AP® CALCULUS AB/CALCULUS BC
2015 SCORING GUIDELINES
Question 1
 t2 
The rate at which rainwater flows into a drainpipe is modeled by the function R, where R( t ) = 20sin   cubic
 35 
feet per hour, t is measured in hours, and 0 ≤ t ≤ 8. The pipe is partially blocked, allowing water to drain out the
other end of the pipe at a rate modeled by D( t ) =
−0.04t 3 + 0.4t 2 + 0.96t cubic feet per hour, for 0 ≤ t ≤ 8.
There are 30 cubic feet of water in the pipe at time t = 0.
(a) How many cubic feet of rainwater flow into the pipe during the 8-hour time interval 0 ≤ t ≤ 8 ?
(b) Is the amount of water in the pipe increasing or decreasing at time t = 3 hours? Give a reason for your
answer.
(c) At what time t, 0 ≤ t ≤ 8, is the amount of water in the pipe at a minimum? Justify your answer.
(d) The pipe can hold 50 cubic feet of water before overflowing. For t > 8, water continues to flow into and out
of the pipe at the given rates until the pipe begins to overflow. Write, but do not solve, an equation involving
one or more integrals that gives the time w when the pipe will begin to overflow.
8
{
1 : integrand
1 : answer
(a)
∫0 R( t ) dt = 76.570
2:
(b)
R ( 3) − D( 3) =
−0.313632 < 0
Since R( 3) < D( 3) , the amount of water in the pipe is
decreasing at time t = 3 hours.
 1 : considers R( 3) and D( 3)
2:
 1 : answer and reason
(c)
The amount of water in the pipe at time t, 0 ≤ t ≤ 8, is
0
 1 : considers R( t ) − D( t ) =

3 :  1 : answer
 1 : justification
30 +
t
∫0 [ R( x ) − D( x )] dx.
R( t ) − D( t ) = 0 ⇒ t = 0, 3.271658
t
0
3.271658
8
Amount of water in the pipe
30
27.964561
48.543686
The amount of water in the pipe is a minimum at time
t = 3.272 (or 3.271) hours.
(d)
30 +
w
50
∫0 [ R( t ) − D( t )] dt =
2:
{
1 : integral
1 : equation
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AP® CALCULUS AB/CALCULUS BC
2014 SCORING GUIDELINES
Question 1
Grass clippings are placed in a bin, where they decompose. For 0 ≤ t ≤ 30, the amount of grass clippings
remaining in the bin is modeled by A( t ) = 6.687 ( 0.931)t , where A(t ) is measured in pounds and t is measured
in days.
(a) Find the average rate of change of A( t ) over the interval 0 ≤ t ≤ 30. Indicate units of measure.
(b) Find the value of A′(15 ) . Using correct units, interpret the meaning of the value in the context of the
problem.
(c) Find the time t for which the amount of grass clippings in the bin is equal to the average amount of grass
clippings in the bin over the interval 0 ≤ t ≤ 30.
(d) For t > 30, L( t ) , the linear approximation to A at t = 30, is a better model for the amount of grass
clippings remaining in the bin. Use L( t ) to predict the time at which there will be 0.5 pound of grass
clippings remaining in the bin. Show the work that leads to your answer.
(a)
A( 30 ) − A( 0 )
=
−0.197 (or −0.196) lbs/day
30 − 0
(b) A′ (15 ) = −0.164 (or −0.163)
The amount of grass clippings in the bin is decreasing at a rate
of 0.164 (or 0.163) lbs/day at time t = 15 days.
