2017
AP Calculus AB
Sample Student Responses
and Scoring Commentary
Inside:
• Free Response Question 4
• Scoring Guideline
•
Student Samples
•
Scoring Commentary
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AP
®
CALCULUS AB/CALCULUS BC
2017 SCORING GUIDELINES
Question 4
(a)
H
′
(
0
)
= −
1
4
(
91 − 27
)
= −16
H
(
0
)
= 91
An equation for the tangent line is
y = 91 − 16 t .
The internal temperature of the potato at time
t = 3
minutes is
approxi
mately
91 − 16 ⋅ 3 = 43
degrees Celsius.
3 :
1 : slope
1 : tangent line
1 : approximation
(
b)
d
2
H 1 dH 1 1 1
= − =
2
(
−
)
(
−
)
(
H − 27
)
=
(
H − 27
)
dt
4 dt 4 4 16
dH
2
1
H > 27
for
t > 0
⇒
=
(
2
H − 27
)
> 0
for
t > 0
dt
16
Therefore, the graph of H is concave up for
t > 0.
Thus, the
answer in part (a) is an underestimate.
1 : underestimate with reason
(c)
dG
= −
dt
(
G
−
27
)
23
⌠
dG
=
(
−
1
)
dt
⌡
(
G
−
23
27
)
∫
3
(
G
−
27
)
13
=
−
tC
+
3
(
91
−
27
)
13
=
0
+
C
⇒
C
=
12
3
(
G
−
27
)
13
=
12
−
t
Gt
(
12
−
t
3
)
3
( )
=
27
+
fo
r 0
≤<
t
10
The internal temperature of the potato at time
t = 3
minutes is
(
12 − 3
27 + = 54
degrees Celsius.
3
)
3
5
1 : separation of variables
1 : antiderivatives
1 : constant of integration and
:
uses initial condition
1 : equation involving G and t
1 : Gt
( )
and G
(
3
)
Note: max
2 5
[1-1-0-0-0] if no constant
of integration
Note:
0 5
if no separation of variables
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© 2017 The College Board.
Visit the College Board on the Web: www.collegeboard.org.
© 2017 The College Board.
Visit the College Board on the Web: www.collegeboard.org.
© 2017 The College Board.
Visit the College Board on the Web: www.collegeboard.org.
© 2017 The College Board.
Visit the College Board on the Web: www.collegeboard.org.
© 2017 The College Board.
Visit the College Board on the Web: www.collegeboard.org.
AP
®
CALCULUS AB/CALCULUS BC
2017 SCORING COMME
NTARY
Question 4
Overview
The context for this problem is the internal temperature of a boiled potato that is left to cool in a kitchen. Initially
at time
t = 0,
the potato’s internal temperature is 91 degrees Celsius, and it is given that the internal temperature
of the potato exceeds 27 degrees Celsius for all times
t > 0
. The internal temperature of the potato at time t
minutes
is modeled by the function H that satisfies the differential equation
dH 1
= −
(
H − 27
)
,
dt 4
where
Ht
( )
is
measured in degrees Celsius and
H
(
0
)
= 91.
In part (a) students were asked for an equation of the line tangent to
the graph of H at
t = 0,
and to use this equation to approximate the internal temperature of the potato at time
t = 3.
Using the initial value and the differential equation, students needed to find the slope of the tangent line to
be
1
H
′
(
0
)
= −
(
91 − 27
)
= −16
4
and report the equation of the tangent line to be
y = 91 − 16 t.
Students needed
to find the approximate temperature of the potato at
t = 3
to be
91 − 16 ⋅ 3 = 43
degrees Celsius. [LO 2.3B/EK
2.3B2] In part (b) stude
nts were asked to use
dH
2
2
dt
to determine whether the approximation in part (a) is an
underestimate or overestimate for the p otato’s internal temperature at time
t = 3.
Students needed to use the
given differential equation to calculate
dH
2
1 dH 1
= − =
(
H −
27
)
d
t
2
4
d
t
16
.
Then using t he given information that the
temperature always exceeds 27 degrees Celsius, students needed to conclude that
dH
2
>
2
0
dt
for all times t. Thus,
the graph of H is concave up, and the line tangent to the graph of H at
t = 0
lies below the graph of H (except
at the point of tangency), so the approximation found in part (a) is an underestimate. [LO 2.1D/EK 2.1D1, LO
2.2A/EK 2.2A1] In part (c) an alternate model, G, is proposed for the internal temperature of the potato at times
t < 10.
