5223_Test_100-137 04/10/06 11:24 AM Page 100 Sample Tests for The Advanced Placement Examinations The Advanced Placement Calculus Examinations are three hours and 15 minutes long and evaluate how well students have mastered the concepts and techniques in either AB or BC Calculus course. Each examination consists of (1) a multiple-choice section that tests proficiency over a broad spectrum of topics and (2) a problem section that requires students to demonstrate their ability in solving problems requiring a more extensive chain of reasoning. Students taking the BC test will receive both an AB and a BC grade to better assist colleges in giving credit for scores. The examination will be based on the Advanced Placement Course Description. MULTIPLE CHOICE (Section I) Part A Part B 28 questions 17 questions 55 minutes 50 minutes No calculator allowed Graphing calculator required (students will have to decide whether a calculator is appropriate since not all problems need a calculator) 45 minutes 45 minutes Graphing calculator required No calculator allowed FREE RESPONSE (Section II) Part A Part B 3 questions 3 questions The problem section (6 questions completed in 90 minutes) is designed to utilize the graphing calculator for some problems. During the timed portion for Part B, students may work on Part A questions without the use of a calculator. Both the multiple-choice and free-response sections are given equal weight in determining the grade for the examination. The following tests are intended to give students practice with the type of exam they will take. You may wish to simulate the conditions of the exam, providing the exact amount of time for each section and using the multiple choice answer sheet that is provided on page 137. 100 Sample AB Test © 2007 Pearson Education, Inc. 5223_Test_100-137 04/10/06 11:24 AM Page 101 Advanced Placement Calculus AB test Section I—Part A (55 minutes) Choose the best answer for each question. Your score is determined by subtracting one-fourth of the number of wrong answers from the number of correct answers. Calculators are not permitted. 1. dy d2y 6 0 and 2 , 6 0? For the graph shown, at which point is it true that dx dx (A) A (B) B (C) C (D) D (E) E 2. Find the area of the region bounded by the x-axis and the graph of y = 1 x + 1 2 1 x - 2 2 2. (A) 5 4 (B) 2 3 4 (C) 5 1 4 (D) 6 1 4 (E) 6 3 4 3. Which of the following is an antiderivative of x2 sec2 x3? (A) 2x sec2 x3 6x4 sec2 x3 tan x3 (B) 2x sec2 x3 6x3 sec x3 (C) 1 tan x3 - 5 3 (D) 3 tan x3 1 (E) - cot x 3 + 4 3 © 2007 Pearson Education, Inc. Sample AB Test 101 5223_Test_100-137 04/10/06 11:24 AM Page 102 4. Line L is tangent to the curve defined by 2xy2 3y 18 at the point (3, 2). The slope of line L is (A) 21 8 (B) 32 3 (C) - 10 21 (D) 8 21 (E) - 8 21 5. A bicyclist rides along a straight road starting from home at t 0. The graph above shows the bicyclist’s velocity as a function of t. How far from home is the bicyclist after 2 hours? (A) 13 miles (B) 16.5 miles (C) 17.5 miles 6. Find the value of x at which the graph of y = (B) 4 2>3 (A) 2 7. Find lim xSq (A) (D) 18 miles (E) 20 miles 1 + 1x has a point of inflection. x (C) 4 (D) 6 (C) 1 (D) - (E) 8 2x - 4x3 8x3 + 4x2 - 3x 2 3 102 Sample AB Test (B) 3 2 1 2 (E) - 3 4 © 2007 Pearson Education, Inc. 5223_Test_100-137 04/10/06 11:24 AM Page 103 8. Which of the following is a slope field for the differential equation dy>dx = - 2x + y? (A) (B) y (E) y y x x x (D) (C) y y x x 1 9. Let f (x) cos (3x2). Find f¿ a b . 3 (A) - 13 (B) 13 (C) 0 (D) - 13p 2 (E) 10. Assume that f (x) is a one-to-one function. The area of the shaded region is equal to which of the following definite integrals? I. II. III. 4 3 f1x2 - 3 4 dx 0 3 f1x2 - 3 4 dx 3 f -1 1y2 dy 30 34 31 (A) I only (B) II only © 2007 Pearson Education, Inc. (C) III only (D) I and III (E) II and III Sample AB Test 103 5223_Test_100-137 04/10/06 11:24 AM Page 104 11. The graph of a function y f(x) is shown above. Which of the following are true for the function f? I. f¿ 1 22 is defined. II. lim+ f1x 2 = lim- f1x2 xS2 xS2 III. f¿ 1x 2 6 0 for all x in the open interval (1, 2). (A) I only (B) II only 12. Let f 1x 2 = sin-1x. Find f¿ a (A) p 4 13. Evaluate (B) 3 12 2 12 2 (C) III only (D) II and III (E) I, II and III (D) 12 (E) Undefined b. (C) 1 2 1cos - e2x 2 dx. (A) -sinx - 1 2x e + C 2 1 (B) sinx - e2x + C 2 (C) - sinx - 2e2x + C (D) sinx - 2e2x + C (E) - cosx - 104 Sample AB Test 1 2x e + C 2 © 2007 Pearson Education, Inc. 5223_Test_100-137 04/10/06 11:24 AM Page 105 3 14. Let f(x) ex -2x2 -4x +5 (A) 2 . Then f has a local minimum at x (B) - 2 3 (C) 2 3 (D) 1 (E) 2 15. The acceleration of a particle moving along the x-axis is a(t) 12t 10. At t 0, the velocity is 4. At t 1, the position is x 8. Find the position at t 2. (A) 5 (B) 4 (C) 10 (D) 11 (E) 7 16. Let f be differentiable for all real numbers. Which of the following must be true for any real numbers a and b? a I. 32 b II. b f1x2 dx = 3a 32 a f1x2 dx + 3a f1 x 2 dx ¢ 3 f1x2 4 2 + f¿ 1x2 ≤ dx = 3 f1 b 2 4 2 - 3f 1 a 2 4 2 b III. 3b b 3 f1x2 dx = 3 (A) I only 3a f1x 2 dx (B) II only (C) I and II (D) I and III 3x at x 3. x -6 (A) 5x y 18 (B) 5x y 12 (C) 5x 3y 24 (D) x 5y 12 17. Find an equation of the line normal to the graph of y = © 2007 Pearson Education, Inc. (E) I, II, and III 2 (E) x y 6 Sample AB Test 105 5223_Test_100-137 04/10/06 11:24 AM Page 106 1x + h2 2 - x2 . For what value of x does g(x) 2? hS0 h 18. Let g 1x2 = lim (A) x 1 (B) x 2 (C) x 3 (D) x 4 (E) x 5 19. Let f be a differentiable function of x that satisfies f(1) 7 and f(4) 3. Which of the following conditions would guarantee that the tangent line at x c is parallel to the secant line joining (1, f(1)) to (4, f(4))? (A) f1c 2 = 3 2 (B) f1c 2 = 5 (C) f¿ 1c2 = - 3 4 (D) f¿ 1 c2 = - 4 3 (E) f¿ 1 c 2 = - 4 3 20. Let f(x) x3 12x. Which statement about this function is false? (A) The function has a relative minimum at x 2. (B) The function is increasing for values of x between 2 and 2. (C) The function has two relative extrema. (D) The function is concave upward for x > 0. (E) The function has one inflection point. 3 21. 32 (A) 8x 1x2 - 52 dx 74 3 22. Let f 1x2 = (B) 30 (C) 90 370 3 (D) 112 (E) (D) 4 (E) 5 x d 1t2 + 16 dt. What is f1 -3 2 ? dx 30 (A) 5 (B) 4 (C) 3 dy 1 = xy2 and y = - when y 2, what is y when x 4? dx 3 1 1 1 1 (A) (B) (C) (D) 3 5 9 3 23. If 106 Sample AB Test (E) 1 9 © 2007 Pearson Education, Inc. 5223_Test_100-137 04/10/06 11:24 AM Page 107 24. Use the Trapezoidal Rule with n 3 to approximate the area between the curve y x2 and the x-axis for 1 x 4. (A) 14 (B) 21 (C) 21.5 (D) 29 (E) 30 25. Let f(x) be a continuous function that is defined for all real numbers x. x2 - x - 6 If f1x2 = 2 when x2 - 5x + 6 Z 0, what is f (3)? x - 5x + 6 (A) 5 (B) 4 (C) 2 (D) 1 (E) 0 26. Find the derivative of cos3 2x. (A) sin3 2x (B) 6 cos2 2x. (C) 6 cos2 2x sin 2x (D) 3 cos2 2x sin 2x (E) 6 cos2 2x sin 2x 27. Let f be a twice-differentiable function whose derivative f ¿ (x) is increasing for all x. Which of the following must be true of all x? I. f(x) 0 II. f ¿ (x) 0 III. f – (x) 0 (A) I only (B) II only (C) III only (D) I and II (E) II and III 28. The function f (x) x3 6x2 9x 4 has a local maximum at (A) x 0 © 2007 Pearson Education, Inc. (B) x 1 (C) x 2 (D) x 3 (E) x 4 Sample AB Test 107 5223_Test_100-137 04/10/06 11:24 AM Page 108 Section I—Part B (50 minutes) Choose the best answer for each question. (If the exact answer does not appear among the choices, choose the best approximation for the exact answer.) Your score is determined by subtracting one-fourth of the number of wrong answers from the number of correct answers. You may use a graphing calculator. 