AP CALCULUS AB/CALCULUS BC 2017 SCORING GUIDELINES · 2018-04-24 · AP® CALCULUS AB 2007 SCORING...

AP® CALCULUS AB/CALCULUS BC 2017 SCORING GUIDELINES

© 2017 The College Board. Visit the College Board on the Web: www.collegeboard.org.

Question 4

(a) ( ) ( )10 91 27 164H ′ = − − = −

( )0 91H = An equation for the tangent line is 91 16 .y t= − The internal temperature of the potato at time 3t = minutes is approximately 91 16 3 43− = degrees Celsius.

1 : slope3 : 1 : tangent line

1 : approximation

(b) ( )( )( ) ( )2

21 1 1 127 274 4 4 16

d H dH H Hdtdt= − = − − − = −

27H for 0t ( )2

21 27 016

d H Hdt

= − for 0t

Therefore, the graph of H is concave up for 0.t Thus, the answer in part (a) is an underestimate.

1 : underestimate with reason

(c) ( )

( )( )

( )( )( )

( ) ( )

2 3

2 3

1 3

1 3

1 3

3

27

127

3 27

3 91 27 0 12

3 27 12

1227 for 103 0

dG dtG

dG dtG

G t C

C C

G t

tG t t

= −−

= −−

− = − +

− = + =

− = −

−= + <

∫

The internal temperature of the potato at time 3t = minutes is

( )312 327 543−+ = degrees Celsius.

( ) ( )

1 : separation of variables 1 : antiderivatives 1 : constant of integration and

5 : uses initial condition

1 : equation involving and 1 : and 3

G tG t G

Note: max 2 5 [1-1-0-0-0] if no constant

of integration Note: 0 5 if no separation of variables

AP® CALCULUS AB 2016 SCORING GUIDELINES


Question 4

Consider the differential equation 2

.1dy ydx x=

−

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

(b) Let ( )y f x= be the particular solution to the given differential equation with the initial condition ( )2 3.f = Write an equation for the line tangent to the graph of ( )y f x= at 2.x = Use your equation to approximate ( )2.1 .f

(c) Find the particular solution ( )y f x= to the given differential equation with the initial condition ( )2 3.f =

(a)

{ 1 : zero slopes2 :

1 : nonzero slopes

(b) ( ) ( )

2

, 2, 3

3 92 1x y

dydx =

= =−

An equation for the tangent line is ( )9 2 3.y x= − + ( ) ( )9 2.1 2 92. 3 3.1f − + =

{ 1 : tangent line equation2 :

1 : approximation

(c)

( )

2

2

1 11

1 11

1 ln 1

1 1ln 2 13 31 1ln 1 3

11 ln 13

dy dxxy

dy dxxy

x Cy

CC

xy

yx

=−

=−

− = − +

− = − +

− = − −

⇒

=

=

−

−

−

Note: This solution is valid for 1 31 1 .x e< < +

1 : separation of variables 2 : antiderivatives

5 : 1 : constant of integration and uses initial condition 1 : solves for y

Note: max 3 5 [1-2-0-0] if no constant of integration Note: 0 5 if no separation of variables


Question 6


Consider the curve given by the equation 3 2.y xy− = It can be shown that 2 .3

dy ydx y x

=−

(a) Write an equation for the line tangent to the curve at the point ( )1, 1 .−

(b) Find the coordinates of all points on the curve at which the line tangent to the curve at that point is vertical.

(c) Evaluate 2

2d ydx

at the point on the curve where 1x = − and 1.y =

(a)

( ) ( ) ( ) ( ), 1, 12

1 143 1 1x y

dydx = − − −

= =

An equation for the tangent line is ( )1 1 1.4 xy + +=

1 : slope2 :

1 : equation for tangent line

(b) 2 203 3x x yy − = = So, ( )( )3 3 22 3 2 1y x y yyy y− = − = = −

( ) ( )31 1 2 3x x− − − = = The tangent line to the curve is vertical at the point ( )3, 1 .−

2 1 : sets 3 : 1 : equation in one variable

1 : coordinates

3 0y x− =

(c) ( ) ( )

( )

22

2 22

3 6 1

3

dy dyy x y yd y dx dxdx y x

− − −=

−

( ) ( )

( )( ) ( )( )( )

22

2 22, 1, 1

1 13 1 1 1 6 1 14 4

3 1 1

11 1216 32

x y

d ydx = −

− − − −=

− −

−= =

· · · · ·

·

2 : implicit differentiation

4 : 1 : substitution for

1 : answer

dydx


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

AP® CALCULUS AB 2007 SCORING GUIDELINES (Form B)

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


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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 11 dy dxy x=+

ln 1 lny x K+ = + ln1 x 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


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

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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7

Question 6

Consider the differential equation 2

.2dy 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) 21

2xdy dx

y= −

214x Cy− = − +

1 1 1;2 4 4C C− = − + = −

2 21 4

1 14 4

yx x

= =++

6 :


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

°°®°°¯


62


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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.


(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

15

15

0

15

12

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

°°°®°°°¯


59


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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.


(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


Copyright © 2002 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.

6

Question 5

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$07+)*+-*)*+').%#') !! ! " )%$)*-#8'#*)*")*+')8(-6+)",)!1):%#&)*+')";7""(&%#-*')",)*+')6"%#*)",)

*-#8'#7<=)-#&)&'*'(>%#')?+'*+'()!)+-$)-)."7-.)>-@%>0>=)."7-.)>%#%>0>=)"()#'%*+'()-*)*+%$)

6"%#*1)A0$*%,<)<"0()-#$?'(1)

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69

Copyright © 2000 by College Entrance Examination Board and Educational Testing Service. All rights reserved.AP is a registered trademark of the College Entrance Examination Board.

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68

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 by3x2 + 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 equationdy

dx=

3x2 + 12y

with the initial

condition f(1) = 4.(d) Use your solution from part (c) to find f(1.2).

(a)dy

dx=

3x2 + 12y

dy

dx

�� x = 1y = 4

=3 + 12 · 4

=48

=12

1: answer

(b) y � 4 =12(x� 1)

f(1.2)� 4 ⇡ 12(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 = (3x2 + 1) dxZ

2y dy =Z

(3x2 + 1) dx

y2 = x3 + x + C

42 = 1 + 1 + C

14 = C

y2 = x3 + x + 14

y =p

x3 + x + 14 is branch with point (1, 4)

f(x) =p

x3 + 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 onefunction out of a family of functions

1: solves for y

0/1 if solving a linear equation in y0/1 if no constant of integration

Note: max 0/5 if no separation of variablesNote: max 1/5 [1-0-0-0-0] if substitutes

value(s) for x, y, or dy/dx beforeantidi↵erentiation

(d) f(1.2) =p

1.23 + 1.2 + 14 ⇡ 4.114 1: answer, from student’s solution tothe given di↵erential equation in (c)

Copyright 1998 College Entrance Examination Board. All rights reserved.Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.

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AP CALCULUS AB/CALCULUS BC 2017 SCORING GUIDELINES · 2018-04-24 · AP® CALCULUS AB 2007 SCORING...

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