Gt
( )
is measured in degrees Celsius and satisfies the differential
equation
dG
= −
(
G − 27
)
23
dt
with
G
(
0
)
= 91.
Students were asked to find an expression for
Gt
( )
and to find the internal temperature of the potato
at time
t = 3
based on this model. Students needed to employ the method of separation
of variables, using the
initial condition
G
(
0
)
= 91
to resolve the constant of integration, and arrive at the particular solution
(
12 − t
3
Gt
( )
= 27 +
3
)
.
Students should then have reported that the model gives an internal temperature of
G
(
3
)
= 54
degrees Celsius for the potato at time
t = 3.
[LO 3.5A/EK 3.5A2] This problem incorporates the
following Mathematical Practices for AP Calculus (MPACs): reasoning with definitions and theorems, connecting
concepts, implementing algebraic/computational processes, building notational fluency, and communicating.
Sample: 4A
Score: 9
The response earned all 9 points: 3 points in part (a), 1 point in part (b), and 5 points in part (c). In part (a) the
student earned the first point for the slope with
1
−
(
91 − 27
)
4
.
The second point was earned for the tangent line
At
( )
=−16
(
t − 0
)
+ 91.
Note that the student names the tangent line
A
(
t
)
and is not penalized. The third point
was earned for the approximation 43. Either of the numerical expressions
−16
(
3
)
+ 91
or
− 48 + 91
would have
earned the third point. The student chooses to simplify a nd does so correctly. In part (b) the student has the correct
answer of “underestimate” and supports the answer with correct reasoning. The student has the correct second
derivative in line 2 on the left and states that the second derivative is always positive in line 2 on the right. The
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AP
®
CALCULUS AB/CALCULUS BC
2017 SCORING COMMENTARY
Question 4 (continued)
student earned the point. In part (c) the student earned the first point for separation in line 2. The antiderivatives
are correct in line 4 and earned the second point. The third point was earned for the constant of integration and use
of the initial condition in lines 8 and 9. The fourth point was earned in line 12 for an equation involving G and t
and a correct numerical value for C. As an aside, the fourth point could have been earned with an implicit
equation such as
(
13
12 − t
G − 27
)
= .
3
The fifth point was earned for
(
(
t
3
Gt
)
= −+4
)
+ 27
3
in line 12 along
with 54 in line 4 on the right. Any of the three numerical expressions in lines 1, 2, and 3 on the right together with
(
t
3
Gt
( )
= −+4
)
+ 27
3
in line 12 would have earned the fifth point. The student chooses to simplify and does so
correctly.
Sample: 4B
Score: 6
The response earned 6 points: 3 poi
nts in part (a), no point in part (b), and 3 points in part (c). In part (a) the
student earned the first point for the slope with
1
−
(
91 − 27
)
4
.
The second point was earned for the tangent line
y − 91 = −16 t .
The third point was earned f or the approximation 43. Either of the numerical expressions
−16
(
3
)
+ 91
or
−+48 91
would have earned the third point. The student chooses to simplify and does so
correctly. In part (b) the student’s answer of “underestimate” is correct, but the reason is based on an incorrect
second derivative in line 3 on the left. The student did not earn the point. In part (c) the student earned the first
point for separation in line 1. In line 2 the student drops a negative sign, so the second point for correct
antiderivatives was not earned. Line 2 should have been:
−3
(
G − 27
)
13
= t+ C.
The student’s equation
involving antiderivatives is eligible for the remaining 3 points. In line 3 the third point was earned for the constant
of integration and use of the initial condition. In line 7 the fourth point was earned f or an equation involving G
and t together with the consistent numerical value for C. The fifth point was not earned because, although the
answer is consistent with the work, the student’s value of
G
(
3
)
is out of the context of the problem because
152 > 91.
Sample: 4C
Score:
3
The response earned 3 points:
3 po
ints in part (a), no point
in part (b), and no points in part
(c). In part (a)
the
student earned the first
point
for the slope
with
1
−
(
91 − 27
)
4
in line 1 on the left. The second point was earned
for the tangent line
H − 91 =−16
(
t
)
.
Note that the student uses H in place of y in the tangent line and is not
penalized. The third point was earned for the approximation 43. Either of the numerical expressions
−16
(
3
)
+ 91
or
−+48 91
would have earned the third point. In part (b) the student’s answer of “underestimate” is correct, but
the reason is based on an incorrect second derivative of
1
−
4
.
The student did not earn the point. In part (c) there is
no separation of variables, and thus, no points were earned.
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