2017 Free-Response Solutions - skylit.com

1 BBee PPrreeppaarreedd for the CCaallccuulluuss EExxaamm Mark Howell Gonzaga High School, Washington, Martha Montgomery Fremont City Schools, Fremont, Ohio Practice exam contributors: Benita Albert Oak Ridge High School, Oak Ridge, Tennessee Thomas Dick Oregon State University Joe Milliet St. Mark's School of Texas, Dallas, Texas Skylight Publishing Andover, Massachusetts Third Edition * AP and Advanced Placement Program are registered trademarks of the College Entrance Examination Board, which was not involved in the production of and does not endorse this book. Copyright 2005- 2017 by Skylight Publishing Chapter 10. Annotated Solutions to Past Free-Response Questions This material is provided to you as a supplement to the book Be Prepared for the AP Calculus Exam. You are not authorized to publish or distribute it in any form without our permission. However, you may print out one copy of this chapter for personal use and for face-to-face teaching for each copy of the Be Prepared book that you own or receive from your school.

2 Skylight Publishing 9 Bartlet Street, Suite 70 Andover, MA 01810 web: e-mail: 3 2017 AB AP Calculus Free-Response Solutions and Notes Question AB-1 (a) The left Riemann sum approximation is 2 3 5 cubic feet. 1 (b) This is an overestimate of the volume of the tank since ()Ah is decreasing. (c) The volume is 100fhdh cubic feet. 2 (d) When the height of the water is h feet, the volume of the water 0hVfudu . The volume is changing at the rate of dVdhfhdtdt . When h = 5, cubic feet per minute. Notes: 1. You can leave the answer unsimplified. For the record, it is cubic feet. 2. Use the given function names in your answers; do not repeat their formulas to avoid transcription errors. 4 Free-Response Solutions ~ 2017 AB Question AB-2 (a) 20ftdt pounds. (b) 7f . 1 The rate at which customers remove bananas from the display table is decreasing at the rate of pounds of bananas per hour per hour 7 hours after the store opened.

3 (c) 5f and 5g . Since (5) (5)fg , the bananas are being removed by customers faster than store employees add them at t = 5, so the number of bananas is decreasing at that time. (d) 880350ftdtgtdt . 2 Notes: 1. Use your calculator no need for symbolic differentiation. 2. The fact that f and g are not defined at the lower limits of integration does not affect the values of the two integrals. Free-Response Solutions ~ 2017 AB 5 Question AB-3 (a) 27xfxftdt , so 621677 4232fftdt , and 52211577 2 3222fftdt . 1 (b) f is increasing on the intervals 62x and 25x since f is continuous on [6,5] and 0fx on (6, 2) and (2,5). (c) The only points where f could have an absolute minimum are 6,2, 2xxx , and 5x . 63f , 27f , 272f , 5102f . The absolute minimum is 72 . 2 (d) 152f ; 3f does not exist, because 33lim23xfx fx and 33lim13xfx fx . Notes: 1. You can leave the answer unsimplified.

4 For the record, it is 10 2 . 2. The candidate test is the easiest way to justify absolute extrema for a continuous function on a closed interval. 6 Free-Response Solutions ~ 2017 AB Question AB-4 (a) 0191 27 164tdHdt . 1 An equation of the tangent line is 91 16yt . At t = 3, y = 91 48 = 43. The temperature of the potato is approximately 43 C. (b) 2211127274416dH ddHHHdtdtdt . Since H > 27 for all t > 0, 220dHdt for all t > 0. Therefore, the tangent line approximation is an underestimate of the temperature. (c) Separating variables, we get 2/327 GdGdt . 2/31/327327 GdGdt G tC . From the initial condition (0) 91G , 1/339127 12C . So 1/3327 12Gt 31/327427433ttGG . At t = 3, 27 27 54G C. Notes: 1. ()Ht models the temperature for 0t , but we can assume that dHdt is defined on a larger interval including t = 0. Free-Response Solutions ~ 2017 AB 7 Question AB-5 (a) Particle P is moving to the left when 2220210 Ptxttt 220 0 1tt.

5 (b) 35 Qvt tt so 0 Qvt for 3 < t < 5. The particles never move left at the same time, but for 1 < t < 3 and 58t , both particles are moving to the right. (c) 282 40 QQQat v t ta . 21330Qv . 2Qa and 2Qv have opposite signs, so the speed of the particle is decreasing at t = 2. (d) Particle Q first changes direction at t = 3. Its position at that time is 333220058 15 54 155 9 36 453ttt dtt t . 1 Notes: 1. You can leave the answer unsimplified. For the record, it is 23. 8 Free-Response Solutions ~ 2017 AB Question AB-6 (a) sin2sin 2cos1xfxx exf . The slope of the tangent line is 1 . (b) kx hfx f xkhff . sin() cos(2)2fe ; 123h ; 1f 11133k . (c) 222mx g x hxgx hx 122422425 21 333mgh gh . 1 (d) 358435 2gg . Since g is differentiable, it is also continuous. So by the Mean Value Theorem applied to g on 5, 3 , there must be a c in 5, 3 such that 4gc.

6 Notes: 1. You can leave the answer unsimplified. 9 2017 BC AP Calculus Free-Response Solutions and Notes Question BC-1 See AB Question 1. Question BC-2 (a) Area is 22012fd . 1 (b) 2222201122kkgfd gfd . (c) wg f . The average distance 2012 Awwd . (d) .. 2 Since 0w , w is decreasing. Notes: 1. Use the given function names in the setups for parts (a), (b), and (c). Doing so saves time and is less prone to transcription errors. 2. Use your calculator to calculate the derivative no need for symbolic differentiation. Question BC-3 See AB Question 3. 10 Free-Response Solutions ~ 2017 BC Question BC-4 See AB Question 4. Question BC-5 (a) 22234 715318 21 5275xfxfxx . 1 (b) 704fxx . Since fx changes sign from positive to negative at 74x , f has a relative maximum at 74x . (c) 15521limlim ln 2 5 ln 125 1bbbbf x dxdxxxxx 52525 55lim lnlim lnlnln 2 ln1144bbbxbxb.

7 2 (d) Since fx is continuous, positive, and decreasing for x > 5, and 5fxdx converges, the series converges by the Integral Test. Notes: 1. You can leave the answer unsimplified. For the record, it is 154 . 2. 8ln5 . Free-Response Solutions ~ 2017 BC 11 Question BC-6 (a) 0101ff ; 0202ff ; 403 06ff . So the first four nonzero terms for the Maclaurin series for f are 234 234262! 3! 4!2 3 4xxxxxxxx . The general term is 11nnxn . (b) At x = 1, the series is which is the alternating harmonic series. It converges by the Alternating Series Test. The series is the harmonic series, which diverges. Therefore, the series converges conditionally at x = 1. (c) 234511( ).. 6 12 201nnxxxxxgxnn . (d) 41122Pg is bounded by the magnitude of the first omitted term, which is 5111220 640 500.