1 30
(c) A( t ) =
=
A( t ) dt ⇒ t 12.415 (or 12.414)
30 ∫0
t ) A( 30 ) + A′( 30 ) ⋅ ( t − 30 )
(d) L(=
A′( 30 ) = −0.055976
A( 30 ) = 0.782928
1 : answer with units
 1 : A′(15 )
2:
 1 : interpretation
 1 : 1 30 A( t ) dt
2: 
30 ∫0
 1 : answer
 2 : expression for L ( t )

4 :  1 : L( t ) = 0.5
 1 : answer
L( t )= 0.5 ⇒ t = 35.054
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AP® CALCULUS BC
2013 SCORING GUIDELINES
Question 1
On a certain workday, the rate, in tons per hour, at which unprocessed gravel arrives at a gravel processing plant
 2
is modeled by G ( t=
) 90 + 45cos  t  , where t is measured in hours and 0 ≤ t ≤ 8. At the beginning of the
 18 
workday ( t = 0 ) , the plant has 500 tons of unprocessed gravel. During the hours of operation, 0 ≤ t ≤ 8, the
plant processes gravel at a constant rate of 100 tons per hour.
(a) Find G′( 5 ) . Using correct units, interpret your answer in the context of the problem.
(b) Find the total amount of unprocessed gravel that arrives at the plant during the hours of operation on this
workday.
(c) Is the amount of unprocessed gravel at the plant increasing or decreasing at time t = 5 hours? Show the
work that leads to your answer.
(d) What is the maximum amount of unprocessed gravel at the plant during the hours of operation on this
workday? Justify your answer.
−24.588 (or − 24.587)
(a) G′( 5 ) =
The rate at which gravel is arriving is decreasing by 24.588
(or 24.587) tons per hour per hour at time t = 5 hours.
(b)
8
∫0 G ( t ) dt = 825.551 tons
2:
=
(c) G ( 5 ) 98.140764 < 100
At time t = 5, the rate at which unprocessed gravel is arriving
is less than the rate at which it is being processed.
Therefore, the amount of unprocessed gravel at the plant is
decreasing at time t = 5.
(d) The amount of unprocessed gravel at time t is given by
A(t ) =
500 +
 1 : G′( 5 )
2: 
 1 : interpretation with units
t
∫0 ( G( s ) − 100 ) ds.
{
1 : integral
1 : answer
 1 : compares G ( 5 ) to 100
2: 
 1 : conclusion
0
 1 : considers A′( t ) =

3 :  1 : answer
 1 : justification
A′( t ) = G ( t ) − 100 = 0 ⇒ t = 4.923480
t
0
4.92348
8
A( t )
500
635.376123
525.551089
The maximum amount of unprocessed gravel at the plant during
this workday is 635.376 tons.
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AP® CALCULUS BC
2011 SCORING GUIDELINES (Form B)
Question 1
A cylindrical can of radius 10 millimeters is used to measure rainfall in Stormville. The can is initially empty,
and rain enters the can during a 60-day period. The height of water in the can is modeled by the function S,
where S ( t ) is measured in millimeters and t is measured in days for 0 ≤ t ≤ 60. The rate at which the height
of the water is rising in the can is given by S ′( t ) = 2sin ( 0.03t ) + 1.5.
(a) According to the model, what is the height of the water in the can at the end of the 60-day period?
(b) According to the model, what is the average rate of change in the height of water in the can over the
60-day period? Show the computations that lead to your answer. Indicate units of measure.
(c) Assuming no evaporation occurs, at what rate is the volume of water in the can changing at time t = 7 ?
Indicate units of measure.
(d) During the same 60-day period, rain on Monsoon Mountain accumulates in a can identical to the one in
Stormville. The height of the water in the can on Monsoon Mountain is modeled by the function M, where
1
3t 3 − 30t 2 + 330t . The height M ( t ) is measured in millimeters, and t is measured in days
M (t ) =
400
for 0 ≤ t ≤ 60. Let D( t ) = M ′( t ) − S ′( t ) . Apply the Intermediate Value Theorem to the function D on
the interval 0 ≤ t ≤ 60 to justify that there exists a time t, 0 < t < 60, at which the heights of water in the
two cans are changing at the same rate.
(
60
∫0
)
(a)
S ( 60 ) =
S ′( t ) dt = 171.813 mm
(b)
S ( 60 ) − S ( 0 )
= 2.863 or 2.864 mm day
60
(c) V ( t ) = 100π S ( t )
V ′( 7 ) = 100π S ′( 7 ) = 602.218
⎧ 1 : limits
⎪
3 : ⎨ 1 : integrand
⎪⎩ 1 : answer
1 : answer
2:
{
1 : relationship between V and S
1 : answer
The volume of water in the can is increasing at a rate of
602.218 mm3 day.
