CALCULUS AB 2004 SCORING GUIDELINES (Form B)...equation with the initial condition f ()03.= (a) 2 :...

AP® CALCULUS AB 2004 SCORING GUIDELINES (Form B)

Copyright © 2004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents).

6

Question 5

Consider the differential equation ( )4 2 .dy x ydx = −

(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) While the slope field in part (a) is drawn at only twelve points, it is defined at every point in the xy-plane. Describe all points in the xy-plane for which the slopes are negative.

(c) Find the particular solution ( )y f x= to the given

differential equation with the initial condition ( )0 0.f =

(a)

2 :

( )

( )

( )

1 : zero slope at each point ,

where 0 or 2

positive slope at each point ,

where 0 and 2

1 :

negative slope at each point ,

where 0 and 2

x yx y

x yx y

x yx y

° = =°°°° ® ° ≠ >° °°° ®° °° °° ≠ <°¯ ¯

(b) Slopes are negative at points ( ),x y

where 0x ≠ and 2.y <

1 : description

(c)

5

5

5

4

5

1

5

1

5

0

1

5

1

2

1ln 2

5

2

2 ,

2

2 2

xC

x C

x

dy x dxy

y x C

y e e

y Ke K eKe K

y e

=−

− = +

− =

− = = ±− = =

= −

6 :

1 : separates variables

2 : antiderivatives

1 : constant of integration

1 : uses initial condition

1 : solves for

0 1 if is not exponential

yy

°°°®°°°¯

Note: max 3 6 [1-2-0-0-0] if no

constant of integration

Note: 0 6 if no separation of variables

59

AP® CALCULUS AB 2004 SCORING GUIDELINES

Copyright © 2004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents).

7

Question 6

Consider the differential equation ( )2 1 .dy x ydx = −

(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 pink test booklet.)

(b) While the slope field in part (a) is drawn at only twelve points, it is defined at every point in the xy-plane. Describe all points in the xy-plane for which the slopes are positive.

(c) Find the particular solution ( )y f x= to the given differential equation with the initial condition ( )0 3.f =

(a)

2 :

( )

( )

( )

1 : zero slope at each point , where 0 or 1

positive slope at each point , where 0 and 1

1 : negative slope at each point ,

where 0 and 1

x yx y

x yx y

x yx y

° = =°°°° ® ° ≠ >° °°° ®° °° °° ≠ <°¯ ¯

(b) Slopes are positive at points ( ),x y where 0x ≠ and 1.y >

1 : description

(c) 211dy x dxy =−

3

3

3

3

13

13

0

13

1ln 1 3

1

1 ,2

1 2

xC

x C

x

y x C

y e e

y Ke K eKe K

y e

− = +

− =

− = = ±= =

= +

6 :

1 : separates variables 2 : antiderivatives 1 : constant of integration 1 : uses initial condition 1 : solves for 0 1 if is not exponential

yy

°°°®°°°¯

Note: max 3 6 [1-2-0-0-0] if no constant of

integration Note: 0 6 if no separation of variables

60


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7

Question 6

Consider the differential equation 2

.2

dy xydx

−= Let

( )y f x= be the particular solution to this differential

equation with the initial condition ( )1 2.f − =

(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 1.x = −

(c) Find the solution ( )y f x= to the given differential equation with the initial condition ( )1 2.f − =

(a)

2 : 1 : zero slopes1 : nonzero slopes

®¯

(b) Slope ( )1 4

22

− −= =

( )2 2 1y x− = +

1 : equation

(c) 2

12xdy dx

y= −

214x Cy− = − +

1 1 1;2 4 4

C C− = − + = −

2 21 4

1 14 4

yx x

= =++

6 :

1 : separates variables 2 : antiderivatives1 : constant of integration1 : uses initial condition

1 : solves for y

°°°®°°°̄

Note: max 3 6 [1-2-0-0-0] if no constant of integration

Note: 0 6 if no separation of variables

61


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7

Question 6

Consider the differential equation 2 .dy xdx y= −

(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 pink test booklet.)

