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Unformatted text preview: This page intentionally left blank 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 © 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 CONTENTS Chapter 1 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 1 1 8 13 16 23 27 Chapter 2 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 3 DIFFERENTIATION 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 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 91 101 112 119 126 132 138 147 157 165 Chapter 4 APPLICATIONS OF THE DERIVATIVE 174 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 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 Chapter 5 THE INTEGRAL 5.1 5.2 5.3 Approximating and Computing Area The Definite Integral The Fundamental Theorem of Calculus, Part I 174 181 191 198 206 220 236 242 250 260 260 274 284 5.4 5.5 5.6 The Fundamental Theorem of Calculus, Part II Net Change as the Integral of a Rate Substitution Method Chapter Review Exercises Chapter 6 APPLICATIONS OF THE INTEGRAL 6.1 6.2 6.3 6.4 6.5 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 Chapter 8 TECHNIQUES OF INTEGRATION 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 370 370 378 383 393 398 401 407 415 424 431 446 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 9 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS 9.1 9.2 9.3 9.4 317 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 7 EXPONENTIAL FUNCTIONS 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 290 296 300 307 Arc Length and Surface Area Fluid Pressure and Force Center of Mass Taylor Polynomials Chapter Review Exercises 555 555 564 569 577 593 iii iv C A L C U L U S CON TEN T S 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 742 759 766 780 789 801 1 PRECALCULUS REVIEW 1.1 Real Numbers, Functions, and Graphs Preliminary Questions 1. Give an example of numbers a and b such that a < b and |a| > |b|. solution Take a = −3 and b = 1. Then a < b but |a| = 3 > 1 = |b|. 2. Which numbers satisfy |a| = a? Which satisfy |a| = −a? What about |−a| = a? solution The numbers a ≥ 0 satisfy |a| = a and | − a| = a. The numbers a ≤ 0 satisfy |a| = −a. 3. Give an example of numbers a and b such that |a + b| < |a| + |b|. solution Take a = −3 and b = 1. Then |a + b| = | − 3 + 1| = | − 2| = 2, but |a| + |b| = | − 3| + |1| = 3 + 1 = 4. Thus, |a + b| < |a| + |b|. 4. What are the coordinates of the point lying at the intersection of the lines x = 9 and y = −4? solution The point (9, −4) lies at the intersection of the lines x = 9 and y = −4. 5. In which quadrant do 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. 6. 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 = 9 has radius 3. 7. The equation f (x) = 5 has a solution if (choose one): (a) 5 belongs to the domain of f . (b) 5 belongs to the range of f . solution The correct response is (b): the equation f (x) = 5 has a solution if 5 belongs to the range of f . 8. 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 1. Use a calculator to find a rational number r such that |r − π 2 | < 10−4 . solution r must satisfy π 2 − 10−4 < r < π 2 + 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 3. [−2, solution (d) 3a <|x| 3b≤ 2 (b) |a| < |b| (e) −4a < −4b (c) ab > 0 1 1 (f) < a b 5. (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| < 2. 7. [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. 1 (−2, 8) June 7, 2011 LTSV SSM Second Pass 2 CHAPTER 1 PRECALCULUS REVIEW In Exercises 9–12, write the inequality in the form a < x < b. 9. |x| < 8 solution −8 < x < 8 11. |2x + 1| < 5 |x − 12| < 8 solution −5 < 2x + 1 < 5 so −6 < 2x < 4 and −3 < x < 2 In Exercises |3x −13–18, 4| < 2express the set of numbers x satisfying the given condition as an interval. 13. |x| < 4 solution (−4, 4) 15. |x − 4| < 2 |x| ≤ 9 solution The expression |x − 4| < 2 is equivalent to −2 < x − 4 < 2. Therefore, 2 < x < 6, which represents the interval (2, 6). 17. |4x − 1| ≤ 8 |x + 7| < 2 solution The expression |4x − 1| ≤ 8 is equivalent to −8 ≤ 4x − 1 ≤ 8 or −7 ≤ 4x ≤ 9. 