(d) D( 0 ) = −0.675 < 0 and D( 60 ) = 69.37730 > 0
Because D is continuous, the Intermediate Value Theorem
implies that there is a time t, 0 < t < 60, at which D( t ) = 0.
At this time, the heights of water in the two cans are changing
at the same rate.
⎧ 1 : considers D( 0 ) and D( 60 )
2: ⎨
⎩ 1 : justification
1 : units in (b) or (c)
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AP® CALCULUS BC
2010 SCORING GUIDELINES
Question 1
There is no snow on Janet’s driveway when snow begins to fall at midnight. From midnight to 9 A.M., snow
accumulates on the driveway at a rate modeled by f ( t ) = 7tecos t cubic feet per hour, where t is measured in
hours since midnight. Janet starts removing snow at 6 A.M. ( t = 6 ) . The rate g ( t ) , in cubic feet per hour, at
which Janet removes snow from the driveway at time t hours after midnight is modeled by
for 0 ≤ t < 6
⎧0
⎪
g ( t ) = ⎨125 for 6 ≤ t < 7
⎪⎩108 for 7 ≤ t ≤ 9 .
(a) How many cubic feet of snow have accumulated on the driveway by 6 A.M.?
(b) Find the rate of change of the volume of snow on the driveway at 8 A.M.
(c) Let h( t ) represent the total amount of snow, in cubic feet, that Janet has removed from the driveway at time
t hours after midnight. Express h as a piecewise-defined function with domain 0 ≤ t ≤ 9.
(d) How many cubic feet of snow are on the driveway at 9 A.M.?
(a)
6
∫ 0 f ( t ) dt = 142.274 or 142.275 cubic feet
2:
{
1 : integral
1 : answer
(b) Rate of change is f ( 8 ) − g ( 8 ) = −59.582 or −59.583 cubic feet per hour.
1 : answer
(c) h( 0 ) = 0
⎧ 1 : h( t ) for 0 ≤ t ≤ 6
⎪
3 : ⎨ 1 : h( t ) for 6 < t ≤ 7
⎪⎩ 1 : h( t ) for 7 < t ≤ 9
t
t
t
t
For 0 < t ≤ 6, h( t ) = h( 0 ) +
∫ 0 g ( s ) ds = 0 + ∫ 0 0 ds = 0.
For 6 < t ≤ 7, h( t ) = h( 6 ) +
∫ 6 g ( s ) ds = 0 + ∫ 6125 ds = 125 ( t − 6 ) .
For 7 < t ≤ 9, h( t ) = h( 7 ) +
∫ 7 g ( s ) ds = 125 + ∫ 7108 ds = 125 + 108 ( t − 7 ).
t
t
for 0 ≤ t ≤ 6
⎧0
⎪
Thus, h( t ) = ⎨125 ( t − 6 )
for 6 < t ≤ 7
⎪⎩125 + 108 ( t − 7 ) for 7 < t ≤ 9
(d) Amount of snow is
9
∫ 0 f ( t ) dt − h( 9 ) = 26.334 or 26.335 cubic feet.
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⎧ 1 : integral
⎪
3 : ⎨ 1 : h( 9 )
⎪⎩ 1 : answer
AP® CALCULUS BC
2009 SCORING GUIDELINES (Form B)
Question 2
A storm washed away sand from a beach, causing the edge of the water to get closer to a nearby road. The
rate at which the distance between the road and the edge of the water was changing during the storm is
modeled by f ( t ) = t + cos t − 3 meters per hour, t hours after the storm began. The edge of the water
was 35 meters from the road when the storm began, and the storm lasted 5 hours. The derivative of f ( t )
1
− sin t.
is f ′( t ) =
2 t
(a) What was the distance between the road and the edge of the water at the end of the storm?
