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The North Carolina Association Of Advanced Placement* Mathematics Teachers Newsletter * * * * * * * * * * * * * * * *

Volume 15

Board Members Deborah Britt ­ Executive Secretary Jeff Lucia ­ Treasurer/Membership Chair Stephen Davis ­ Web Master Rhea Caldwell ­ Western Region Gloria Dupree ­ President-Elect, Western Region Ray Jernigan ­ President, Eastern Region Emogene Kernodle ­ Central Region Patricia Morris ­ Immediate Past President, Central Region Martha Ray ­ Central Region David Royster ­ College Representative, Western Region Dan Teague ­ Central Region Sue Wall ­ Eastern Region Notes from the President's Desk

The purpose of NCA2PMT is to foster and encourage the association of advanced placement mathematics teachers by keeping them informed about advanced placement programs, teaching strategies, education directives, and other mathematical topics of general interest. I encourage you to be active member and let us hear from you. How can we better support you in the teaching of mathematics? I would also like to take this opportunity to express appreciation to those who have contributed to our newsletter. Your time and efforts in writing articles and supporting teachers is greatly appreciated. As a member of NCA2PMT, please consider sharing an activity in our newsletter. I would like to take this opportunity to welcome Dr. David Royster to the Board of Directors for NCA2PMT. David is the Director of the Center for Mathematics, Science and Technology Education, University of North Carolina at Charlotte. He has been a strong supporter of teachers and education in our state. I would also like to thank Dr. Ben Klein for his service to our board. Ben has worked hard as your college representative and NC was lucky to have him on the Test Development Committee. Dr. Stephen Davis, our webmaster is going to be the new chair of the TDC, so NC is keeping a strong place in AP. Both Stephen and Ben are at Davidson College. Visit our website for more information, and do let us hear from you. I hope you have time to enjoy a few days of relaxation before the final days of summer wind down and you begin gearing up for the fall. Ray Jernigan

Summer 2007

Issue No. 2

Web Address www.ncaapmt.org/calculus

*Advanced Placement is a registered trademark of The College Board, which was not involved in the production of, and does not endorse, this publication. 1

Notes From The Secretary's Desk

Once again, the summer newsletter is jam-packed with important information and discussions of the latest AP exams. It has been a busy spring for me and things are changing. I completed my doctoral coursework and was able to get Chapters 1 and 2 of my dissertation in draft form. Then during the past 2 months I was able to make a nice start on Chapter 3. After completing my written and oral comprehensive exams in June, I feel like there is an end in sight. I am now officially a full-time doctoral student at Ohio University on a nice scholarship. I have been selected to be the editor of the Rural Mathematics Educator (Go to http://www.acclaim-math.org/ and click on Newsletter Archives on the left). The biggest news is that I have resigned my teaching position at East Tennessee State University and plan to work full-time on my dissertation during the upcoming year. I will also be conducting workshops, editing and searching for a job. Please write and let me know your comments about our newsletter. We want to serve you. If you have items of interest, pass them on to me. Deb Britt, Mars Hill, NC, [email protected] Please remember to renew your membership to receive the two newsletters. You can send your $5.00 check, payable to NCA 2 PMT, to Jeff Lucia, 718 Landsdowne Road, Charlotte, NC 28270. His email address is [email protected]

Exciting News ­ Problem solving activity for 2007-08 coming soon to our website

We are working on the details. Beginning the first week of September, we will present a challenge problem or series of problems for teams of students to work on. The problems will focus on problem solving using the content of calculus, so the problems will be somewhat non-standard, but we will attempt to keep them within the abilities of a group of students working together. Student groups will receive partial credit for partial results and we encourage classrooms to participate as a team. The teacher can serve as an advisor, but cannot contribute to the student's solutions. Solutions will be given two weeks after the problems appear along with new problems. Particularly interesting student approaches will also be shared on the site. Teachers should look at the NCAAPMT website around September 4 for information on where to find the site hosting the problems.

Table of Contents Overall 2007 Question Comments (Prior to Grading) 2007 AP Questions ­ Commentary The Alternate Exam Unofficial Means 2007 Reading Events The Water Skier Problem The Big Bang and the Age of the Universe Items of Interest 2007 AB Free Response Frequency Table 2007 BC Free Response Frequency Table Treasurer and Membership Report

Pages 3­4 4 ­ 14 14 ­ 15 16 16 ­ 21 22 ­ 23 24 ­ 26 27 ­ 29 30 ­ 32 33 ­ 36 37

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**** Please visit the AP Central Website for copies of the AP Exam questions. **** The 2007 AP Calculus Examinations ­ Commentary written prior to the grading Lin Mc Mullin - Peekskill, NY, www.LinMcMullin.net

