Results from the 2012 AP Calculus AB and BC Exams

Results from the 2012 AP Calculus AB and BC Exams Stephen Kokoska Bloomsburg University Chief Reader, AP Calculus [email protected]

March, 2013

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AP Calculus Outline 1 Exams: background information 2 The Reading (a) Leadership Structure (b) The Flow and Question Teams (c) Logistics and Numbers 3

Three Free Response Questions (a) The Standard (b) Statistics (c) The good, bad, and some suggestions

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Exams AP Calculus Exams US Main: United States, Canada, Puerto Rico, US Virgin Islands Form A: US Alternate Exam: late test Form I: International Main Exam Form J: International Alternate Exam Parts Section I: Multiple Choice. Section II: Free Response. Calculator and Non-Calculator Sections AB and BC Exams. 3 / 31

The Reading Leadership Structure Chief Reader (CR) Chief Reader Associate (CRA) Assistant Chief Reader (ACR) Chief Aides (CA) Exam Leaders (EL) (2 → 5) Question Leaders (QL) (9 → 20) Question Team Members (QTM) Table Leaders (TL) Readers 4 / 31

The Reading: 2012 Grading Flow AB 11-28

BC 61-66

AB-3/BC-3 Tara Smith Dale Hathaway Carolyn Williamson

AB-2 Vicki Carter Stan Perrine Queen Harris

BC-4 Tracy Bibelnieks Guy Mauldin Alex McAllister

AB-6 Frances Rouse Stephen Greenfield Nancy Stephenson

BC-2 Rose Gundacker Ruth Dover Roger Roger Nelsen Nelsen

AB-3/BC-3 Tara Smith Dale Hathaway Carolyn Williamson

BC 71-76

AB 31-49

AB-1/BC-1 Virge Cornelius Jacquelyn Reinschmidt Tom Dick

AB-5/BC-5 Ben Klein Mark Kannowski Pam Hooten

BC-6 John Jensen Dwight Pierre Georgia Weickel

AB-4 Pario-Lee Law Stephanie Ogden Mark Kiraly

AB-1/BC-1 Virge Cornelius Jacquelyn Reinschmidt Tom Dick 5 / 31

The Reading Logistics and Participants Kansas City: Convention Center (versus college campus) Westin Hotel (versus college dorms) Total Participants: 853 High School: 55% College: 45% 50 states, DC, and other countries

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The Reading Number of (all) AP Calculus Exams 2012 Totals AB: 268,396 BC: 94,882 Total: 363,278

350000

Exams

300000 250000 200000 150000 100000 50000 0 1955

1960

1970

1980

1990

2000

2012

Year 7 / 31

The Reading 2012 Scores (US Main Exam)

Score 5 4 3 2 1

AB 24.9% 17.0% 17.4% 10.3% 30.5%

US Main BC 50.6% 16.2% 16.2% 5.4% 11.8%

AB subscore 60.3% 16.7% 9.0% 5.8% 8.2%

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The Reading General Comments We awarded points for good calculus work, if the student conveyed an understanding of the appropriate calculus concept. Students must show their work (bald answers). Students must communicate effectively, explain their reasoning, and present results in clear, concise, proper mathematical notation. Practice in justifying conclusions using calculus arguments. Decimal presentation errors, use of intermediate values. 9 / 31

2012 Free Response General Information Six questions on each exam (AB, BC). Three common questions: AB-1/BC-1, AB-3/BC-3, AB-5/BC-5. 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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2012 Free Response Statistics Question Mean St Dev % 9s % 0s AB-1 3.96 2.88 5.6 16.6 BC-1 5.88 2.55 14.7 4.1 AB-2