29. Which of the following functions has the fastest rate of growth as x S q ? (A) y x18 5x (B) y 5x2 (C) y ln x2 (D) y (ln x)2 (E) y e0.01x 30. The velocity of a particle moving along a straight line is given by v(t) 3x2 4x. Find an expression for the acceleration of the particle. (A) x3 4 (B) x3 2x2 (C) 3x2 4 (D) 3x 4 (E) 6x 4 31. Find the average value of the function y x3 4x on the closed interval [0, 4]. (A) 8 (B) 12 (C) 24 (D) 32 (E) 48 32. A region is enclosed by the x-axis and the graph of the parabola y 9 x2. Find the volume of the solid generated when this region is revolved about the x-axis. (A) 36 (B) 40.5 (C) 129.6 (D) 194.4 (E) 259.2 33. Which of the following is an antiderivative of x 1x2 + 3? 1 1 1 2 (A) x3>2 (B) x3 (C) 1 x2 + 3 2 3>2 (D) 1 x2 + 3 2 3>2 3 3 3 3 108 Sample AB Test (E) (x2 3)3/2 © 2007 Pearson Education, Inc. 5223_Test_100-137 04/10/06 11:24 AM Page 109 34. x f(x) 3.3 3.69 3.4 3.96 3.5 4.25 3.6 4.56 3.7 4.89 Let f be a differentiable function that is defined for all real numbers x. Use the table above to estimate f¿ 13.62 . (A) 0.3 (B) 1.8 (C) 2.7 (D) 3.0 (E) 3.2 35. The weight in pounds of a certain bear cub t months after birth is given by w(t). If w(2) 36, dw w(7) 84, and was proportional to the cub’s weight for the first 15 months of his life, how dt much did the cub weigh when he was 11 months old? (A) 125 pounds 36. Let f1x2 = b (B) 135 pounds (C) 145 pounds (D) 155 pounds (E) 165 pounds 3x2 - 4, for x 1 . 6x - 5, for x 7 1 Which of the following are true statements about this function? I. lim f1x2 exists. xS1 II. f¿ 112 exists. III. lim f¿ 1x2 exists. xS1 (A) None (B) II only (E) III only (D) II and III (E) I, II, and III 37. Two particles are moving along the x-axis. Their positions are given by x1(t) 2t2 5t 7 and x2(t) sin 2t, respectively. If a1(t) and a2(t) represent the acceleration functions of the particles, find the numbers of values of t in the closed interval [0, 5] for which a1(t) a2(t). (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 or more (D) 3 (E) 4 or more 38. The function f(x) ex x3 has how many critical points? (A) 0 (B) 1 © 2007 Pearson Education, Inc. (C) 2 Sample AB Test 109 5223_Test_100-137 04/10/06 11:24 AM Page 110 39. A dog heading due north at a constant speed of 2 meters per second trots past a fire hydrant at t 0 sec. Another dog heading due east at a constant speed of 3 meters per second trots by the hydrant at t 1 sec. At t 9 sec, the rate of change of the distance between the two dogs is (A) 3.2 m/sec (B) 3.6 m/sec 40. Let f (x) x5 x. Find the value of (A) - 1 6 (B) 1 6 (C) 4.0 m/sec (D) 4.4 m/sec (E) 4.8 m/sec (D) 6 (E) 81 d 1 f (x) at x 2. dx (C) 1 81 1 ln t 2 2 3 ft >sec for t 1 sec. If the t volume of the balloon is 1.3 ft3 at t 1 sec, what is the volume of the balloon at t 5 sec? 41. Suppose air is pumped into a balloon at a rate given by r1 t 2 = (A) 2.7 ft3 (B) 3.0 ft3 (C) 3.3 ft3 (D) 3.6 ft3 (E) 3.9 ft3 42. Find the approximate value of x where f1 x 2 = x2 - 3 1x + 2 has its absolute minimum. (A) 4.5 (B) 2 (C) 0 (D) 0.5 (E) 2.5 43. The graph of y = f¿ 1 x2 is shown. Which of the following statements about the function f(x) are true? I. f(x) is decreasing for all x between a and c. II. The graph of f is concave up for all x between a and c. III. f(x) has a relative minimum at x a. (A) I only 110 Sample AB Test (B) II only (C) III only (D) I and III (E) I, II, and III © 2007 Pearson Education, Inc. 5223_Test_100-137 04/10/06 11:24 AM Page 111 44. Suppose f and g are even functions that are continuous for all x, and let a be a real number. Which of the following expressions must have the same value? a I. 3-a 3 f1x2 + g1x 2 4 dx a II. 2 30 3 f1x2 + g1x 2 4 dx a III. 3-a a f1x2 dx + (A) I and II only 3-a g1 x2 dx (B) I and III only (C) II and III only (D) I, II, and III (E) None 45. Let f(x) g(h(x)), where h(2) 3, h¿ 1 2 2 = 4, g(3) 2, and g¿ 1 32 = 5. Find f¿ 1 2 2 . (A) 6 (B) 8 (C) 15 (D) 20 (E) More information is needed to find f¿ 1 2 2 . © 2007 Pearson Education, Inc. Sample AB Test 111 5223_Test_100-137 04/10/06 11:24 AM Page 112 Advanced Placement Calculus AB test Section II (90 minutes) Show your work. In order to receive full credit, you must show enough detail to demonstrate a clear understanding of the concepts involved. You may use a graphing calculator. Where appropriate, you may give numerical answers in exact form or as decimal approximations correct to three decimal places. For Problems 1–3, a graphing calculator may be used. (45 minutes) 1. For t 0, a particle moves along the x-axis with a velocity given by v(t) 2t 5 sin t. At t 0, the particle is located at x 0 (a) Write an expression for the acceleration a (t) of the particle. (b) Write an expression for the position x(t) of the particle. (c) For what values of t (t 0) is the particle moving to the left? (d) For t 1, find the position of the particle the first time the velocity of the particle is zero. 112 Sample AB Test © 2007 Pearson Education, Inc. 5223_Test_100-137 04/10/06 11:24 AM Page 113 2. A function y f(x) is defined by 5 1x + xy + y3 + 11. (a) Find an expression for f' 1 x2 = dy in terms of x and y. dx (b) Find the equation of the line that is tangent to the graph of y f (x) at the point (0.25, 2). (c) Use the tangent line from part (b) to estimate f (0.6). (d) Write an equation whose solution is the exact value of f(0.6). To the nearest thousandth, what is f(0.6)? (e) Would it be appropriate to use the tangent line from part (b) to estimate f(0.1)? Explain. © 2007 Pearson Education, Inc. Sample AB Test 113 5223_Test_100-137 04/10/06 11:24 AM Page 114 3. The figure above shows a pump connected by a flexible tube to a spherical balloon. The pump consists of a cylindrical container of radius 8 inches, with a piston that moves up and down accord24 + ln 1 t + 1 2 for 0 t 100, where t is measured in seconds and ing to the equation h1 t2 = t +1 h (t) is measured in inches. As the piston moves up and down, the total volume of air enclosed in the pump and the balloon remains constant, and r (t) 0 at t 0. ¢ The volume of a sphere with 4 radius r is r 3. ≤ 3 (a) Write an expression in terms of h (t) and r(t) for the total volume of the air enclosed in the pump and the balloon. (Do not include the air in the flexible tube.) (b) Find the minimum volume of air in the pump and when it occurs. (c) Find the rate of change of the volume of the air enclosed in the pump at t 3 sec. (d) At t 3 sec, find the radius of the balloon and the rate of change of the radius of the balloon. 114 Sample AB Test © 2007 Pearson Education, Inc. 5223_Test_100-137 04/10/06 11:24 AM Page 115 No calculator may be used for Problems 4, 5, & 6. Students may continue working on Problems 1–3, but may not use a calculator. (45 minutes) 4. The shaded region is enclosed by the graphs of y x3 and y 4 14x. (a) Find the coordinates of the point in the first quadrant where the two curves intersect. 