(b) Using correct units, interpret the value f ′( 4 ) = 1.007 in terms of the distance between the road and
the edge of the water.
(c) At what time during the 5 hours of the storm was the distance between the road and the edge of the
water decreasing most rapidly? Justify your answer.
(d) After the storm, a machine pumped sand back onto the beach so that the distance between the road
and the edge of the water was growing at a rate of g ( p ) meters per day, where p is the number of
days since pumping began. Write an equation involving an integral expression whose solution would
give the number of days that sand must be pumped to restore the original distance between the road
and the edge of the water.
(a) 35 +
5
∫ 0 f ( t ) dt = 26.494 or 26.495 meters
(b) Four hours after the storm began, the rate of change of
the distance between the road and the edge of the water
is increasing at a rate of 1.007 meters hours 2 .
(c)
f ′( t ) = 0 when t = 0.66187 and t = 2.84038
The minimum of f for 0 ≤ t ≤ 5 may occur at 0,
0.66187, 2.84038, or 5.
2:
{
1 : integral
1 : answer
⎧ 1 : interpretation of f ′( 4 )
2: ⎨
⎩ 1 : units
⎧ 1 : considers f ′( t ) = 0
⎪
3 : ⎨ 1 : answer
⎪⎩ 1 : justification
f ( 0 ) = −2
f ( 0.66187 ) = −1.39760
f ( 2.84038 ) = −2.26963
f ( 5 ) = −0.48027
The distance between the road and the edge of the
water was decreasing most rapidly at time t = 2.840
hours after the storm began.
(d) −
5
x
∫ 0 f ( t ) dt = ∫ 0 g ( p ) dp
2:
{
1 : integral of g
1 : answer
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AP® CALCULUS BC
2009 SCORING GUIDELINES
Question 2
The rate at which people enter an auditorium for a rock concert is modeled by the function R given by
R( t ) = 1380t 2 − 675t 3 for 0 ≤ t ≤ 2 hours; R( t ) is measured in people per hour. No one is in the
auditorium at time t = 0, when the doors open. The doors close and the concert begins at time t = 2.
(a) How many people are in the auditorium when the concert begins?
(b) Find the time when the rate at which people enter the auditorium is a maximum. Justify your answer.
(c) The total wait time for all the people in the auditorium is found by adding the time each person waits,
starting at the time the person enters the auditorium and ending when the concert begins. The function
w models the total wait time for all the people who enter the auditorium before time t. The derivative
of w is given by w′( t ) = ( 2 − t ) R( t ) . Find w( 2 ) − w(1) , the total wait time for those who enter the
auditorium after time t = 1.
(d) On average, how long does a person wait in the auditorium for the concert to begin? Consider all people
who enter the auditorium after the doors open, and use the model for total wait time from part (c).
(a)
2
∫ 0 R( t ) dt = 980 people
2:
(b) R′( t ) = 0 when t = 0 and t = 1.36296
The maximum rate may occur at 0, a = 1.36296, or 2.
R( 0 ) = 0
R( a ) = 854.527
R( 2 ) = 120
{
1 : integral
1 : answer
⎧ 1 : considers R′( t ) = 0
⎪
3 : ⎨ 1 : interior critical point
⎪⎩ 1 : answer and justification
The maximum rate occurs when t = 1.362 or 1.363.
(c) w( 2 ) − w(1) =
2
2:
{
1 : integral
1 : answer
2:
{
1 : integral
1 : answer
2
∫ 1 w′( t ) dt = ∫ 1 ( 2 − t ) R( t ) dt = 387.5
The total wait time for those who enter the auditorium after
time t = 1 is 387.5 hours.
(d)
1
1 2
w( 2 ) =
( 2 − t ) R ( t ) dt = 0.77551
980
980 0
On average, a person waits 0.775 or 0.776 hour.
∫
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AP® CALCULUS BC
2008 SCORING GUIDELINES (Form B)
Question 2
(
For time t ≥ 0 hours, let r ( t ) = 120 1 − e −10t
2
) represent the speed, in kilometers per hour, at which a
car travels along a straight road. The number of liters of gasoline used by the car to travel x kilometers is
modeled by g ( x ) = 0.05 x 1 − e− x 2 .
