2018
AP Calculus BC
Sample Student Responses
and Scoring Commentary
Inside:
Free Response Question 5
R Scoring Guideline
R
Student Samples
R
Scoring Commentary
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®
CALCULUS BC
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2018 SCORING GUIDELINES
Question 5
(a)
1
53
π
Area =
(
2
∫
4 −
(
3 + 2cos
θ
)
2
d
θ
2
π
3
)
:
{
1 : constant and limits
3
2 : integrand
(b)
dr dr
=−2sin
θ
⇒ =−2
d
θ
d
θ
r
( )
θ
=
π
2
π
= 3 + 2cos
2
(
π
)
= 3
2
dy
dr
y = r sin
θ
⇒ = sin
θ
+ r cos
θ
d
θ
d
θ
dx dr
x = r cos
θ
⇒ = cos
θ
− r sin
θ
d
θ
d
θ
π
−2sin
dy
dy d
θ
=
(
2
=
)
π
+ 3cos
(
2
)
2
=
dx
θ
=
π
dx d
θ
2
θ
=
π
(
π
) (
π
2
−2cos
)
3
− 3sin
2 2
The slope of the line tangent to the graph of
r = 3 + 2cos
θ
at
π
θ
=
2
is
2
.
3
— OR —
dy
y = r sin
θ
=
(
3 + 2cos
θ
)
sin
θ
⇒ = 3cos
θ
+2cos
2
θ
− 2sin
2
θ
d
θ
dx
x = r cos
θ
=
(
3 + 2cos
θ
)
cos
θ
⇒= −3sin
θ
− 4sin
θ
cos
θ
d
θ
π
3c
2
π
os +2cos − 2sin
2
π
dy
dy d
θ
2 2 2
2
= =
( ) ( ) ( )
=
dx
θ
=
π
2
dx d
θ
θ
=
π
(
π
) (
π
) (
π
2
−3sin − 4sin cos
2 2 2
)
The slope of the line tangent to the graph of
r = 3 + 2cos
θ
at
π
θ
=
2
is
2
.
3
dy
dr
1 : = sin
θ
+
r
cos
θ
d
θ
d
θ
dx dr
or c= os
θ
−
r
sin
θ
d
θ
d
θ
3 :
dr
sin
θ
+
r
cos
θ
dy
d
θ
1 : =
dx dr
cos
θ
−
r
sin
θ
d
θ
1 : answer
(c)
dr dr d
θ
d
θ
d
θ
dr 1
=⋅ = −2sin
θ
⋅ ⇒ = ⋅
dt d
θ
dt dt dt dt −2sin
θ
d
θ
1 3
= 3 ⋅
( )
= =−3 radians p er second
dt
θ
=
π
3
π
− 3
−2sin
3
3
dr dr d
θ
1 : = ⋅
dt d
θ
dt
3 :
d
θ
dr 1
1 : = ⋅
dt dt −2sin
θ
1 : answer with units
6A
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2018 SCORING COMMENTARY
Question 5
Overview
In this problem a polar graph is provided for polar curves r 4 and r 32 cos
. It was given that the curves
intersect at
3
and
5
.
3
In part (a) students were asked for an integral expression that gives the area of the
region
R
that is insid e the graph of r 4 and outside the graph of r 3 2cos
. A correct response should
resource the formula for the area of a simple polar region as half of a definite integral of the square of the radius
function. The area of
R
is given by
1
53
2
1
53
1
5
3
4 d
3 2cos
2
d
4
2
3 2cos
2
d
.
2
3
2
3
2
3
In part (b) students were asked for the slope of the line tangent to the graph of r
3
2cos
at
.
2
A correct
response should deal with the conversion between polar and rectangular coordinate systems given by
y r
sin
and
x r
cos
, differentiate these with respect to
using the product rule, and find the slope of the line tangent to
the graph as the value of
dy
dy d
dx dx d
at
.
