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AP Calculus BC
2015 Scoring Guidelines
®
© 2015 The College Board. College Board, Advanced Placement Program, AP, AP Central, and the acorn logo
are registered trademarks of the College Board.
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AP® CALCULUS AB/CALCULUS BC
2015 SCORING GUIDELINES
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Question 1
 t2 
The rate at which rainwater flows into a drainpipe is modeled by the function R, where R( t ) = 20sin   cubic
 35 
feet per hour, t is measured in hours, and 0 ≤ t ≤ 8. The pipe is partially blocked, allowing water to drain out the
other end of the pipe at a rate modeled by D( t ) =
−0.04t 3 + 0.4t 2 + 0.96t cubic feet per hour, for 0 ≤ t ≤ 8.
There are 30 cubic feet of water in the pipe at time t = 0.
(a) How many cubic feet of rainwater flow into the pipe during the 8-hour time interval 0 ≤ t ≤ 8 ?
(b) Is the amount of water in the pipe increasing or decreasing at time t = 3 hours? Give a reason for your
answer.
(c) At what time t, 0 ≤ t ≤ 8, is the amount of water in the pipe at a minimum? Justify your answer.
(d) The pipe can hold 50 cubic feet of water before overflowing. For t > 8, water continues to flow into and out
of the pipe at the given rates until the pipe begins to overflow. Write, but do not solve, an equation involving
one or more integrals that gives the time w when the pipe will begin to overflow.
8
{
1 : integrand
1 : answer
(a)
∫0 R( t ) dt = 76.570
2:
(b)
R ( 3) − D( 3) =
−0.313632 < 0
Since R( 3) < D( 3) , the amount of water in the pipe is
decreasing at time t = 3 hours.
 1 : considers R( 3) and D( 3)
2:
 1 : answer and reason
(c)
The amount of water in the pipe at time t, 0 ≤ t ≤ 8, is
0
 1 : considers R( t ) − D( t ) =

3 :  1 : answer
 1 : justification
t
30 + ∫ [ R( x ) − D( x )] dx.
0
R( t ) − D( t ) = 0 ⇒ t = 0, 3.271658
t
0
3.271658
8
Amount of water in the pipe
30
27.964561
48.543686
The amount of water in the pipe is a minimum at time
t = 3.272 (or 3.271) hours.
(d)
w
30 + ∫ [ R( t ) − D( t )] dt =
50
0
2:
{
1 : integral
1 : equation
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At time t ≥ 0, a particle moving along a curve in the xy-plane has position ( x( t ) , y ( t ) ) with velocity vector
)
v( t ) = cos t 2 , e0.5t . At t = 1, the particle is at the point ( 3, 5 ) .
(a) Find the x-coordinate of the position of the particle at time t = 2.
(b) For 0 < t < 1, there is a point on the curve at which the line tangent to the curve has a slope of 2.
At what time is the object at that point?
(c) Find the time at which the speed of the particle is 3.
(d) Find the total distance traveled by the particle from time t = 0 to time t = 1.
(a) x( 2 )= 3 +
dy
(b)=
dx
∫1 cos ( t ) dt = 2.557 (or 2.556)
2
dy dt
=
dx dt
2
 1 : integral

