Section II: Free Response ... The Reading. Number of AP Calculus Exams. 1955.
1960. 1970. 1980. 1990. 2000 ... Don't give up on a free response problem.
Results from the 2011 AP Calculus AB and BC Exams Stephen Kokoska Bloomsburg University Chief Reader, AP Calculus
[email protected]
July, 2011
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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
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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.
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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
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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
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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
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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
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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
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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
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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.
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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?
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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.
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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.
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(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 atraveled vt the 1.949 velocity dtby (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 vvtt dtdt14.134
(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 vvtt dtdt14.134
(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
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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.
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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
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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 100C 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?
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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 5560 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 B10 8.817 B10t 100minutes. 1 : integrand time 010 : integrand B10 100 B t dt 34.18275; H 10 B10 8.817 3 : 1 uses B 0 100 010 The biscuits are 8.817 degrees Celsius cooler than the tea. 1 : integrand B10 100 B t dt 34.18275; H 10 B10 8.817 uses B 0 100 3 : 1 : answer 010 The biscuits are 8.817 degrees Celsius cooler than the tea. B10 100 B t dt 34.18275; H 10 B10 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.
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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.
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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).
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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.
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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
.
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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 x1 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 x1 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
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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.
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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).
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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.
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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
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4 x 3. There is3no point c, 4 c 3, for which f c is equal to that average rate of change. (d) Find average off change on9the 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 xstatement does not contradict the Mean Value Theorem. 1 : g 3 0 4 3 3 x 3 : 11 :: gg 9 (a) g x33222 f3f x 3 f2 t dt 6 1 : g 3 4 0 3 g 1 : 3 3 : 1 : g x 9 (a) gg x33222 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 f3x 2 52. 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 52.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 rateof 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 atx7 . 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.
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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.
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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.
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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 tothe (a) Find the particular 300differential W 0solution 300 W 1W1400 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 onW theinterval 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 theinterval 1400 0 Wt . (b) Therefore, 300 and 0 onW 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 300WdW 300 25 dt 25 1 1 1 ln t C dt dWW 300 1 dW (c) W 300W 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 1e 0 C ln 1100 C W 300300 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.
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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
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So, lim f xf iscontinuous 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 11f1x 2sin 0 e dx 0 x 0 1 4 x x 1 2sin x x dx0 e e 4 xdx x112cos 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 11 2sin x dx and 1 : answer 4: 2 : antiderivatives 1 : answer
1 4x
0 e dx 1 4x 0 e dx
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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.
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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.
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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.
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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
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or 4,to5.467 (b) Find the slopeof 4, the5.466 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 yx3 3 y 3 13.006 or 13.007 2 (b) Slope or 0.032 0.031 1 :: answer3. (d) Find the total x 3distance 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 yx3 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
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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.
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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.
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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).
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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
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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
xk 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
xk x 0
4
4
(c)
(c)
dV dk e 4k dt dt 1 dV 1 e2 . When k , 2 dt 3
xk 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).
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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 .
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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.
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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
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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 xx 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 xxx 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.
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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
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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.
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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
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