the theorem on the slope of the tangent line gives dy dy sin t dt = = dx dx − cos t dt dy The points where the tangent line has slope 12 are those where dx = 12 We solve for t: dy = dx sin t = − cos t (1) sin t = − cos t We let u = sin t Then cos t = ± − sin2 t = ± − u2 Hence 2u = ± − u2 We transfer sides and square to obtain ± − u2 = 2u − 1 − u2 = 4u2 − 4u + 5u2 − 4u = u(5u − 4) = u = 0, u = We find t by the relation u = sin t: u = 0: sin t = ⇒ t = 0, t = π u= 4 : sin t = ⇒ t ≈ 0.93, t ≈ 2.21 5 These correspond to the points (0, 1), (π, 2), (0.13, 0.40), and (1.41, 1.60), respectively, for < t < 2π 19 Find the equation of the Bézier curve with control points Find the points on (t + sin t, t − sin t) where the tangent is vertical or horizontal P0 = (−1, −1), P1 = (−1, 1), P2 = (1, 1), P3 (1, −1) solution We substitute the given points in the appropriate formulas in the text to find the parametric equations of the Bézier curve We obtain x(t) = −(1 − t)3 − 3t (1 − t)2 + t (1 − t) + t = −(1 − 3t + 3t − t ) − (3t − 6t + 3t ) + (t − t ) + t = (−2t + 4t − 1) y(t) = −(1 − t)3 + 3t (1 − t)2 + t (1 − t) − t = −(1 − 3t + 3t − t ) + (3t − 6t + 3t ) + (t − t ) − t = (2t − 8t + 6t − 1) 21 Find the speed (as a function of t) of a particle whose position at time t seconds is c(t) = (sin t + t, cos t + t) What Find themaximal speed at speed? t = π4 of a particle whose position at time t seconds is c(t) = (sin 4t, cos 3t) is the particle’s solution We use the parametric definition to find the speed We obtain ds = dt = ((sin t + t) )2 + ((cos t + t) )2 = (cos t + 1)2 + (1 − sin t)2 cos2 t + cos t + + − sin t + sin2 t = + 2(cos t − sin t) We now differentiate the speed function to find its maximum: d 2s = dt + 2(cos t − sin t) = √ − sin t − cos t + 2(cos t − sin t) www.elsolucionario.net June 15, 2011 LTSV SSM Second Pass Chapter Review Exercises 805 We equate the derivative to zero, to obtain the maximum point: d 2s =0 dt − sin t − cos t =0 √ + 2(cos t − sin t) − sin t − cos t = − sin t = cos t sin(−t) = cos(−t) π −t = + πk π t = − + πk Substituting t in the function of speed we obtain the value of the maximal speed: + cos − π π − sin − 4 √ 2 − − 2 √ = 3+2 = √ 3+2 −t cos t, e−t sin t) In Exercises 24,of let(3e c(t) =3,(e4e t − t + 7) for ≤ t ≤ Find 23 theand length 23 Show that c(t) for ≤ t < ∞ has finite length and calculate its value solution We use the formula for arc length, to obtain: s= = = = = ∞ ((e−t cos t) )2 + ((e−t sin t) )2 dt ∞ (−e−t cos t − e−t sin t)2 + (−e−t sin t + e−t cos t)2 dt ∞ e−2t (cos t + sin t)2 + e−2t (cos t − sin t)2 dt ∞ ∞ e−t cos2 t + sin t cos t + sin2 t + cos2 t − sin t cos t + sin2 tdt ∞ √ √ √ e−t 2dt = 2(−e−t ) =− √ √ = − 2(0 − 1) = lim e−t − e0 t→∞ Plotfirst c(t)positive = (sinvalue 2t, of cost t)such for 0that ≤the t ≤ π Express of theand curve as a definite integral, 25 Find the tangent line to the c(t0 length ) is vertical, calculate the speed at t = tand 0 approximate it using a computer algebra system solution We use a CAS to plot the curve The resulting graph is shown here y −2 x −1 −1 −2 Plot of the curve (sin 2t, cos t) To calculate the arc length we use the formula for the arc length to obtain s= π (2 cos 2t)2 + (−2 sin t)2 dt = π cos2 2t + sin2 t dt We use a CAS to obtain s = 6.0972 www.elsolucionario.net Convert the points (x, y) = (1, −3), (3, −1) from rectangular to