This preview shows page 1. Sign up to view the full content.
Unformatted text preview: n 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: a (2) = v′(2) ≈ −0.133 . 2. Give a reason for full credit. 3. Justification is required for full credit. 5 FREERESPONSE SOLUTIONS ~ 2004 AB Question 4
(a) Using implicit differentiation,
2 x + 8 yy′ = 3 y + 3 xy′ ⇒ y′ ( 8 y − 3 x ) = 3 y − 2 x ⇒ y′ = (b) dy
= 0 ⇒ 3 y − 2 x = 0 . x = 3 and y = 2 satisfy this equation, 8 ⋅ 2 − 3 ⋅ 3 ≠ 0 , and
dx
32 + 4 ⋅ 22 = 7 + 3 ⋅ 3 ⋅ 2 , so the point P = (3, 2) is indeed on the curve. d y d 3y − 2x =
=
dx 2 dx 8 y − 3 x 2 (c) 3 y − 2x
.
8 y − 3x
1 (8 y − 3x ) 3 dy dy − 2 − (3 y − 2 x) 8 − 3 dx
dx . At P = (3, 2),
2
(8 y − 3x ) 3 y − 2 x = 0, 8 y − 3 x = 7, and dy
d2y
=0 ⇒
dx
dx 2 x = 3, y = 2 2
= − . The first derivative is 0
7 and the second derivative is negative, so the curve has a local maximum at P.
Notes:
1. You must verify that the point is on the curve. Question 5
0 2 +1
. g ′(0) = f (0) = 1 .
2 (a) g (0) = ∫ f ( x) dx = 3 ⋅ (b) At a relative maximum, g ′( x) = f ( x) must change from positive to negative. There
is only one such point, at x = 3 . (c) The absolu...
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.
 Fall '11
 Mr.Snickles
 Calculus

Click to edit the document details