Results from the 2012 AP Calculus AB and BC Exams

Transcription

1 Results from 2012 AP Calculus AB and BC Exams Stephen Kokoska Bloomsburg University Chief Reader, AP Calculus March, / 31

2 AP Calculus Outline 1 Exams: background informion 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) Stistics (c) The good, bad, and some suggestions 2 / 31

3 Exams AP Calculus Exams US Main: United Stes, Canada, Puerto Rico, US Virgin Islands Form A: US Alterne Exam: le test Form I: Internional Main Exam Form J: Internional Alterne Exam Parts Section I: Multiple Choice. Section II: Free Response. Calculor and Non-Calculor Sections AB and BC Exams. 3 / 31

4 The Reading Leadership Structure Chief Reader (CR) Chief Reader Associe (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

5 The Reading: 2012 Grading Flow AB BC BC AB AB-3/BC-3 Tara Smith Dale Hhaway Carolyn Williamson AB-1/BC-1 Virge Cornelius Jacquelyn Reinschmi Tom Dick AB-2 Vicki Carter Stan Perrine Queen Harris BC-4 Tracy Bibelnieks Guy Mauldin Alex McAllister AB-5/BC-5 Ben Klein Mark Kannowski Pam Hooten AB-6 Frances Rouse Stephen Greenfield Nancy Stephenson BC-2 Rose Gundacker Ruth Dover Roger Nelsen BC-6 John Jensen Dwight Pierre Georgia Weickel AB-4 Pario-Lee Law Stephanie Ogden Mark Kiraly AB-3/BC-3 Tara Smith Dale Hhaway Carolyn Williamson AB-1/BC-1 Virge Cornelius Jacquelyn Reinschmi Tom Dick 5 / 31

6 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 stes, DC, and or countries 6 / 31

7 The Reading Number of (all) AP Calculus Exams Exams Totals AB: 268,396 BC: 94,882 Total: 363, Year 7 / 31

8 The Reading 2012 Scores (US Main Exam) US Main Score AB BC AB subscore % 50.6% 60.3% % 16.2% 16.7% % 16.2% 9.0% % 5.4% 5.8% % 11.8% 8.2% 8 / 31

9 The Reading General Comments We awarded points for good calculus work, if student conveyed an understanding of approprie calculus concept. Students must show ir work (bald answers). Students must communice effectively, explain ir reasoning, and present results in clear, concise, proper mamical notion. Practice in justifying conclusions using calculus arguments. Decimal presention errors, use of intermedie values. 9 / 31

10 2012 Free Response General Informion 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. 10 / 31

11 2012 Free Response Stistics Question Mean St Dev % 9s % 0s AB BC AB AB BC AB AB BC / 31

12 2012 Free Response Stistics Question Mean St Dev % 9s % 0s AB BC BC BC / 31

13 AP Calculus AB-1/BC-1 Version Free Response: AB-1/BC-1 t (minutes) Wt (degrees Fahrenheit) The temperure of wer in a tub time t is modeled by a strictly increasing, twice-differentiable function W, where Wt is measured in degrees Fahrenheit and t is measured in minutes. At time t 0, temperure of wer is 55 F. The wer is heed for 30 minutes, beginning time t 0. Values of Wt selected times t for first 20 minutes are given in table above. (a) Use da in table to estime W 12. Show computions th lead to your answer. Using correct units, interpret meaning of your answer in context of this problem. 20 (b) Use da in table to evalue Wt. Using correct units, interpret meaning of 0 W t 0 in context of this problem (c) For 0 t 20, average temperure of wer in tub is. 20 W t Use a left Riemann sum with four subintervals indiced by da in table to approxime. 20 W t Does this 0 approximion overestime or underestime average temperure of wer over se 20 minutes? Explain your reasoning. (d) For 20 t 25, function W th models wer temperure has first derivive given by Wt 0.4 t cos0.06 t. Based on model, wh is temperure of wer time t 25? / 31