AB 1 / BC 1 This is the standard area-volume problem that has appeared on the exam every year since Newton. The new twist was not printing the graph. Good idea. If you want a graph you use your calculator. I liked that. We expect students to use a graphing calculator as a tool -- not just to draw a graph when a graph is asked for. So here they can use the tool to draw the graph if they want to (or not, it's easy enough to sketch by hand -- even, x-axis asymptote, max at x = 0). AB 2/ BC 2 This is really a combination of two "standard" type questions: (1) the rate in - rate out idea (Amusement Park, Sandy Beach) and (2) From the graph of the derivative, tell me about the function. Part (a) was the usual integrate the rate to get the amount. Part (b) was pretty obvious (a closed interval and a half-open interval or all open may be required, x = 3 should not be in any interval). Part (c) - What clicked for me here was the Amusement Park problem. When I first saw the Calculus in Motion animation of that question, I had one of those Duh moments. Of course you find the critical points by solving E(t) - L(t) = 0 or E(t) = L(t) , but the animation included a graph on the same axes of E and L. The maximum number of people in the park occurs when the rate leaving, L, goes above the rate entering, E. Same thing here, when g(t) jumps above f(t) at t = 3 this must be location of the (local) maximum. I am assuming an endpoint analysis will be required for the justification here. AB 3 This is not a new format for a problem about the various important theorems. See in the multiple choice 1998 AB 26, AB 85 (= BC 85) and 1993 AB 18, BC 79 and BC 83 (both the BC questions are suitable for AB). Also similar is last year's FR AB 6. For variations and practice change to h(x) = f (x) ! g(x) or h(x) = f (x) / g(x) or h(x) = f (2x) ! (g(x) ) etc. Another way to change the problem is to give the

3

information in a graph rather than a table (probably line segments - see D&S Marketing's AB review book page 11 #24, and page 86 #31 for examples). Can part (b) be done by calculating h '(1) = 10 and

h '(3) = !8 and using the IVT? Why or why not?

AB 4 No problem with (a) except you need to know the trig values of special angles. (b) Two things: First, I assume that solving a separable DEq was in the multiple choice, so having no DEq here did not bother me. Second, one of the things readers noticed a few years back is that many students didn't seem to understand that the solution of a DEq was function (not a number) that, when substituted with its derivative(s), satisfied the given DEq. That seemed to be what was being hinted at here. AB 5 / BC 5 All basic calculus topics here: (a) tangent line approximation and its relationship to concavity; (b) related rate problem; (c) Riemann sum and meaning of integral; (d) graphical understanding of how a right Riemann sum is formed. All standard stuff, but it seemed like a lot in one question. AB 6 (a) and (b) were hardly extreme. (c) perhaps a little different. And no, I do not mind a little algebra now and then.

3

BC 3 This seemed a complete coverage of the little BC students are supposed to know about polar curves. The pre-calculus information was given (graph and points of intersection). Find the area, find and interpret dr/dt and dy/dt. Notice that BC students also had to work with the trig functions of the special angles (AB 4 was not a common question), but BC students could use their calculator for them. BC 4 Find first and second derivative and integration by parts. All standard stuff. BC 6 Generate a power series by substitution, followed by using the series in two application situations. I like that -- kids should see some of the ways powers series are used. And no, I do not mind a little algebra now and then for BC also. I think that if any of these nine questions with slightly different format were slipped into, say, last year's test no one would have minded much. Using several in one year made things look different and hence superficially more difficult.

The 2007 AP Calculus Examinations ­ Individual Questions AB/BC 1 "The Mustache Problem" Eugene Cloutier - Southlake Christian Academy, Huntersville, NC

As usual, the first problem on the free response dealt with finding the area and volume of a region. Below is a list of the criteria used in grading this problem. This problem was nicknamed the mustache problem since the various regions considered by the students resembled different types of mustaches. This is not meant to be an exhaustive list, but I hope it helps you to see what we as readers focused on while reading your students' responses. As you can see, how a student set up these problems was more important than finding the answer. Part (a): Rationale ­ The student is able to find the area between two curves. 1st Point: This point was awarded for the student's determination of the correct region. A point was awarded if the student chose the correct limits in either parts a, b, or c. To be considered, the limits were required to be placed in an integral. Due to the symmetric properties of these equations, we also awarded points for considering the area of the region in Quadrant I and multiplying the answer by 2. If the student chose incorrect limits or decided to only consider Quadrant I, the student would lose this point, but we would read with the student for the rest of the points. If the student chose to work with the wrong shaded region, such as an unbounded region, the student was ineligible for the rest of the answer points (because of the improper integral), yet could still earn some integrand points. ! ! 20 " " # 2 % dx 2nd Point: This point was awarded for the correct integrand: $ $ 2 % && 1+ x ' ' We did our best to ignore offensive notations (such as leaving off the dx or incorrect use of grouping symbols). Even a reversal was ignored for this point, but addressed in the next. The students were also allowed to find the area of the region by integrating with respect to y.

(

3rd Point: To be eligible for this point, the student must have previously earned the integrand point. This point was awarded for the correct answer either rounded or truncated correctly to the thousandths place. Many students carelessly lost answer points by rounding to another digit. In the case of a reversal of the integrand, a student would earn this point only if there was clear evidence that the student acknowledged that, although the answer to the integral was negative, the answer to the problem was positive.

