Results from the 2011 AP Calculus AB and BC Exams
Stephen Kokoska
Bloomsburg University
Chief Reader, AP Calculus
skokoska@bloomu.edu
July, 2011
1 / 60
APDC
AP Calculus Development Committee
Co-Chair: Thomas Becvar, St. Louis University High School, St.
Louis, MO
Co-Chair: Tara Smith, University of Cincinnati, Cincinnati, OH
Robert Arrigo, Scarsdale High School, Scarsdale, NY
Vicki Carter, West Florence High School, Florence, SC
Donald King, Northeastern University, Boston, MA
James Sellers, Penn State University, State College, PA
Chief Reader: Stephen Kokoska, Bloomsburg University, PA
College Board Advisor: Kathleen Goto, Iolani School, Honolulu, HI
ETS Consultants: Fred Kluempen, Craig Wright
2 / 60
APDC
Committee Responsibilities
Plan, develop, and approve each exam.
Participate in the Reading.
Consider statistical analysis associated with the exam.
Review, revise, and update the Course Description
Consider the following items.
1
2
3
Special Focus topics.
Advise the Chief Reader on reading issues.
Participate in Outreach Events.
Four meetings each year. Additional virtual meetings.
3 / 60
Exams
AP Calculus Exams
Operational: North, Central, and South America (includes Alaska and
Hawaii)
Form A: Alternate Exam, late test
Form I: International, operational
Form J: International, alternate
Section I: Multiple Choice. Section II: Free Response
Calculator and Non-calculator Sections
AB and BC Exams
Common Problems
4 / 60
The Reading
Leadership
Chief Reader:
Michael Boardman, Pacific University, OR
Chief Reader Designate:
Steve Kokoska, Bloomsburg University, PA
Assistant to the Chief Reader:
Tom Becvar, Saint Louis University, MO
Chief Aide:
Craig Turner, Georgia College and State University, GA
5 / 60
The Reading
Leadership - Exam Leaders
AB Exam Leader:
Jim Hartman, College of Wooster, OH
BC Exam Leader:
Peter Atlas, Concord Carlisle Regional High School, MA
Alternate Exam Leader:
Stephen Davis, Davidson College, NC
Overseas Exam Leader:
Deanna Caveny, College of Charleston, SC
6 / 60
The Reading
Leadership - Question Leaders
AB1: Particle Motion
Frances Rouse, UMS-Wright Preparatory School, AL
AB2/BC2: Model Numerically/Analytically (Tea and Biscuits)
Ben Klein, Davidson College, NC
AB3: Tangent Line, Area, Volume
Pario-Lee Law, D’Evelyn Junior Senior High School, CO
Dale Hathaway, Olivet Nazarene University, IL
AB4/BC4: Analysis of a Graph, FTC
Tara Smith, University of Cincinnati, OH (NSF)
AB5/BC5: Landfill, Differential Equation
Virge Cornelius, Lafayette High School, MS
AB6: Piecewise Defined Function
Mike Koehler, Blue Valley North High School, KS
7 / 60
The Reading
Leadership - Question Leaders
BC1: Parametric Particle Motion
John Jensen, Rio Salado College, AZ
BC3: Perimeter, Volume
Tracy Bibelnieks, Augsburg College, MN
BC6: Taylor Series
Mark Kannowski, DePauw University, IN
Alternate Exam TQL:
Julie Clark, Hollins University, VA
Overseas Exam TQL:
Chris Lane, Pacific University, OR
8 / 60
The Reading
Logistics and Participants
Kansas City:
Convention Center (versus college campus)
Westin Hotel (versus college dorms)
Total Participants: 857
High School: 49%
College: 51%
50 states, DC, and other countries
9 / 60
The Reading
350000
Number of AP Calculus Exams
300000
350000
250000
AB Exams
200000
300000
150000
Exams
250000
BC Exams
100000
Total Exams
50000
200000
0
19551960
150000
1970
1980
1990
2000
2011
2011 Totals
AB: 257,054
BC: 85,647
100000
50000
0
1955
1960
1970
1980
1990
Year
2000
2011
Since 1955:
> 4, 913, 300
10 / 60
The Reading
2011 Scores (Operational Exam)
Operational
Score
AB
BC
5
21.0%
47.0%
4
16.3%
16.3%
3
18.5%
17.2%
2
10.8%
5.9%
1
33.4%
13.6%
Total 257,054 85,647
11 / 60
The Reading
General Comments
Students must show their work (bald answers).
Students must communicate effectively, explain their reasoning, and
present results in clear, concise, proper notation.
Practice in justifying conclusions using calculus arguments.
More practice with functions represented in graphical and tabular
form. Consider piecewise-defined functions.
Difficulty with application problems, or problems in context.
Correct units.
Don’t give up on a free response problem. Try all parts.
12 / 60
The Reading
Some Calculator Tips
Use radian mode.
Report final answers accurate to three digits to the right of the
decimal. Do not round or truncate intermediate values.
Show the mathematical setup, then use the calculator.
Examples:
R6
|v (t)|dt = 12.573
dy
Derivative at a point:
= 7.658
dx x =1.2
Definite integral:
0
Do not use calculator syntax.
fnint(x,x,0,5)
Which calculator?
13 / 60
2011 Free Response
General Information
Six questions on each exam (AB, BC).
Three common questions: AB2/BC2, AB4/BC4, AB5/BC5.
Scoring: 9 points for each question.
Complete and correct answers earn all 9 points.
The scoring standard is used to assign partial credit.
14 / 60
2011 Free Response
AP Calculus AB–1
Version 1.0
201
AB1: Particle Motion
For 0  t  6, a particle is moving along the x-axis. The particle’s position, x t  , is not explicitly given.