14 (d) Explain For 20 your t reasoning. 25, function W th models wer temperure has first derivive given by (d) For 20 25, function th models wer temperure has first derivive given by (d) WFor 20 t 25, function W th models wer has first derivive given by t 0.4 t cos 0.06 t. Based on model, wh is temperure of wer time t 25? Based on model, wh is temperure of wer time 25 Based on model, wh is temperure of wer time t 25? W t 2012 Free t cos cos t Response:. AB-1/BC-1 W W (a) W (a) W15 15 W (a) W (or (or ) 1.016) 6 The wer temperure (or 1.016) is increasing a re of approximely The wer temperure is increasing re of approximely The wer F per per temperure minute minute is time time increasing t 12 minutes. 12 minutes. a re of approximely F per minute time t 12 minutes. 1 : est ima te 2 : estime estima 2 : 1 : interpre tion w ith units interpretion ith units 1 : interpretion with units W t W 20 W W t W 20 W : value value 0 2 : 1 : value 1 : interpretion with units The wer has warmed by 16 F over interval from t 0 to interpretion with units The wer has warmed by 16 over interval from to 1 : interpretion with units The t 20 wer minutes. has warmed by 16 F over interval from t 0 to 20 minutes. t 20 minutes W t W W W W : left Riemann sum left Riemann sum 20 left Riemann sum : 1 : approximion approximion : 20 1 : approximion underestime with reason underestime with reason 20 1 : underestime with reason This approximion This approximion 20 is an underestime since a left Riemann sum is an underestime since left Riemann sum This used used approximion and function and function is an W underestime is strictly increasing. is strictly increasing. since a left Riemann sum is used and function W is strictly increasing. 25 W W W t W t 20 2 : integral integral 20 2 : 1 : integral 1 : answer answer : answer The College Board. Visit College 2012 The College Board. Visit College Board Board 2012 on on The College Web: Web: Board. 14 / 31 (b) (b) (b) (c) (c) (c) (d) (d) (d)

15 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. Interpretion tougher. Needed correct units. Average re of change (your answer). Part (b): Good job recognizing FTC. Interpretion: change, units, and interval. 15 / 31

16 2012 Free Response: AB-1/BC-1 Results Part (c): Left Riemann sum and compution good. Explanion: inadeque or incorrect reasons. Part (d): Students did fairly well. 16 / 31

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

18 2012 Free Response: AB-1/BC-1 To Help Students Improve Performance In general, most students were able to apply approprie concepts and compute correct numerical answers. Interpretion and communicion of results. Clearly indice mamical steps to a final solution. 18 / 31

19 2012 AP Calculus Free AB-2 Response: Version AB Let R be region in first quadrant bounded by x-axis and graphs of y ln x and y 5 x, as shown in figure above. (a) Find area of R. (b) Region R is base of a solid. For solid, each cross section perpendicular to x-axis is a square. Write, but do not evalue, an expression involving one or more integrals th gives volume of solid. (c) The horizontal line y k divides R into two regions of equal area. Write, but do not solve, an equion involving one or more integrals whose solution gives value of k. ln x 5 x x Therefore, graphs of y ln x and y 5 x intersect in first quadrant point AB, , B y (a) Area 5 y e dy or : integrand 3 : 1 : limits 19 / 31

20 gives involving Therefore, ln x 5 volume one x or graphs more of x of integrals solid. y lnwhose x and solution y 5 gives x intersect value in of k. (c) The first horizontal quadrant line y point k divides AB, R into , two regions of equal. area. Write, but do not solve, an equion involving one or more integrals whose solution gives value of k. Therefore, graphs of y ln x and y 5 x intersect in ln x 5 x x first Free quadrant Response: point AB, , AB-2. Therefore, graphs of y ln x and y 5 x intersect in ln x 5 x x first quadrant B Area y point AB, , y e dy 0 Therefore, graphs of y ln x and y 5 x intersect in B Area first y quadrant 5 or y e point dy AB, , (a) (a) OR or y e dy B (a) Area y 5 0 OR A 5 Area B or y (a) Area x 5 A 0A 5 OR Area x or or ln y e dy 5 x 1 ln 5 x 1 A A or OR Area ln x 5 1 x A A A ln or (b) Volume Area x ln x 5 x 5 x 1 1 A A A 5 (b) Volum e2.986 ln or x x A 1 : integrand 3 : 1 : limits integrand 1 : answer 3 : 1 : limits 1 : integrand answer 3 : 1 : limits 1 : integrand answer 3 : 1 : limits 1 : answer : 2 : integrands 3 : 2 : integrands A : 2 : integrands (b) Volume ln x 5 x 1 A expression for total volume (c) 5 A (b) Volume or 1 1 : integrand k y e y dy : ln x 5 x 1 3 : A 2 1 : integrands limits k y (c) expression for total volume or 1 integrand y e dy 1 : equion 3 : 1 : limits k y 5 y e dy or 1 1 : integrand equion : 1 : limits (c) 1 : expression for total volume 1 : expression for total volume 20 / 31