(
)
(a) How many kilometers does the car travel during the first 2 hours?
(b) Find the rate of change with respect to time of the number of liters of gasoline used by the car when
t = 2 hours. Indicate units of measure.
(c) How many liters of gasoline have been used by the car when it reaches a speed of 80 kilometers per
hour?
(a)
(b)
2
∫ 0 r ( t ) dt = 206.370 kilometers
dg
dg dx
dx
=
⋅ ;
= r(t )
dt
dx dt
dt
dg
dg
=
⋅ r( 2)
dt t = 2
dx x = 206.370
2:
{
1 : integral
1 : answer
3:
{
2 : uses chain rule
1 : answer with units
= ( 0.050 )(120 ) = 6 liters hour
(c) Let T be the time at which the car’s speed reaches
80 kilometers per hour.
Then, r (T ) = 80 or T = 0.331453 hours.
⎧ 1 : equation r ( t ) = 80
⎪
4 : ⎨ 2 : distance integral
⎪⎩ 1 : answer
At time T, the car has gone
x( T ) =
T
∫ 0 r ( t ) dt = 10.794097 kilometers
and has consumed g ( x(T ) ) = 0.537 liters of gasoline.
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AP® CALCULUS BC
2007 SCORING GUIDELINES (Form B)
Question 3
The wind chill is the temperature, in degrees Fahrenheit ( °F ) , a human feels based on the air temperature, in
degrees Fahrenheit, and the wind velocity v, in miles per hour ( mph ) . If the air temperature is 32°F, then the
wind chill is given by W ( v ) = 55.6 − 22.1v 0.16 and is valid for 5 ≤ v ≤ 60.
(a) Find W ′( 20 ) . Using correct units, explain the meaning of W ′( 20 ) in terms of the wind chill.
(b) Find the average rate of change of W over the interval 5 ≤ v ≤ 60. Find the value of v at which the
instantaneous rate of change of W is equal to the average rate of change of W over the interval 5 ≤ v ≤ 60.
(c) Over the time interval 0 ≤ t ≤ 4 hours, the air temperature is a constant 32°F. At time t = 0, the wind
velocity is v = 20 mph. If the wind velocity increases at a constant rate of 5 mph per hour, what is the rate of
change of the wind chill with respect to time at t = 3 hours? Indicate units of measure.
(a)
W ′( 20 ) = −22.1 ⋅ 0.16 ⋅ 20−0.84 = −0.285 or −0.286
When v = 20 mph, the wind chill is decreasing at
0.286 °F mph.
(b) The average rate of change of W over the interval
W ( 60 ) − W ( 5 )
5 ≤ v ≤ 60 is
= −0.253 or −0.254.
60 − 5
W ( 60 ) − W ( 5 )
when v = 23.011.
W ′( v ) =
60 − 5
(c)
dW
dt
t =3
=
( dWdv ⋅ dvdt )
t =3
= W ′( 35 ) ⋅ 5 = −0.892 °F hr
OR
W = 55.6 − 22.1( 20 + 5t )0.16
dW
= −0.892 °F hr
dt t = 3
Units of °F mph in (a) and °F hr in (c)
2:
{
1 : value
1 : explanation
⎧ 1 : average rate of change
⎪
3 : ⎨ 1 : W ′( v ) = average rate of change
⎪⎩ 1 : value of v
⎧ 1 : dv = 5
⎪
dt
⎪ 1 : uses v( 3) = 35,
⎪
3: ⎨
or
⎪
uses v( t ) = 20 + 5t
⎪
⎪⎩ 1 : answer
1 : units in (a) and (c)
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AP® CALCULUS BC
2007 SCORING GUIDELINES
Question 2
The amount of water in a storage tank, in gallons, is modeled
by a continuous function on the time interval 0 ≤ t ≤ 7,
where t is measured in hours. In this model, rates are given
as follows:
(i) The rate at which water enters the tank is
f ( t ) = 100t 2 sin ( t ) gallons per hour for 0 ≤ t ≤ 7.