3.09

3.10

6.9

41.0

AB-3 BC-3

2.67 4.29

2.58 2.61

1.3 3.8

30.9 11.3

AB-4

4.09

2.61

2.8

14.3

AB-5 BC-5

2.87 4.75

2.24 2.55

1.4 7.1

16.3 5.5 11 / 31

2012 Free Response Statistics Question Mean St Dev % 9s % 0s AB-6 3.59 2.82 5.5 19.1 BC-2

5.07

2.66

10.7

6.0

BC-4

5.43

2.84

18.0

7.6

BC-6

4.23

2.70

5.3

11.5

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2012 Free Response: AB-1/BC-1 AP Calculus AB-1/BC-1

Version 1.0

2012

t (minutes)

0

4

9

15

20

W  t  (degrees Fahrenheit)

55.0

57.1

61.8

67.9

71.0

The temperature of water in a tub at time t is modeled by a strictly increasing, twice-differentiable function W, where W  t  is measured in degrees Fahrenheit and t is measured in minutes. At time t  0, the temperature of the water is 55F. The water is heated for 30 minutes, beginning at time t  0. Values of W  t  at selected times t for the first 20 minutes are given in the table above. (a) Use the data in the table to estimate W12 . Show the computations that lead to your answer. Using correct units, interpret the meaning of your answer in the context of this problem. (b) Use the data in the table to evaluate

20

0

W  t  dt. Using correct units, interpret the meaning of

20

0

W  t  dt

in the context of this problem. 1 20 (c) For 0  t  20, the average temperature of the water in the tub is W  t  dt. Use a left Riemann sum 20  0 1 20 with the four subintervals indicated by the data in the table to approximate W  t  dt. Does this 20  0 approximation overestimate or underestimate the average temperature of the water over these 20 minutes? Explain your reasoning.

(d) For 20  t  25, the function W that models the water temperature has first derivative given by W  t   0.4 t cos  0.06t . Based on the model, what is the temperature of the water at time t  25 ? 13 / 31

t  reasoning. 25, the function W that models the water temperature has first derivative given by (d) Explain For 20 your (d) For 20  t  25, the function W that models the water temperature has first derivative given by 0.4 t  t25, (d) W For t20 function W on thatthemodels water temperature hasoffirst given model,the what is the temperature thederivative water at time  cos the 0.06 t . Based t by25 ?  W  t   0.4 t cos  0.06t . Based on the model, what is the temperature of the water at time t  25 ? W  t   0.4 t cos  0.06t . Based on the model, what is the temperature of the water at time t  25 ?

2012 Free Response: AB-1/BC-1     W 15  W 9 67.9  61.8 (a) W 12   W 15  W  9   67.9  61.8 (a) W 12   W 15 15  W 9  9   67.9  6 61.8 15  9 6 (a) W 12    1.01715(or 1.016)  9 6  1.017 (or 1.016)  1.017 (or 1.016) The water temperature is increasing at a rate of approximately The water temperature is increasing at a rate of approximately t  12 minutes. minute at is time 1.017water F per The temperature increasing at a rate of approximately t  12 minutes. minute at time 1.017 F per 1.017 F per minute at time t  12 minutes.

(b) (b) (b)

(c) (c) (c)

(d) (d) (d)

2: 2: 2:

 

1 : estimate 1 : estimate estimatetation with units 1 : interpre 1 : interpretation with units 1 : interpretation with units

2: 2: 2:

 

1 : value 1 : value value 1 : interpretation with units 1 : interpretation with units 1 : interpretation with units

3: 3: 3:

: left Riemann sum 1  1 : left Riemann sum  left Riemann sum  : approximation 1  1 : approximation    approximationwith reason : underestimate 1   1 : underestimate with reason    1 : underestimate with reason

2: 2: 2:

 

20

 t  dt  W  20   W  0   71.0  55.0  16  002020W W  t  dt  W  20   W  0   71.0  55.0  16 W  t  dt  W  20   W  0   71.0  55.0  16 The The water has warmed by 16 F over the interval from t  0 to 0 water has warmed by 16 F over the interval from t  0 to