4

Part (b): Rationale ­ The student is able to find the volume of the solid generated by rotating a region about the x-axis. Students are expected to use the Washer Method. ! ! 20 "2 " th th 2 4 and 5 Points: These points were awarded for the correct integrand: # # $ % 2 $ dx . To be # & 1+ x2 ' $ & ' eligible, the student's integrand must be in the form of the difference of two squares of the correct functions. The only exception would be a copy error of the original function. Any errors in constants (such as omitting ! ) were taken off the answer (6th) point. A student who chose to use the shell method and did so correctly was eligible for all points.

(

6th Point: This point was awarded for the correct answer either rounded or truncated correctly to the thousandths place. The same rules used for grading the 3rd point apply. Part (c): Rationale ­ The student is able to find the volume of the solid using semi-circular cross sections. 7

th

and 8

th

Points:

2

! 10 " # 1% dx or These points were rewarded only for the correct integrand: $ 2 & 1+ x '

(

2

! 20 " # 2 % dx . To be eligible, the student's integrand must be the square of either the radius or even $ 2 & 1+ x ' diameter of the cross section. The only exception would be a copy error of the original function. Reversals were ignored since the difference in functions is squared. Any errors in constants were taken off the answer (9th) point.

(

9th Point: This point was awarded for the correct answer either rounded or truncated correctly to the thousandths place. Common mistakes students made were choosing to square the diameter instead of the radius or choosing to find the area of the circles instead of the semi-circles.

AB/BC 2 Norman Allen ­ Lejeune High School, Camp Lejeune NC

This is a typical rates problem where there is one increasing rate, one decreasing rate and a graph showing the relationship. This was the second calculator active problem that required students to give the answer to the nearest gallon, but the AP Board did accept the normal answer to the thousands. The points on this problem were distributed as 2 ­ 2 ­ 5. Part (a) (2 Points) The first point was for the integral having limits 0 ­ 7. The second point was for the answer.

"

7

0

f (t)dt ! 8.264 gallons

You could miss the answer, but still get the integral point, or the bald answer would be worth one point. Most students got these two points. Part (b) (2 Points) The amount of water in the tank is decreasing on the intervals 0 t 1.617 and 3 t 5.076 because f(t) < g(t) for 0 t < 1.617 and 3 < t < 5.076. The AP Board accepted open or closed intervals and the reason point could be obtained without the first interval point. Most students got the interval point, but missed the reason point. Part (c) (5 Points) Since f (t) ­ g (t) changes sign from positive to negative only at t = 3, the candidates for the absolute maximum are at t = 0, 3, and 7.

5

t (hours) 0 3 7

gallons of water 5000

5000 +

" (f (t) ! g(t) )dt =5126.591

0

3

5126.591 +

"

7

3

(f (t) ! g(t) )dt =4513.807

The amount of water in the tank is greatest at 3 hours. At that time, the amount of water in the tank, rounded to the nearest gallon, is 5127 gallons. The first point was for recognizing that 3 was a candidate. Many students received this point. The second point, which most students missed, was for the integrand,

5000 +

" (f (t) ! g(t) )dt =5126.591 .

0

3

The third point was for finding the amount of water for t = 3 , which

7

most students missed since they missed the second point. The fourth point was for finding the amount of water at t = 7 , which is 5126.591 +

"

3

(f (t) ! g(t) )dt =4513.807 .

Very few students received this point.

The fifth point was the conclusion point. The conclusion was that the greatest amount of water was 5127 gallons at t = 3 hours. Very few students received this point. The mean score for this problem was 3.02, but if you take away the students that received a 0 on this problem, then the mean was 4.38.

AB 3 Gloria Nan Dupree, PhD ­ CD Owen High School, Black Mountain, NC

AB 3 proved to be an interesting problem. It has been several years since the Calculus Exam included a table of this nature. Most students were at a loss as to what to do with it, contributing to the lowest mean that I can remember. The unofficial mean (with most of the books scored) was .95. The scoring rubrics were explicit in guiding the readers to an appropriate score for the students' work. Part (a) (2 points): The first point was awarded for correctly evaluating h(1) = 3 and h(3) = !7 . Any computational errors lost this point. Unlabeled values of 3 and !7 did not earn this point. The second point was for the student's conclusion based on the Intermediate Value Theorem (IVT). The student could not earn this point without using his/her h(1) and h(3) . The student could state that the conclusion was true by stating that h or f was continuous, that h or f was differentiable, or by the IVT. The students who stated the IVT with no supporting work earned no points on Part (a). The student who lost the first point because of one computational error was still eligible for the second point. The student could not earn the second point by appealing to a wrong theorem. Students were allowed to reason using the interval 2 < r < 3 (since it is a subinterval of the correct interval) and earn both points ­ the first point for h(2) = 4 and h(3) = !7 . Again, evaluation errors came off this point, and unlabeled values of 4 and !7 did not earn the first point. As with the other students, they were eligible for the second point if they used their computed values as stated and drew the correct conclusion using the IVT. There was an alternate solution that was accepted. The student who reasoned that h(r) = !5 implies that f (g(r) ) = 1 could earn the first point if f (g(1) ) = 9 and f (g(3) ) = !1 . The student could earn the second point if he/she reached the Value Theorem with no evidence of any calculations. Unfortunately, they needed to compute the values and connect them with the IVT to receive any credit. correct conclusion with the IVT and f (g(1) ) and f (g(3) ) . Many students simply stated the Intermediate

h(3) ! h(1) . The 3 !1 student could lose this point with an incorrect evaluation of the difference quotient. The second point was earned by a correct conclusion using the Mean Value Theorem (MVT). The student could state that h was

