•
•
•
2022
AP
®
Calculus AB
Sample Student Responses
and Scoring Commentary
Inside:
Free-Response Question 4
• Scoring Guidelines
• Student Samples
• Scoring Commentary
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AP® Calculus AB/BC 2022 Scoring Guidelines
Part
B
(AB or BC):
Graphing calculator not allowed
Question
4
9 points
General Scoring Notes
The model solution is presented using standard mathematical notation.
Answers (numeric or algebraic) need not be simplified. Answers given as a decimal approximation should be
correct to three places after the decimal point. Within each individual free-response question, at most one
point is not earned for inappropriate rounding.
t
(days)
0 3 7 10 12
rt
′
(
)
(centimeters per d
ay)
− 6.1 5.0 − 4.4 − 3.8 − 3.5 −
An ice sculpture melts in such a way that it can be modeled as a cone that maintains a conical shape as it
decreases in size. The radius of the base of the cone is given by a twice-differentiable function
r,
where
rt
( )
is measured in centimeters and
t
is measured in days. The table above gives selected values of
rt
′
( )
,
the rate
of change of the radius, over the time interval
0 t 12. ≤≤
Model Solution Scoring
(a)
Approximate
r
′′
(
8.5
)
using the average rate of change of
r
′
over the interval
7 t 10 ≤≤
. Show the
computations that lead to your answer, and indicate units of measure.
r
′
(
10
)
− r
′
(
7
)
−3.8 −−
(
4.4
)
r
′′
(
8.5
)
≈ =
10 − 7 10 − 7
r
′′
(
8.5
)
with
supporting work
1 point
0.6
3
= = 0.2
centimeter per day per day
Units
1 point
Scoring notes:
• To earn the first point the supporting work must include at least a difference and a quotient.
• Simplification is not required to earn the first point. If the numerical value is simplified, it must be
correct.
• The second poi
nt can be earned with an incorrect approximation for
r
′′
(
8.5
)
but cannot be earned
without some value for
r
′′
(
8.5
)
presented.
•
Units may be written in any equivalent form (such as
cm
day
2
).
Total for part (a)
2
points
© 2022 College Board
AP® Calculus AB/BC 2022 Scoring Guidelines
(b)
Is there a time
t,
0 t ≤≤3,
for which
rt
′
( )
= −6?
Justify your answer.
rt
( )
is twice-differentiable.
⇒ rt
′
( )
is differentiable.
⇒ rt
′
( )
is continuous.
r
′
( )
0 =−6.1 <−6 <−5.0 = r
′
( )
3
Therefore, by the Intermediate Value Theorem, there is a time
t,
0 t 3, <<
such that
rt
′
( )
= −6.
r
′
( )
0 <−6 < r
′
( )
3
1 point
Conclusion using
Intermediate Value
Theorem
1 point
Scoring notes:
• To earn the first point, the response must establish that
−6
is between
r
′
( )
0
and
r
′
( )
3
(or
−6.1
and
−5
). This statement may be represented symbolically (with or without including one or both
endpoints in an inequality) or verbally. A response of “
rt
′
( )
= −6
because
r
′
( )
0 = −6.1
and
r
′
( )
3 = −5
” does not state that
−6
is between
−6.1
and
−5.
Thus this response does not earn the
first point.
• To earn the second point:
o The response must state that
rt
′
( )
is continuous because
rt
′
( )
is differentiable (or because
rt
( )
is twice differentiable).
o The response must have earned the first point.
Exception: A response of “
rt
′
( )
= −6
because
r
′
( )
0 = −6.1
and
r
′
( )
3 = −5
” does not earn
the first point because of imprecise communication but may nonetheless earn the second
point if all other criteria for the second point are met.
o The response must conclude that there is a time
t
such that
rt
′
( )
= −6.
(A statement of “yes”
would be sufficient.)
• To earn the second point, the response need not explicitly name the Intermediate Value Theorem,
but if a theorem is named, it must be correct.
Total for part (b)
2 points
© 2022 College Board
AP® Calculus AB/BC 2022 Scoring Guidelines
(c) Use a right Riemann sum with the four subintervals indicated in the table to approximate the value of
∫
12
r
′
(
t
)
dt.
0
∫
12
r
′
(
t
)
dt≈ 3 r
′
(
3
)
+ 4r
′
(
7
)
+ 3r
′
(
10
)
+ 2r
′
(
12
)
0
= 3
(
−5.0
)
+ 4
(
−4.4
)
+ 3
(
−3.8
)
+ 2
(
−3.5
)
= −51
Form of right
Riemann sum
1 point
Answer 1 point
Scoring notes:
• To earn the first point, at least seven of the eight factors in the Riemann sum must be correct. If
there is any error in the Riemann sum, the response does not earn the second point.