The water has warmed by 16 F over the interval from t  0 to minutes. t  20 t  20 minutes. t  20 minutes. 1 20 1 1  20W  t  dt  1  4  W  0   5  W  4   6  W  9   5  W 15   20 1  0020W  t  dt  20 1  4  W  0   5  W  4   6  W  9   5  W 15   20 20 W  t  dt   4  W  0   5  W  4   6  W  9   5  W 15   1 20  0 20  1  4  55.0  5  57.1  6  61.8  5  67.9   20  4  55.0  5  57.1  6  61.8  5  67.9  1  20 1  4  55.0  5  57.1  6  61.8  5  67.9   20 1  1215.8  60.79  20 1  1215.8  60.79  20is an 1215.8  60.79 since a left Riemann sum This approximation underestimate This approximation20is an underestimate since a left Riemann sum is used and the function is strictly increasing. This approximation is anW underestimate since a left Riemann sum is used and the function W is strictly increasing. is used and the function W is strictly increasing. 25 W  25  71.0   25W  t  dt W  25  71.0   20 25W  t  dt W  25  71.0   20 W  t  dt  71.0  2.043155  73.043 20  71.0  2.043155  73.043  71.0  2.043155  73.043 © 2012 The College Board.

1 : integral 1 : integral integral 1 : answer 1 : answer 1 : answer

2012on The College Board. Visit the College© Board the Web: www.collegeboard.org. 2012on The College Board. Visit the College©Board the Web: www.collegeboard.org.

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2012 Free Response: AB-1/BC-1 Results In general, students performed well. Multiple entry points. No surprises. Part (a): Most students set up a difference quotient. Interpretation tougher. Needed correct units. Average rate of change (your answer). Part (b): Good job recognizing the FTC. Interpretation: change, units, and interval.

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2012 Free Response: AB-1/BC-1 Results Part (c): Left Riemann sum and computation good. Explanation: inadequate or incorrect reasons. Part (d): Students did fairly well.

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2012 Free Response: AB-1/BC-1 Common Errors Interpretation of the answer in the context of the problem. Correct units. Part (b): The meaning of the definite integral. Part (c): Incorrectly assumed the width of each subinterval was the same. Explanation associated with an underestimate. Part (d): Use of 0 as a lower bound on the definite integral. 17 / 31

2012 Free Response: AB-1/BC-1 To Help Students Improve Performance In general, most students were able to apply appropriate concepts and compute correct numerical answers. Interpretation and communication of results. Clearly indicate the mathematical steps to a final solution.

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2012 Free Response: AB-2 AP Calculus AB-2

Version 1.0

2012

Let R be the region in the first quadrant bounded by the x-axis and the graphs of y  ln x and y  5  x, as shown in the figure above. (a) Find the area of R. (b) Region R is the base of a solid. For the solid, each cross section perpendicular to the x-axis is a square. Write, but do not evaluate, an expression involving one or more integrals that gives the volume of the solid. (c) The horizontal line y  k divides R into two regions of equal area. Write, but do not solve, an equation involving one or more integrals whose solution gives the value of k.

ln x  5  x  x  3.69344 Therefore, the graphs of y  ln x and y  5  x intersect in the first quadrant at the point  A, B    3.69344, 1.30656.

(a) Area 

y  0 5  y  e  dy B

 2.986  or 2.985

 1 : integrand  3 :  1 : limits

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gives ofx the solid. involving one or more integrals the value Therefore, the graphs in of k. y  5  gives x intersect y  ln whose x and solution ln x the 5 volume x  of 3.69344 the first quadrant 1.30656  A, BR into  3.69344, . area. Write, but do not solve, an equation (c) The horizontal lineat ythe point two regions of equal k divides Therefore, the graphs of y  ln x and y  5  x intersect in involving or more ln x  5 one x  x  integrals 3.69344 whose solution gives the value of k. the first quadrant at the point  A, B    3.69344, 1.30656.