The velocity of the particle is given by v t   2sin et
1
a  t   et
2
4

cos e
t 4
4
  1. The acceleration of the particle is given by
 and x 0   2.
(a) Is the speed of the particle increasing or decreasing at time t  5.5 ? Give a reason for your answer.
(b) Find the average velocity of the particle for the time period 0  t  6.
(c) Find the total distance traveled by the particle from time t  0 to t  6.
(d) For 0  t  6, the particle changes direction exactly once. Find the position of the particle at that time.
(a) v 5.5   0.45337, a 5.5   1.35851
2 : conclusion with reason
The speed is increasing at time t  5.5, since velocity and
acceleration have the same sign.
15 / 60
(b) Find the average velocity of the particle for the time period 0  t  6.
6
1 : integral
1 5.5
(a)
v 5.5the
0.45337,
1.35851
  total
(b)
2 ::6.conclusion with reason
 atraveled
vt 
 the
1.949
velocity
 dtby
(c) Average
Find
distance
particle from time t  0 to t 2
6 0
1 : answer
(d) For 0  t  6, the particle changes direction exactly once. Find the position of the particle at that time.
The speed is increasing at time t  5.5, since velocity and
acceleration have the same sign.
1 : integral
1 6   1.35851
(a)
22 :: conclusion
with reason
v 5.5   velocity
0.45337,
(b) Average
 a 5.5
v t  dt  1.949
6 0
1 : answer
The speed is increasing
at time t  5.5, since velocity and
6
1 : integral
2:
(c) acceleration
Distance  have
v tthe
 12.573
 dtsame
sign.
0
1 : answer
1 : integral
1 6
(b) Average velocity   v t  dt  1.949
2:
6 0
1 : answer

2011 Free Response

(c) Distance 
6
0
v t  dt  12.573
1 6
(b) Average
v t  Let
dt b1.949
(d)
v t   0 velocity
when t  5.19552.
 5.19552.
6 0
v t  changes sign from positive to negative at time t  b.
(c) Distance
x b   2 
6b
12.573or 14.135
 0 vvtt dtdt14.134
(d) v t   0 when t  5.19552. Let b  5.19552.
v t  changes sign from positive to negative at time t  b.
(c) Distance
x b   2 
6b
12.573or 14.135
 0 vvtt dtdt14.134
(d) v t   0 when t  5.19552. Let b  5.19552.
v t  changes sign from positive to negative at time t  b.
x b   2 
b
2:
2:
3:
2:
3:
2:






1 : integral
1 : answer
1 : integral
 1 : considers v t   0
 1 : answer
 1 : integral
 11 :: integral
answer
1 : answer
 1 : considers v t   0

 1 : integral
 11 :: integral
answer
1 : answer

 1 : considers v t   0

3 :  1 : integral
 1 : answer
14.135
The College Board. All rights reserved.
 0 v t  dt  14.134 or© 2011
16 / 60
2011 Free Response
AB1: Particle Motion
Exam
AB
Mean
2.79
St Dev
2.44
% 9s
2.5
Overall, student performance was poor.
In general, students did not do well on part (a). Many did not earn
either point.
Most students correctly set up and evaluated the average value
integral to earn both points in part (b).
In part (c), most students failed to earn either point. However, those
who successfully set up the integral for total distance usually earned
the answer point also.
In part (d), few students earned all three points.
17 / 60
2011 Free Response
AB1: Particle Motion
Commons errors:
Problems entering the correct functions for velocity and acceleration
into the calculator, leaving off the +1 or misplacing the parentheses
on the sine function.
Considered the sign of one function, velocity or acceleration.
v (6) − v (0)
Use of average acceleration,
rather than average velocity.
6−0
Z 6
Ignored the change in direction:
Z 6
v (t) dt instead of
0
|v (t)| dt
0
Did not use the time t = 5.196 to find position.
Did not use the initial condition x (0) = 2.
18 / 60
2011 Free Response
AB1: Particle Motion
To help students improve performance:
Understand the difference between:
Speed and velocity.
Average velocity and average acceleration.
Total distance traveled and displacement.
Careful definitions of calculator functions.
Finding the position of a particle using the Fundamental Theorem of
Calculus.
Z b
x (b) = x (a) +
v (t) dt
a
19 / 60
2011 Free Response
AP Calculus AB-2/BC-2
Version 1.0
2011
AB2/BC2: Model Numerically/Analytically
t
(minutes)
0
2
5
9
10
H t 
(degrees Celsius)
66
60
52
44
43
As a pot of tea cools, the temperature of the tea is modeled by a differentiable function H for 0  t  10, where
time t is measured in minutes and temperature H  t  is measured in degrees Celsius. Values of H  t  at selected
values of time t are shown in the table above.
(a) Use the data in the table to approximate the rate at which the temperature of the tea is changing at time
t  3.5. Show the computations that lead to your answer.
1 10
H  t  dt in the context of this problem. Use a trapezoidal
10  0
1 10
H  t  dt.
sum with the four subintervals indicated by the table to estimate
10 0
(b) Using correct units, explain the meaning of

(c) Evaluate
10
0
H  t  dt. Using correct units, explain the meaning of the expression in the context of this
problem.
(d) At time t  0, biscuits with temperature 100C were removed from an oven. The temperature of the
biscuits at time t is modeled by a differentiable function B for which it is known that
B t   13.84e 0.173t . Using the given models, at time t  10, how much cooler are the biscuits than
the tea?
20 / 60
B ttea?
. Using the given models, at time t  10, how much cooler are the biscuits than
  13.84e
the
t
biscuits
time et 0.173
is modeled
a differentiable
it is known
13.84
given models,function
at time tB for
10,which
B ttea?
. Usingbythe
how much
cooler that
are the biscuits than
  at
the
B ttea?
the
  13.84e 0.173t . Using the given models, at time t  10, how much cooler are the biscuits than
H  5  H  2 
H tea?
3.5  
1 : answer
the
H  55  2H  2 
H  3.5  
1 : answer

52
60
H  55  2H 22.666

or  2.667 degrees Celsius per minute
H  3.5  

1 : answer
3
52
 2H 22.666
H 
5560

or  2.667 degrees Celsius per minute
H  3.5  
 52 
1 : answer
3560

2

 2.666 or  2.667 degrees Celsius per minute
3
1 10  52  60  2.666 or  2.667 degrees Celsius per minute
H  t  dt3is the average temperature of the tea, in degrees Celsius,
 1 : meaning of expression
10
1  010

H  t  dt is the average temperature of the tea, in degrees Celsius,
meaning of sum
expression
3
:
 1 : trapezoidal

10
over
the 10 minutes.
1 010
 1 : meaning of expression
is
the
average
temperature
of
the
tea,
in
degrees
Celsius,
H
t
dt