3 :  1 : uses initial condition

 1 : answer
e0.5t
cos t 2
2:
{
1 : slope in terms of t
1 : answer
( )
2:
{
1 : speed in terms of t
1 : answer
2:
{
1 : integral
1 : answer
( )
e0.5t
=2
cos t 2
( )
t = 0.840
(c) Speed = cos 2 t 2 + et
( )
cos 2 t 2 + et =
3
t = 2.196 (or 2.195)
Distance
(d)=
∫0 cos ( t ) + e dt = 1.595 (or 1.594)
1
2
2
t
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Question 2
( ( )
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AP® CALCULUS BC
2015 SCORING GUIDELINES
© 2015 The College Board.
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AP® CALCULUS AB/CALCULUS BC
2015 SCORING GUIDELINES
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Question 3
t
(minutes)
v( t )
(meters per minute)
0
12
20
24
40
0
200
240
–220
150
Johanna jogs along a straight path. For 0 ≤ t ≤ 40, Johanna’s velocity is given by a differentiable function v.
Selected values of v( t ) , where t is measured in minutes and v( t ) is measured in meters per minute, are given in
the table above.
(a) Use the data in the table to estimate the value of v′(16 ) .
(b) Using correct units, explain the meaning of the definite integral ∫
40
0
Approximate the value of ∫
40
0
table.
v( t ) dt in the context of the problem.
v( t ) dt using a right Riemann sum with the four subintervals indicated in the
(c) Bob is riding his bicycle along the same path. For 0 ≤ t ≤ 10, Bob’s velocity is modeled by
B( t ) =t 3 − 6t 2 + 300, where t is measured in minutes and B( t ) is measured in meters per minute.
Find Bob’s acceleration at time t = 5.
(d) Based on the model B from part (c), find Bob’s average velocity during the interval 0 ≤ t ≤ 10.
(a) v′(16 ) ≈
(b)
240 − 200
=
5 meters/min 2
20 − 12
40
∫0 v( t ) dt is the total distance Johanna jogs, in meters, over the
time interval 0 ≤ t ≤ 40 minutes.
40
1 : approximation
 1 : explanation

3 :  1 : right Riemann sum
 1 : approximation
∫0 v( t ) dt ≈ 12 · v(12 ) + 8 · v( 20 ) + 4 · v( 24 ) + 16 · v( 40 )
= 12 · 200 + 8 · 240 + 4 · 220 + 16 · 150
= 2400 + 1920 + 880 + 2400
= 7600 meters
(c) Bob’s acceleration is B′(=
t ) 3t 2 − 12t.
B′( 5 ) = 3 ( 25 ) − 12 ( 5 ) = 15 meters/min 2
(d) Avg vel =
1 10 3
t − 6t 2 + 300 dt
10 ∫0
(
)
10

1  t4
=
− 2t 3 + 300t 

10  4
0
=
 1 : uses B′( t )
2:
 1 : answer
 1 : integral

3 :  1 : antiderivative
 1 : answer
1  10000
 350 meters/min
− 2000 + 3000
=
10  4

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Question 4
dy
Consider the differential equation = 2 x − y.
dx
(a) On the axes provided, sketch a slope field for the given differential equation at the six points indicated.
d2y
in terms of x and y. Determine the concavity of all solution curves for the given differential
dx 2
equation in Quadrant II. Give a reason for your answer.
(b) Find
(c) Let y = f ( x ) be the particular solution to the differential equation with the initial condition f ( 2 ) = 3.
Does f have a relative minimum, a relative maximum, or neither at x = 2 ? Justify your answer.
y mx + b is a solution to the differential equation.
(d) Find the values of the constants m and b for which =
(a)
(b)
2:
d2y
dy
=2 −
=2 − (2 x − y ) =2 − 2 x + y
2
dx
dx
In Quadrant II, x < 0 and y > 0, so 2 − 2 x + y > 0.
Therefore, all solution curves are concave up in Quadrant II.
(c)
dy
= 2 ( 2) − =
3 1 =/ 0
dx ( x, y ) = ( 2, 3)
Therefore, f has neither a relative minimum nor a relative
maximum at x = 2.
(d) y = mx + b ⇒
dy
=
dx
d
m
( mx + b ) =
dx
2x − y =
m
2 x − ( mx + b ) = m
0
( 2 − m) x − ( m + b) =
2−m =0 ⇒ m =2
b =−m ⇒ b =−2
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AP® CALUCLUS AB/CALCULUS BC
2015 SCORING GUIDELINES
{
1 : slopes where x = 0
1 : slopes where x = 1