• A response of
3
(
−5.0
)
+ 4
(
−4.4
)
+ 3
(
−3.8
)
+ 2
(
−3.5
)
earns both the first and second points, unless
there is a subsequent error in simplification, in which case the response would earn only the first
point.
• A response that presents t h
e correct answer, with accompanying work that shows the four products
in the Riemann sum (without explicitly showing all of the factors and/or the sum process) does not
earn the first point but earns the second point. For example,
−15 + 4
(
−4.4
)
+ 3
(
−3.8
)
+−7
does not
earn the first point but earns the second point. Similarly,
−15, −17.6, −11.4, −7 → −51
does not
earn the first point but earns the second point.
• A response that presents the correct answer (
−51
) with no supporting work earns no points.
• A response that provides a completely correct left Riemann sum and approximation
∫
12
0
r
′
(
t
)
dt
(i.e.,
3r
′
(
0
)
+ 4r
′
(
3
)
+ 3r
′
(
7
)
+ 2r
′
(
10
)
= 3
(
−6.1
)
+ 4
(
−5.0
)
+ 3
(
−4.4
)
+ 2
(
−3.8
)
= −59.1
)
earns
1 of the 2 points. A response that has any error in a left Riemann sum or evaluation for
∫
12
r
′
(
t
)
dt
0
earns no points.
• Units are not required or read in this part.
Total for part (c) 2 points
© 2022 College Board
AP® Calculus AB/BC 2022 Scoring Guidelines
(d)
The height of the cone decreases at a rate of
2
centimeters per day. At time
t = 3
days, the radius is
100
centimeters and the height
is
50
centimeters. Find the rate of change of the volume of the cone with
respect to time, in cubic centimeters per
day, at time
t = 3
days. (The volume
V
of a cone with radius
r
and height
h
is
1
V
=
π
rh
2
.
)
3
dV 2 dr 1
2
dh
=
π
rh
+
π
r
dt 3 dt 3 dt
Product rule 1 point
Chain rule 1 point
dV 2 1
70,000
π
=
π
(
100
)(
50
)(
−+5
)
π
(
100
)
2
(
−2
)
=−
dt
t = 3
3 3 3
The rate of change of the volume of the sculpture at
t = 3
is
approximately
70,000
π
−
3
cubic centimeters per day.
Answer 1 point
Scoring notes:
•
The first 2
points could be earned in either order.
•
A response with a completely correct product rule, missing one or both of the correct differentials,
earns
the product rule point, but
not the chain rule point.
For example,
dV
dt
=
2
3
π
rh +
1
3
π
r
2
earns
the first point, but not the second.
•
A response that treats
r
or
h
(but
not both)
as a constant
is eligible for
the
chain rule point but not
the product rule point. For example,
dV 2 dr
=
π
rh
dt 3 dt
is correct if
h
is constant, and thus
earns the
chain rule point.
•
Note: Neither
dV 2 dh
dV 2 dr dh
=
π
r
nor
=
π
rh
dt 3 dt
dt 3 dt dt
earns
any points.
•
A response that assumes a functional relationship
between
r
and
h
(such as
r = 2h
), and uses this
relationship to create a function for volume in terms of one variable, is eligible for at most the chain
rule point. For example,
r = 2h → V =
1
3
π
(
2hh
)
2
=
4
3
π
h
3
→
dV
dt
= 4
π
h
2
dh
dt
earns only the
chain rule point.
•
A response that mishandles the constant
1
3
π
cannot earn the third point but is eligible for the first
2 points.
•
The third point cannot be earned without both of the first
2
points.
•
dV 2 1
=
π
(
100
)(
50
)(
−+5
)
π
(
100
)
2
(
−2
)
dt 3 3
earns all 3 points.
• Units are not required or read in this part.
Total for part (d)
3
points
Total for question
4
9 points
© 2022 College Board
Sample 4A1 of 2
Sample 4A2 of 2
Sample 4B1 of 2
Sample 4B2 of 2
Sample 4C1 of 2
Sample 4C2 of 2
AP
®
Calculus AB/BC 2022 Scoring Commentary
Question 4
Note: Student samples are quoted verbatim and may contain spelling and grammatical errors.
Overview
In this problem the melting of an ice sculpture can be modeled as a cone that maintains a conical shape as it
decreases in size. The radius of the base of the cone is a twice-differentiable function
rt
( )
measured in centimeters,
with time
t,
0 ≤≤t 12,
in days. Selected values of
rt
′
( )
are provided in a table.
In part (a) students were asked to ap
proximate using the
average rate of change of
r
′
r
′′
(
8.5
)
over the interval
7 ≤≤t 10
and to provide correct units. A correct response should estimate the value using a difference quotient,
drawing from the data in the table that most tightly bounds
t = 8.5.
The response should include units of
centimeters per day per day.
In part (b) students were asked to justify whether there is a time
t,
0 ≤≤t 3,
for which the rate of change of
r
is
equal to
−6.
