_ap05_sg_calculus_ab__46570

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Unformatted text preview: 1 : solves for y ′ y 1 = 2y − x 2 2y = 2y − x x=0 y=± 2 y 1 ⎧1: = ⎪ 2y − x 2 2: ⎨ ⎪ 1 : answer ⎩ y =0 2y − x y=0 The curve has no horizontal tangent since ⎧1: y = 0 2: ⎨ ⎩ 1 : explanation ( 0, (c) 2 ) , ( 0, − 2 ) 02 ≠ 2 + x ⋅ 0 for any x. (d) When y = 3, 32 = 2 + 3 x so x = 7 . 3 ⎧ 1 : solves for x ⎪ 3 : ⎨ 1 : chain rule ⎪ 1 : answer ⎩ dy dy dx y dx = ⋅ = ⋅ 2 y − x dt dt dx dt 3 9 dx dx At t = 5, 6 = ⋅ = ⋅ 7 dt 11 dt 6− 3 dx 22 = dt t = 5 3 Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 6 AP® CALCULUS AB 2005 SCORING GUIDELINES (Form B) Question 6 dy − xy 2 . Let = dx 2 y = f ( x ) be the particular solution to this differential equation with the initial condition f ( −1) = 2. Consider the differential equation (a) On the axes provided, sketch a slope field for the given differential equation at the twelve points indicated. (Note: Use the axes provided in the test booklet.) (b) Write an equation for the line tangent to the graph of f at x = −1. (c) Find the solution y = f ( x ) to the given differential equation with the initial condition f ( −1) = 2. (a) ⎧ 1 : zero slopes 2: ⎨ ⎩ 1 : nonzero slopes − ( −1) 4 =2 2 y − 2 = 2 ( x + 1) (b) Slope = (c) 1 : equation 1 x dy = − dx 2 2 y ⎧ 1 : separates variables ⎪ 2 : antiderivatives ⎪ 6 : ⎨ 1 : constant of integration ⎪ 1 : uses initial condition ⎪ ⎪ ⎩ 1 : solves for y 1 x2 =− +C 4 y 1 1 1 − = − + C; C = − 2 4 4 1 4 y= 2 =2 1 x x +1 + 4 4 − Note: max 3 6 [1-2-0-0-0] if no constant of integration Note: 0 6 if no separation of variables Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 7...
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This note was uploaded on 10/01/2009 for the course OC 9876 taught by Professor Dq during the Spring '09 term at UC Merced.

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