3
:
estimate sum
 1 : trapezoidal
 01010the 10 minutes.
10
over

1
 44 Celsius,
1
1
66  temperature
60
60 of52the tea,52
44  43
H  tt  dt
meaning of sum
expression
3 :  1 : trapezoidal
estimate
dt is the average
 3
 4  in degrees
 1
2
 minutes.
over
theH10
10  0010

10
10
2 60
2 52
2 43
1
1
66 
60 
52 2 44
44 
3
:
trapezoidal
sum
1
:
estimate


dt 
 3
 4
 1
2
 t minutes.
over
theH10
10
10
2 60
2 52
2 43
 52.95
1  010
1
66 
60 
52 2 44
44 
 1 : estimate
H  t  dt 
 3
 4
 1
2

10
0
10
10
2
2
2
2 43

52.95
1
1
66  60
60  52
52  44
44 
H  t  dt 
 3
 4
 1
2
 52.95
1010 0
10
2
2
2
2
1 : value of integral
10   H  0   43  66  23
 010 H  t  dt  H 52.95
2:
 t  dt  H 10   H  0   43  66  23
value of integral
1 : meaning
of expression
 010 Htemperature
The
of the tea drops 23 degrees Celsius from time t  0 to
2:
1 : meaning
value of integral
H  t  dt  H 10   H  0   43  66  23
of expression
The
t

10
minutes.
time
10
0 temperature of the tea drops 23 degrees Celsius from time t  0 to
2:

value of integral
 H 10   H  0   43  66  23
1 : meaning
of expression
 0 Htemperature
t t 10dt minutes.
time
The
of the tea drops 23 degrees Celsius from time t  0 to
2:
1 : meaning of expression
t  10 minutes.
time temperature
The
of
the
tea
drops
23
degrees
Celsius
from
time
to
t

0
10
 10
  B t  dt  34.18275; H 10   B10   8.817
B10t 
100minutes.
 1 : integrand
time
010
 : integrand
B10   100   B t  dt  34.18275; H 10   B10   8.817
3
:
 1 uses B  0   100
010
The biscuits are 8.817
degrees Celsius cooler than the tea.
 1 : integrand
B10   100   B t  dt  34.18275; H 10   B10   8.817
uses B  0   100
3
:
 1 : answer
010
The biscuits are 8.817
degrees Celsius cooler than the tea.

B10   100   B t  dt  34.18275; H 10   B10   8.817
uses B  0   100
3 :  1 : integrand
answer
The biscuits are 8.817
degrees
Celsius
cooler
than
the
tea.
0

© 2011 The College Board. All rights reserved.
uses B  0   100
3 :  1 : answer
The biscuits are 8.817 degrees
Celsius
cooler
than
the
tea.
Visit
the
College
Board
on
the
Web:
www.collegeboard.org.
© 2011 The College Board. All rights reserved.
 1 : answer
Visit the College Board on the Web: www.collegeboard.org.
2011
Free Response
(a)
(a)
(a)
(a)
(b)
(b)
(b)
(b)
(c)
(c)
(c)
(c)
(d)
(d)
(d)
(d)








© 2011 The College Board. All rights reserved.
Visit the College Board on the Web: www.collegeboard.org.
© 2011 The College Board. All rights reserved.



21 / 60
2011 Free Response
AB2/BC2: Model Numerically/Analytically
Exam
AB
BC
Mean
3.31
5.48
St Dev
2.87
2.84
% 9s
5.5
18.0
Overall performance was average.
Part (a): most students earned the point.
Parts (b) and (c): many students had difficulty explaining the
meaning of the expressions.
Part (b): sufficient presentation to convey the trapezoidal rule.
Part (d): use of a definite integral.
22 / 60
2011 Free Response
AB2/BC2: Model Numerically/Analytically
Common errors:
Explanation of expressions involving definite integrals.
Failure to recognize the time intervals were of unequal lengths.
Application of the Fundamental Theorem of Calculus.
Z 10
H 0 (t) dt = H(10) − H(0)
0
Part (d): found the temperature of the biscuits, B(10). Did not give
a final answer (difference).
23 / 60
2011 Free Response
AB2/BC2: Model Numerically/Analytically
To help students improve performance:
The Fundamental Theorem of Calculus in various forms.
Z b
f (b) = f (a) +
f 0 (t) dt
a
Application of computational skills to contextual problems.
Appropriate use of units.
24 / 60
2011 Free Response
AP Calculus AB–3
Version 1.0
2011
AB3: Tangent Line/Area/Volume
Let R be the region in the first quadrant enclosed by the graphs of f  x   8 x3
and g  x   sin  x  , as shown in the figure above.
(a) Write an equation for the line tangent to the graph of f at x 
1
.
2
(b) Find the area of R.
(c) Write, but do not evaluate, an integral expression for the volume of the solid
generated when R is rotated about the horizontal line y  1.
(a)
f
 12   1

 1 : f  1
2
2: 
 1 : answer
 12   6
f  x   24 x 2 , so f 
An equation for the tangent line is y  1  6 x 
1
.
25 / 60
 12   6
 1 : answer
f  x   24 x 2 , so f 
(a) f    1
2 Free Response
2011
1
An equation for the tangent line is y  1  6  x   .
1

2
1
6
f  x   24 x , so f 
2
1
(a) f
1
1
An 2equation for the tangent line is y  1  6 x  .
2
1
2
6
f  x   24 x , so f 
2
2




1 2


 1 : f  1
2
2: 
 1 : answer

 1 : f  1
2
2: 
 1 : answer
(b) Area 
 g  x   f  x   dx
0 for the tangent line is y  1  6 x  1 .
An equation
2
1 2
sin  x   8 x3 dx

 1 : integrand

4 :  2 : antiderivative
 1 : answer
(b) Area   1  g  x   f  x  4 dxx  1
 0 cos  x   2 x
 
 x  0
1 2
3

x
8
x
dx
 1 sin



1
 0 
18 2 
x1
1 g  x   f  x  4 dx
(b) Area 
  0 cos
 x   2 x 
 
 x 0
2
 1 : integrand

4 :  2 : antiderivative
 1 : answer
2
 1 : integrand

4 :  2 : antiderivative
 1 : answer
0 
1 2

 



1 2



  1 sin
1  x   8 x3 dx
 0 
8 
x1 2
1
   cos  x   2 x 4 



 x 0
1 2 
(c)  
11 f 1x  2  1  g  x  2 dx
0
 
 2
1 28

 
1  8 x3  1  sin  x  2 dx

(c) 