(b) Let ( )y f x= be the particular solution to the differential equation with the initial condition ( )1 1.f = − Write an equation for the line tangent to the graph of f at ( )1, 1− and use it to approximate ( )1.1 .f

(c) Find the particular solution ( )y f x= to the given differential equation with the initial condition ( )1 1.f = −

(a) 2 : { 1 : zero slopes

1 : nonzero slopes

(b) The line tangent to f at ( )1, 1− is ( )1 2 1 .y x+ = − Thus, ( )1.1f is approximately 0.8.−

2 : ( )1 : equation of the tangent line1 : approximation for 1.1f

®¯

(c) 2dy xdx y= −

2y dy x dx= − 2

22y x C= − +

1 1 ;2 C= − + 32C =

2 22 3y x= − + Since the particular solution goes through ( )1, 1 ,−

y must be negative. Thus the particular solution is 23 2 .y x= − −

5 :


1 : solves for y

°°®°°¯

Note: max 2 5 [1-1-0-0-0] if no constant of integration Note: 0 5 if no separation of variables

62


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6

Question 5

Consider the differential equation ( ) ( )21 cos .dy y xdx π= −

(a) On the axes provided, sketch a slope field for the given differential equation at the nine points indicated. (Note: Use the axes provided in the exam booklet.)

(b) There is a horizontal line with equation y c= that satisfies this differential equation. Find the value of c.

(c) Find the particular solution ( )y f x= to the differential equation with the initial condition ( )1 0.f =

(a)

2 : { 1 : zero slopes1 : all other slopes

(b) The line 1y = satisfies the differential equation, so 1.c =

1 : 1c =

(c) ( )

( )21 cos

1dy x dx

yπ=

−

( ) ( )1 11 siny x Cππ−− − = +

( )1 1 sin1 x Cy ππ= +−

( )11 sin C Cππ= + =

( )1 1 sin 11 xy ππ= +−

( )sin1 xyπ π π= +−

( )1 siny xπ

π π= − + for x− < <∞ ∞

6 :


1 : answer

°°®°°¯

Note: max 3 6 [1-2-0-0-0] if no constant of integration Note: 0 6 if no separation of variables

63


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6

Question 5

Consider the differential equation 1

,dy ydx x

+= where 0.x ≠

(a) On the axes provided, sketch a slope field for the given differential equation at the eight points indicated.

(Note: Use the axes provided in the pink exam booklet.)

(b) Find the particular solution ( )y f x= to the differential equation with the initial condition ( )1 1f − = and

state its domain.

(a)

2 : sign of slope at each point and relative

steepness of slope lines in rows and

columns

(b) 1 1

1dy dxy x=+

ln 1 lny x K+ = +

ln

1x Ky e ++ =

1 y C x+ =

2 C=

1 2y x+ =

2 1y x= − and 0x <

or

2 1y x= − − and 0x <

7 : [ ]

1 : separates variables

2 : antiderivatives

6 : 1 : constant of integration

1 : uses initial condition

1 : solves for

Note: max 3 6 1-2-0-0-0 if no

constant of integration

Note:

y

°°®°°¯

0 6 if no separation of variables

1 : domain

°°°°°°®°°°°°°¯

64


Question 5

© 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents).

Consider the differential equation 1 1.2dy x ydx � �

(a) On the axes provided, sketch a slope field for the given differential equation at the nine points indicated. (Note: Use the axes provided in the exam booklet.)

(b) Find 2

2d ydx

in terms of x and y. Describe the region in the xy-plane in

which all solution curves to the differential equation are concave up. (c) Let � �y f x be a particular solution to the differential equation with the

initial condition � �0 1f . Does f have a relative minimum, a relative maximum, or neither at Justify your answer. 0 ?x

(d) Find the values of the constants m and b, for which y mx b � is a solution to the differential equation.

(a)

2 : Sign of slope at each

point and relative steepness of slope lines in rows and columns.

(b) 2

21 12 2

d y dy x ydxdx � � � 1

2

Solution curves will be concave up on the half-plane above the line 1 1 .2 2y x � �

3 :

2

2 2 :

1 : description

d ydx

°®°̄

(c) � �0, 1

0 1 1 0dydx � � and

� �

2

20, 1

10 1 02d ydx

� � !