A correct response will use the Intermediate Value Theorem, first noting that the conditions for
applying this theorem are met—specifically that
rt
′
( )
is continuous because
rt
( )
is twice-differentiable and that
−6
is bounded between the values of
r
′
(
0
)
and
r
′
(
3
)
given in the table. Therefore, by the Intermediate Value
Theorem, there is a time
t
such that
0 <<t 3,
with
rt
′
( )
= −6
.
In part (c) students
were
asked to use
a right Riemann sum and the subintervals
indicated by the table
to approximate
the value of
∫
12
r
′
(
t
)
dt
0
.
A correct response should present the sum of the four products ∆t⋅ rt
′
i
(
i
)
drawn from the
table and obtain an approximation value of
−51.
In part (d) students were told that the height of the cone decreases at a rate of
2
centimeters per day and that at time
t = 3
the radius of the cone is
100
cm and the height is
50
cm. They are asked to find the rate of change of the
volume of the cone with respect to time at time
t = 3
days. A correct response will use the product and chain rules
to differentiate the given function for the volume of a cone,
V =
1
3
π
rh
2
,
and then evaluate the resulting derivative
using values
dh
dr
r = 100,
h = 50,
= −2,
and
= −5
dt
t =3
dt
t =3
(from the table) to obtain a rate of
−
70,000
π
3
cubic centimeters per day.
Sample: 4A
Score: 9
The response earned 9 points: 2 points in part (a), 2 points in part (b), 2 points in part (c), and 3 points in part (d).
In part (a) the respons
e would have earned the first point with the expression
−3.8 −−
(
4.4
)
3
in line 1
, with no
simplification. In this case, correct simplification to the boxed answer of
0.2
in line 2 earned the first point. The
response earned the second point for the correct units presented in the boxed answer in line 2.
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AP
®
Calculus AB/BC 2022 Scoring Commentary
Question 4 (continued)
In part (b) the response earned the first point with the statements given in the last three lines, which place the value
−6
between the values of
−5.0
and
−6.1.
The response earned the second point with the statements in lines 1
through 5, concluding that
rt
′
( )
is continuous because
rt
′
( )
is differentiable. The response names the Intermediate
Value Theorem, which is not required but is correct.
In part (c) the response earned the first point with the sum of products expression given in line 1. The response
would have also earned the second point for this expression in line 1, with no further simplification. In this case,
correct simplification to the boxed answer of
−51
in line 6 earned the second point.
In part (d) the response earned the first and second points with the correct
dV
dt
expression given in line 2. The
response would have also earned the third point
for the correct evaluation of this expression given in line 3, with no
further simplification. In this case, correct simplification presented in the boxed answer in line 6 earned the third
point.
Sample: 4B
Score: 5
The response earned 5 points: 2 points in part (a), no points in part (b), 1 point in part (c), and 2 points in part (d).
In part (a) the response would have earned the first point with the expression
−3.8 −−4.4
10 − 7
at the beginning of line
1, with no simplification. In
this case, correct simplification to the boxed answer of
1
5
at the end of line 1 earned
the first point. The response earned the second point for the correct units presented in the boxed answer in line 1.
In part (b) the response does not establish that
−6
is between
r
′
(
0
)
and
r
′
(
3
)
;
thus, it did not earn the first point.
The response did not earn the second point because the conclusion of “NO” is incorrect.
In part (c) the response earned the first point with the sum of products given in line 1. The response would have
also earned the second point for this expression in line 1, with no further simplification. In this case, the response
did not earn the second point because of an incorrect simplification to
−37.0.
In part (d) the response earned the first and second points with the correct expression for
dV
dt
given on line 3. The
response did not earn the third point because this expression was never evaluated.
Sample: 4C
Score: 3
The response earned 3 points: 1 point in part (a), 1 point in part (b), 1 point in part (c), and no points in part (d).
In part (a) the response would have earned the first point with the expression
−3.8 − −
(
4.4
)
10 − 7
in line 1,
with no
simplification. In this case, correct simplification to the final answer of
0.2
at the end of line 1 earned the first
point. The response does not present correct units, so it did not earn the second point.
© 2022 College Board.
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AP
®
Calculus AB/BC 2022 Scoring Commentary
Question 4 (continued)
In part (b) the response earned the first point for the statement given in line 4: “... since
−6
is between
−6.1
and
−5.0.
” The response does not establish that the continuity condition of the Intermediate Value Theorem has been
met and incorrectly names the theorem as the Mean Value Theorem, so it did not earn the second point.
In part (c) the response earned the first point with the sum of products given in line 1. The response would have
also earned the second point for this expression in line 1, with no further simplification. In this case, incorrect
simplification leads to a final answer of
−52,
so the response did not earn the second point.
In part (d) the response earned no points. The response does not present either a correct product rule or correct
chain rule; thus, it did not earn either of the first two points and is not eligible for the third point.
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