3 (b) Use an integral with respect to x to find the area of the shaded region. (c) Set up an integral with respect to y that could be used to find the area of the shaded region. (d) Without using absolute values, write an integral expression that gives the volume of the solid generated by revolving the shaded region about the line x 1. Do not evaluate. © 2007 Pearson Education, Inc. Sample AB Test 115 5223_Test_100-137 04/10/06 11:24 AM Page 116 5. The graph of a differentiable function g is shown. The area of the shaded region is 8 square units. Let f be a differentiable function such that f(0) 3 and f¿ 1 x 2 = g 1 x2 for 1 x 5. (a) Find f(4) (b) For what values of x is the graph of y f(x) concave upward? Explain. (c) Write an expression for f(x). Your answer should involve a definite integral and should be expressed in terms of the function g. (d) Sketch a possible graph for y f(x). 116 Sample AB Test © 2007 Pearson Education, Inc. 5223_Test_100-137 04/10/06 11:24 AM Page 117 dy (y 5) (x 2). dx (a) On the axes provided below, sketch a slope field for the given differential equation at the nine points indicated. 6. Consider the differential equation given by y 3 2 1 –3 –2 –1 x (b) Find the particular solution y f(x) to the given differential equation with the initial condition f(0) 2. (c) Find the domain and range of your particular function. © 2007 Pearson Education, Inc. Sample AB Test 117 5223_Test_100-137 04/10/06 11:24 AM Page 118 Advanced Placement Calculus BC test Section 1—Part A (55 minutes) Choose the best answer for each question. Your score is determined by subtracting one-fourth of the number of wrong answers from the number of correct answers. Calculators are not permitted. dy = -x with initial condition y(0) 1 dx (A) is always concave up 1. The solution to (B) is always concave down (C) is undefined at x 0 (D) is always decreasing (E) is always increasing 2. Which of the following is a term in the Taylor series about x 0 for the function f(x) cos 2x? 1 (A) - x2 2 3. Evaluate (A) (B) (C) (D) (E) 118 3 4 (B) - x3 3 (C) 2 4 x 3 (D) 1 5 x 60 (E) 4 6 x 45 x cos 2x dx. 1 1 x cos 2x - sin 2x + C 2 4 1 1 x sin 2x - cos 2x + C 2 4 1 1 x sin 2x - sin 2x + C 2 4 1 1 x cos 2x + sin 2x + C 2 4 1 1 x sin 2x + cos 2x + C 2 4 Sample BC Test © 2007 Pearson Education, Inc. 5223_Test_100-137 04/10/06 11:24 AM Page 119 dy = 1x + 3 2e -2y, then which of the following is a possible expression for y? dx 1 (A) ln 1 x2 + 6x + 52 2 (B) ln 1x2 + 6x - 4 2 1 (C) ln 1 x2 + 6x2 - 3 2 3 1 1 (D) ln a x2 + x b 2 4 2 1 (E) ln 1 x2 + 3x2 2 4. If 5. Let f1x2 = b 2x - 5, 1x + 1, for x 3 . for x 7 3 8 Find 30 f1x2dx. (A) 24 (B) 45 2 (C) 52 3 (D) 20 3 (E) 32 12 - 2 13 3 6. The line tangent to the graph of y x3 3x2 2x 1 at x 1 will also intersect the curve at which of the following values of x? (A) x 4 7. lim tan a hS0 (B) x 5 (C) x 6 (D) x 7 (E) x 8 p p + h b - tan 3 3 = h (B) 13 (A) 4 (C) 1 13 (D) 13 2 (E) 1 2 8. A curve in the xy-plane is defined by the parametric equations x t3 2 and y t2 4t. Find the slope of the line tangent to the curve at the point where x 6. (A) - 3 2 (B) 2 3 (C) - 2 3 (D) 1 2 (E) 2 9. Assume that g¿ 1x2 h(x) and f(x) x2. Which of the following expressions is equal to (A) 2x g (x) (B) 2x h(x) © 2007 Pearson Education, Inc. (C) 2 g(x) h(x) (D) f¿ 1 x2 g 1 x2 h1 x2 d f 1 g 1x 2 2 ? dx (E) x2 h(x) 2x g(x) Sample BC Test 119 5223_Test_100-137 04/10/06 11:24 AM Page 120 10. Let f 1x 2 = b for x 6 1 2x, . 2x - 3, for x 6 1 Let g (x) ln [(x 1)2]. Which of the following functions are continuous at x 1? I. g (x) II. f¿ 1x 2 x III. 30 f1t 2 dt (A) I only (B) II only (C) III only (D) I and II (E) I and III q 1x - 22n 11. Find the values of x for which the series a n converges. n =1 n1 -3 2 (A) x 2 only (B) 1 x 5 (C) 1 x 5 (D) 1 x 5 (E) All real numbers 12. The slope field for a certain differential equation is shown below. Which of the following could be a specific solution to that differential equation? y 5 O 5 x (A) y = 1x (B) y = x2 (C) y = 1>x (D) y = ln x (E) y = ex 120 Sample BC Test © 2007 Pearson Education, Inc. 5223_Test_100-137 04/10/06 11:24 AM Page 121 13. A particle is moving along the x-axis according to the equation x (t) 4t2 sin 3t where x is given p in feet and t is given in seconds. Find the acceleration at t . 2 (A) 1 ft/sec2 (B) 5 ft/sec2 (C) 11 ft/sec2 (D) 17 ft/sec2 (E) 2 ft/sec2 14. If the derivative of f is f¿ x (x 1)2 (x 2)3 (x 3)4, find the number of points where f has a local maximum. (A) None (B) One (C) Two (D) Three (E) Four 1 -4 e 2 1 (D) - e-4 2 (E) q q 15. Evaluate 2 32 xe -x dx. 1 (A) e -2 2 1 (B) - e-2 2 (C) 16. Let f and g be functions that are differentiable for all real numbers, with lim f1 x 2 = 3 and lim g 1x 2 = 5. xS0 xS0 f1 x2 f1 x 2 ? ¢ You may assume that lim exists. ≤ xS0 g 1 x2 xS0 g 1 x 2 Which of the following must be equal to lim I. II. 3 5 f102 g102 f¿ 102 xS0 g¿ 102 III. lim (A) None © 2007 Pearson Education, Inc. (B) I and II (C) I and III (D) II and III (E) I, II, and III Sample BC Test 121 5223_Test_100-137 04/10/06 11:24 AM Page 122 q 2p b. 17. Let f (x) a 1cosx2 3n. Evaluate f a 3 n =1 1 7 1 9 1 7 8 9 The series diverges. (A) (B) (C) (D) (E) 18. Let f 1x 2 = 2 (A) ex x2 2 30 et +t dt. Find f¿ 1x2. (B) 2x ex 1x 2 +2x 2 +12 4 (C) ex +x2 (D) 2e1x +x2 2 2 (E) 2x ex +2x 19. A particle is moving along the graph of the curve y ln (3x 5). At the instant when the particle crosses the y-axis, the y-coordinate of its location is changing at the rate of 15 units per second. Find the rate of change of the x-coordinate of the particle’s location. (A) (B) (C) (D) (E) 5 ln 3 units per second 9 units per second 25 units per second 45 units per second 3 ln 5 units per second 20. Find lim a xSq 2x + 1 3x b 2x (A) 1.5 122 Sample BC Test (B) 6 (C) e1.5 (D) e6 (E) q © 2007 Pearson Education, Inc. 5223_Test_100-137 04/10/06 11:24 AM Page 123 21. Use implicit differentiation to find 1 (A) - exy 4 (B) y x + 4e-xy 3 22. Which of following is equal to 3 (A) 1 u3 du 4 31 dy for the equation 4y exy 7. dx 31 (C) - y (D) - exy + 7 4 (E) 7 - yexy 4 + xexy 1 2x2 - 5 2 3 x dx? 13 (B) yexy xexy - 4 1 u3 du 4 3-3 13 (C) 3-3 3 u3 du (D) 4 31 13 u3 du (E) 4 3-3 23. Find the area of the region above the x-axis and beneath one arch of the graph of y (A) 2p + 13 3 (B) 2p +1 3 (C) 13 - p 3 (D) 13 + 4p 3 (E) u3 du 1 sin x. 2 7p 13 + +1 12 2 24. A curve is defined parametrically by x t3 5 and y e2t for 0 t 4. Which of the following is equal to the length of the curve? 4 (A) 19t4 + 4e4t dt 30 4 (B) 16t2e2t + 1 dt 30 4 (C) 2 4 (D) 1t4 + e4t dt 30 30 1 1t3 - 52 2 + e4t dt 4 (E) 2p 30 1t3 - 52 19t4 + 4e4t dt © 2007 Pearson Education, Inc. Sample BC Test 123 5223_Test_100-137 04/10/06 11:24 AM Page 124 25. Find the values of x for which the graph of y xex is concave upward. (A) x 2 (B) x 2 (C) x 1 26. Find the sum of the geometric series (A) 3 5 (B) 5 8 (D) x 1 (E) x 0 3 1 1 9 - + - + ... 8 4 2 3 (C) 13 24 (D) 27 8 (E) 27 40 27. The graph of f(x) x3 x2 has a point of inflection at (A) x = 1 3 (B) x = - 1 3 (C) x = - 2 3 (D) x = 2 27 (E) x = 0 5 28. Use partial fractions to evaluate (A) ln 3 ln 5 124 Sample BC Test 4x - 9 dx 33 2x - 9x + 10 (B) 2 ln 3 ln 5 2 (C) ln 3 2 ln 5 (D) ln 5 ln 3 (E) 2 ln 5 ln 3 © 2007 Pearson Education, Inc. 5223_Test_100-137 04/10/06 11:24 AM Page 125 Section I—Part B (50 minutes) Choose the best answer for each question. (If the exact answer does not appear among the choices, choose the best approximation for the exact answer.) Your score is determined by subtracting one-fourth of the number of wrong answers from the number of correct answers. You may use a graphing calculator. 