Unformatted text preview: (1n −1 ) ( x − 1) ⇒ y = 1 + n( x − 1) . Its xintercept 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 FreeResponse
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 FREERESPONSE 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.
 Fall '11
 Mr.Snickles
 Calculus

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