APcal 2004-Solutions

# 2 1 2004 bc form b ap calculus free response

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Unformatted text preview: (1n −1 ) ( x − 1) ⇒ y = 1 + n( x − 1) . Its x-intercept is x = 1 − . The area of n 1 1 1 the triangle is (1) 1 − 1 − = . 2 n 2n (c) Area of S = ∫ 1 0 x n dx − area of T = g (n) = 1 1 1 1 − . g ′(n) = − + 2. 2 n + 1 2n ( n + 1) 2n g ′(n) = 0 and changes sign from positive to negative when 1 2 . Therefore, the area of S reaches ( n + 1) = 2n 2 ⇒ n + 1 = 2n ⇒ n = 2 −1 1 maximum at n = .1 2 −1 Notes: 1. Do not despair if your answer is 2 + 1: 1 = 2 + 1. 2 −1 2004 BC (Form B) AP Calculus Free-Response Solutions and Notes Question 1 2 (a) 4t 3 a(t ) = (b) 2 dx dy Speed s (t ) = + . s (0) = dt dt dy dx 2 t +9 4 = , 2et − 5e− t t4 + 9 ∫ 3 Distance = (d) x(3) = x(0) + ∫ 0 = t =0 s (t ) dt = ∫ (c) 3 0 4 + 9 ) + ( 2et + 5e − t ) 2 = 9 + 49 . t =0 = 0, −3 . t =0 2et + 5e − t t =0 (t 7 7 . An equation for the tangent line is y − 1 = ( x − 4) . 3 3 3 0 (t 4 + 9 ) + ( 2et + 5e− t ) dt 3 dx dt = 4 + ∫ t 4 + 9 dt 0 dt 2 ≈ 45.227 . ≈ 4 + 13.931 . 17 18 FREE-RESPONSE SOLUTIONS ~ 2004 BC (FORM B) Question 2 (a) f (2) = 7 . f ′′(2) = −9 ⇒ f ′′(2) = −18 . 2! (b) Yes. f ′(2) = 0 and f ′′(2) < 0 . Therefore f...
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## This note was uploaded on 07/07/2013 for the course MATH AP taught by Professor Mr.snickles during the Fall '11 term at Benjamin N Cardozo High School.

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