29. Find the average value of the function y = x 1cos x on the closed interval [5, 7]. (A) 4.4 (B) 5.4 30. The series x + x3 + (A) x ln (1 x2) (C) 6.4 (D) 7.4 (E) 10.8 x7 x2n +1 x5 + +p + + p is the Maclaurin series for 2! 3! n! (B) x ln (1 x2) (C) x2ex 2 (D) xex 2 (E) ex 31. Find the area, in terms of k, for the region enclosed by the graphs of y x4 and y k. (Assume k 0.) (A) 1 2 + k2 1k 4 (B) 2k a k - k2 b 5 4 (C) 2 1 1 + k2 1k (D) 1.6k5>4 (E) 1.8k5>4 (D) 12.5 (E) 25 32. The area enclosed by the graph of r 5 cos 4 is (A) 5 (B) 10 (C) 6.25 33. A region is enclosed by the graphs of the line y 2 and the parabola y 6 x2. Find the volume of the solid generated when this region is revolved about the x-axis. (A) 76.8 © 2007 Pearson Education, Inc. (B) 107.2 (C) 167.6 (D) 183.3 (E) 241.3 Sample BC Test 125 5223_Test_100-137 04/10/06 11:24 AM Page 126 34. Let f(x) be a differentiable function whose domain is the closed interval [0, 5], and let x F(x) = 30 f1t2 dt. If F(5) 10, which of the following must be true? I. F(x) 2 for some value of x in [0, 5]. II. f(x) 2 for some value of x in [0, 5]. III. f ¿ (x) 2 for some value of x in [0, 5]. (A) I only 35. Let g 1x2 = (B) II only x (C) III only (D) I and II (E) I, II and III 1t + 22 1 t - 32e-t dt. 30 For what values of x is g decreasing? (A) x 1.49 (B) x 0.37 (C) –2 x 3 (D) x 2.72, x 0 (E) Nowhere 36. Use the Trapezoidal Rule with the indicated subintervals to estimate the area of the shaded region. (A) 48 126 Sample BC Test (B) 50 (C) 51 (D) 52 (E) 54 © 2007 Pearson Education, Inc. 5223_Test_100-137 04/10/06 11:24 AM Page 127 37. The velocity of a particle moving along the x-axis is given by v(t) t sin t2. Find the total distance traveled from t 0 to t 3. (A) 1.0 (B) 1.5 (C) 2.0 (D) 2.5 (E) 3.0 38. A 15-foot ladder is leaning against a building as shown, so that the top of the ladder is at (0, y) and the dy = -12 feet per bottom is at (x, 0).The ladder is falling because the ground is slippery; assume that dt dx second at the instant when x 9 feet. Find at this instant. dt (A) 6 feet per second (B) 9 feet per second (C) 12 feet per second (D) 16 feet per second (E) 20 feet per second 5 in the first quadrant is revolved about the x-axis x +1 to generate a solid. The volume of this solid is 39. The infinite region beneath the curve y = (A) 5 © 2007 Pearson Education, Inc. (B) 5 (C) 25 (D) 25 (E) q Sample BC Test 127 5223_Test_100-137 04/10/06 11:24 AM Page 128 40. Let f (t) sin t 2 cos t2, where 0 t 4. For what value of t is f(t) increasing most rapidly? (A) 1.76 (B) 2.81 (C) 3.32 (D) 3.56 (E) 3.77 41. 2 A rectangle is inscribed under the curve y = e -x as shown above. Find the maximum possible area of the rectangle. (A) 0.43 (B) 0.61 (C) 0.71 (D) 0.86 (E) 1.77 42. Let fn(x) denote the nth-order Taylor polynomial at x 0 for cos x (that is, the sum of the terms up to and including the xn term). For what values of n is fn(0.8) cos x? (A) 0, 2, 4, 6, 8, 10, … (B) 1, 3, 5, 7, 9, 11, … (C) 1, 2, 5, 6, 9, 10, … (D) 2, 3, 6, 7, 10, 11, … (E) 3, 4, 7, 8, 11, 12, … 43. Find the average rate of change of y with respect to x on the closed interval [0, 3] if (A) 1 ln 10 6 128 Sample BC Test (B) 1 ln 3 6 (C) 1 ln 10 2 (D) 1 10 dy x = 2 . dx x +1 (E) 3 10 © 2007 Pearson Education, Inc. 5223_Test_100-137 04/10/06 11:24 AM Page 129 44. The position vector of a particle moving in the xy-plane is given by r 1 t 2 = 8sin-1 t, 1t + 42 2 9 9 for 1 t 1. The velocity vector at t 0.6 is (A) 8sin -1 0.6, 21.169 (B) 81. 25, 9.29 5 (C) h , 1.2 i 3 5 (D) h , 9.2 i 3 (E) h 75 , 2i 64 45. The base of a solid is the region in the xy-plane beneath the curve y sin kx and above the x-axis for p 0 x . Each of the solid’s cross-sections perpendicular to the x-axis has the shape of a rectangle 2k with height cos2 kx. If the volume of the solid is 1 cubic unit, find the value of k. (Assume k 0.) (A) 3 (B) 3 © 2007 Pearson Education, Inc. (C) 1 3p (D) p 3 (E) 1 3 Sample BC Test 129 5223_Test_100-137 04/10/06 11:24 AM Page 130 Advanced Placement Calculus BC test Section II Show your work. In order to receive full credit, you must show enough detail to demonstrate a clear understanding of the concepts involved. You may use a graphing calculator. Where appropriate, you may give numerical answers in exact form or as decimal approximations correct to three decimal places. For Problems 1–3, a graphing calculator may be used. (45 minutes) 1. The figure above shows a pump connected by a flexible tube to a spherical balloon. The pump consists of a cylindrical container of radius 8 inches, with a piston that moves up and according to 24 + ln 1t + 12 for 0 t 100, where t is measured in seconds and h(t) is the equation h1t 2 = t +1 measured in inches. As the piston moves up and down, the total volume of air enclosed in the pump and the balloon remains constant, and r(t) = 0 at t = 0 a The volume of a sphere with radius r 4 is pr3. b 3 (a) Write an expression in terms of h(t) for the total volume of the air enclosed in the pump and the balloon. (Do not include the flexible tube.) (b) Find the rate of change of the volume of the air enclosed in the pump at t 3 sec. (c) Find the rate of change of the radius of the ballon at t 3 sec. (d) Find the maximum volume of the balloon and when it occurs. 130 Sample BC Test © 2007 Pearson Education, Inc. 5223_Test_100-137 04/10/06 11:24 AM Page 131 2. Let f be a function that has derivatives of all orders on the interval (1, 1). Assume that f102 = 6, f¿ 1 0 2 = 8, f– 1 0 2 = 30, f– 1 02 = 48, and 0 f142 1 x2 0 75 for all x in the interval (0, 1). (a) Find the third-order Taylor series about x 0 for f(x). (b) Use your answer to part (a) to estimate the value of f(0.2). What is the maximum possible error in making this estimate? (c) Let g(x) = x f(x2). Find the Maclaurin series for g(x). (Write as many nonzero terms as possible.) (d) Let h(x) be a function that has the properties h(0) 5 and h¿ (x) f(x). Find the Maclaurin series for h(x). (Write as many terms as possible.) © 2007 Pearson Education, Inc. Sample BC Test 131 5223_Test_100-137 04/10/06 11:24 AM Page 132 3. Consider the family of polar curves defined by r 2 cos ku, where k is a positive integer. (a) Show that the area of the region enclosed by the curve does not depend on the value of k. What is the area? (b) Write an expression in terms of k and u for the slope (c) Find the value of 132 Sample BC Test dy of the curve. dx dy p at u = , if k is a multiple of 4. dx 4 © 2007 Pearson Education, Inc. 5223_Test_100-137 04/10/06 11:24 AM Page 133 No calculator may be used for Problems 4–6. Students may continue working on Problems 1–3, but may not use a calculator. (45 minutes) 4. A particle travels in the xy-plane according to the equations x(t) t3 5 and y(t) 4t2 3 for t 0. (a) For t 5, find the velocity vector and its magnitude. (b) Find the total distance traveled (i.e., the length of the path traced) by the particle during the interval 0 t 5. (c) Find dy as a function of t. dx (d) Find d2y as a function of t. dx2 © 2007 Pearson Education, Inc. Sample BC Test 133 5223_Test_100-137 04/10/06 11:24 AM Page 134 5. 