The 2018 AP Calculus AB International Practice Exam FRQ scoring guidelines emphasize precise decimal answers (three places), clear justification for claims (e.g., nonincreasing rates), and accurate use of technology for accumulation and derivative problems. Key topics include differential equations, particle motion, and average rates of change.

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AP® Calculus AB
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AP® Calculus AB
295
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© 2018 The College Board.
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(a)
( )
( ) ( )
51
20.5 15.1
3 1.35
51 4
gg
g
−
−
≈
′
= =
−
At time
3t
=
minutes, the rate at which grain is being added to the silo is
increasing at a rate of 1.35 cubic feet per minute per minute.
1 : approximation
2 :
1 : interpretation with units
(b)
The total amount of grain added to the silo from time
t 0=
to time
8t =
is
( )
8
0
g t dt
∫
cubic feet.
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
8
0
110 5 51 6 65 8 86
15.1 1 20.5 4 18.3 1 22.7 2 160.8
g t dt g g g g≈ ⋅− + ⋅−+ ⋅ − + ⋅−
= ⋅+ ⋅ + ⋅+ ⋅ =
∫
1 : integral expression
3 :
1 : right Riemann sum
1 : approximation
(c)
( )
8
0
99.051497w t dt =
∫
The approximate
amount of unspoiled grain remaining in the silo at
time
8t =
is
( )
8
0
160.8 61.749w t dt−=
∫
(or 61.748) cubic feet.
1 : integral
2 :
1 : answer
(d)
( ) ( )
6 6 18.3 16.063173 2.236827 0gw−=− = >
Because
( ) ( )
6 6 0gw− ,
the amount of unspoiled grain is increasing at
>
time
6t = .
( ) ( )
1 : considers 6 6
2 :
1 : answer
gw−
AP
®
CALCULUS AB
2018 SCORING GUIDELINES
Question 1
AP
®
CALCULUS AB
2018 SCORING GUIDELINES
© 2018 The College Board.
Visit the College Board on the Web: www.collegeboard.org.
Question 2
(a)
( )
5 0.538462v
′
=
The accelerati
on of the snail at time
t = 5
minutes is 0.538 inches per
minute per minute.
1 : answer
(b)
( )
15
0
76.043074v t dt =
∫
The d
isplacement of the snail over the
interval
0 15t≤≤
minutes is
76.043 inches.
{
1 : integral
2 :
1 : answer
(c)
( )
15
0
1
5.069538
15
v t dt =
∫
( )
2
1.4ln 1 5.069538 6.031tt+ = ⇒=
minutes
{
1 : average value expression
2 :
1 : answer
(d)
The velocity of the ant at time t,
12 15t≤≤ ,
is
2dt t c= +
∫
2
inches
per minute for some constant c.
For
12 15t≤≤ ,
the displacement of the ant is
( )
( )
15
15
2
12
12
2 81 3
t
t
t c dt t ct c
=
=
+=+ =+
∫
inches.
Thus,
81 3 76.043074 1.652309c+ = ⇒=−c.
The velocity
of the ant at time
12t =
is
B =2⋅−12 1.652309 82 =24.3
(or 22.347) inches per minute.
— OR —
The velocity of the ant at time t,
12 15t≤≤ ,
is
( )
2 12t− B+
inches per
minute.
For
12 15t≤≤ ,
the displacement of the ant is
( )
( )
( )
( )
2
15
2
12
15
1
2 2 91
t
t
t B dt Bt Bt
=
=
− −+ =+=
∫
1 2 3+
inches.
9 3 76.043074 22.348BB+ = ⇒=
(or 22.347) inches per minute
1 : ant’s velocity
1 : ant’s displacement
4 :
1 : equation
1 : answer
© 2018 The College Board.
Visit the College Board on the Web: www.collegeboard.org.
AP
®
CALCULUS AB
2018 SCORING GUIDELINES
Question 3
(a)
( ) ( )
7
0
9
7 3 7 21 3 24
2
f g t dt
π
=+= =⋅ −+ −
∫
9
2
π
( ) ( )
73 733fg
′
=+ =+=6
( )
( )
1 : 7
2 :
1 : 7
f
f
′
(b)
On the interval
4 3,x−≤ ≤
( ) ( )
3f x gx
′
= + .
Because
( )
0fx
′
≥
for
4 3x−≤ ≤,
f is nondecreasing over
the entire interval, and the maximum must occur when
3x = .
2 : answer with justification
(c)
( )
0
1
lim
2
x
gx
−
→
′
= −
( )
0
lim
x
gx
+
→
′
does not exist.
{
1 : left-hand limit
2 :
1 : right-hand limit
(d)
( )
( )
( )
2
0
2
lim 7 6 7 0
x
f x g t dt
−
→−
+ =−+ + =
∫
( )
36
2
lim 1 0
x
x
e
+
→−
−=
Using L’Hospital’s Rule,
( ) ( ) ( )
36 36
22
7 32
31
lim lim
3 3
13
xx
xx
fx f x g
ee
++
→− →−
′
+ +−
+
= = = =
4
3
−
.
1 : limits equal 0
3 :
1 : applies L’Hospital’s Rule
1 : answer
Note: max
13
[1-0-0] if no limit notation
attached to a ratio of derivatives
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End of Document
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