polar coordinates June 15, 2011 LTSV SSM Second Pass 806 C H A P T E R 12 PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS 27 Convert the points (r, θ) = 1, π6 , 3, 5π from polar to rectangular coordinates solution We convert the points from polar coordinates to cartesian coordinates For the first point we have √ π x = r cos θ = · cos = π y = r sin θ = · sin = For the second point we have √ 5π x = r cos θ = cos =− √ 5π y = r sin θ = sin =− 2 cos θ 29 Write r =(x + y)2 = xy as an equation in rectangular coordinates Write cos θ − sin θ + as an equation in polar coordinates solution We use the formula for converting from polar coordinates to cartesian coordinates to substitute x and y for r and θ: cos θ r= cos θ − sin θ 31 x2 + y2 = 2r cos θ r cos θ − r sin θ x2 + y2 = 2x x−y Convert the equation Show that r = is the polar equation of a line cos θ − sin θ 9(x + y ) = (x + y − 2y)2 to polar coordinates, and plot it with a graphing utility solution We use the formula for converting from cartesian coordinates to polar coordinates to substitute r and θ for x and y: 9(x + y ) = (x + y − 2y)2 9r = (r − 2r sin θ)2 3r = r − 2r sin θ = r − sin θ r = + sin θ The plot of r = + sin θ is shown here: r = + 2sin −1 −2 −4 −3 −2 −1 Plot of r = + sin θ 33 Calculate the area of one petal of r = sin 4θ (see Figure 1) Calculate the area of the circle r = sin θ bounded by the rays θ = π3 and θ = 2π y y y x n = (4 petals) x n = (8 petals) x n = (12 petals) FIGURE Plot of r = sin(nθ) www.elsolucionario.net June 15, 2011 LTSV SSM Second Pass Chapter Review Exercises 807 solution We use a CAS to generate the plot, as shown here 0.8 0.6 0.4 0.2 −0.2 −0.4 −0.6 −0.8 −1 −1 −0.8 −0.4 r = 4sin 0.4 0.8 Plot of r = sin 4θ We can see that one leaf lies between the rays θ = and θ = obtain A= = θ We now use the formula for area in polar coordinates to π/4 sin 8θ π/4 π/4 θ− sin 4θ dθ = (1 − cos 8θ) dθ = 4 π π − (sin 2π − sin 0) = 16 32 16 sin θ 35 Calculate the totalr area enclosed by the r = cos 2) (Figure 1) Compute the total area of the The equation = sin(nθ), where n ≥curve is even, is a θe “rose” (Figure of 2n petals flower, and show that it does not depend on n y x −1 FIGURE Graph of r = cos θesin θ solution to obtain: Note that this is defined only for θ between −π/2 and π/2 We use the formula for area in polar coordinates A= π/2 π/2 r dθ = cos θesin θ dθ −π/2 −π/2 We evaluate the integral by making the substitution x = sin θ dx = cos θ dθ : A= π/2 1 cos θesin θ dθ = ex = e − e−1 −π/2 −1 37 Find the area enclosed by the cardioid r = a(1 + cos θ), where a > Find the shaded area in Figure solution The graph of r = a (1 + cos θ) in the rθ-plane for ≤ θ ≤ 2π and the cardioid in the xy-plane are shown in the following figures: y r θ= 2a π ,r=a θ = π, r = a θ= π π 3π r = a (1 + cos θ) 3π ,r=a θ=0 r = 2a x 2π The cardioid r = a (1 + cos θ), a > As θ varies from to π the radius r decreases from 2a to 0, and this gives the upper part of the cardioid www.elsolucionario.net June 15, 2011 LTSV SSM Second Pass 808 C H A P T E R 12 PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS The lower part is traced as θ varies from π to 2π and consequently r increases from back to 2a We compute the area enclosed by the upper part of the cardioid and the x-axis, using the following integral (we use the identity cos2 θ = 12 + 12 cos 2θ): π π