2 1 2004 bc form b ap calculus free response

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Ask a homework question - tutors are online