Part (b) (2 points): The first point was earned by using a correct difference quotient,

6

continuous and differentiable, that h or f was differentiable, or by the MVT. The student was eligible for the second point only if he/she set up a correct difference quotient. The student could not earn the second point by appealing to a wrong theorem. As in Part (a), many students lost both points by correctly stating the Mean Value Theorem without the supporting difference quotient. It is important for students to understand that these theorems need supporting evidence. Part (c) (2 points): The first point was earned by applying the chain rule, either explicitly or implicitly. Any of w '(x) = f (g(x) )! g '(x) , w '(3) = f (g(3) )! g '(3) , or w '(3) = f (4) ! 2 was accepted. The second point was the answer point and could only be earned if the first point was earned. A bald answer received 0 of 2 points. There were students that knew Part (c) involved the Fundamental Theorem of Calculus but did not apply the chain rule. Lack of the use of the chain rule cost the students both points. Part (d) (3 points): The first point was earned when the student declared that g ! 1 (2) = 1 or had a tangent line that used the point (2,1) . There were students who lost this point because they declared the point

(2,1) based on a statement that g '(2) = 1 . The second point was earned when the student declared that

1 . The third point was earned if the 5 ' 1 student presented the correct equation, or an equation with a correct g ! 1 (2) = 1 or g ! 1 (2) = and a 5 declared value for the one that was incorrect. Examples: 1 g ! 1 (2) = 3 and y ! 3 = (x ! 2) was scored 0 ­ 1 ­ 1. 5

!1 '

(g ) (2) = 1 5

or had a tangent line equation that used the slope

( )

(g ) (2) = 5 and y ! 1 = 5(x ! 2) was scored 1 ­ 0 ­ 1.

!1 '

Acceptable declarations for the first point were g ! 1 (2) = # and g(#) = 2 . An ordered pair ( 2, # ) was not an acceptable declaration. Many students did not remember how to take the derivative of an inverse function. Some declarations for the second point were: ACCEPTABLE

(g ) (2) = #

!1 '

UNACCEPTABLE

g! = #

dy =# dx y '(2) = #

derivative of g derivative = #

!1

m=# slope = #

=#

Bald equations were scored as follows: 1 The correct equation, y ! 1 = (x ! 2) , in any correct form, earned all 3 points. 5 A tangent line equation with incorrect slope that used the correct point (2,1) was scored 1 ­ 0 ­ 0. Example: y ! 1 = 5(x ! 2) A tangent line equation with correct slope that used an incorrect point was scored 0 ­ 1 ­ 0. 1 Example: y ! 3 = (x ! 2) 5 The students who earned only one point on this question generally earned one computation point somewhere on the question. If teachers will peruse the earlier exams, they will find questions similar to this one.

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AB 4 Ann Carroll ­ Ravenscroft School, Raleigh, NC

This problem was the first of the three non-calculator questions. It was clearly stated and surprisingly easy to grade. Part (a) required that students be able to use both the product and chain rule to differentiate a position function, solve a trigonometric equation on a defined domain, and justify where the minimum value occurs on that domain. Part (b) asked students to find a value of A that would satisfy a second order differential equation. Though the points for the two parts were distributed 5 ­ 4, students could attain 2 of the points for Part (a) by differentiating the function in part (b). There were quite a few students who omitted or earn a zero on this question. Part (a) (5 points): Two points were awarded for the correct derivative of x(t) . Students who forgot to use the product rule earned neither of these two points. There was a 1 point deduction for each sign error, since it was difficult to know if the incorrect sign was a result of a mistake in differentiating trigonometric functions or e ! t . Students could earn these two points by expressing the derivative as x '(t), v(t) or as an unlabeled correct expression for x '(t) in either Parts (a) or (b). The third point was awarded for setting the derivative equal to 0. If it was unclear whether students had set the derivative equal to 0, they could earn ! 5! this third point for stating the correct critical values t = , t = . Students could earn the third point 4 4 even if they missed the first two by setting their derivative equal to 0. The fourth point was awarded for the 5! correct answer t = . The fourth point could again be earned even if the derivative was wrong. Graders 4 were asked to read with the student. The fifth point was the justification point and very few students successfully earned this point. It required that students evaluate the original function at all possible candidates, the two endpoints and the two critical values. Students could eliminate two of the possibilities, ! " ! 5! # t = and t = 2! , by explaining that the function was decreasing on the interval $ , % and increasing 4 &4 4 '

5! " 5! # on the interval $ , 2! % . Students still needed to evaluate x(t) at t = 0 and t = . Many students 4 & 4 ' were successful on the first three points of this part. Many students were unable to solve the trigonometric ! equation without the aid of a calculator, some stated only the first critical value t = , and some listed too 4 ! 3! 5! 7 ! many critical values: t = , . There were students who tried to use integration instead of , , 4 4 4 4 differentiation to answer this question.