(ii) The rate at which water leaves the tank is
⎧ 250 for 0 ≤ t < 3
gallons per hour.
g (t ) = ⎨
⎩2000 for 3 < t ≤ 7
The graphs of f and g, which intersect at t = 1.617 and t = 5.076, are shown in the figure above. At
time t = 0, the amount of water in the tank is 5000 gallons.
(a) How many gallons of water enter the tank during the time interval 0 ≤ t ≤ 7 ? Round your answer to
the nearest gallon.
(b) For 0 ≤ t ≤ 7, find the time intervals during which the amount of water in the tank is decreasing.
Give a reason for each answer.
(c) For 0 ≤ t ≤ 7, at what time t is the amount of water in the tank greatest? To the nearest gallon,
compute the amount of water at this time. Justify your answer.
(a)
2:
{
1 : integral
1 : answer
2:
{
1 : intervals
1 : reason
7
∫ 0 f ( t ) dt ≈ 8264 gallons
(b) The amount of water in the tank is decreasing on the
intervals 0 ≤ t ≤ 1.617 and 3 ≤ t ≤ 5.076 because
f ( t ) < g ( t ) for 0 ≤ t < 1.617 and 3 < t < 5.076.
(c) Since f ( t ) − g ( t ) changes sign from positive to negative
only at t = 3, the candidates for the absolute maximum are
at t = 0, 3, and 7.
t (hours) gallons of water
0
5000
3
5000 +
7
5126.591 +
⎧ 1 : identifies t = 3 as a candidate
⎪ 1 : integrand
⎪
5 : ⎨ 1 : amount of water at t = 3
⎪ 1 : amount of water at t = 7
⎪
⎩ 1 : conclusion
3
∫ 0 f ( t ) dt − 250 ( 3) = 5126.591
7
∫3
f ( t ) dt − 2000 ( 4 ) = 4513.807
The amount of water in the tank is greatest at 3 hours. At
that time, the amount of water in the tank, rounded to the
nearest gallon, is 5127 gallons.
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AP® CALCULUS BC
2006 SCORING GUIDELINES
Question 2
At an intersection in Thomasville, Oregon, cars turn
t
cars per hour
left at the rate L( t ) = 60 t sin 2
3
over the time interval 0 ≤ t ≤ 18 hours. The graph of
y = L( t ) is shown above.
(a) To the nearest whole number, find the total
number of cars turning left at the intersection
over the time interval 0 ≤ t ≤ 18 hours.
(b) Traffic engineers will consider turn restrictions
when L( t ) ≥ 150 cars per hour. Find all values
of t for which L( t ) ≥ 150 and compute the
average value of L over this time interval.
Indicate units of measure.
(c) Traffic engineers will install a signal if there is any two-hour time interval during which the product of the
total number of cars turning left and the total number of oncoming cars traveling straight through the
intersection is greater than 200,000. In every two-hour time interval, 500 oncoming cars travel straight
through the intersection. Does this intersection require a traffic signal? Explain the reasoning that leads to
your conclusion.
()
18
(a)
∫0
(b)
L ( t ) = 150 when t = 12.42831, 16.12166
Let R = 12.42831 and S = 16.12166
L ( t ) ≥ 150 for t in the interval [ R, S ]
S
1
L ( t ) dt = 199.426 cars per hour
S − R ∫R
L ( t ) dt ≈ 1658 cars
2:
{
1 : setup
1 : answer
⎧ 1 : t -interval when L ( t ) ≥ 150
⎪
3 : ⎨ 1 : average value integral
⎪⎩ 1 : answer with units
(c) For the product to exceed 200,000, the number of cars
turning left in a two-hour interval must be greater than 400.
15
∫13 L ( t ) dt = 431.931 > 400
⎧ 1 : considers 400 cars
⎪ 1 : valid interval [ h, h + 2]
⎪
h+2
4: ⎨
⎪ 1 : value of ∫ h L ( t ) dt
⎪
⎩ 1 : answer and explanation
OR
OR
The number of cars turning left will be greater than 400
on a two-hour interval if L( t ) ≥ 200 on that interval.