3 The shaded region is enclosed by the graphs of y x3 and y = 4 14x. (a) Find the coordinates of the point in the first quadrant where the two curves intersect. (b) Use an integral with respect to x to find the area of the shaded region. (c) Write an integral with respect to y that could be used to confirm your answer to part (b). (d) Without using absolute values, write an integral expression that gives the volume of the solid generated by revolving the shaded region about the line x 1. Do not evaluate. 134 Sample BC Test © 2007 Pearson Education, Inc. 5223_Test_100-137 04/10/06 11:24 AM Page 135 6. The graph of a differentiable function g is shown. Assume that the area of the shaded region is 8 x>3 +2 square units. Let f1x2 = 32 g 1 t 2 dt. (a) Find f(12). (b) Find f ¿ (6). (c) Write an expression for f– 1 x 2 in terms of the functions g. © 2007 Pearson Education, Inc. Sample BC Test 135 5223_Test_100-137 04/10/06 11:24 AM Page 136 (d) For what values of x is the graph of y f(x) concave downward? Explain. (e) Sketch a possible graph for y f(x). 136 Sample BC Test © 2007 Pearson Education, Inc. 5223_Test_100-137 04/10/06 11:24 AM Page 137 DATE NAME SAMPLE AB TEST SAMPLE BC TEST 1. A B C D E 26. A B C D E 1. A B C D E 26. A B C D E 2. A B C D E 27. A B C D E 2. A B C D E 27. A B C D E 3. A B C D E 28. A B C D E 3. A B C D E 28. A B C D E 4. A B C D E 29. A B C D E 4. A B C D E 29. A B C D E 5. A B C D E 30. A B C D E 5. A B C D E 30. A B C D E 6. A B C D E 31. A B C D E 6. A B C D E 31. A B C D E 7. A B C D E 32. A B C D E 7. A B C D E 32. A B C D E 8. A B C D E 33. A B C D E 8. A B C D E 33. A B C D E 9. A B C D E 34. A B C D E 9. A B C D E 34. A B C D E 10. A B C D E 35. A B C D E 10. A B C D E 35. A B C D E 11. A B C D E 36. A B C D E 11. A B C D E 36. A B C D E 12. A B C D E 37. A B C D E 12. A B C D E 37. A B C D E 13. A B C D E 38. A B C D E 13. A B C D E 38. A B C D E 14. A B C D E 39. A B C D E 14. A B C D E 39. A B C D E 15. A B C D E 40. A B C D E 15. A B C D E 40. A B C D E 16. A B C D E 41. A B C D E 16. A B C D E 41. A B C D E 17. A B C D E 42. A B C D E 17. A B C D E 42. A B C D E 18. A B C D E 43. A B C D E 18. A B C D E 43. A B C D E 19. A B C D E 44. A B C D E 19. A B C D E 44. A B C D E 20. A B C D E 45. A B C D E 20. A B C D E 45. A B C D E 21. A B C D E 21. A B C D E 22. A B C D E 22. A B C D E 23. A B C D E 23. A B C D E 24. A B C D E 24. A B C D E 25. A B C D E 25. A B C D E © 2007 Pearson Education, Inc. Sample BC Test 137 5223_ans_138-156 04/10/06 11:18 AM Page 138 Answers Concepts Worksheets (g) Sections 1.2–1.6 1. (a) (h) (b) (i) (c) (j) (d) (k) (e) (l) (f) 138 Answers © 2007 Pearson Education, Inc. 5223_ans_138-156 04/10/06 11:18 AM Page 139 Symmetry with respect to y-axis or origin Even or Odd Function— f(x) f(x) or f(x) f(x) Is the function periodic? If so, state the period. Is f(x) a one-toone function? (For each f(x) only one x exists) Function Domain Range y f(x) Zeros (Find x when f(x) 0) (a) f(x) x2 R y0 x0 y-axis Even No No (b) f(x) x3 R R x0 Origin Odd No Yes (c) f(x) 0 x 0 R y0 x0 y-axis Even No No (d) f(x) sin x R 0y0 1 k Origin Odd Yes, 2 No (e) f(x) cos x R 0y0 1 p x = 12k + 12 , 2 y-axis k HJ Even Yes, 2 No (f) f(x) tan x x H R, p x Z 1 2k + 1 2 , 2 k H J, R Origin Odd Yes, No (g) f(x) sec x x H R, p x Z 1 2k + 1 2 , 2 k H J, 0y0 1 None y-axis Even Yes, 2 No (h) f(x) 2x R y0 None Neither Neither No Yes (i) f(x) log2 x x0 R x1 Neither Neither No Yes x H R, x0 x H R, y0 None Origin Odd No Yes (k) f(x) 1x x0 y0 x0 Neither Neither No Yes (l) f(x) 1a2 - x2 0 x 0 a 0ya x a y-axis Even No No (j) f(x) 1 x © 2007 Pearson Education, Inc. x k, HJ x k, k HJ Answers 139 5223_ans_138-156 04/10/06 11:18 AM Page 140 3. Yes. Even functions have y-axis symmetry, while odd functions are symmetric about the origin. 4. (a) (b) 9. 10. x g(t), y f(t); 3 11. x e2t, y et [–, ] by [–1.5, 1.5] (c) [–4, 4] by [–3, 3] 5. (a) (b) 12. (a) One possible answer: x t2, y t (b) One possible answer: 1 1 x = 2, t 0t0 [–, ] by [–1.5, 1.5] (c) Sections 2.2–2.3 [–4, 4] by [–3, 3] Section 1.4 1. 2. 3. 4. 5. 6. 7. t<0 t>0 t=0 8. 1. 2. 3. 4. 5. Both Neither Right end behavior model Left end behavior model (a) Continuous at x c (b) Discontinuous at x c; removable discontinuity (c) Discontinuous at x c; removable discontinuity (d) Discontinuous at x c; jump discontinuity (e) Discontinuous at x c; infinite discontinuity (f) Discontinuous at x c; infinite discontinuity 6. Possible answers: 1 1 (a) f1 x2 = (b) f1x 2 = 2 x x -4 (c) f(x) tan x 7. Possible answers: (a) f(x) 3x2 5 0, if x is rational (b) f1 x 2 = e 1, if x is irrational [–4, 4] by [–3, 3] 140 Answers © 2007 Pearson Education, Inc. 5223_ans_138-156 04/10/06 11:18 AM Page 141 Sections 3.1–3.3 1. (a) 3p 2 p +1 2 13 (d) 36 p (f) 8 113 (h) 2 (b) (c) - 4 3 4 (e) -p (g) 2 9 2. (a) At t 1 sec and at t L 4.25 sec (b) 5 t 6 (c) t 3 sec (d) 2. (a) (b) x = -2, x = 1 3. (a) None (b) x 0 (c) x 0 (d) x 0 (e) x 0, x 1 (f) x 0 4. Corner points, cusp points, very sharply changing points, as well as points at which the function is discontinuous or the tangent line is vertical. Section 3.4 1. (a) 5 t 7 (b) 0 t 1, 3 t 4 (c) 1 t 3, 4 t 5 (d) Speed(in./sec) 3. (a) 108 cm (b) 18 cm/sec 1 2 (c) v(t) t + 4t; a 1 t 2 = t + 4 2 (d) none 4. (a) (b) (c) (d) 5. (a) (b) 10 ft/sec; 10 ft/sec; 29 ft/sec t L 1 sec, t L 2 sec t L 3.5 sec t 3, x 228; t 7, x 272 6 4 [–400, 400] by [–1, 2] 2 1 2 3 4 5 6 7 t(sec) Section 3.7 –2 dy dx 2. (a) (b) (c) 3. –4 1. –6 Speed(in./sec) 6 4 2 1 2 3 4 5 6 7 y2 - 5x4 - 4x3y x4 - 2xy - 3y2 1 1 4 = t(sec) –2 –4 –6 (e) 3 t 4 © 2007 Pearson Education, Inc. 4. No; The expression is undefined at (0, 0). 5. 0 6. 1 Answers 141 5223_ans_138-156 04/10/06 11:18 AM Page 142 7. No; f ¿ (0) does not exist because the right- and left-hand derivatives at x 0 are not equal. 8. 2; 3 9. f(x) 0 Section 3.8 1. 2. 4. 1eln 2 2 x = ex ln 2; ex ln 2 1 ln 2 2 = 2 x 1 ln 2 2 ; ln 2 5. 1 eln 3 2 x = ex ln 3; ex ln 3 1 ln 3 2 = 3x 1 ln 3 2 ; ln 3 6. 1 eln x 2 x = ex ln x; 1 ex ln x a x # + ln x b = xx 1 1 + ln x 2 x Section 4.1 1. (a) x a, c, g, k (b) x h 3. 4. (c) x b, d, f, h x i, x j, k, m (d) [b, f ], [ j, k], [m, q ) (e) ( q , b], [f, h), [i, j], [k, m] (f) x f, h x i, x k 5. a, c, e, f 6. e, f 7. Each is the inverse function of the other. 1 8. g¿ 1x2 = f¿ 1g1x2 2 9. (g) One possible answer: x f (Answer may depend on assumptions made regarding the end behavior of the function.) (h) x b, h x i, x j, x m (i) None (j) (d, e), (l, n) (k) ( q , b), (b, d), (e, h), ( j, l), (n, q ) (l) x d, e, l, n 2. (a) y 0 (and possibly y p for some positive constant p) (b) x = h 3. x = b, h, i, j 1 7 x 2 2 11 1 11. y = x + 4 4 10. y = 12. y = -2x - 2 Section 3.9 3x +h - 3x 3x # 3h - 3x 3x 1 3h - 1 2 , lim , lim hS0 hS0 hS0 h h h 1. lim 2. 3. 0.693; 1.099 142 Answers © 2007 Pearson Education, Inc. 5223_ans_138-156 04/10/06 11:18 AM Page 143 4. x = b, d, e, h x j, x = l, n 5. Section 4.2 1. (a) (b) The Mean Value Theorem does not apply, because the function is not differentiable at x d. (c) (b) (d) (e) (c) 3. (a) (b) (d) The variation of Rolle’s Theorem does not apply because f1a2 Z f1 b2 . 