Part (b) (4 points): Students performed much better on part (b) than in part (a). Two points were awarded for the correct second derivative. A student who forgot the product rule earned 0 points and the sign errors resulted in a 1 point deduction. Students who forgot to use the chain rule when differentiating e ! t ultimately lost both points since they were required to differentiate e ! t twice. One point was awarded for substituting the first and second derivative expressions into the differential equation. This point could be earned even if the student's derivatives were incorrect. Students still earned the substitution point even if they copied their derivatives incorrectly. The final point was awarded for solving correctly for the value of A. Most derivative mistakes made this task impossible. Students could earn the final point by substituting in a value for t in the given interval.

AB/BC 5 "The Balloon Problem" Sarah Raynor ­ Wake Forest University, Winston-Salem, NC

The final common problem (the only shared AB/BC non-calculator problem) had several different parts covering three major concepts from three different portions of the course. Students are presented with the

8

information that a spherical hot air balloon is expanding, that the radius is twice-differentiable, and that the graph of the radius is concave down on the time interval 0 < t < 12. They are given data for r' at 6 points on the interval (including both endpoints), in units of feet/min. Finally they are told that r = 30 when t = 5. Part (a) (2 points): Students are asked to use a tangent line approximation to estimate r (5.4), and then state whether the result is smaller or greater than the actual value, with a reason. The correct answer, 30.8 feet, was worth 1 point. This was easy and straightforward to grade because there were no decimal presentation issues. However, correct use of the tangent line approximation, in absence of the correct answer, was worth nothing, which penalized a number of students with minor arithmetic errors. That the estimation was too large, with justification, was worth a second point. Students were required to say that the estimation was too large because r is concave down on the interval t = 5 to 5.4. Equivalent statements, such as "r' is decreasing," or "the tangent line is above the graph," were accepted. Students had to give an acceptable time interval (containing 5 to 5.4 and contained in 0 to 12), or use an "invisible" time interval. However, point-based arguments, such as "r' (5.4) is less than r' (5)" lost the point. Part (b) (3 points): Students are asked to find

dV at t = 5 , with units. The correct answer was dt

dV dr = 4!r 2 = 7200! ft 3 / min . Correct use of the chain rule was worth 2 points, and the answer was dt dt worth one point. Units counted separately (more on that later). If a student had an arithmetic error (such as 8 dV getting !r 2 for ) they lost one point for the chain rule but could still receive the answer point. 3 dr Students also did not have to state the chain rule explicitly. For example, the answer 4 (30)2(2) was worth 3 points. However, a bald answer of 7200 was worth 0. The burden was on the students to convince us that they had used the chain rule correctly to receive any points. This portion of the question was definitely the easiest for students to receive full credit on.

Part (c) (2 points plus 1 point for units ): Students are asked to estimate r' (t) dt from t=0 to t=12

!

12

0

r '(t)dt using a Riemann sum with the given tabular data and to interpret their answer with correct units.

The correct answer, 19.3 ft, was worth 1 point. The explanation was worth 1 point. Correct units in both Parts (b) and (c) earned students a "units point." Students struggled with the Riemann sum because the partition was not even. Additionally, as in Part (a), correct set up of the Riemann sum was worth nothing in the absence of the correct answer. Therefore, it was advisable for students not to simplify their answer. A student who simplified incorrectly could lose as much as 3 points (1 each on (a), (b), and (c)), whereas unsimplified answers were accepted in all cases. For the interpretation point, students had to state that this integral represented the change in the radius of the balloon on the time interval t = 0 to 12. Radius or r had to be stated explicitly, as did the time interval. Statements like "the twelve minutes" cost the students the point. Part (d) (1 point): Students were asked whether their answer for (c) was too large or too small, and why. Students were required to state that it was too small because r' is decreasing. Other, equivalent statements, such as "r' ' < 0" or "r is concave down," were not accepted. The burden was on the student to convince us that they were referring to r' in particular. This cost many students the single point awarded in (d). Overall, this was an interesting question that tested several fundamental concepts. It was obviously difficult to allocate only nine points to these four parts. As a consequence, the rubric was very strict and focused on correct numerical answers and certain exact phrases in the explanation sentences. It is crucial to remind students to be as precise and explicit as possible, especially in naming functions and time intervals, so as to avoid losing points unnecessarily on this type of question.