1 02



2
2
 0  1  f  x    1  g  x    dx

1 2
3 2
2


3:

1 : limits and constant
2 : integrand
3:

1 : limits and constant
2 : integrand
26 / 60
2011 Free Response
AB3: Tangent Line/Area/Volume
Exam
AB
Mean
4.64
St Dev
2.69
% 9s
7.7
Overall student performance was good, especially in parts (a) and (b).
Part (a): most students earned both points.
Part (b): most students earned the setup point and at least one
antiderivative point.
Part (c): the most difficult for students.
27 / 60
2011 Free Response
AB3: Tangent Line/Area/Volume
Commons errors:
Antiderivative of sin(πx )
Z 1/2
Reversal:
(f (x ) − g(x )) dx
0
Negative answer to positive area.
Part (c): incorrect placement of parentheses and square of the wrong
term(s).
28 / 60
2011 Free Response
AB3: Tangent Line/Area/Volume
To help students improve performance:
In addition to formulas for the volume of a solid of revolution,
consider the volume of a solid with known cross-sectional area.
Mathematical notation: missing, misplaced symbols, poor
presentation.
Present intermediate work.
29 / 60
2011 Free Response
AP Calculus AB–4/BC–4
Version 1.0
2011
AB4/BC4: Graphical Analysis
The continuous function f is defined on the interval  4  x  3.
The graph of f consists of two quarter circles and one line
segment, as shown in the figure above.
Let g  x   2 x 
x
0
f  t  dt.
(a) Find g  3 . Find g  x  and evaluate g 3 .
(b) Determine the x-coordinate of the point at which g has an
absolute maximum on the interval  4  x  3.
Justify your answer.
(c) Find all values of x on the interval  4  x  3 for which
the graph of g has a point of inflection. Give a reason for
your answer.
(d) Find the average rate of change of f on the interval
 4  x  3. There is no point c,  4  c  3, for which f  c  is equal to that average rate of change.
Explain why this statement does not contradict the Mean Value Theorem.
(a)
g  3  2  3 
g  x   2  f  x 
3
0
f  t  dt  6 
9
4
 1 : g  3

3 :  1 : g  x 
30 / 60
 4  x  3. There is3no point c,  4  c  3, for which f  c  is equal to that average rate of change.
(d) Find
average
off change
on9the interval
(a)
g  3the
2  this
3 rate
t  dt
6f contradict
of
 why
Explain
statement
does
not
the Mean Value Theorem. 1 : g  3
0 no point c,  4  4c  3, for which f  c  is equal to
 that average rate of change.
 4  x  3. There is
3
9
3 :  1 : g  x 
(a) Explain
g x3 why
f
t
dt
6





22 this
f3 xstatement
does not contradict
the Mean Value Theorem. 1 : g  3
0
4

 3
3
x
3 :  11 :: gg  
9
(a) g x33222 
f3f x 3   f2 t  dt  6 
 1 : g  3
4
0
 3
g
1
:


3
3 :  1 : g  x 
9
(a) gg x33222 
f3f x 3   f2 t  dt  6 
 1 : g  3
4
0
 1 : g  3
3 :  1 : g  x 
g  
3

2

f

3

2



x  2  f  x
5
(b) g  x   0 when f  x   2. This occurs at x  .
 11 :: considers
g  3 g  x   0
2

3 :  1 : identifies interior
candidate
g  
 f  f3x 2 52. This occurs at x 5 5 .
(b) gg
1
:
considers
g  x   0
 xx 3 002when



for  4  x  and g  x   0 for 2 x  3.
 1 : answer with justification
2
25
3 :  1 : identifies interior candidate
5
(b) gg  xx  
 00 when
f xx  52.and
This
 1 : considers g  x   0
g occurs
x   0atforx 5 . x  3.
for
 4an
Therefore,
g has
absolute
 1 : answer with justification
2 maximum at x  22 2.
3 :  1 : identifies interior candidate
5
5
5
(b) gg  xx  
0
0 when
f xx  2.and
This
occurs
atforx 5 . x  3.
 1 : considers g  x   0

g
x

0
for

4


Therefore, g has an absolute
 1 : answer with justification
2 maximum at x  22 2.
3 :  1 : identifies interior candidate
5
5
5  x  3.


g
x

0
g
x

0
for
and
for

4

x





Therefore, g has an absolute
maximum at x  2 .
with justification
 1 : answer
2 only
1 : answer
with reason
(c) g  x   f  x  changes sign
at x  0. Thus,
2 the graph
5
x

0.
of
g
has
a
point
of
inflection
at
 . the graph
an absolute
maximum
1 : answer with reason
(c) Therefore,
g  x   f g xhas
sign only
at x at0.x Thus,
 changes
2
of g has a point of inflection at x  0.
1 : answer with reason
(c) g  x   f  x  changes sign only at x  0. Thus, the graph
rate of
of inflection
change ofatf on
(d) The
1 : average rate of change
x the
0. interval  4  x  3 is
of g average
has a point
12 : answer with reason
(c) g  x   f  x  changes sign only at x  0. Thus, the graph
1 : explanation
f
3

f
(

4)


2
rateof
of
change
ofatf on
(d) The
1 : average rate of change
.
 inflection
x the
0. interval  4  x  3 is
of g average
has a point
2:
3  ( 4)
7
1 : explanation
f  3  f ( 4)
2
.
 of
 Value
average
change
of f on the
interval
 4  x  3 is
(d) The
1 : average rate of change
To apply
the rate
Mean
Theorem,
f must
be differentiable
3  ( 4)
7
2:
atf each
point
in
the
interval

4

x

3.
However,
f
is
not
1 : explanation
3  f ( 4)
2
To apply
the rate
Mean
Theorem,
f must
be differentiable
.
 of
 Value
average
change
of f on the
interval
 4  x  3 is
(d) The
1 : average rate of change
differentiable
at x 7 3 and x  0.
3

(

4)
2
:
atf each
 4  x  3. However, f is not
1 : explanation
3 point
f ( 4)in the interval
2
To
apply
the
Mean
Value
Theorem,
f
must
be
differentiable
differentiable
atx7 . 3 and x  0.
3