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 − 4 > 2 or x − 4 < −2 ⇒ x > 6 or x < 2 ⇒ (−∞, 2) ∪ (6, ∞) 21. {x : |x 2 − 1| > 2} {x : |2x + 4| > 3} √ √ solution x 2 − 1 > 2 or x 2 − 1 < −2 ⇒ x 2 > 3 or x 2 < −1 (this will never happen) ⇒ x > 3 or x < − 3 ⇒ √ √ (−∞, − 3) ∪ ( 3, ∞). 23. Match (a)–(f) with (i)–(vi). {x : |x 2 + 2x| > 2} (a) a > 3 ! ! ! 1 !! ! (c) !a − ! < 5 3 (e) |a − 4| < 3 (i) (ii) (iii) (iv) (v) (vi) (b) |a − 5| < 1 3 (d) |a| > 5 (f) 1 ≤ a ≤ 5 a lies to the right of 3. a lies between 1 and 7. The distance from a to 5 is less than 13 . The distance from a to 3 is at most 2. a is less than 5 units from 13 . a lies either to the left of −5 or to the right of 5. solution (a) On the number line, numbers greater than 3 appear to the right; hence, a > 3 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 | < 5 is satisfied by those numbers less than 5 units from 1 : (v). 3 (d) The inequality |a| > 5 is equivalent to a > 5 or a < −5; that is, either a lies to the right of 5 or to the left of −5: (vi). (e) The interval described by the inequality |a − 4| < 3 has a center at 4 and a radius of 3; that is, the interval consists of those numbers between 1 and 7: (ii). (f) The interval described by the inequality 1 < x < 5 has a center at 3 and a radius of 2; that is, the interval consists of those numbers less than 2 units from 3: (iv). # an interval. Hint: Plot y = x 2 + 2x − 3. 25. Describe {x :"x 2 + 2x < 3} as x < 0 as an interval. Describe x : solution The inequality x + 1x 2 + 2x < 3 is equivalent to x 2 + 2x − 3 < 0. In the figure below, we see that the graph of 2 y = x + 2x − 3 falls below the x-axis for −3 < x < 1. Thus, the set {x : x 2 + 2x < 3} corresponds to the interval −3 < x < 1. June 7, 2011 LTSV SSM Second Pass S E C T I O N 1.1 y = x2 + 2x − 3 −4 −3 −2 Real Numbers, Functions, and Graphs 3 y 10 8 6 4 2 −2 1 2 x 27. Show that if a > b, then b−1 > a −1 , provided that a and b have the same sign. What happens if a > 0 and b < 0? Describe the set of real numbers satisfying |x − 3| = |x − 2| + 1 as a half-infinite interval. solution Case 1a: If a and b are both positive, then a > b ⇒ 1 > ab ⇒ b1 > a1 . Case 1b: If a and b are both negative, then a > b ⇒ 1 < ab (since a is negative) ⇒ b1 > a1 (again, since b is negative). Case 2: If a > 0 and b < 0, then a1 > 0 and b1 < 0 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| 2 and − 5||(a <+ 1?b) − 13| < 1. Hint: Use the triangle inequality. solution |a + b − 13| = |(a − 5) + (b − 8)| ≤ |a − 5| + |b − 8| < (by the triangle inequality) 1 1 + = 1. 2 2 Suppose that |a − 6| ≤ 2 and |b| ≤ 3. Suppose that |x − 4| ≤ 1. What is the largest possible value of |a + b|? (a) What is the maximum possible value of |x + 4|? What is the smallest possible value of |a + b|? (b) Show that |x 2 − 16| ≤ 9. solution |a − 6| ≤ 2 guarantees that 4 ≤ a ≤ 8, while |b| ≤ 3 guarantees that −3 ≤ b ≤ 3. Therefore 1 ≤ a + b ≤ 11. It follows that (a) the largest possible value of |a + b| is 11; and (b) the smallest possible value of |a + b| is 1. 31. (a) (b) 33. Express r1 = 0.27 as a fraction. Hint: 100r1 − r1 is an integer. Then express r2 = 0.2666 . . . as a fraction. Prove that |x| − |y| ≤ |x − y|. Hint: Apply the triangle inequality to y and x − y. solution Let r1 = 0.27. We observe that 100r1 = 27.27. Therefore, 100r1 − r1 = 27.27 − 0.27 = 27 and r1 = 27 3 = . 99 11 Now, let r2 = 0.2666. Then 10r2 = 2.666 and 100r2 = 26.666. Therefore, 100r2 − 10r2 = 26.666 − 2.666 = 24 and r2 = 4 24 = . 