d2y
1:
2:
dx 2
 1 : concave up with reason

 1 : considers dy

dx ( x, y ) =( 2, 3)
2:
 1 : conclusion with justification
 1 : d ( mx + b ) = m

dx
3 :  1 : 2x − y = m

 1 : answer
Therefore, m = 2 and b = −2.
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AP® CALCULUS BC
2015 SCORING GUIDELINES
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Question 5
Consider the function f ( x ) =
f ′( x ) =
(
k − 2x
x 2 − kx
)
2
1
, where k is a nonzero constant. The derivative of f is given by
x − kx
2
.
(a) Let k = 3, so that f ( x ) =
whose x-coordinate is 4.
1
. Write an equation for the line tangent to the graph of f at the point
x − 3x
2
1
. Determine whether f has a relative minimum, a relative maximum, or
x − 4x
neither at x = 2. Justify your answer.
(b) Let k = 4, so that f ( x ) =
2
(c) Find the value of k for which f has a critical point at x = −5.
(d) Let k = 6, so that f ( x ) =
Find ∫ f ( x ) dx.
(a)
f ( 4) =
1
. Find the partial fraction decomposition for the function f.
x − 6x
2
1
1
=
4 − 3· 4 4
=
f ′(4)
2
(
3 − 2·4
2
4 − 3· 4
)
2
= −
5
16
2:
{
An equation for the line tangent to the graph of f at the point
5
1
whose x-coordinate is 4 is y = − ( x − 4 ) + .
16
4
=
(b) f ′( x )
(
4 − 2x
x2 − 4 x
)
2
=
f ′( 2 )
4 − 2·2
=
0
2
22 − 4 · 2
(
)
f ′( x ) changes sign from positive to negative at x = 2.
Therefore, f has a relative maximum at x = 2.
(c)
(d)
f ′( −5 ) =
k − 2 · ( −5 )
( ( −5)2 − k · ( −5))
2
=
0 ⇒ k =−10
A
B
1
1
=
=
+
⇒ 1 = A ( x − 6 ) + Bx
x x−6
x ( x − 6)
x2 − 6 x
1
x =0 ⇒ 1 =A · ( −6 ) ⇒ A =−
6
1
x = 6 ⇒ 1 = B · ( 6) ⇒ B =
6
1 6
−1 6
1
=
+
x
x−6
x ( x − 6)
1 6 
 −1 6
∫ f ( x ) dx = ⌠⌡  x + x − 6  dx
1
1
1
x−6
ln
+C
= − ln x + ln x − 6 +=
C
6
6
6
x
1 : slope
1 : tangent line equation
 1 : considers f ′( 2 )
2:
 1 : answer with justification
1 : answer
4:
{
2 : partial fraction decomposition
2 : general antiderivative
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AP® CALCULUS BC
2015 SCORING GUIDELINES
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Question 6
∞
The Maclaurin series for a function f is given by ∑
n =1
( −3)n −1
n
xn = x −
( −3)
3 2
x + 3 x3 −  +
2
n
n −1
x n +  and
converges to f ( x ) for x < R, where R is the radius of convergence of the Maclaurin series.
(a) Use the ratio test to find R.
(b) Write the first four nonzero terms of the Maclaurin series for f ′, the derivative of f. Express f ′ as a
rational function for x < R.
(c) Write the first four nonzero terms of the Maclaurin series for e x . Use the Maclaurin series for e x to write
the third-degree Taylor polynomial for g ( x ) = e x f ( x ) about x = 0.
(a) Let an be the nth term of the Maclaurin series.
( −3)n x n +1
an +1
an
lim
−3n
n
=
·
·x
n −1 n
+1
n +1
n
( −3) x
−3n
n →∞ n + 1
·x = 3 x
3 x <1 ⇒ x <
1
3
The radius of convergence is R =
1
.
3
(b) The first four nonzero terms of the Maclaurin series for f ′ are
1 − 3 x + 9 x 2 − 27 x 3 .
=
f ′( x )
 1 : sets up ratio

3 :  1 : computes limit of ratio

 1 : determines radius of convergence
3:
{
2 : first four nonzero terms
1 : rational function
1
1
=
1 − ( −3 x ) 1 + 3 x
(c) The first four nonzero terms of the Maclaurin series for e x are
x2
x3
1+ x +
+
.
2!
3!
The product of the Maclaurin series for e x and the Maclaurin
series for f is
(


3
x 2 x3
1
+
x
+
+
+   x − x 2 + 3 x3 − 

2!
3!
2


1 2
=−
x
x + 2 x3 + 
2
 1 : first four nonzero terms

of the Maclaurin series for e x
3:

 2 : Taylor polynomial
)
The third-degree Taylor polynomial for g ( x ) = e x f ( x )
1 2
x
x + 2 x3 .
about x = 0 is T3 ( x ) =−
2
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