21 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 = e y. 21 / 31

22 2012 Free Response: AB-2 Results Part (b): Students did well. Two distinct, separe integrals to find total volume. Part (c): Those working in terms of y more successful. Some compliced yet correct solutions in terms of x. 22 / 31

23 2012 Free Response: AB-2 Common Errors No calculor use to find (A, B). Point of intersection reported as (4, 1). Solving for x in terms of y (inverse functions). Part (a): 5 1 (5 x ln x) Incorrect limits: 0 as a lower bound, 4 as an upper bound. Part (b): 5 1 (5 x ln x) 2 (constant π) Part(c): Equion in terms of x. 23 / 31

24 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 re are distinct right and left boundaries. Using best, most efficient method for finding area of a region. Computing volume of a solid with known cross sectional area. Computing area of certain known geometric regions an arbitrary point x. 24 / 31

25 2012 Free Response: BC-2 AP Calculus BC 2 Version For For t 0, 0, a is is moving moving along along a curve curve so so th th its its position position time time t is is x t, y t. At At time time t 2, 2, t 2 dy dy 2 is is position position 1, 5. It It is is known known th th and and sin sin t. t e (a) (a) Is Is horizontal horizontal movement movement of of to to left left or or to to right right time time t 2? 2? Explain Explain your your answer. answer. Find Find slope slope of of ph ph of of time time t (b) (b) Find Find x-coordine x-coordine of of s s position position time time t (c) (c) Find Find speed speed of of time time t Find Find accelerion accelerion vector vector of of time time t (d) (d) Find Find distance distance traveled traveled by by from from time time t 2 to to t (a) (a) 2 2 t2 e Since Since 0, 0, is is moving moving to to right right t2 time time t dy dy dy dy t or t t2 2 t 1 : moving moving to to right right with with reason reason dy dy 3 : 1 : considers considers 1 : slope slope t 2 25 / 31

26 dy t2 t2 dy t or t2 2 (b) t 2 x or t 2 e 2012 Free 4 Response: BC-2 (b) t 2 x or t 2 e 4 (b) t 2 x or t 2 (c) Speed x2 4e y or (c) Spe Accelerion ed x 4x 4, yy or , (c) Spe Accelerion ed x 4x 4, yy or , Accelerion x 4, y , (d) Distance xt yt or (d) Distance xt yt or (d) Distance xt yt or : 1 : integral 1 : answer 2 : 1 : integral 1 : answer 2 : 1 : integral answer 2 : 1 : speed 1 : accelerion 2 : 1 : speed 1 : accelerion 2 : 1 : speed 1 : accelerion 2 : 1 : integral 1 : answer 2 : 1 : integral 1 : answer 2 : 1 : integral 1 : answer 2012 The College Board. Visit College Board on Web: The College Board. Visit College Board on Web: The College Board. 26 / 31

27 2012 Free Response: BC-2 Results Student performance generally very good. Understanding of parametrically defined curve. Arithmetic, algebra, decimal presention errors. Correct presentions followed by incorrect answers (from calculor). Inapproprie use of initial condition. 27 / 31

28 2012 Free Response: BC-2 Common Errors Part (a): Horizontal movement using dy or x(2). Arithmetic, algebra errors in computing slope. Part (b): Initial condition not used. Symbolic antiderivive of (abandoned, restart with calculor) 28 / 31

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

30 2012 Free Response: BC-2 To Help Students Improve Performance Decimal presention instructions and intermedie values. More practice with use of Fundamental Theorem of Calculus. x(b) = x(a) + b a x (t) Concepts of speed and total distance traveled. 30 / 31

31 AP Calculus Reading Or Informion AP Teacher Community: A new online collaborion space and professional learning network for AP Educors. Discussion boards, resource library, member directory, digests, notificions, etc. Each community is modered Reading: June 11-17, Kansas City, Session II 31 / 31