90 15 35. The text states: If the decimal expansions of numbers a and b agree to k places, then |a − b| ≤ 10−k . Show that the Represent 1/7 and 4/27 as repeating decimals. converse is false: For all k there are numbers a and b whose decimal expansions do not agree at all but |a − b| ≤ 10−k . solution Let a = 1 and b = 0.9 (see the discussion before Example 1). The decimal expansions of a and b do not agree, but |1 − 0.9| < 10−k for all k. 37. (a) (b) Find the equation of the circle with center (2, 4): Plot each pair of points and compute the distance between them: with radius r = 3. (a) (1, 4) and (3, 2) (b) (2, 1) and (2, 4) that passes through (1, −1). (c) (0, 0) and (−2, 3) (d) (−3, −3) and (−2, 3) solution (a) The equation of the indicated circle is (x − 2)2 + (y − 4)2 = 32 = 9. (b) First determine the radius as the distance from the center to the indicated point on the circle: $ √ r = (2 − 1)2 + (4 − (−1))2 = 26. Thus, the equation of the circle is (x − 2)2 + (y − 4)2 = 26. 39. Determine the domain and range of the function Find all points with integer coordinates located at a distance 5 from the origin. Then find all points with integer coordinates located at a distance 5 fromf (2, 3).s, t, u} → {A, B, C, D, E} : {r, defined by f (r) = A, f (s) = B, f (t) = B, f (u) = E. solution The domain is the set D = {r, s, t, u}; the range is the set R = {A, B, E}. June 7, 2011 LTSV SSM Second Pass 4 CHAPTER 1 PRECALCULUS REVIEW In Exercises find the and range of the D function. Give 41–48, an example of adomain function whose domain has three elements and whose range R has two elements. Does a exist whose domain D has two elements and whose range R has three elements? 41. function f (x) = −x solution D : all reals; R : all reals 43. f (x) = x 3 4 g(t) = t solution D : all reals; R : all reals 45. f (x) = |x|√ g(t) = 2 − t solution D : all reals; R : {y : y ≥ 0} 1 47. f (x) = 2 1 h(s) = x s solution D : {x : x ) = 0}; R : {y : y > 0} In Exercises 49–52,1determine where f (x) is increasing. g(t) = cos 49. f (x) = |x + 1|t solution A graph of the function y = |x + 1| is shown below. From the graph, we see that the function is increasing on the interval (−1, ∞). y 2 1 −3 −2 −1 1 x 51. f (x) = x 4 3 f (x) = x solution A graph of the function y = x 4 is shown below. From the graph, we see that the function is increasing on the interval (0, ∞). y 12 8 4 −2 −1 1 2 x In Exercises 53–58, find 1 the zeros of f (x) and sketch its graph by plotting points. Use symmetry and increase/decrease f (x)where = 4 appropriate. information x + x2 + 1 2 53. f (x) = x − 4 solution Zeros: ±2 Increasing: x > 0 Decreasing: x < 0 Symmetry: f (−x) = f (x) (even function). So, y-axis symmetry. y 4 2 −2 −1 1 −2 2 x −4 55. f (x) = x 3 − 4x f (x) = 2x 2 − 4 solution Zeros: 0, ±2; Symmetry: f (−x) = −f (x) (odd function). So origin symmetry. y 10 5 −2 −1 −5 −10 June 7, 2011 1 2 x LTSV SSM Second Pass Real Numbers, Functions, and Graphs S E C T I O N 1.1 5 57. f (x) = 2 − x33 f (x) = x √ solution This is an x-axis reflection of x 3 translated up 2 units. There is one zero at x = 3 2. y 20 10 −2 −1−10 −20 1 x 2 59. Which of the curves in Figure 26 is the graph of a function? 1 f (x) = (x − 1)2 + 1 y y x x (A) (B) y y x x (C) (D) FIGURE 26 solution (B) is the graph of a function. (A), (C), and (D) all fail the vertical line test. 61. Determine whether the function is even, odd, or neither. Determine1 whether the 1function is even, odd, or neither. 3− 2−t (a) f(a) (t)f=(x)4= x 5 − (b)(b)g(t) == 2t t− g(t) t2 t + t + 1 t4 − t + 1 1 (c) G(θ) (d) H (θ) = sin(θ 2 ) (c) F= (t)sin = θ 4+ cos2 θ t +t solution (a) This function is odd because f (−t) = 1 1 − (−t)4 + (−t) + 1 (−t)4 − (−t) + 1 1 1 = 4 − 4 = −f (t). t −t +1 t +t +1 (b) g(−t) = 2−t − 2−(−t) = 2−t − 2t = −g(t), so this function is odd. (c) G(−θ ) = sin(−θ ) + cos(−θ ) = − sin θ + cos θ which is equal to neither G(θ) nor −G(θ), so this function is neither odd nor even. (d) H (−θ ) = sin((−θ )2 ) = sin(θ 2 ) = H (θ), so this function is even. 1 3 + 12x 2 −=3x + 4 as 63. Determine f (x) is increasing or decreasing. Write f the (x) interval = 2x 4 −on5xwhich x − 4 the sum of an even and an odd function. solution A graph of the function is shown below. From this graph we can see that f (x) is decreasing on (−∞, 4) and also decreasing on (4, ∞). y 6 4 2 0 x 2 −2 4 6 8 10 −2 −4 −6 June 7, 2011 LTSV SSM Second Pass 6 CHAPTER 1 PRECALCULUS REVIEW In Exercises f (x) be theisfunction shown in Figureor27. State 65–70, whetherletthe function increasing, decreasing, neither. (a) (b) (c) (d) Surface area of a sphere as a function of its radius y Temperature at a point on the equator as a function of time 4 price of oil Price of an airline ticket as a function of the 3 of volume Pressure of the gas in a piston as a function 2 1 0 1 2 3 4 x FIGURE 27 65. Find the domain and range of f (x)? solution D : [0, 4]; R : [0, 4] %1 & 67. Sketch the the graphs of fof(2x), , and 2f (x). 