2. (a) It appears that the graph is not differentiable at x = d. The Mean Value Theorem does not apply. Every point between a and b would be an appropriate c in the Mean Value Theorem, since the tangent line coincides with the secant line for every c. The Mean Value Theorem does not apply, because the function is not differentiable at x = 0. Applying the Mean Value Theorem with a = 0 and b = 4, there is at least one value of c in (0, 4) such that f 14 2 - f 1 02 5 -3 1 f¿ 1 c2 = = = . 4 -0 4 -0 2 Since f¿ 1 x2 is continuous and differentiable, we may apply the Mean Value Theorem to f¿ 1 x2 with a = 1 and b = 3. There is at least one value of c in (1, 3) such that f– 1 c2 = © 2007 Pearson Education, Inc. f¿ 1 4 2 - f¿ 10 2 -1 - 3 = = -1. 4 -0 4 -0 Answers 143 5223_ans_138-156 04/10/06 11:18 AM Page 144 2. Section 4.3 1. 3. 2. One possible answer: 4. 5. 3. 4. (a) Origin; -f (x) (c) 0 (e) 2a, 0 (b) 0 (d) 5a (f) (2a, 6a) 6. c 8. a 10. (a) 0 (c) 0 x, 1 (e) 7. d 9. b (b) Increasing (d) None 11. (a) 0 (b) 0 1 (d) 2 Section 5.2 1 1. 1 2 x dx 30 2. 1 30 2 1x + x2 2 dx 1 1 1 dx dx or 2 30 1 + x 30 31 2x 2 2 1 1 5. dx 6. dx 30 1 + 2x 30 1 + 2x 1 4 1 1 51 dx 8. dx 7. 2 30 1 + x 2 31 x 9. Yes, because they can all be represented by the same limit of Riemann sums. 3. 11 + 2x dx 4. Sections 5.3–5.4 1. 144 Answers (c) 1 (e) 3 2 1 3 , 6 x 6 2 2 2 1 3 (g) 6 x 6 2 2 (h) (f) 0 6 x 6 © 2007 Pearson Education, Inc. 5223_ans_138-156 04/10/06 11:18 AM Page 145 12. (a) 0 (c) 2b 13. (b) 2b b -a . The inscribed rectangular n area approximation is 4. Let ¢x = I = 3f1x0 2 + f1x1 2 + f1x2 2 + p + f1 xn-12 4 ¢x, and the circumscribed rectangular area approximation is C = 3f1x1 2 + f1x2 2 + f1x3 2 + p + f1xn 2 4 ¢x. Section 5.4 1 1. 1 + x2 Therefore, 1 1 +C = B f1 x0 2 + 2f 1 x1 2 + 2f 1x2 2 + p 2 2 2. 2 14x2 + 1 + 2f 1 xn-1 2 + f 1 xn 2 R ¢x sin x 4. 2 1x 3. 0 2 x 7. 16 6. 115 1 8. 2 5. - Section 5.5 = b -a B f1 x0 2 + 2f 1 x1 2 + 2f 1x2 2 + p 2n + 2f1 xn - 1 2 + f 1xn 2 R . Sections 6.1–6.2 1. 2 1. x3 dx 4. The trapezoidal approximation 30 is greater since y x3 is always concave up for 0 x 2 and the top side of each trapezoid lies above the curve. 20 2. 10 x 99 x dx . The trapezoidal dx 1 2 32 4 31 approximation would equal the integral 2. answer since the function being integrated represents a straight line which the trapezoids would exactly fit. 4 3. (a) 30 1x dx = 16 b + 211 2 + 2 a b + 21 12 2 2 2 110 114 + 2a b + 21 132 + 2 a b + 2R 2 2 (b) B0 + 2 a 1 4 16 3 12 (c) Note that f1x2 = 1x is concave down. Since the Trapezoidal Rule approximation is based on straight segments which lie below the graph of y f(x) and the Simpson’s Rule approximation is based on curves which nearly match the graph of y f(x), expect the Simpson’s Rule approximation to be larger. To confirm, observe that S L 5.3046 and T L 5.2650. © 2007 Pearson Education, Inc. x 1 dt = ln x 31 t 1 1 (b) dt = 0 31 t 1 1 (c) dt = -ln x 3x t 4. (a) 3. (a) Answers 145 5223_ans_138-156 04/10/06 11:18 AM Page 146 (b) Sections 7.1–7.5 60 R1 t 2 dt (b) 842.5 cm 30 2. (a) L 4.6; 4.6 inches of rain fell during the 24 hours beginning at midnight. (b) L 1.2; 1.2 inches of rain fell between 4 A.M. and noon. (c) L 2.5; 2.5 inches of rain fell between 8 A.M. and 8 P.M. 3. $2,500,000 4. 3200 ft-lb 1. (a) (c) 5. (a) Section 7.2 b 1. 3a f1 x 2 dx c 2. (b) 3a b 3. 3a c 4. f1 x 2 dx + 3a + 6. (a) ln 0 x 0 (b) ln 11 + x2 (c) -ln 1 1 – x2 7. If t = au, then dt = adu. Therefore, when t = a, u = 1 and when t = ax, u = x. ax x x 1 1 # du adu = dt = 3a t 31 u 31 au ax ax 1 dt = ln t ` = ln 1ax 2 - ln 1 a 2 a 3a t x x du = ln u ` = ln x - ln 1 = ln x 1 31 u Therefore, ln (ax) ln (a) ln x, so ln (ax) ln x ln (a). 1 dy = 8. (a) (b) y = ln x + C dx x 1 dy 9. (a) = dx x (b) The equations are the same. Since ln 2x = ln 2 + ln x, the general solution y = 1 dy ln x + C of the equation = includes dx x the particular solution y = ln 2x. 146 Answers 3c 6. 2 30 32 7. 3 f1 x 2 dx 3g 1 x 2 - f1 x2 4 dx 3c d 5. 3c 3 f1 x2 - g 1 x2 4 dx b (c) b 3f 1 x2 - g 1 x 2 4 dx 3g 1 y 2 - f 1 y2 4 dy a f1 y 2 dy or 8. (a) D1 t 2 = a 3-a f1 y 2 dy t 1 R1 x2 - E1 x2 2 dx 30 (b) - 105.5 or a deficit of 105.5 muds (c) The deficit disappears in 2039. Section 7.3 1. (a) (b) © 2007 Pearson Education, Inc. 5223_ans_138-156 04/10/06 11:18 AM Page 147 (c) px4 (d) px 4dx 2 (e) p x4 dx 30 (f) 2. (a) (b) (d) p 1 4 –y2 dy (c) 4p - py 4 (e) p 30 14 - y2 dy 3. (a) (c) 32p 5 (f) 8p (b) 1 4 - 2y 2 a b = 12 - y2 2 2 12 2 (d) (2 y)2 dy (f) 4. (a) (b) (c) (d) (e) 30 1 2 - y 2 2 dy b 3a xf1x2 dx Sections 9.1–9.5 1 1 6 x 2 2 1 1 1 1 (b) f a b = + - p 4 2 8 24 1. (a) - + + 5 x2n-1 d + p 1 2n - 1 2 ! Note that, if n is replaced by n + 2 in the general term for f (x), we obtain the opposite of the general term for f(x). This means that f (x) = -f(x), so y = f(x) solves y + y = 0. (b) All real numbers (c) f1 x2 = 3 cosx + sinx 8 3 f(x) 2px 2px f 1x2 2px f1x 2 dx (e) 2p 3 (c) ln a b 2 (d) f¿ 1 x2 = 2 - 4x + 8x2 - p + 1 -1 2 n + 1 1 2 2 1 2x 2 n - 1 + p ; 1 4 f¿ a b = 4 3 2. (a) 7 - 3 (x - 3) + 6 (x - 3)2 (b) 8.44; 0.162 3 (c) -2 + 7 (x - 3) - (x - 3)2 + 2 (x - 3)3 2 2 3 4 (d) 7x - 3x + 6x 5 3 5 3. (a) f¿ 1 x2 = 5 - 3x - x2 + x3 + x4 2! 3! 4! 3 5 5 6 - x - x + p 5! 6! 3 + 1 -12 n c x2n - 1 1 2n - 1 2! 5 + x2n d + p 1 2n2 ! 3 5 3 f– 1x 2 = -3 - 5x + x2 + x3 - x4 2! 3! 4! 5 5 - x + p 5! 3 + 1 -12n c x2n-2 1 2n - 22 ! Section 10.2 1. (a) 1 - 12, 2 2 ,1 12, 2 2 (b) (0, 0) 16 3 16 3 (c) (0, 0), a , b, a , b 2 2 2 2 (d) 1 - 1 2 n+1 1 n a b + p n 2 © 2007 Pearson Education, Inc. Answers 147 5223_ans_138-156 04/10/06 11:18 AM Page 148 p p 13 3 b = r¿ a b = h ,- i 6 6 2 2 p p 1 3 13 a a b = r– a b = h , i 6 6 2 2 p 3 3 13 i (b) r a b = h , 6 2 2 2. (a) v a 2. (a) (b) (c) (c) minimum t = 0, p, 2p p 3 maximum t = , p 2 2 2 3 3. (a) 8t , t – 19 (b) (c) r¿ 11 2 = 82, 39 (d) r– 1t 2 = 82, 6t 9, r– 102 = 82, 0 9 (e) There exist none since the horizontal component of the vector is always nonzero. 4. (a) r 1t 2 = 8C1et, C2et 9 where C1, C2 are arbitrary constants. (b) r 1t2 = 8C1e-t, C2e-t 9 where C1, C2 are arbitrary constants. Section 10.3 1. (a) 148 3. (a) (b) 4. 5. (a) (b) (b) (c) (c) Answers © 2007 Pearson Education, Inc. 5223_ans_138-156 04/10/06 11:18 AM Page 149 6. (a) (b) 7. (a) x = (1 + cos u) cos u, y = (1 + cos u) sin u dx = - sin u - 2sin u cos u (b) du = - sin u - sin 2u; dy = cos u + cos2u - sin2u du = cos u + cos 2u cos u + cos 2u (c) - sin u - sin 2u (d) y = x + 1 (e) y = 1 - x 8. (a) x = -cos2 u, y = -cos u sin u dx = 2 sin u cos u = sin 2u, (b) du dy = -cos2 u + sin2 u = -cos 2u du - cos 2u = -cot 2u (c) sin 2u p 3p (d) , 4 4 p (e) 0, , p 2 9. To find point of intersection: 2 cos u = 2 sin u p or at the pole. u = 4 For r = 2 cos u, dy 2 cos2 u - 2 sin2 u 2 cos 2u = = = -cot 2u dx -4 sin u cos u -2 sin 2u dy p = 0 at u = . dx 4 For r = 2 sin u, 4 sin u cos u dy = = tan 2u dx 2 cos 2u p dy is undefined (vertical slope) at u = . dx 4 Therefore, the slopes of the tangents at p u = indicate the lines are perpendicular. 4 © 2007 Pearson Education, Inc. At the pole: r = 2 cos u and r = 2 sin u intersect for different values at u, namely 0 = p>2 and u = 0, respectively. Slope of the tangent to r = 2 cos u at 0 = p>2 is undefined. Slope of the tangent to r = 2 sin u at u = 0 is 0. Therefore, the slopes indicate the tangent are perpendicular at the pole. p 3p 3 3 , sin-1 , p - sin-1 10. (a) u = , 2 2 5 5 (b) 11. The point closest to the pole is at (1, p), and the line tangent to r = 2 + cos u at this point is vertical. For the “dimpling effect” we are interested in any other vertical tangents symmetrically situated on the curve about u = p. The only other vertical tangent occurs when u = 0. Therefore, there is no dimple around u = p. 12. 