(

4)
at each point in the interval  4  x  3. However, f is not
To
apply the Mean
Theorem,
differentiable
at x Value
3 and
x  0. f must be differentiable
at each point in the interval  4  x  3. However, f is not
31 / 60
2011 Free Response




2011 Free Response
AB4/BC4: Graphical Analysis
Exam
AB
BC
Mean
2.44
4.13
St Dev
2.26
2.29
% 9s
0.4
1.6
Student performance was below expectation.
Part (a): many students earned only one of three points.
Part (b): Lack of justification for an absolute maximum.
Part (c): few students earned the point.
Part (d): many correctly found the average rate of change of f . Few
could explain why the statement does not contradict the Mean Value
Theorem.
32 / 60
2011 Free Response
AB4/BC4: Graphical Analysis
Commons errors:
Part (a):
Z
−3
Z
0
f (t) dt = −
0
f (t) dt
−3
Ignored 2x when finding g 0 (x ).
Used 3 instead of −3.
Part (b): Justification for a local maximum rather than a global
maximum.
Part (c):
Listed x = −3 as an inflection point.
Rejected x = 0 (because g 00 (x ) = f 0 (x ) DNE).
Part (d): application of the Mean Value Theorem.
33 / 60
2011 Free Response
AB4/BC4: Graphical Analysis
To help students improve performance:
Understand the difference between a local extrema and a global
extrema. And, be able to use a variety of methods to justify.
Communicate mathematics using clear, mathematically correct,
precise language.
Practice with functions whose definitions include an integral.
Be able to reason from the graph of a function, the graph of a
derivative, or a function whose definition involves a presented graph.
34 / 60
2011 Free Response
AP Calculus AB–5/BC–5
Version 1.0
2011
AB5/BC5: Modeling, Differential Equation
At the beginning of 2010, a landfill contained 1400 tons of solid waste. The increasing function W models
the total amount of solid waste stored at the landfill. Planners estimate that W will satisfy the differential
dW
1

equation
W  300  for the next 20 years. W is measured in tons, and t is measured in years from
dt
25
the start of 2010.
(a) Use the line tangent to the graph of W at t  0 to approximate the amount of solid waste that the landfill
1
contains at the end of the first 3 months of 2010 (time t  ).
4
d 2W
d 2W
in terms of W. Use
to determine whether your answer in part (a) is an underestimate or
dt 2
dt 2
1
an overestimate of the amount of solid waste that the landfill contains at time t  .
4
(b) Find
(c) Find the particular solution W  W  t  to the differential equation
condition W  0   1400.
(a)
dW
dt
t 0

1
W  0   300   1 1400  300   44
25
25
The tangent line is y  1400  44t.
dW
1

W  300  with initial
25
dt
 1 : dW at t  0
2: 
dt
 1 : answer
35 / 60
condition W  0   1400.
dW
1
(c)
 t  tothe
(a) Find the particular
300differential

W  0solution
  300 W 1W1400
  44 equation
dt t  0 25
25
2:
condition
W
0

1400.


The
dW tangent 1line is y  1400  44
1 t.
(a)

W  0   300   1400  300   44
dt 1 t  0 25
25
2:
1
W
 1400  44
 1411 tons
4
4
The
dW tangent 1line is y  1400  44
1 t.
(a)

W  0   300   1400  300   44
1
dt 1
25
25
2:
W t  0 1400  44
 1411 tons
4
4
The tangent line is y  1400  44t.
2011 Free Response




1 dW 44  11   1411 tons
d W1   1400
W
 4 W  300  and W  1400
4 
25 dt
625
dt
2
(b)
2
2
1 ddW
1
d 2W
W
1
(b) Therefore,
300  and

 0 onW
theinterval
 1400
0 Wt 
.
25 dt
625
dt2
4
dt 2
2
The
part (a)1 is an underestimate. 1
W
1 din
d 2Wanswer
dW
theinterval
 1400
0 Wt 
.
(b) Therefore,
300  and

 0 onW
4
25 dt
625
dt2
dt 2
The answer in2 part (a) is an underestimate.
d W
1
 0 on the interval 0  t  .
Therefore,
2
4
dW
1 dt
(c) The answer

W  300 
dt
25 in part (a) is an underestimate.
1

dW 1 1
  dt
(c)  W  300WdW 300
 25
dt
25
1
1
1
ln
  t C

dt
dWW  300
1 dW
(c)  W  300W 25
300
 25
dt
25
1
1
ln
1400
 300
ln W
1 300
   t25
 C
10   C  ln 1100   C

dW25

dt
 W  300
1 25
1t
25
ln 
1400
 1e  0   C  ln 1100   C
W
300300
1100
ln W  300 
t25 C
25 1 1 t
W  t   300  1100
et 25 , 0  t  20
251
dW
 1 : dW1 at
Wt 300
0  with initial
dt dt25
 1 : answer
 1 : dW at t  0
dt

 1 : answer
dW
at t  0
 1 :
dt

 1 : answer

d 2W
1:
2: 
dt 2
 1 : answer
d 2W with reason
1:
2: 
dt 2
 1 : answer
with reason

d 2W
1:
2: 
dt 2
 1 : answer with reason
 1 : separation of variables
 1 : antiderivatives

5 :  1 : separation
constant ofof
integration
variables
 1 : antiderivatives
uses initial condition

5 :  1
1 :: separation
solves
forofW
constant
integration
of
variables
 1 : uses initial condition
1
:
antiderivatives

