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APcal 2004-Solutions

# APcal 2004-Solutions - Be Prepared for the AP Calculus Exam...

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B B e e P P r r e e p p a a r r e e d d for the A A P P C C a a l l c c u u l l u u s s E E x x a a m m Mark Howell Gonzaga High School, Washington, D.C. 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

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Copyright © 2005 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 (ISBN 0-9727055-5-4). 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. Skylight Publishing 9 Bartlet Street, Suite 70 Andover, MA 01810 web: http://www.skylit.com e-mail: [email protected]
3 2004 AB AP Calculus Free-Response Solutions and Notes Question 1 (a) 30 0 82 4sin 2 t dt + ­ 2474 . (b) 7 t dF dt = ­ 1.873 0 ≈ − < ( ) F t is decreasing. (c) 15 10 1 82 4sin 5 2 t dt + ­ 81.899 cars/min. (d) [ ] 1 1 15 10 (15) (10) 4sin 4sin 5 5 2 2 F F = ­ 1.518 cars/min 2 . Question 2 (a) Area = [ ] 1 1 0 0 ( ) ( ) 2 (1 ) 3( 1) f x g x dx x x x x dx = ­ 1.133 . (b) Volume= ( ) ( ) 1 2 2 0 2 ( ) 2 ( ) g x f x dx π = ( ) ( ) 2 1 2 0 2 3( 1) 2 2 (1 ) x x x x dx π ­ 16.179 . (c) ( ) ( ) 2 1 1 2 0 0 ( ) ( ) (1 ) 3( 1) 15 h x g x dx kx x x x dx = = . ± 1 ± Notes: 1. Do not solve.

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4 F REE -R ESPONSE S OLUTIONS ~ 2004 AB Question 3 (a) ( ) 2 2 2 4 ( ) (2) 1 1 1 t t t t dv e e e a t a dt e e e = = − = − = − + + + . ± 1 (b) ( ) 1 2 (2) 1 tan 0 v e = < and, from Part (a), (2) 0 a < . Since v (2) and a (2) have the same sign, the speed is increasing. ± 2 (c) ( ) ( ) 1 ( ) 0 tan 1 tan(1) ln tan(1) t t v t e e t = = = = ­ 0.443 . There is a local max at t = .443 since v is positive to the left and negative to the right. t = .443 is the only critical number in the domain, therefore y reaches an absolute maximum at this time. ± 3 (d) ( ) ( ) 2 1 0 (2) 1 1 tan t y e dt = − + ­ 1.360 ≈ − . The particle is moving away from the origin because y (2) < y (0) and, v (2) < 0 (from Part (b)). ± Notes: 1. Leave it at that to save time and avoid mistakes. Or just use your calculator to evaluate the derivative: (2) (2) a v = ­ 0.133 ≈ − . 2. Give a reason for full credit. 3. Justification is required for full credit.
F REE -R ESPONSE S OLUTIONS ~ 2004 AB 5 Question 4 (a) Using implicit differentiation, ( ) 3 2 2 8 3 3 8 3 3 2 8 3 y x x yy y xy y y x y x y y x + = + = = .

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