Solutions to the2009 AP Calculus ABFree Response Questions
Louis A. Talman, Ph.D.
Department of Mathematical & Computer SciencesMetropolitan State College of Denver
Problem 1.
ü
a.
At time t = 7.5, the acceleration of Caren's bicycle is

1
ÅÅÅÅÅÅÅ
10
miles per minute per minute.
ü
b.
The integral
Ÿ
012
»
v
H
t
L »
„
t
gives, in miles, the total distance that Caren traveled during the period 0
§
t
§
12. The value of this integral is
9
ÅÅÅÅ
5
miles.
ü
c.
Her turnaround time corresponds to the point on the graph where the sign of her velocity changes from positive to negative.That's
t
=
2 minutes.
ü
d.
Caren lives
Ÿ
512
v
H
t
L
„
t
=
7
ÅÅÅÅ
5
miles from school because she left home at
t
=
5, arrived at school at
t
=
12, traveled in onedirection only, and the distance she traveled during that time is given by the integral. Larry's distance is the integral of hisvelocity over the interval
@
0,12
D
, or
AB2009.nb
1
p ÅÅÅÅÅÅÅ
15
‡
012
Sin
A
p ÅÅÅÅÅÅÅ
12
t
E
„
t
8
ÅÅÅÅ
5
At
8
ÅÅÅÅ
5
miles, Larry lives farther than Caren, at
7
ÅÅÅÅ
5
miles from the school. Thus, Caren lives closer.
Problem 2.
ü
a.
At time
t
=
2, the auditorium contains
Ÿ
02
H
1380
t
2

675
t
3
L
„
t
people.
‡
02
H
1380 t
2

675 t
3
L
„
t
980
That's 980 people.
ü
b.
We are given
R
H
t
L
=
1380
t
2

675
t
3
, so that
R
'
H
t
L
=
2760
t

2025
t
2
=
15
t
H
184

135
t
L
. Thus,
R
H
t
L
is increasing onthe interval
@
0,
184
ÅÅÅÅÅÅÅÅÅ
135
D
and decreasing on the interval
@
184
ÅÅÅÅÅÅÅÅÅ
135
,2
D
, because
R
'
H
t
L
is positive on the first of these intervals andnegative on the second. It follows that the maximal rate at which people enter the auditorium is at
t
=
184
ÅÅÅÅÅÅÅÅÅ
135
hours.
ü
c.
We have
R
@
t_
D
=
1380 t
2

675 t
3
1380 t
2

675 t
3
Dw
@
t_
D
=
H
2

t
L
R
@
t
D
H
2

t
L H
1380 t
2

675 t
3
L
By the Fundamental Theorem of Calculus, the difference
w
H
2
L

w
H
1
L
is given by
‡
12
Dw
@
t
D
„t
775
ÅÅÅÅÅÅÅÅÅÅ
2
Total wait time for those who enter the auditorium after
t
=
1 is
775
ÅÅÅÅÅÅÅÅÅ
2
hours.
AB2009.nb
2
ü
d.
From part (a) of this problem, above, we know that there are 980 people in the auditorium at time
t
=
2. We also know thatthe total wait time for these 980 people is
‡
02
Dw
@
t
D
„t
760
760 hours. Consequently, average waiting time is
760
ÅÅÅÅÅÅÅÅÅ
980
=
38
ÅÅÅÅÅÅÅ
49
hours.
Problem 3.
ü
a.
Mighty's profit on a cable of length
k
meters is given by
P
H
k
L
=
120
k

Ÿ
0
k
6
è!!!
x
„
x
, in dollars.
P
@
k_
D
=
120 k

6
‡
0k
è!!!!
x
„
x
120 k

4k
3
ê
2
P
@
25
D
2500
Thus, profit on a 25meter cable is $2500
ü
b.
The integral
Ÿ
2530
6
è!!!
x
„
x
gives the cost, in dollars, to Mighty for building the last five meters of a 30meter cable.
ü
c.
As we saw in part (a) of this problem, above, Mighty's profit, in dollars, on the sale of a cable that is
k
meters long is120
k

6
Ÿ
0
k
è!!!
x
„
x
, or 120
k

4
k
3
ê
2
=
4
k
I
30

è!!!
k
M
.
AB2009.nb
3
ü
d.
We first observe that profit for a cable
k
meters long, which is given by 4
k
I
30

è!!!
k
M
, is nonnegative only when0
§
k
§
900 and is zero at the endpoints of this interval. Thus, the maximum does not occur at an endpoint of the interval (oroutside the interval). However, the profit function is continuous, so there must be a maximum ocurring at some
k
œ
@
0,900
D
.As we have observed, we must actually have
k
œ
H
0,900
L
, so
k
must be a critical number for the profit function. Noting thatthe derivative
P'
@
k
D
120

6
è!!!
k
of the profit function vanishes only when
k
=
400, we conclude that the maximum profit must come at
k
=
400. Themaximal profit is therefore
P
H
400
L
=
16,000 dollars.We also notice that if cables longer than 900 meters ever become very popular, Mighty should restructure something.
Problem 4.
ü
a.
The desired area is
‡
02
H
2 x

x
2
L
„
x
4
ÅÅÅÅ
3
ü
b.
The desired volume is
‡
02
Sin
A
p ÅÅÅÅ
2
x
E
„
x
4
ÅÅÅÅp
ü
c.
Solving both equations for
y
in terms of
x
gives us
x
=
1
ÅÅÅÅ
2
y
for the first and
x
=
è!!!
y
for the second. The length of the baseof the square corresponding to
y
=
y
0
is therefore
è!!!!!
y
0

1
ÅÅÅÅ
2
y
0
, and the desired volume is therefore
‡
04
Iè!!!
y

1
ÅÅÅÅ
2
y
M
2
„
y
.(For those who are unable to comply with the instruction no to evaluate the integral,, this is
8
ÅÅÅÅÅÅÅ
15
.)
AB2009.nb
4