Note:
2 5for
[1-1-0-0-0]
if no constant of
:: solves
5 :  1
1max
constant
ofWintegration
 1 integration
: uses initial condition
 max
0 5 2if no
separation of
Note:
5 [1-1-0-0-0]
if variables
no constant of
 1 : solves for W
36 / 60
integration
2011 Free Response
AB5/BC5: Modeling, Differential Equation
Exam
AB
BC
Mean
1.63
3.53
St Dev
2.32
2.91
% 9s
0.5
2.4
Overall student performance was very poor.
For those who earned points, generally both in part (a) - tangent line
approximation.
Most students failed to earn any points in part (b).
Part (c) was a traditional separable DE. Most common - all 5 or
middle 3.
37 / 60
2011 Free Response
AB5/BC5: Modeling, Differential Equation
Commons errors:
Algebra and arithmetic errors throughout the problem.
Part (a): string of equalities.
Example: (1/25)(1100) = 44(1/4) = 11 + 1400 = 1411
d 2W
1
Part (b):
=
(incorrect)
2
dt
25
d 2W
> 0 at a single point rather than on the entire interval.
dt 2
Part (c): errors in separating variables.
38 / 60
2011 Free Response
AB5/BC5: Modeling, Differential Equation
To help students improve performance:
Solving DEs in contextual problems.
The behavior of a solution to a DE without actually solving the DE.
Solving DEs in which the independent variable does not appear.
Determination of an underestimate or an overestimate requires
knowledge of the behavior of the function and its derivatives on an
entire interval, not at a single point.
39 / 60
2011 Free Response
AP Calculus AB–6
AP Calculus AB–6
Version 1.0
Version 1.0
2011
2011
AB6: Analysis of a Piecewise Defined Function
1  2sin x for x  0
Let f be a function defined by f  x   1 4 x2sin x for x  0
Let f be a function defined by f  x   e  4 x
for x  0.
for x  0.
e
(a) Show that f is continuous at x  0.
(a) Show that f is continuous at x  0.
(b) For x  0, express f  x  as a piecewise-defined function. Find the value of x for which f  x   3.
(b) For x  0, express f  x  as a piecewise-defined function. Find the value of x for which f  x   3.
(c) Find the average value of f on the interval  1, 1.
(c) Find the average value of f on the interval  1, 1.
(a)
(a)
lim 1  2sin x   1
x  1
xlim
 0 1  2sin
x  0  4 x
lim e  4 x  1
1
xlim
 0 e
x  0
2 : analysis
2 : analysis
f  0  1
f  0  1
So, lim f  x   f  0  .
0 f  x   f  0 .
So, xlim
x 0
Therefore, f is continuous at x  0.
Therefore, f is continuous at x  0.
(b)
2cos x for x  0
x for x  0
f  x   2cos
4x
 2 : f  x 
40 / 60
So,
lim f  xf iscontinuous
f  0 .
Therefore,
at x  0.
x 0
Therefore, f is continuous at x  0.
2011
Free Response
2cos x for x  0
f  x   
4x
for x  0
  4e
2cos x for x  0
(b) f  x   
4x
x  0of
2cos x 43e for allfor
values
1
  of
ln

4e  4 xx  33 when
2cos
for allxvalues
4
1
4x
 4e
 3 when x   ln
4
Therefore, f  x   3 for x 
 2 : f  x 
3: 
 1 : value of x
 2 : f  x 
3: 
 1 : value of x
(b)
x  0.
3
x4  0.0.
3
 0.
41
3
 ln
.
4
4
1
3
Therefore, f  x   3 for x   ln
.
4
4


(c)
(c)
1
0


1
 1 f  x  dx   1 f  x  dx   0 f  x  dx
1
0
1
0
1 4x
dx x  dxf x  dx
 1 f  x  dx   11f1x 2sin
 0 e dx
0
x 0
1  4 x  x 1

 2sin
x  x  dx0 e e  4 xdx
  x112cos

 x 1  4
 x  0
x 1
x 0
1 1 4e  4 x1
 x  2cos x 


 3  2cos  1x
 1  e4   x  0
4
4
1
1
  3  2cos  1    e  4 
4
4
1 1
Average value   f  x  dx
2 1
1 1
Average value  13  cos
f  x  dx 1  4
28 1  1  8 e
0
1




 1 : 0 1  2sin x  dx and
 1

4: 
0
 21 :: antiderivatives

 11  2sin x  dx and
 1 : answer


4: 
2 : antiderivatives

 1 : answer
1 4x
 0 e dx
1 4x
 0 e dx
41 / 60
2011 Free Response
AB6: Analysis of a Piecewise Defined Function
Exam
AB
Mean
3.03
St Dev
2.32
% 9s
0.5
Students did reasonably well, given a piecewise defined function.
Part (a): most students earned one of the two points.
Part (b): many students differentiated the expressions but did not
express f 0 (x ) as a piecewise defined function. Few students earned
the third point.
Part (c): many students earned the first point and one of the two
antiderivative points.
42 / 60
2011 Free Response
AB6: Analysis of a Piecewise Defined Function
Commons errors:
Part (a): attempt to show continuity by evaluating the function at
each piece (1/2). Need to use limits to demonstrate continuity.
Part (b): Failure to express f 0 (x ) as a piecewise defined function.
Attempt to solve −2 cos x = −3 and presentation of an answer.
f (1) − f (−1)
Part (c): average rate of change of f :
2
rather than the average value of f .
Few students earned the fourth point.
43 / 60
2011 Free Response
AB6: Analysis of a Piecewise Defined Function
To help students improve performance:
Correctly use the definition of continuity.
lim f (x ) = f (a)
x →a
Piecewise defined functions.
Computation of derivatives and antiderivatives analytically.
44 / 60
2011 Free Response
AP Calculus BC–1
Version 1.0
2011
BC1: Parametric Particle Motion
At time t, a particle moving in the xy-plane is at position  x t  , y  t   , where x t  and y  t  are not explicitly
dy
dx
 4t  1 and
 sin t 2 . At time t  0, x 0   0 and y  0    4.
dt
dt
 
given. For t  0,
(a) Find the speed of the particle at time t  3, and find the acceleration vector of the particle at time t  3.
(b) Find the slope of the line tangent to the path of the particle at time t  3.
(c) Find the position of the particle at time t  3.
(d) Find the total distance traveled by the particle over the time interval 0  t  3.
(a) Speed   x 3 2   y 3 2  13.006 or 13.007
Acceleration  x 3 , y 3
 4, 5.466 or 4, 5.467
(b) Slope 
y 3
 0.031 or 0.032
2:

1 : speed
1 : acceleration
1 : answer
45 / 60
or 4,to5.467
(b) Find the slopeof 4,
the5.466
line tangent
the path of the particle at time t  3.