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AB 6 Jack Self ­ North Davidson High School, Lexington, NC

The student was presented with a function f(x) involving an unknown positive constant k. In Part (a) the student was asked to find f '(x) and f "(x) , necessitating the ability to handle a square root radical as well as the natural logarithm function. In Part (b) the student was asked to find the value of k for which f(x) had a critical point at x = 1 and to determine whether there was a relative maximum, minimum or neither and to justify the answer. In Part (c) the student was asked to find the value of k for which the graph of f had an inflection point on the x-axis. There were relatively few blank answer booklets for AB 6, indicating that most students felt they had some understanding of the requirements of this question. Although they were not always successful, most students at least attempted to answer all three parts. Part (a) (2 points): This was fairly straightforward with each correct derivative scoring 1 point. Most students differentiated f(x) correctly, with minor errors in handling the derivatives of the square root function. Many students made errors in finding f "(x) in not only differentiating the square root function

1 with its antiderivative. x Part (b) (4 points): The student was expected to set f '(1) equal to 0 and solve for k, and most students were able to do this correctly. In determining whether there was a local maximum or minimum or neither, the student had the option of using either the first or the second derivative test. Setting f '(x) = 0 was awarded 1 point, solving for k a second point, stating whether there was a maximum, minimum, or neither was given a third point, and the justification was a fourth point. If the student used the first derivative test for the justification, he or she was required to describe the sign change of f '(x) at x = 1. Many students failed to earn the justification point because they inadequately described the behavior of f '(x) . Some created a sign chart without a verbal explanation, which is not in line with the College Board's policy on the use of sign charts. Some described a change in sign but did not refer to the derivative function. Use of such expressions as "it changes from negative to positive" or "the function changes from negative to positive" fail to adequately describe the fact that f '(x) is what is changing sign. It was also necessary to reference where the sign change occurred. Merely saying that f '(x) changed from negative to positive was not sufficient. The student needed to state that the sign change was occurring at x = 1. If the student used the second derivative test in the justification, he or she was expected to find the value of the second derivative at x = 1 and note that since f "(1) was positive, there was a relative minimum. Merely finding

but also confusing the derivative of the value of the second derivative was insufficient. Without referencing the fact that f "(1) > 0, the student did not show sufficient understanding of the significance of the sign of the second derivative. Part (c) (3 points): The student was expected to set f(x) = 0 and f "(x) = 0 and to combine the results to produce an equation in one variable, ultimately solving for k. Setting either function equal to 0 earned the first point, producing a combined equation in one variable earned a second point, and solving for k earned a final point. As the equation involved a natural logarithm and calculators were not allowed, the correct solution was an expression involving a power of e. Many students erroneously were still working under the assumption that x = 1 (apparently still mentally in Part (b) ), and that resulted in loss of all three points. While most students correctly understood that they were supposed to set the second derivative equal to zero, many were unable to generate a correct single variable equation in order to solve for k. It is clear from the reading of this problem that many students are still unaware that a sign chart alone is not sufficient justification for a calculus conclusion. The student must state in words what is significant about the sign chart, i.e., describe the sign change and the resulting conclusion. It is also clear that students continue to be lax in using proper terminology. They must clearly refer to the first derivative or the second derivative if that is what they are referencing rather than a nebulous "it" or "the function". Working with an unknown constant within a function continues to be a popular format for AP test questions, and this question does a good job of tying the students' knowledge of calculus concepts (first derivative, second

10

derivative, relative extrema, first and second derivative test, inflection points) with the algebraic manipulation necessary to solve for the unknown constant.

BC 3 Al Letarte ­ WG Enloe High School, Raleigh, NC

This problem combined computation of area in polar coordinates and motion of a particle along a polar curve via parametric equations. Part (a) (4 points): Area =

2 1 19! 11 3 2 ! " 22 + & (3 + 2 cos # ) d# = $ % 10.370 3 2 2! 3 3 2

1 2 2 d# of a "relevant" 2$ "

!

4! 3

The first point, the circular-region point, was earned by computing the area circular sector--i.e., with limits

" 2# 2# % $ 2# % , - , , 0, - , (0, 2# ), (0, # )+ or the equivalent. The 0. 3 3 / . 3 / 1 sector area could be computed either via a polar-area integral or by multiplying the circular area by an

(&, ' )( *$ ) ,

!

appropriate fraction.

Although there was reason to doubt whether students who wrote

$ 2d#

"

!

really

understood computation of areas in polar, the circular-region point was awarded in this case as well. The second point, the integrand point, was earned for writing any integral, with or without limits, in which the integrand contained either an expression of the form ± (3 + 2 cos ! ) or ± 22 ± (3 + 2 cos ! ) . Students

2

(

2

)

had to include an integral symbol in order to qualify for the integrand point. The third point, the limits-and-constants point, was earned for writing either

where ( ) involved the expression 3 + 2 cos ! . If both integrand and limits-and-constants points were to be earned, they had to be earned with the same integral. For the fourth point, the answer point, three-place accuracy was expected: 10.370 or 10.37. Because some TI calculators have a factory default of six significant figures and would give the approximation 10.3705, we accepted any approximation in the

1 #( 2 2! 3

4! 3

)d"

or

2! 3

# ( )d" ,

!