L( t ) ≥ 200 on any two-hour subinterval of
[13.25304, 15.32386].
⎧ 1 : considers 200 cars per hour
⎪⎪ 1 : solves L( t ) ≥ 200
4: ⎨
⎪ 1 : discusses 2 hour interval
⎪⎩ 1 : answer and explanation
Yes, a traffic signal is required.
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3
AP® CALCULUS BC
2005 SCORING GUIDELINES (Form B)
Question 2
A water tank at Camp Newton holds 1200 gallons of water at time t = 0. During the time interval
0 ≤ t ≤ 18 hours, water is pumped into the tank at the rate
W ( t ) = 95 t sin 2
( 6t ) gallons per hour.
During the same time interval, water is removed from the tank at the rate
R( t ) = 275sin 2
( 3t ) gallons per hour.
(a) Is the amount of water in the tank increasing at time t = 15 ? Why or why not?
(b) To the nearest whole number, how many gallons of water are in the tank at time t = 18 ?
(c) At what time t, for 0 ≤ t ≤ 18, is the amount of water in the tank at an absolute minimum? Show the
work that leads to your conclusion.
(d) For t > 18, no water is pumped into the tank, but water continues to be removed at the rate R( t )
until the tank becomes empty. Let k be the time at which the tank becomes empty. Write, but do not
solve, an equation involving an integral expression that can be used to find the value of k.
(a) No; the amount of water is not increasing at t = 15
since W (15 ) − R (15 ) = −121.09 < 0.
(b) 1200 +
18
∫0
(W ( t ) − R( t ) ) dt = 1309.788
⎧ 1 : limits
⎪
3 : ⎨ 1 : integrand
⎪⎩ 1 : answer
1310 gallons
(c)
W ( t ) − R( t ) = 0
t = 0, 6.4948, 12.9748
⎧ 1 : interior critical points
⎪⎪ 1 : amount of water is least at
3: ⎨
t = 6.494 or 6.495
⎪
⎩⎪ 1 : analysis for absolute minimum
t (hours) gallons of water
0
6.495
12.975
18
1 : answer with reason
1200
525
1697
1310
The values at the endpoints and the critical points
show that the absolute minimum occurs when
t = 6.494 or 6.495.
(d)
k
∫18 R( t ) dt = 1310
⎧ 1 : limits
2: ⎨
⎩ 1 : equation
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3
AP® CALCULUS BC
2004 SCORING GUIDELINES
Question 1
Traffic flow is defined as the rate at which cars pass through an intersection, measured in cars per minute.
The traffic flow at a particular intersection is modeled by the function F defined by
F ( t ) = 82 + 4sin
( 2t ) for 0 ≤ t ≤ 30,
where F ( t ) is measured in cars per minute and t is measured in minutes.
(a) To the nearest whole number, how many cars pass through the intersection over the 30-minute
period?
(b) Is the traffic flow increasing or decreasing at t = 7 ? Give a reason for your answer.
(c) What is the average value of the traffic flow over the time interval 10 ≤ t ≤ 15 ? Indicate units of
measure.
(d) What is the average rate of change of the traffic flow over the time interval 10 ≤ t ≤ 15 ? Indicate
units of measure.
30
 1 : limits

3 :  1 : integrand
 1 : answer
(a)
∫0
(b)
F ′ ( 7 ) = −1.872 or −1.873
Since F ′ ( 7 ) < 0, the traffic flow is decreasing
at t = 7.
1 : answer with reason
(c)
1 15
F ( t ) dt = 81.899 cars min
5 ∫10
 1 : limits

3 :  1 : integrand
 1 : answer
(d)
F (15 ) − F (10 )
= 1.517 or 1.518 cars min 2
15 − 10
1 : answer
F ( t ) dt = 2474 cars
1 : units in (c) and (d)
Units of cars min in (c) and cars min 2 in (d)
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2
AP® CALCULUS BC
2002 SCORING GUIDELINES (Form B)
Question 2
The number of gallons, 2 J , of a pollutant in a lake changes at the rate 2 J !A 0.2
J
gallons per day, where J is measured in days. There are 50 gallons of the pollutant in the lake at
time J = 0. The lake is considered to be safe when it contains 40 gallons or less of pollutant.