2 the particle
2 at time t  3.
(c)
of
(a) Find
Speedthe position
1 : speed
yx3 3    y 3   13.006 or 13.007
2


(b)
Slope
or
0.032
0.031
1 :: answer3.
(d) Find the total
x 3distance traveled by the particle over the time interval 0 1t : acceleration
Acceleration  x 3 , y 3
 4, 5.466 or 4, 5.467
(a) Speed  yx3 3 2   y 3 2  13.006 or 13.007
1 : speed
2
 0.031 or 0.032
(b) Slope 
1 :: answer
1 : acceleration
x 3
Acceleration  x 3 , y 3
3 4, 5.466 or 4, 5.467
dx
(c) x 3  0  
dt  21
 2 : x-coordinate
0 dt

1 : integral
y 3
 0.031 or 0.032
(b) Slope 
1 : answer

x 3 3 dy
1 : answer


y  3   4 
dt  3.226
4: 
30 dt
y -coordinate
2
:

dx
(c) x 3  0  
dt  21
 2 : 1x-coordinate
: integral
0 dt
At time t y 33, the particle is at position  21, 3.226  .

1 :: answer
integral
1

3 0.031 or 0.032
(b) Slope 
1 : answer
x 3 dy
1 : answer


y  3   4 
dt  3.226
4: 
0 dt
 2 : y -coordinate
3
dx
: integral
(c) x 3  0  
dt  21
 2 : 1x-coordinate
dt particle is at position  21, 3.226  .
At time t  
3,0 the

1 : answer

integral

3
1 : answer

dy

y  3   4  3
dt  3.226
4: 
12 :: integral
dy 2
30 dt
dx 2
y -coordinate
(d) Distance  

dt  21.091
dx
2 :  2 : x-coordinate
(c) x 3  0  
21
dt

dt
dt
 1 : answer
1 : integral
00 dt


At time t  3, the particle is at position  21, 3.226  .
1 :: answer
integral
1

3
1 : answer

dy

y  3   4  3
dt  3.226
4: 
12 :: integral
dy 2
 dt
dx 2
y -coordinate
(d) Distance   0

dt  21.091
2: 
dt
dt
 1 : answer
1 : integral
0
At time t  3, the particle is at position  21, 3.226  .

1 : answer
2011 Free Response

   

   

46 / 60
2011 Free Response
BC1: Parametric Particle Motion
Exam
BC
Mean
5.24
St Dev
2.64
% 9s
13.9
Student performance was very good.
Most students earned all points in parts (a) and (b).
Part (c): many students earned the two points for the x -coordinate,
but did not earn points for the y -coordinate.
Part (d): if correct setup, then correct answer.
47 / 60
2011 Free Response
BC1: Parametric Particle Motion
Commons errors:
y 0 (3)
Part (a): Speed = 0
x (3)
Part (b): speed as the slope of the tangent line.
Part (c): many students tried to find an atiderivative for sin t 2 rather
than use a calculator to evaluate the definite integral.
Failure to use y (0) = −4.
48 / 60
2011 Free Response
BC1: Parametric Particle Motion
To help students improve performance:
Practice with particle motion, including computations of speed,
velocity, and distance traveled.
The Fundamental Theorem of Calculus written as:
Z b
f (b) = f (a) +
f 0 (t) dt
a
Show intermediate work.
Careful attention to writing complex expressions (including
parentheses).
49 / 60
2011 Free Response
AP Calculus BC–3
Version 1.0
2011
BC3: Perimeter, Volume
Let f  x   e2 x . Let R be the region in the first quadrant bounded by the
graph of f, the coordinate axes, and the vertical line x  k , where
k  0. The region R is shown in the figure above.
(a) Write, but do not evaluate, an expression involving an integral that
gives the perimeter of R in terms of k.
(b) The region R is rotated about the x-axis to form a solid. Find the
volume, V, of the solid in terms of k.
(c) The volume V, found in part (b), changes as k changes. If
determine
(a)
dk 1
 ,
dt
3
dV
1
when k  .
dt
2
f  x   2e 2 x
Perimeter  1  k  e
2k

k
0

1  2e

2x 2
dx
 1 : f  x 

3 :  1 : integral
 1 : answer
50 / 60
dk 1
k
2
(c) The
volume 1V,found
Perimeter
k  e 2ink part
  (b),
1 changes
2e 2 x asdxk changes. If dt  3 ,
0
dV
1
when k  .
determine
dt
2


2011 Free Response
 0 e 
(b) Volume  2 x
(a) f  x   2e
k
2x 2
Perimeter  1  k  e
(b) Volume  
 0 e 
k
dx   
2k
2x 2

k
0
dx   
k 4x
e
0

dx 
1  2e
k 4x
e
0

4

2x 2
dx 
e
4x
xk
x 0


4
e
4k
e
4k


4
dx

4
e
4x
dV
dk
  e 4k
dt
dt
1 k dV2 x 2 2 1 k 4 x
e   . e dx   e4 x
k

When
e  dx
(b) Volume  2, dt
3 0
4
0
xk
x 0


4


4
(c)
 
(c)
dV
dk
  e 4k
dt
dt
1 dV
1
  e2  .
When k  ,
2 dt
3
xk
x 0


4
e
4k


4

3 :  1 : integral
 1 : answer
 1 : integrand
 1 : limits
4: 
f  x 
 1 : antiderivative

3 :  1 : integral
answer
 1 : answer
 1 : integrand
 1 : limits
4: 
 1 : antiderivative
 1 : answer

1 : applies chain rule
2 :  1 : integrand
1 : answer
 1 : limits
4: 
 1 : antiderivative
 1 : answer
2:

1 : applies chain rule
1 : answer
© 2011 The College Board. All rights reserved.
Visit the College Board on the Web: www.collegeboard.org.
51 / 60
2011 Free Response
BC3: Perimeter, Volume
Exam
AB Part
BC Part
Total
Mean
3.84
1.67
5.51
St Dev
1.91
1.27
2.74
% 9s
19.5
Students performed well, mode score 9.
Part (a): almost half of all students earned all three points.
Part (b): most students earned the first two points. The
antiderivative and answer were more difficult.
Part (c): errors in the computation of the derivative.
Could use integral setup or answer in part (b).
52 / 60
2011 Free Response
BC3: Perimeter, Volume
Commons errors:
Part (a): incorrect integral for arc length, no integral for arc length,
or leaving out other boundaries for R.
Part (b): integral expression, but no evaluation.
Incorrect antiderivative.
Part (c): incorrect application of the chain rule.
Derivative of e 4x .
53 / 60
2011 Free Response
BC3: Perimeter, Volume
To help students improve performance:
Differentiation and integration of functions of the form e ax .
Intermediate steps and communication of the final answer.
Inclusion of the terminating differential (dx ).
Separates the integral from the remaining expression.
Indicates the variable of integration.
54 / 60
2011 Free Response
AP Calculus BC–6
Version 1.0
2011
BC6: Taylor Series
 