2012 Free Response: AB-2 Therefore, the graphs of y  ln x and y  5  x intersect in ln x  5 Bx  x  3.69344 first Area  quadrant 5  yatthe e y point dy  A, B    3.69344, 1.30656. (a) the Therefore,0 the graphs of y  ln x and y  5  x intersect in B  2.986 2.985 dy  A, B    3.69344, 1.30656. Area  quadrant 5  oryat e y point (a) the first the

  0 

OR (a) Area OR Area (a) Area Area OR

 

 2.986  or 2.985 B   5  y  e y dy





0 A 5 2.986 dx 2.985  5  B ln x or 1 5  y  e y A dy 0A 2.986 or 2.9855

 x  dx        ln x dx    5  x  dx  2.986  or 2.985 1 A  2.986  or 2.9855  A OR Area   ln x dx    5  x  dx





1

A

A  or 2.985 5 2 5 2 A Volume2.986 (b) Area   ln1 x ln dxx dx 5  Ax5dx x  dx

1

A

A

   ln (b) Volume2.986 2 dx  or x2.985 1

(c) (b) (c) (c)

A

5

2

5

2

 A 5  x 

dx

1  ln x  dx   A 5  x  dx 1 1 5  y  eAy  dy   2.986 or  2.985  0 lum 5  22 2 2 Vo e  ln x dx  5  x     dx 1 y A k 1 1 5  y  e dy   2.986 or    2  2.985 0 2

(b) Volume 

2

k

1 1 y  5  y  e  dy   2.986  or  2.985 k

 1 : integrand  3 :  1 : limits : integrand  1  1 : answer 3 :  1 : limits  1 : answer  1 : integrand  3 :  1 : limits answer  1 : integrand  3 :  1 : limits  1 : answer

3: 3:

   

2 : integrands 1 : expression for total volume 2 : integrands 1 : expression for total volume

2 : integrands 3 :  1 : integrand  1 : expression for total volume 3 :  1 : limits 2 :: integrands integrand 3 :  1  1 : equation for total volume limits 3 :  11 :: expression  1 : equation  1 : integrand  20 / 31 3 :  1 : limits

2012 Free Response: AB-2 Results Area / volume problem with two regions: difficult for students. Working in x: OK in parts (a) and (b). Part (c) was very challenging for students working with respect to x. Part (a): Common solution in terms of x, 2 regions. For those in y: OK if correctly found x = ey .

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2012 Free Response: AB-2 Results Part (b): Students did well. Two distinct, separate integrals to find total volume. Part (c): Those working in terms of y more successful. Some complicated yet correct solutions in terms of x.

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2012 Free Response: AB-2 Common Errors No calculator use to find (A, B). Point of intersection reported as (4, 1). Solving for x in terms of y (inverse functions). Z 5 Part (a): (5 − x − ln x) dx 1

Incorrect limits: 0 as a lower bound, 4 as an upper bound. Z 5 Part (b): (5 − x − ln x)2 dx (constant π) 1

Part(c): Equation in terms of x. 23 / 31

2012 Free Response: AB-2 To Help Students Improve Performance Practice in solving for x in terms of y (inverse functions). Computing areas in which there are distinct right and left boundaries. Using the best, most efficient method for finding the area of a region. Computing the volume of a solid with known cross sectional area. Computing the area of certain known geometric regions at an arbitrary point x. 24 / 31

2012 Free Response: BC-2 AP AP Calculus Calculus BC–2 BC–2

Version Version 1.0 1.0

2012 2012

For  0, 0, aa particle  2, 2, the For tt  particle is is moving moving along along aa curve curve so so that that its its position position at at time time tt is is  xx tt  ,, yy tt  .. At At time time tt  the dy dx t  2 2 dx  t  2 and dy  sin 2 t. particle is at position It is known that 1, 5 .   particle is at position 1, 5. It is known that dt  ett and dt  sin t. dt dt e (a) (a) Is Is the the horizontal horizontal movement movement of of the the particle particle to to the the left left or or to to the the right right at at time time tt  Explain your your answer. answer. 2 2 ?? Explain Find Find the the slope slope of of the the path path of of the the particle particle at at time time tt   2. 2. (b) Find Find the the x-coordinate x-coordinate of of the the particle’s particle’s position position at at time time tt   4. 4. (b) (c) (c) (d) (d)