2 xand Sketch graphs f (x f+ 2) f (x) + 2. solution The graph of y = f (2x) is obtained by compressing the graph of y = f (x) horizontally by a factor of 2 (see the graph below on the left). The graph of y = f ( 12 x) is obtained by stretching the graph of y = f (x) horizontally by a factor of 2 (see the graph below in the middle). The graph of y = 2f (x) is obtained by stretching the graph of y = f (x) vertically by a factor of 2 (see the graph below on the right). y y y 4 4 8 3 3 6 2 2 4 1 1 1 2 3 4 x 2 2 f (2x) 4 6 8 x f (x/2) 1 2 3 4 x 2 f(x) 69. Extend the graph of f (x) to [−4, 4] so that it is an even function. Sketch the graphs of f (−x) and −f (−x). solution To continue the graph of f (x) to the interval [−4, 4] as an even function, reflect the graph of f (x) across the y-axis (see the graph below). y 4 3 2 1 −4 −2 2 4 x 71. Suppose that f (x) has domain [4, 8] and range [2, 6]. Find the domain and range of: Extend the graph of f (x) to [−4, 4] so that it is an odd function. (a) f (x) + 3 (b) f (x + 3) (c) f (3x) (d) 3f (x) solution (a) f (x) + 3 is obtained by shifting f (x) upward three units. Therefore, the domain remains [4, 8], while the range becomes [5, 9]. (b) f (x + 3) is obtained by shifting f (x) left three units. Therefore, the domain becomes [1, 5], while the range remains [2, 6]. (c) f (3x) is obtained by compressing f (x) horizontally by a factor of three. Therefore, the domain becomes [ 43 , 83 ], while the range remains [2, 6]. (d) 3f (x) is obtained by stretching f (x) vertically by a factor of three. Therefore, the domain remains [4, 8], while the range becomes [6, 18]. 73. Suppose that the graph of f (x) = sin x is compressed horizontally by a factor of 2 and then shifted 5 units to the 2 right. Let f (x) = x . Sketch the graph over [−2, 2] of: (a) f is (xthe + 1) (b) f (x) + 1 (a) What equation for the new graph? (c) f (5x) (d)by5f (b) What is the equation if you first shift by 5 and then compress 2?(x) (c) Verify your answers by plotting your equations. June 7, 2011 LTSV SSM Second Pass Real Numbers, Functions, and Graphs S E C T I O N 1.1 7 solution (a) Let f (x) = sin x. After compressing the graph of f horizontally by a factor of 2, we obtain the function g(x) = f (2x) = sin 2x. Shifting the graph 5 units to the right then yields h(x) = g(x − 5) = sin 2(x − 5) = sin(2x − 10). (b) Let f (x) = sin x. After shifting the graph 5 units to the right, we obtain the function g(x) = f (x − 5) = sin(x − 5). Compressing the graph horizontally by a factor of 2 then yields h(x) = g(2x) = sin(2x − 5). (c) The figure below at the top left shows the graphs of y = sin x (the dashed curve), the sine graph compressed horizontally by a factor of 2 (the dash, double dot curve) and then shifted right 5 units (the solid curve). Compare this last graph with the graph of y = sin(2x − 10) shown at the bottom left. The figure below at the top right shows the graphs of y = sin x (the dashed curve), the sine graph shifted to the right 5 units (the dash, double dot curve) and then compressed horizontally by a factor of 2 (the solid curve). Compare this last graph with the graph of y = sin(2x − 5) shown at the bottom right. y y 1 1 −6 −4 −2 2 4 x 6 −6 −4 −2 −1 2 4 6 2 4 6 x −1 y y 1 1 −6 −4 −2 2 4 x 6 −6 −4 −2 x −1 −1 % & 75. Sketch the28 graph of fthe (2x) andof f f12(x) x ,= where (x)Match = |x|the + 1functions (Figure 28). Figure shows graph |x| +f1. (a)–(e) with their graphs (i)–(v). solution by a factor of 2 (a) f (x The − 1)graph of y = f (2x) is obta...
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