Let f  x   sin x 2  cos x. The graph of y  f  5  x  is
shown above.
(a) Write the first four nonzero terms of the Taylor series for
sin x about x  0, and write the first four nonzero terms
 
of the Taylor series for sin x 2 about x  0.
(b) Write the first four nonzero terms of the Taylor series for
cos x about x  0. Use this series and the series for
 
sin x 2 , found in part (a), to write the first four nonzero
terms of the Taylor series for f about x  0.
(c) Find the value of f  6   0  .
(d) Let P4  x  be the fourth-degree Taylor polynomial for f about x  0. Using information from the graph of
1
1
1
y  f  5  x  shown above, show that P4
 f

.
4
4
3000
 
(a) sin x  x 
x3 x5 x 7



3! 5! 7!
 1 : series for sin x
3: 
2
55 / 60
 
 
5! above,
7! show that P
3: 
y  f  x3!
 f

.
 shown
4
4
4
3000
sinthe
x 2graph of
10
(d) Let P24  x  be
the 6fourth-degree
from
 2 : series for
x14 Taylor polynomial for f about x  0. Using information
2x3 x x5 x x 7
sin
x

x





1 : series for sin x
(a) sin x  x5
3! 5!  

1
1
1

7!
5! above,
7! show that P4
y  f  x3!
 f

.
 shown
3: 
2
4
4
3000
6
10
 2 : series for sin x
x14
2
2x3 x x5 x x 7

1
:
series
for
sin
x
(a) sin
sin xx  xx  3!  5!  


3! 5! 7! 7!
3: 
2
6
10
14
 2 : series for sin x
23 x x54 x x 76
x
2
2x
x
x   x    
sin
x

x

1
:
series
for
sin
x
(a)
sin
x

x


(b) cos x  1  3! 3!5! 5!7!  
 1 : series for cos x
2! 4! 6! 7!
3: 
f  x x 2
64
10
14
 2 : series for sin
2
6
x
x
x
2
x 6 x  
sin
 x 2xx 2  xx 4  121

f  xxx 
 11 
 2 3!4! 5! 6! 
(b) cos
1
:
series
for
cos
x
7!

2! 4! 6!
3: 
2
:
series
for
f
x



22
44
6
x
x
121
x6 x
f  xx  1 
 2  4!  6!  
(b) cos
1 : series for cos x

2!
6!
3: 
 2 : series for f  x 
6 x6
x 22 x 44 121
 6
x
f
0
xx  1 



6 
cos



(b)
 1 : series for cos x
is the
of x in the Taylor series for f about x  0.
(c)
13 :: answer
2 coefficent
4! 6!6!
2!

6!
 2 : series for f  x 
2
4
6
x
x
121x
Therefore,
f  6x 0 1  f (6) 0   121. 6 
2 coefficent
4!
6!
is the
of x in the Taylor series for f about x  0.
(c)
1 : answer
6!
 
2011 Free Response
 
 
 
 
 
 
 
(c)
(d)
(c)
(d)
(d)
(d)
f (6)  0   121.
Therefore,
f  6  0 
is the coefficent of x 6 in the Taylor series for f about x  0.
1 : answer
6!
(5)
 5
1 : form of the error bound
The
graph off (6)
y 0 f 121.
x  indicates that max f ( x)  40.
Therefore,
f  6  0 
 
1
2
0  x series
for f about x  0.
is the coefficent of x 6 in the Taylor
1 :: answer
4
1 : analysis
6!
Therefore, (6)
(5)
 5
1 : form of the error bound
The
graph
of
y

f
x
indicates
that
max
f
(
x
)

4
0.


Therefore, f  0   121.(5)
2:
0  x 1
max f  x 
4
1 : analysis
5
1

x

0
1
1
1
40
1
1
Therefore,
 fof y  f  5 4x  indicates
 that  max f 5(5)( x)  4
P4 graph
.
1 : form of the error bound
The
4
4
5!(5)
4
3072 0. 3000

120
4
1
2:
0  x
max f  x 
4
1 : analysis
5
1
0  x
1
1
1
40
1
1
Therefore,
 fof y  f  5 4x  indicates
 that  max f 5(5)( x)  4
P4 graph
1 : form of the error bound
The
0. 3000 .
4
4
5!(5)
4
307
2
x  1 4 Board. All rights reserved. 2 :
max f  x  © 2011 The0 120
College
4
1 : analysis
5
0  x 1
Visit
the
College
Board
on
the
Web:
www.collegeboard.org.
1
1
1
40
1
1
4
Therefore,
 f





P4
.
56 / 60
 
 
 







5
2011 Free Response
BC6: Taylor Series
Exam
BC
Mean
3.97
St Dev
2.26
% 9s
1.2
Better on this series question than in previous years. Mode score 5.
Part (a): most students earned both points.
Part (b): most students earned the first point, and at most 1 of the
next 2.
Part (c): most did not earn the point.
Part (d): 0 or 2. Understood the error bound, or not.
57 / 60
2011 Free Response
BC6: Taylor Series
Commons errors:
Series for sin x and cos x .
Series manipulation: substituting x 2 for x .
Part (b): combining of like terms in order to present the first four
nonzero terms of the Taylor series for f .
Part (c): use of the coefficient on x 6 to determine f (6) (0).
Part (d): many students did not know the Lagrange
error
bound.
(5)
Difficulty in finding a correct value for max f (x )
0≤x ≤1/4
58 / 60
2011 Free Response
BC6: Taylor Series
To help students improve performance:
Know the power series expansions for common functions, including
sin x and cos x .
Creating new series from old series using substitution and algebraic
manipulation.
Practice with error bounds.
59 / 60
AP Calculus Reading
Other Information
AP Central:
Course Description and Teacher’s Guide
Newsletter, Course Perspective, and Audit information
Released Exams, Free Response Questions, Special Focus Material,
Curriculum Modules, Lesson Plans, Teaching Strategies, and Course
Content Related Articles
2008 released exam, for sale
Free response questions: 2 calculator, 4 non-calculator
Rights-only scoring on the multiple choice quesitons
2012 Reading: June 10-16, Kansas City, Session II
60 / 60