19! 11 3 8! + 1.99289 , etc.. , " 3 3 2 The answer point could not be earned unless both integrand points were earned. Mistakes in combining regions came off the answer point. Copy errors were handled as follows: Students who miscopied 3 + 2 cos ! in (a) did not earn integrand points, but could earn the answer point if their answer was the correct one. We read with students who imported copy errors to (b) and / or (c) (in practice, this seldom happened).

interval [ .370,10.371]. Exact or mixed answers were also accepted: 10 Part (b) (2 points):

dr dt

=

!=" 3

dr dr # $1.732 . The particle is moving closer to the origin, since <0 d! !=" 3 dt

! and r > 0 when " = . 3

For the first point, the answer point, bald answers were accepted. The second point, the interpretation point, could be earned in the presence of an incorrect answer, provided a student committed to a numerical dr value and labeled it as such. Students were given some latitude in terms of language they could dt !=" 3 use in interpretation: Words / phrases like "center," "pole," "particle moving inward," "toward," "closer,"

11

etc., were acceptable. Observing only that the particle is moving toward the origin, with no mention of the fact that r > 0 , was enough to earn the interpretation point. In the context of correct wording, the rate could be described as either + 3 or ! 3 . On the other hand, answers like "radius is decreasing," "distance from particle to center is changing at rate ! 3 ," "motion of particle toward origin is changing at

! 3 ," "position of particle relative to origin is changing at rate ! 3 ," or "particle moving down and to the left" were not acceptable for the interpretation point.

Part (c) (3 points): y = r sin ! = (3 + 2 cos ! )sin !

dy dt

=

!=" 3

dy = 0.5 The particle is moving d! !=" 3

dy ! > 0 and y > 0 when " = . away from the x-axis, since dt 3

For the first point in this section, students had to express y in terms of r and ! . Any of the following dy = (3 + 2 cos ! )cos ! + sin ! ("2sin ! ) . assertions earned this point: y = r sin ! , y = (3 + 2 cos ! )sin ! , d!

dy dy , , or y! equals 0.5, with or dt d! ! 1 without the qualification " = . The answer had to be labeled - i.e., a bald did not earn the point. The 3 2 derivative point was awarded for an incorrect answer if a student either (1) imported a copy error from (a) 5 3 or (2) believed, and had stated explicitly, that y = r cos ! (in which case y! = " # "4.33 ). 2

The second point, the derivative point, was earned by observing that The third point, the interpretation point, was earned for assertions like "particle moving away from x-axis," "particle's y-component increasing," or "particle moving upward." We did not expect students to observe dy that y > 0 . Students who arrived at an incorrect value for and then went on to interpret this result dt !=" 3 correctly were eligible for the interpretation point. The interpretation point could not be earned without committing to an explicit y! value.

BC 4 Vicki Carter ­ SCA2PMT President, West Florence High School, Florence, SC

BC4 turned out to be relatively easy for the BC students. It was the first problem the students encountered after putting away their graphing calculators. It was also the problem that had an AB sub-score grading part and a BC grading part. The combined average was 5.78, which was higher than that of BC1, the classic area/volume question. The average score for the AB part was 3.98 out of 5 points. The average score for the BC part was 1.80 out of 4 points. Part (a) (2 points): The student had to produce the equation of a tangent line given the first derivative function, which involved logarithms. The first point was awarded for find the slope at e and the 2nd point for writing the equation of the tangent line. For the most part students successfully completed this task with most of them earning the 2 points. The students who tried to write the equation in slope-intercept form tended to lose points more often as well as the student who attempted to simplify after writing the equation in point-slope form. Part (b) (3 points): The student had to determine the concavity of the function on an open interval. Students were expected to find the 2nd derivative and make a conclusion about concavity over the given interval. Two points were awarded for the 2nd derivative and one point for the conclusion. The 2nd derivative had to include ln x and demonstrate use of the product rule. Algebraic errors with the 2nd

12

derivative came from the 3rd point, the answer point. For the answer point, the student had to indicate that the 2nd derivative was positive on the given interval. Students could use a statement about f ! such as, the derivative of f ! is positive, f ! is increasing, as long as there was a connection to the interval. Additional work was ignored unless it was blatantly wrong. Parts (a) and (b) made up the 5 AB sub-score points. Most students got 2 out of 2 on Part (a) and 2 out of 3 on Part (b). Part (c) (4 points): This was an integration by parts problem that is a BC only topic. The antiderivative was worth 2 points, the use of an initial value to attempt to find the constant of integration was worth 1 point, and finding the correct constant was worth 1 point. Many students found the antiderivative and stopped, since the information to use to find the constant of integration was in the problem stem. The typical student received 2 out of 4 for Part (c) of the question. Some students attempted tabular integration by parts, which was not a good idea for this problem. The students who correctly identified u and dv and then set up an integration by parts usually earned at least 2 points. Students who did recognize the need to use the initial value most often worked out the constant correctly, but many neglected to do this at all. If the student made a single polynomial mistake in attempting to find the antiderivative, they were eligible for the 3rd and 4th points. We had a very generous grading policy to earn the 3rd point by using the initial value in almost any equation involving x, y, and C. The students were only eligible for the last point if they had earned 1 of the first 2 points for the antiderivative. No parts, bad parts, an antiderivative that did not contain ln x, or multiple polynomial mistakes took the students out of the running for 3 of the 4 points in this problem. We have not seen integration by parts in a free response question in quite a while. Make sure your students can correctly demonstrate the process, especially if it also involves finding the constant of integration.