1 a 6 2 Group Activity Explorations Chapter 1 Review Here is one possible solution: A: 4 B: 6 C: 10 D: 18 E: 1 F: y = 4x + 1 1 1 G: H: y = - x + 18 4 4 1 I: 1 J: 4 K: -1 L: Denmark M: 8 N: 1 O: tan P: p Q: 0 R: 0 S: 1 T: 162 U: 9 V: SINE W: U X: 1 Special message: You won! Answers 149 5223_ans_138-156 04/10/06 11:18 AM Page 150 Section 2.3 A 1. When x takes on only irrational values, f(x) 0 and lim 0 = 0. Section 4.4 1. xS0 2. When x takes on only rational values, f(x) x and lim x = 0. xS0 3. lim 0 = 0 and lim a = a. xSa xSa 4. Since lim 0 = lim a only when a 0, f(x) is xSa xSa continuous only at x 0. B 1. When x is any integer multiple of p, sin x 0. sin px, if x is rational 2. f 1x 2 = e 0, if x is irrational 3. f (x) int x (the step function). 2. This makes sense because a car consumes more fuel when traveling very slow or very fast compared to traveling at a moderate speed. 3. The critical point is approximately (36.84, 0.95). The consumption rate C (in gallons/hour) is minimized. y C1 v 2 gallons>hour C 1v 2 = = gallons>mile 4. x v miles>hour v 5. One possible answer: C x2 - 4, 0, x3 - x, 2. f 1x 2 = e 0, px , sin 2 • f 1 x2 = 3. 0, 1. f 1x 2 = e if x is rational if x is irrational if x is rational if x is irrational if x is rational if x is irrational 1 D f (x) is continuous when x = ; , where n is a n positive integer. Section 3.1 f1x + h 2 - f1x 2 h f1x2 + f1h2 + 2xh - f 1 x2 = lim hS0 h f1h2 + 2xh f1h2 2xh = lim = lim + hS0 hS0 h h h = 10 + 2x One possible answer: f(x) = x2 + 10x One possible answer: f(x) = x2 + 10x + 1 f(x) = x2 + 10x + C; C is any real number. f 10 + h 2 = f102 + f1h 2 + 2 # 0 # h f 1h 2 = f102 + f 1h2 + 0 0 = f102 + 0 f10 2 = 0 f(x) = x2 + 10x f(x h) = (x + h)2 + 10(x + h) = x2 + 2xh + h2 + 10x + 10h = x2 + 10x + h2 + 10h + 2xh = f(x) + f(h) + 2xh f (x) = 2x2 + 10x 9. f(x) = x2 + 20x A f 1x2 = x 2 + Bx 2 1. f¿ 1 x2 = lim hS0 2. 3. 4. 5. 6. 7. 8. 10. 150 Answers The slope of this line represents the number of gallons per mile at the chosen speed. 6. Gallons per mile is not minimized at the critical point because the slope of the line through a point on the graph and the origin is not minimized at the critical point. 7. When the slope of the line through a point on the graph and the origin is minimized, fuel consumption will be minimized at that point. This occurs when the line is tangent to the graph. C1 v 2 C¿ 1 v 2 = 8. v 1.5 3 -7 3 4v * 10 - 0.02 = v * 10-7 - 0.02 + v 1.5 4v3 * 10-7 = v3 * 10-7 + v 1.5 3 -7 3v * 10 = v 1.5 4 v = 3 * 10-7 1>4 1.5 b L 47.287 v = a 3 * 10-7 9. One possible answer: A calculator cannot determine the velocity that optimizes fuel consumption given this equation because the minimum value of the graph does not correspond to the optimum velocity. 10. One possible answer: When a shark starts at zero velocity, it must expend extra energy to © 2007 Pearson Education, Inc. 5223_ans_138-156 04/10/06 11:18 AM Page 151 overcome inertia. Once it is moving it requires less energy to stay moving, but it must expend increasingly more energy against the water resistance as it swims faster. 11. One possible answer: The graph shows us that a shark is expending less energy when it is moving slowly than when it is staying still in the water. 12. One possible answer: It is true that the shark has to swim faster to minimize calories per mile than to minimize calories per hour. It would always make sense to minimize calories per hour (requiring less feeding) if that were an option, but the shark does not have that luxury. It must cover great distances to “graze” for food. That is why it depends on the circumstances. When it is hunting, it should minimize calories per mile; when it is resting, it should minimize calories per hour. Section 5.4 1. P = (h,r) 2. 2 r 3. V = p (radius)2(height) = p a xk b ¢x h h n pr2 2 pr2 2 4. lim a 2 xk ¢x = 2 x dx nSq 30 h k =1 h 5. Since we are evaluating the integral with pr2 respect to x, 2 is a constant. It is always safe h to move a constant factor of the integrand outside of the integral. h pr2 pr2 h3 1 x2 dx = 2 . = pr2h 6. 2 h 30 h 3 3 7. radius = r; height = h 8. Since we are adding the same discs, the Riemann sum is unchanged: h n pr2 2 pr2 2 lim a 2 xk ¢x = 2 x dx nSq 30 h k =1 h 1 9. V = pr2h 3 © 2007 Pearson Education, Inc. Section 6.1 I. dy = x + y, the slope should be zero 1. For dx dy = x - y, the slope where y = -x. For dx should be zero where y = x. dy = x + y, 2. Graph A is the slope field of dx dy = x - y. and graph B is the slope field of dx 3. As you move to the right, x increases, so the values of x + y and x - y are both increasing. II. 2 1. The slope is always positive because 1.1x and 2 1.1y are always positive. 2. Graph A shows slopes that depend on x but not on y, and graph B shows slopes that depend on y but not on x. dy 2 = 1.1x , and 3. Graph A is the slope field of dx dy 2 = 1.1y . graph B is the slope field of dx 4. For either differential equation, the solution that passes through (0, 0) would be an odd function, while all other solutions would be neither odd nor even. III. 1. Because both equations depend on y but not on x. dy = sin (y3), and 2. Graph A is the slope field of dx dy = sin (y2). graph B is the slope field of dx You can tell because sin (y2) is an even function of y, meaning that the slope shown at any point (x, y) is the same as the slope shown at (x, y). Since sin (y3) is an odd function of y, the slope shown at any point (x, y) is the opposite of the slope shown at (x, y). 3. No. Both cos (y2) and cos (y3) are even functions of y. IV. dy = cos 1xy 2 , 1. Graph A is the slope field of dx dy = sin 1xy 2 . and graph B is the slope field of dx 2. Since sin (0) = 0, the slopes representing dy = sin 1 xy 2 should be zero along the axes, dx and since cos (0) = 1, the slopes representing dy = cos 1 xy 2 should be 1 along the axes. dx Answers 151 5223_ans_138-156 04/10/06 11:18 AM Page 152 3. Since the sine function is odd, changing the sign of x or y should change the sign of the slope, as shown in graph B. Since the cosine function is even, the slope shown at any point should be the same as the slope at any corresponding point on the other side of either axis, as shown in graph A. Section 8.4 1. Let f1 x2 = e-1x >22 and g 1 x2 = e-x # f and g are 2 continuous functions for all x, 0 f (x) q g(x) for all x 2, and 32 So, by Theorem 6 (Direct Comparison Test), Section 7.3 r1 xk xk = ; r1 = r2 2. r2 h h q e-1x >22 dx also converges. 2 32 q e-1x >22 dx. Since f 1 x2 = e-1x >22 is 2 Let a = 32 2 q an even function, 2 b = 3. Rotate the cone 90º about the x-axis, then use the same similar triangles as in step 2. 4. Since the cross section is scaled both xk horizontally and vertically by a factor of , h xk 2 the area is scaled by a factor of a b . For h example, a triangle of height a and base b which is scaled both horizontally and vertically by a factor of r will have area 1 1 1rb 2 1 ra 2 = a ba b r2, which is the original 2 2 area scaled by a factor of r2. 