(a) Is the amount of pollutant increasing at time J = 9 ? Why or why not?
(b) For what value of J will the number of gallons of pollutant be at its minimum? Justify your
answer.
(c) Is the lake safe when the number of gallons of pollutant is at its minimum? Justify your
answer.
(d) An investigator uses the tangent line approximation to 2 J at J = 0 as a model for the
amount of pollutant in the lake. At what time J does this model predict that the lake
becomes safe?
1 : answer with reason
(a) 2 (9) 1 3A 0.6 0.646 < 0
so the amount is not increasing at this time.
(b) 2 J !A 0.2
J
J = 5 ln 3 = 30.174
2 =J is negative for 0 < J < 5 ln 3 and positive
for J > 5 ln 3 . Therefore there is a minimum at
£
¦
1 : sets 2 a(J ) 0
¦
¦
¦
3 ¤ 1 : solves for J
¦
¦
¦
1 : justification
¦
¦
¥
J = 5 ln 3 .
(c) 2 (30.174) 50 30.174
¨0
1 3A0.2 J @J
= 35.104 < 40, so the lake is safe.
(d) 2 ! . The lake will become safe
when the amount decreases by 10. A linear model
1 : integrand
£
¦
¦
¦
¦
¦ 1 : limits
3 ¦¤
1 : conclusion with reason
¦
¦
¦
¦
¦
based on integral of 2 a(J )
¦
¥
£
¦ 1 : slope of tangent line
2 ¦¤
¦
1 : answer
¦
¥
predicts this will happen when J = 5.
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3
AP® CALCULUS BC 2002 SCORING GUIDELINES
Question 2
The rate at which people enter an amusement park on a given day is modeled by the function E defined by
E t 15600
t 24t 160 .
The rate at which people leave the same amusement park on the same day is modeled by the function L defined by
L t 9890
.
t 38t 370 Both - J and L t are measured in people per hour and time t is measured in hours after midnight. These
functions are valid for 9 > J > 23, the hours during which the park is open. At time J 9, there are no people in
the park.
(a) How many people have entered the park by 5:00 P.M. ( J % )? Round answer to the nearest whole number.
(b) The price of admission to the park is $15 until 5:00 P.M. ( J % ). After 5:00 P.M., the price of admission to
the park is $11. How many dollars are collected from admissions to the park on the given day? Round your
answer to the nearest whole number.
(c)
Let H t J
¨' E x Lx dx
for 9 > J > 23. The value of 0 % to the nearest whole number is 3725.
Find the value of 0 = % and explain the meaning of 0 % and 0 = % in the context of the park.
(d) At what time t, for 9 > J > 23, does the model predict that the number of people in the park is a maximum?
(a)
%
¨'
£
¦
1 : limits
¦
¦
¦
3 ¤ 1 : integrand
¦
¦
¦
1 : answer
¦
¦
¥
- (J ) @J 6004.270
6004 people entered the park by 5 pm.
(b) 15 ¨
%
'
- (J ) @J 11¨
!
%
- (J ) @J 104048.165
1 : setup
The amount collected was $104,048.
or
!
¨%
- (J ) @J 1271.283
1271 people entered the park between 5 pm and
11 pm, so the amount collected was
$15 (6004) $11 (1271) $104, 041.
(c) H (17) E (17) L(17) 380.281
There were 3725 people in the park at t = 17.
The number of people in the park was decreasing
at the rate of approximately 380 people/hr at
time t = 17.
1 : value of 0 (17)
2 : meanings
1 : meaning of 0 (17)
3 1 : meaning of 0 (17)
1 if no reference to J 17
1 : E (t ) L(t ) 0
2 1 : answer
(d) H t E t Lt t = 15.794 or 15.795
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3
1998 Calculus BC Scoring Guidelines
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