2
In part (c) the motion of a particle along the portion of the curve
r 32cos
for
0
2
is such that the distance between the particle and the origin increases at a constant
rate of 3 units per second. Students were asked for the rate at which the angle
changes with respect to time at the
instant when the position of the particle corresponds to
3
and to indicate units of measure. A correct response
should use the chain rule to relate the rates of
r and
with respect to time
dr d
t: 2sin
.
dt dt
Recognizing that
dr
3
dt
from the problem statement, it follows that
d
3
dt
3
radians per second.
For part (a) see LO 3.4D/EK 3.4D1 (BC). For part (b) see LO 2.1C/EK 2.1C7 (BC), LO 2.2A/EK 2.2A4 (BC), LO
2.3B/EK 2.3B1. For part (c) see LO 2.1C/EK 2.1C7 (BC), LO 2.2A/EK 2.2A4 (BC), LO 2.3C/EK 2.3C2. This
problem incorporates the following Mathematical Practices for AP Calculus (MPACs): reasoning with definitions
and theorems, connecting concepts, implementing algebraic/computational processes, connecting multiple
representations, building notational fluency, and communicating.
Sample: 5A
Score: 9
The response earned all 9 points: 3 points in part (a), 3 points in part (b), and 3 points in part (c). In part (a) the
response earned the first point with the constant of
1
2
and the limits of integration of
3
and
5
3
in line 1. The
response would have earned the 2 i
ntegrand points with
4
2
3 2cos
2
in line 1 with no
simp
lification. In this case, the correct simplification in line 2 earned the points. In part (b) the response earned
the first point for one of the following:
dx
d
is computed in line 5 on the left, and
dy
d
is computed in line 4 on the
left. The response earned the second point with the assembly of
dy
dx
in line 1 on the right. The response would
have earned the third point with the expression
21
0 0
21
0
3
1
0
on the right with
no simplification. In this
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®
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case, the correct simplification to
2
3
earned the point. In part (c) the response earned the first point for application
of the chain rule with
dr d
2sin
dt dt
in line 1. Although the response substitutes 3 for
dr
dt
and
3
for
before isolating
d
,
dt
the equation
3
2
d
dt
sin
3
in the last line on the left earned the second point. The response
earned the third point with
3 radians per second in the l
ast line on the right.
Sample: 5B
Score: 6
The response earned 6 points: 1 point in part (a), 3 points in part (b), and 2 points in part (c). In part (a) the
response earned the first point with the constant of
1
2
and the limits of integration of
3
and
5
3
in line 1. The
response did not earn either of the integrand points because the integrand is incorrect. In part (b) the response
earned the first point for one of the following:
dy
d
is computed in the numerator of line 3, and
dx
d
is computed
in the denominator of line 3. The response earned the second point with the assembly of
dy
dx
in line 3. The
response would have earned the third point with the expression
00 2
30
in line 3 with no si
mplification. In this
case, the correct sim
plification to
2
3
earned the point. In part (c) the response earned the first point for application
of the chain rule with
dr
d
2sin
dt dt
in line 2. The response earned the second point because although
sin
3
is incorrectly
evaluated as
1
2
in line 3,
d
dt
is isolated correctly with a consistent result in the equation
d
3
dt
in the last line. The response did not earn the third point because the evaluation of
sin
3
as
1
2
makes the
response not eligible for the third point. The units are correct.
Sample: 5C
Score: 3
The response earned 3 points: 1 point in part (a), no points in part (b), and 2 points in part (c). In part (a) the
response earned the first point with the constant of
1
2
and the limits of integration of
3
and
5
.
3
The response
did not earn either of the integrand points because the integrand is incorrect. In part (b) the response did not earn
any points. The response did not earn the first point because neither
dx
d
nor
dy
d
is presented. Therefore, the
response is not eligible for the second and third points. In part (c) the response earned the first point for
application of the chain rule with
dr d
2sin
dt dt
in line 1 on the right. Although the response substitutes 3 for
AP
®
CALCULUS BC
2018 SCORING COMMENTARY
Question 5 (continued)
dr
dt
and
3
for
before isolating
d
,
dt
the equation
d
3
dt
3
in line 4 on the right earned the second point.
The response did not earn the third point because the a
nswer is incorrect. The units are correct.
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AP
®
CALCULUS BC
2018 SCORING COMMENTARY
Question 5 (continued)