Thus, f has a relative minimum at � �0, 1 .

2 : ^ 1 : answer 1 : justification

2 : ^ 1 : value for 1 : value for

mb

(d) Substituting y m into the differential equation: x b �

� � � � � �1 11 12 2m x mx b m x b � � � � � �

Then 10 2m � and 1:m b � 12m � and 1 .2b

65


Question 5

© 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com.

Consider the differential equation 21,dy y

dx x−= where 0.x ≠

(a)

(b)

(c)

On the axes provided, sketch a slope field for the given differential equation at the nine points indicated. (Note: Use the axes provided in the exam booklet.)

Find the particular solution ( )y f x= to the differential equation with the initial condition ( )2 0.f =

For the particular solution ( )y f x= described in part (b), find ( )lim .

xf x

→∞

(a)

2 : { 1 : zero slopes1 : all other slopes

(b)

( ) ( )

2

1

1

1

12

12

1 12

1 11

1ln 1

1

1

1 , where

1

1 , 0

Cx

C x

Cx

x

dy dxy x

y Cx

y e

y e e

y ke k e

ke

k e

f x e x

− +

−

−

−

−

=−

− = − +

− =

− =

− = = ±

− =

= −

= − >

6 :

1 : separates variables 2 : antidifferentiates1 : includes constant of integration

1 : uses initial condition 1 : solves for y

°°®°°¯

Note: max 3 6 [1-2-0-0-0] if no constant

of integration Note: 0 6 if no separation of variables

(c) ( )1 12lim 1 1x

xe e

−

→∞− = −

1 : limit

66

1998 AP Calculus AB Scoring Guidelines

4. Let f be a function with f(1) = 4 such that for all points (x, y) on the graph of f the slope is

given by

3x

2

+ 1

2y

.

(a) Find the slope of the graph of f at the point where x = 1.

(b) Write an equation for the line tangent to the graph of f at x = 1 and use it to approximate

f(1.2).

(c) Find f(x) by solving the separable di↵erential equation

dy

dx

=

3x

2

+ 1

2y

with the initial

condition f(1) = 4.

(d) Use your solution from part (c) to find f(1.2).

(a)

dy

dx

=

3x

2

+ 1

2y

dy

dx

�� x = 1

y = 4

=

3 + 1

2 · 4

=

4

8

=

1

2

1: answer

(b) y � 4 =

1

2

(x� 1)

f(1.2)� 4 ⇡ 1

2

(1.2� 1)

f(1.2) ⇡ 0.1 + 4 = 4.1

2

(1: equation of tangent line

1: uses equation to approximate f(1.2)

(c) 2y dy = (3x

2

+ 1) dx

Z2y dy =

Z(3x

2

+ 1) dx

y

2

= x

3

+ x + C

4

2

= 1 + 1 + C

14 = C

y

2

= x

3

+ x + 14

y =

px

3

+ x + 14 is branch with point (1, 4)

f(x) =

px

3

+ x + 14

5

8>>>>>>>>>>>>>>>><

>>>>>>>>>>>>>>>>:

1: separates variables

1: antiderivative of dy term

1: antiderivative of dx term

1: uses y = 4 when x = 1 to pick one

function out of a family of functions

1: solves for y

0/1 if solving a linear equation in y

0/1 if no constant of integration

Note: max 0/5 if no separation of variables

Note: max 1/5 [1-0-0-0-0] if substitutes

value(s) for x, y, or dy/dx before

antidi↵erentiation

(d) f(1.2) =

p1.2

3

+ 1.2 + 14 ⇡ 4.114

1: answer, from student’s solution to

the given di↵erential equation in (c)

Copyright 1998 College Entrance Examination Board. All rights reserved.

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67

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6

Question 5

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69

CALCULUS AB 2004 SCORING GUIDELINES (Form B)...equation with the initial condition f ()03.= (a) 2 :...

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Transcript of CALCULUS AB 2004 SCORING GUIDELINES (Form B)...equation with the initial condition f ()03.= (a) 2 :...