Find Find the the speed speed of of the the particle particle at at time time tt  Find the the acceleration acceleration vector vector of of the the particle particle at at time time tt   4. 4. Find  4. 4. Find the distance traveled by the particle from time to t  2 t  4. Find the distance traveled by the particle from time t  2 to t  4.

dx (a) (a) dx dt dt

t 2 t 2

2 2   e22 e

dx dx Since Since dt t  2  the particle particle is is moving moving to to the the right right  0, 0, the dt t  2 at time time tt  at  2. 2.

dy dy dx dx

t 2 t 2

dy dy dt dt   dx dt dx dt

t 2 t 2 t 2 t 2

1 : moving to the right with reason   1 : moving to the right with reason  dy dt  33 ::  1 : considers dy dt  1 : considers dx dx dt dt   1 :: slope slope at at tt  2 2 1   

 3.055  or  3.055 or 3.054 3.054 

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t 2

t 2

dy dt t  2 dy   3.055  or 3.054  dx t  2 dx 4dt tt 2 2 x 4   1   dt  1.253  or 1.252   2 et 4 t2 x 4   1   dt  1.253  or 1.252   2 et 4 t2 x 4   1   dt  1.253  or 1.252   Speed   x2 4 e2t   y 4 2  0.575  or 0.574 

2012 Free Response: BC-2  (b) (b) (b) (c)

Acceleration ed   x4 x24 ,yy442  0.575  or 0.574  (c) Spe  0.041, 0.989 Acceleration ed   x4 x24 ,yy442  0.575  or 0.574  (c) Spe  0.041, 0.989 Acceleration  x 4  , y 4  4 0.98  t  2  (d) Distance   x0.041,  y9 t  2 dt

2

2: 2: 2: 2: 2: 2:

2:

 0.651 4  or 0.650  (d) Distance   x t  2   y t  2 dt

2:

4  0.651  or 0.650  (d) Distance   x t  2   y t  2 dt

2:

2 2

 0.651  or 0.650 

       

1 : integral 1 : answer 1 : integral 1 : answer 1 : integral 1 1 :: answer speed 1 : acceleration 1 : speed 1 : acceleration 1 : speed 1 : acceleration 1 : integral 1 : answer 1 : integral 1 : answer 1 : integral 1 : answer

© 2012 The College Board. Visit the College Board on the Web: www.collegeboard.org. © 2012 The College Board. Visit the College Board on the Web: www.collegeboard.org. © 2012 The College Board.

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2012 Free Response: BC-2 Results Student performance generally very good. Understanding of parametrically defined curve. Arithmetic, algebra, decimal presentation errors. Correct presentations followed by incorrect answers (from calculator). Inappropriate use of initial condition.

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2012 Free Response: BC-2 Common Errors dy or x(2). dx Arithmetic, algebra errors in computing the slope.

Part (a): Horizontal movement using

Part (b): Initial condition not used. dx Symbolic antiderivative of dt (abandoned, restart with calculator)

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2012 Free Response: BC-2 Common Errors Part (c): Incorrect formula for speed. dy dx and analytically. Derivatives of dt dt OK, but quotient rule or algebra errors. dy Part (d): Use of as the integrand in computing dx distance. Use of the formula for arc length assuming y = f (x). 29 / 31

2012 Free Response: BC-2 To Help Students Improve Performance Decimal presentation instructions and intermediate values. More practice with the use of the Fundamental Theorem of Calculus. Z b x0 (t) dt x(b) = x(a) + a

Concepts of speed and total distance traveled.

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AP Calculus Reading Other Information AP Teacher Community: A new online collaboration space and professional learning network for AP Educators. Discussion boards, resource library, member directory, email digests, notifications, etc. Each community is moderated. 2013 Reading: June 11-17, Kansas City, Session II

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