5. The volume of the slab equals the area of its base times its height. Since the area of its base x 2 is A a k b , and its height is x, the volume h x 2 of the slab is A a k b ¢x. h n n x 2 6. lim a Vk = lim a A a k b ¢x nSq nSq h k=1 k=1 h A 2 = 2 x dx 30 h 7. Since we are evaluating the integral with A respect to x, 2 is a constant, so we can move h it outside of the integral by the Constant Multiple Rule for Definite Integrals. h A A x3 h A h3 1 x2 dx = 2 B R = 2 * = Ah 2 h 30 h 3 0 h 3 3 1 2 13 2 9. a. V = s h b. V = sh 3 12 13 2 c. V = sh 2 8. 152 Answers e-x dx converges. e 3-2 q - 1x2>22 e-1x >22 dx = a. Now let 2 32 dx . We have e-1x >22dx = a + b + a = 2a + b. 2 3-q q Therefore 4 2. e-1x >22 dx converges. 2 3-q e-1x >22 dx L 2.50646950 2 3-4 6 e-1x >22 dx L 2.50662827 2 3-6 8 e1x >22 dx L 2.50662828 2 3-8 40 3. e-1x >22 dx L 60 2 3-40 e-1x >22dx 2 3-60 80 e-1x >22 dx L 2.506628275. 2 L 3-80 4. It appears that N e-1x >22 dx L 2.50662875. 2 lim NSq 3-N We cannot conclude that this is an exact value of the integral. 5. No 6. (2.506628275)2 6.283185307 6.283185307 = 3.141592654; 7. 2 q e-1x >22dx = 12p 2 8. 9. 10. 11. 3-q 0 120 units wide 2 e-60 >2 L 0 All 100 of the rectangles will have area L 120 # 0 = 0; The sum is 0. © 2007 Pearson Education, Inc. 5223_ans_138-156 04/10/06 11:18 AM Page 153 12. 150 rectangles would each be 80 units wide. 200 rectangles would each be 60 units wide. The MRAM sum would still be 0. 13. If all of the rectangles have a calculated height of zero, then the sum of all of the rectangle areas will also be zero. Section 9.2 1. TO 1x 2 = x + 2. x3 x5 x7 x 2n+1 x2 x4 x6 x2n + + +p + + p ; TE 1x 2 = 1 + + + +p + + p 3! 5! 7! 12n + 1 2! 2! 4! 6! 12n 2! T 1x2 + T1 -x 2 2 a1 + x + = x2 x3 xn 1 -x2 2 1 -x2 3 1 -x2 n + + p + + p b + a 1 + 1 -x2 + + + p + 1 -12 n + pb 2! 3! n! 2! 3! n! 2 x3 x2 x3 x2 + + pb + a1 - x + + pb 2! 3! 2! 3! 2 a1 + x + = 2x2 2x4 + + p x2 2! 4! x4 x2n = 1 + = + + p + + p = TE 1 x2 2 2! 4! 1 2n2 ! T 1x2 - T1 -x 2 3. 2 2 + a1 + x + = x2 x3 xn 1 -x2 2 1 -x2 3 1 -x2 n + + p + + p b - a 1 + 1 -x2 + + + p + 1 -12 n + pb 2! 3! n! 2! 3! n! 2 x2 x3 x2 x3 + + pb - 1 + x + + p 2! 3! 2! 3! = 2 3 5 2x 2x 2x + + + p 3! 5! x3 x5 x2n+1 = = 1 + + + p + + p = TO 1 x2 2 3! 5! 1 2n + 1 2 ! TE 1x2 is similar to cos x except that every other term in the expansion of cos x is negative. TO 1x2 is similar to sin x except that every other term in the expansion of sin x is negative. ex - e-x sinh x = 2 x -x x e +e e - e-x ex + e-x + ex - e-x 2ex + = = = ex 2 2 2 2 f1x 2 + f1 -x 2 f 1x 2 - f 1 -x 2 f1 x2 + f1 - x2 + f1 x2 - f1 -x2 2f 1 x 2 + = = = f 1x 2 2 2 2 2 f1x 2 - f1 -x 2 f 1 -x2 - f1 x2 Furthermore, fO 1x2 = is odd because fO 1 -x2 = 2 2 f1x 2 - f1 - x 2 f 1 x2 + f 1 -x2 = = -fO 1 x2 , and fE 1 x2 = is even because 2 2 a1 + x + 4. 5. 6. 7. 8. fE 1 -x2 = f 1 -x2 + f1 x2 f 1x2 + f1 -x2 = -fE 1 x2 . = 2 2 © 2007 Pearson Education, Inc. Answers 153 5223_ans_138-156 04/10/06 11:18 AM Page 154 1 1 +x 1 1 -x + + f 1 x 2 + f 1 -x2 2 1 -x 1 +x 1 1 + x2 1 1 - x2 1 1 + x 2 1 1 - x2 = = = 9. fE 1x 2 = 2 2 2 1 - x2 1 1 = 2 1 - x2 # 1 1 1 +x 1 -x f1x2 - f1 -x2 1 -x 1 +x 11 + x2 11 - x2 11 + x2 11 - x2 2x fO 1x2 = = = = 2 2 2 1 - x2 # 1 x = 2 1 - x2 x = x + x3 + x5 + x7 + p + x2n+1 + p 1 - x2 1 = 1 + x2 + x4 + x6 + p + x2n + p 1 - x2 1 = 1 + x + x2 + x3 + x4 + p + xn + p 1 - x x2 x3 x4 xn 11. ln 1x + 12 = x + + p + 1 -12 n-1 +p 2 3 4 n x2 x4 x6 x2n ln 11 - x2 = - p - p 2 4 6 2n 1 + x x3 x5 x2n-1 ln a b = x + + + p + + p A1 - x 3 5 2n - 1 10. Section 10.3 2. If n is odd, then there are n petals. If n is even, then there are 2n petals. 3. If n is odd, then each petal is traced twice over the interval 0 u 2p. So, for all positive integers, n, there are 2n petals traced over the interval 0 u 2p. 4. NINT((2 sin (2x))2, x, 0, 2p)/2 L 6.283185307. We integrate over the interval 0 x 2p to include the entire graph. 5. 2p 6. There are three times as many petals, but the area is the same. 7. NINT ((2 sin (3x))2, x, 0, p)/2 L 3.141592654. We integrate over the interval 0 x p so that we include each petal only once. The exact area is p. 2p, n even 8. A = e p, n odd 9. One petal is traced over the interval p 0 u , so n p>n 1 A = 12 sin 1nu 2 2 2 du 2 30 p>n 1 = 4 sin2 1nu 2 du 2 30 p>n =2 154 30 Answers sin2 1nu 2 du 10. Let u = nu and du = n du. du = du. So, 0 u p and n p>n 2 30 sin 1 nu2 du = 2 2 p sin2 1 u2 du n 30 p 2 = sin2 1 u2 du. n 30 p 2 A sin2 1 u2 du = . 11. The area of 1 petal is n 30 n If n is odd, there are n petals and the area is A n a b = A. If n is even, there are 2n petals n A and the area is 2n a b = 2A. n p p u sin 2u 2 R = p. 12. A = 2 sin 1 u2 du = 2 B 2 4 0 30 So, if n is odd the area is p, and if n is even the area is 2p. Sample Advanced Placement Tests AB: Section I Answers 1. 5. 9. 13. 17. 21. B B A B D B 2. E 6. C 10. E 14. E 18. A 22. E 3. C 7. D 11. D 15. D 19. D 23. C 4. E 8. C 12. D 16. D 20. B 24. C © 2007 Pearson Education, Inc. 5223_ans_138-156 04/10/06 11:18 AM Page 155 25. 29. 33. 37. 41. 45. A E C B A D 26. E 30. E 34. E 38. D 42. D 27. C 31. A 35. E 39. B 43. A 28. 32. 36. 40. 44. (d) B E C B D 6. (a) y AB: Section II Answers 3 1. (a) 2 - 5p cos pt 5 5 (b) t2 + cos pt p p (c) 0 6 t 6 0.885 or 2.418 6 t 6 2.500 (d) L 4.658 5 y + 2 1x 2. (a) x + 3y2 15 4 (b) y = - x + 7 7 (c) 1.8 (d) 5 10.6 + 0.6y + y3 = 11; f1 0.62 L 1.821 (e) No; 0.1 is not in the domain of f (x) because 1 -0.1 is not a real number. 4 3. (a) 64ph 1t2 + pC r1t2 D 3 3 (b) Volume = 64p11 + ln 24 2 L 840.048 in3 at t = 23 sec (c) -80p L -251.327 in3>sec 3 (d) Radius = 1 48118 - ln4 2 L 9.273 in.; 5 rate = L 0.233 in.>sec 3 6118 - ln 4 2 4 2>3 4. (a) 12, 82 (b) (c) 2 3 14 1 4x - x3 2 dx = 8 8 3 a1 y - 30 30 8 (d) p 30 y3 b dy 256 3 B11 y + 12 2 - a 2 or 2p 2 y3 + 1 b R dy 256 3 1x + 12 14 1 4x - x3 2 dx 30 5. (a) 5 (b) -1 6 x 6 2, since g(x) is increasing for these vaues of x. x (c) -3 + 30 g1t2 dt © 2007 Pearson Education, Inc. 2 1 –3 (b) –2 x –1 dy = 1y + 52 1x + 22 dx dy = 1 x + 2 2 dx y +5 dy = 1 x + 2 2 dx 3 3y + 5 x2 ln 0 y + 5 0 = + 2x + C 2 2 ex >2 + 2x + C = y + 5 y = C ex >2 + 2x - 5 f1 0 2 = 2 . . . C = 7 2 y = 7e x >2 + 2x - 5 2 (c) Domain: all real numbers Range: g 5 BC: Section I Answers 1. 5. 9. 13. 17. 21. 25. 29. 33. 37. 41. 45. B D C A B C B B E E D E 2. C 6. B 10. C 14. B 18. B 22. B 26. E 30. D 34. D 38. D 42. D 3. E 7. A 11. C 15. C 19. C 23. A 27. B 31. D 35. E 39. D 43. A 4. A 8. B 12. D 16. B 20. C 24. A 28. A 32. D 36. C 40. E 44. B Answers 155 5223_ans_138-156 04/10/06 11:19 AM Page 156 BC: Section II Answers 4 1. (a) 64ph1t 2 + p3 r1t2 4 3 3 (b) - 80p L - 251.327in3>sec 5 L 0.233 in.>sec (c) 36 118 - ln4 2 4 2>3 (d) Volume = 64p 123 - ln 24 2 L 3985.439 in3 at t = 23 sec 2. (a) (b) (c) (d) 6 + 8x + 15x2 + 8x3 8.264; maximum error is 0.005. 6x + 8x3 + 15x5 + 8x7 + . . . 5 + 6x + 4x2 + 5x3 + 2x4 + . . . 2p 1 r2 du = 4.5p L 14.137 2 30 k sin ku sin u - 12 + cos ku 2 cos u (b) k sin ku cos u + 1 2 + cos ku 2 sin u p (c) 4 3. (a) Area = 5. (a) (2, 8) (b) (c) 2 3 141 4x - x3 2 dx = 8 8 3 a1 y - 30 30 8 (d) p 30 3 c11 y + 122 - a 2 2p y3 b dy 256 2 y3 + 1 b d dy or 256 3 1x + 12 141 4x - x3 2 dx 30 6. (a) 8 (b) 1 x 1 (c) g¿ a + 2 b 9 3 (d) x 6, since this is where g a x + 2 b is 3 decreasing. (e) 4. (a) 875, 409; magnitude 85 (b) 163 units 8 (c) 3t 8 (d) - 4 9t 156 Answers © 2007 Pearson Education, Inc.