a2 π r dθ = a (1 + cos θ)2 dθ = + cos θ + cos2 θ dθ 2 = 1 a2 π + cos θ + + cos 2θ 2 = a2 π 3θ + sin θ + sin 2θ = a2 dθ = a2 π + cos θ + cos 2θ 2 dθ 3π 3π a + sin π + sin 2π − = 4 Using symmetry, the total area A enclosed by the cardioid is A=2· 3πa 3πa = 0.5θ sin θ for ≤ θ ≤ 2π Use a computer algebra system to approximate Figure shows of thethe graph of with r = epolar 39 Calculate the5 length curve equation r = θ in Figure the difference in length between the outer and inner loops y 10 x −6 FIGURE solution We note that the inner loop is the curve for θ ∈ [0, π ], and the outer loop is the curve for θ ∈ [π, 2π ] We express the length of these loops using the formula for the arc length The length of the inner loop is π s1 = (e0.5θ sin θ)2 + ((e0.5θ sin θ) )2 dθ = π eθ sin2 θ + e0.5θ sin θ + e0.5θ cos θ 2 dθ and the length of the outer loop is s2 = 2π eθ sin2 θ + π e0.5θ sin θ + e0.5θ cos θ 2 dθ We now use the CAS to calculate the arc length of each of the loops We obtain that the length of the inner loop is 7.5087 and the length of the outer loop is 36.121, hence the outer one is 4.81 times longer than the inner one In Exercises 41–44, the and conic section thethe vertices and foci r= f2 (θ)Find define same curves in polar coordinates if f1 (θ ) = −f2 (θ + π ) Use Show thatidentify r = f1 (θ) show the following define the same conic section: this y that x to + =1 41 −de de √ , are r= solution This is an ellipse in standardr = position Itsθ foci (±1 +32e − − e cos cos2 θ, 0) = (± 5, 0) and its vertices are (±3, 0), (0, ±2) y = − (x − y)2 43 2x + 2y = x2 − solution We simplify the equation: 2x + y = − (x − y)2 4x + 2xy + y = − x + 2xy − y 2 5x + y = 4 5x 5y + =1 16 ⎛ ⎞2 ⎛ ⎞2 x y ⎝ ⎠ +⎝ ⎠ =1 √ √ www.elsolucionario.net June 15, 2011 LTSV SSM Second Pass Chapter Review Exercises This is an ellipse in standard position, with foci √4 − 0, ± √2 = 0, ± 12 809 and vertices ± √2 , , 0, ± √4 In Exercises 45–50, find the equation of the conic section indicated (y − 3)2 = 2x − √ 45 Ellipse with vertices (±8, 0) and foci (± 3, 0) solution Since the foci of the desired ellipse are on the x-axis, we conclude that a > b We are given that the points √ (±8, 0) are vertices of the ellipse, and since they are on the x-axis, a = We are given that the foci are (± 3, 0) and √ we have shown that a > b, hence we have that a − b2 = Solving for b yields a − b2 = √ a − b2 = 82 − b = b2 = 61 √ b = 61 Next we use a and b to construct the equation of the ellipse: y = √ 61 x + 47 Hyperbola with vertices (±8, 0), asymptotes y = ± 43 x Ellipse with foci (±8, 0), eccentricity 18 solution Since the asymptotes of the hyperbola are y = ± 34 x, and the equation of the asymptotes for a general hyperbola in standard position is y = ± ab x, we conclude that ab = 34 We are given that the vertices are (±8, 0), thus a = We substitute and solve for b: b = a b = b=6 Next we use a and b to construct the equation of the hyperbola: x y − = 49 Parabola with focus (8, 0), directrix x = −8 Hyperbola with foci (2, 0) and (10, 0), eccentricity e = solution This is similar to the usual equation of a parabola, but we must use y as x, and x as y, to obtain x= y 32 51 Find the asymptotes of the hyperbola 3x + 6x − y − 10y = Parabola with vertex (4, −1), directrix x = 15 solution We complete the squares and simplify: 3x + 6x − y − 10y = 3(x + 2x) − (y + 10y) = 3(x + 2x + − 1) − (y + 10y + 25 − 25) = 3(x + 1)2 − − (y + 5)2 + 25 = 3(x + 1)2 − (y + 5)2 = −21 y+5 − √ 21 x+1 =1 √ We obtained a hyperbola with focal axis that is parallel to the y-axis, and is shifted −5 units on the y-axis, and −1 units in the x-axis Therefore, the asymptotes are √ √ x + = ± √ (y + 5) or y + = ± 3(x + 1) 21 www.elsolucionario.net June 15, 2011 LTSV SSM Second Pass 810 C H A P T E R 12 PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS dy 53 Show that the relation dx = (e2 − 1) x holds on a standard ellipse or hyperbola of eccentricity e Show that the “conic section” withyequation x − 4x + y + = has no points solution We differentiate the equations of the standard ellipse and the hyperbola with respect to x: Ellipse: x2 Hyperbola: y2 x2 y2 − =1 a2 b2 2x 2y dy =0 − 2 a b dx + =1 a2 b 2x 2y dy =0 + 2 a b dx b2 x b2 x dy dy =− = dx dx a y a y √ 2 The eccentricity of the ellipse is e = a a−b , hence e2 a = a − b2 or e2 = − √ 2 The eccentricity of the hyperbola is e = a a+b , hence e2 a = a + b2 or dy Combining with the expressions for dx we get: b2 yielding b2 = − e2 a2 a2 2 b e = + , giving b2 = e2 − a a Ellipse: Hyperbola: dy x x = −(1 − e2 ) = (e2 − 1) dx y y dy x = (e2 − 1) dx y dy We, thus, proved that the relation dx = (e2 − 1) xy holds on a standard ellipse or hyperbola of eccentricity e 55 Refer to Figure 25 in Section 12.5 Prove that the product of the perpendicular distances F1 R1 and F2 R2 from the The orbit of Jupiter is an ellipse with the sun at a focus Find the eccentricity of the orbit if the perihelion (closest foci to a tangent line of an ellipse is equal6 to the square b2 of the semiminor axes distance to the sun) equals 740 × 10 km and the aphelion (farthest distance from the sun) equals 816 × 106 km solution We first consider the ellipse in standard position: x2 y2 + =1 a b The equation of the tangent line at P = (x0 , y0 ) is x0 x y y + 02 = a2 b or b2 x0 x + a y0 y − a b2 = The distances of the foci F1 = (c, 0) and F2 = (−c, 0) from the tangent line are |b2 x0 c − a b2 | F1 R1 = b4 x02 + a y02 ; F2 R2 = |b2 x0 c + a b2 | b4 x02 + a y02 We compute the product of the distances: b2 x0 c − a b2 F1 R1 · F2 R2 = b2 x0 c + a b2 b4 x02 + a y02 = b4 x02 c2 − a b4 b4 x02 + a y02 The point P = (x0 , y0 ) lies on the ellipse, hence: x02 y02 + = ⇒ a y02 = a b2 − a b2 x02 a2 b2 We substitute in (1) to obtain (notice that b2 − a = −c2 ) F1 R1 · F2 R2 = = |b4 x02 c2 − a b4 | |b4 x02 c2 − a b4 | = |b4 x02 + a b2 − a b2 x02 | |b2 (b2 − a )x02 + a b2 | |b4 x02 c2 − a b4 | | − b2 x02 c2 + a b2 | = |b2 (x02 c2 − a )| | − (x02 c2 − a )| = | − b2 | = b2 The product F1 R1 · F2 R2 remains unchanged if we translate the standard ellipse www.elsolucionario.net June 15, 2011 LTSV SSM Second Pass (1) ... intentionally left blank www.elsolucionario.net Student’s Solutions Manual to accompany Jon Rogawski’s Single Variable CALCULUS SECOND EDITION BRIAN BRADIE Christopher Newport University ROGER... −π − 2π k The solutions on the interval ≤ θ < 2π are then θ = 0, sin θ = sin 2θ June 7, 2011 6π 8π 2π 4π , , π, , 5 5 www.elsolucionario.net LTSV SSM Second Pass 22 CHAPTER PRECALCULUS REVIEW... of f (x) are therefore x = −3, x = −1.5, x = 1, and x = How many positive solutions does x − 12x + = have? How many solutions does x − 4x + = have? solution The graph of y = x − 12x + shown below