BC6 The Series Problem Vic Levine, Question Leader ­ Memorial High School, Madison, WI

Philosophy: This problem asked the students to work with a Taylor series centered at 0 for f(x) = e ! x Students were expected to modify the series for e x to accomplish this. In part (a), students were asked for the first four non-zero terms of the series, and then the general term of the series. In pat (b) students were asked to find a limit using their series from part (a). In part (c), the students were asked to integrate their series from (a) and write the first four non-zero terms of the antiderivative of e ! x , along with an estimate 1 of the definite integral from 0 to . In part (d), the students were asked to show that the their estimate in 2 1 (c) differs from the exact value of the definite integral by less than , and to explain their reasoning. 200 Part (a) (3 points): Many students attempted to calculate the Taylor series directly by differentiating the function directly. This was a long and tedious process. Some students were able to find the first four nonzero terms by this method, but were unable to come up with the general term. Part (b) (1 point): Students were asked to use their answer from (a) to find the limit. Students who worked directly from the function e ! x did not receive credit, since they did not use their answer from (a). Students who imported their answer from (a) and found a finite answer based upon their work were awarded the point. Part (c) (3 points): Most students were able to integrate term by term their answer from (a) to earn the 1 first two points. The third point was for evaluating the first two non-zero terms of their answer at , and 2 many students were able to successfully accomplish this.

2 2 2

13

Part (d) (2 points): Many students were able to show that the error bound was always less than the absolute value of the first omitted term of the convergent alternating series, and that this number was less 1 than . The first point was earned by stating that the error bound was their third non-zero term 200 1 evaluated at . Very few students, however, were able to successfully state the necessary conditions for 2 the error bound being less than the first omitted term of the convergent alternating series. Teacher Notes: Students are expected to know the Taylor series centered at 0 for the following functions: 1 ex, sin x, cos x, and , and that these series can be differentiated and integrated at any point inside the 1! x interval of convergence. Trying to work directly from e ! x was usually fruitless. In (b), there were too many students who inserted their answer from (a) correctly into the fraction, but then could not distribute the negative sign, and had a sign error in the final answer. Since this problem was only worth 1 point, nothing was earned in this case. In (c), most students who had earned at least two points in (a) were able to earn all of the points in (c). Some students only earned the third point in (c) for evaluating their first two 1 nonzero terms at . Very few students, however, earned the explanation point in (d). Students need to 2 know what hypotheses need to be satisfied to apply a particular theorem. The Alternating Series Test has the same hypotheses needed for the Remainder Estimation Theorem for Alternating Series. But, there are convergent alternating series that do not pass the Alternating Series Test. Thus, just stating that the alternating series is convergent is not a sufficient explanation. Students also need to recognize the importance of looking at the absolute value of the terms of the series, and that they have to be monotonically decreasing to zero. Also, there were many students who indicated that any series, whether convergent or not, alternating or not, has an error bound less than the absolute value of the first omitted term.

2

The Alternate Exam ­ Five Years in an Alternate Universe Tom Polaski ­ Winthrop University, Rock Hill, SC

At the AP Calculus reading, almost all of the readers concentrate on the exam that most students take. This exam is called the "operational exam" in ETS jargon. Approximately 269,000 students took the operational exam this year (207,000 took the AB Exam and 62,000 took the BC Exam). However, there are other versions of the AP Calculus exam which must also be read. The goal of this article is to provide a glimpse into the "reading within the reading" that is used to grade these tests. Two exams other than the operational exam are graded at the reading. The Form A Exam, also known as the Alternate Exam, is given to students who have an approved conflict with the exam schedule ­ a list of approved conflicts may be found at AP Central. The Form A Exam is given two weeks after the operational exam. The Form B Exam is given to students outside the Western Hemisphere within a day of the operational exam. The number of students taking these exams is quite small relative to the operational exam. This year approximately 4600 students took the Form A Exam (3400 AB and 1200 BC) and 4300 took the Form B Exam (3100 AB and 1200 BC). The free response questions from the Form B Exam are released to the public, but the questions from the Form A Exams are not released. Thus readers in Form A rooms must sign a statement stipulating that they will neither disclose the content of the exam nor discuss the questions and student responses outside the reading room. However, I would not be breaching my agreement to assure you that you wouldn't be any more or less shocked by the content of the Form A Exam than you are by each successive year's operational and Form B Exams. Since 2003, two rooms of readers have read the Form A Exams. This year an additional room was assigned to read the Form B Exams. The readers of the Form A and Form B Exams are chosen from readers with several years' experience. Many readers are former table leaders and exam leaders, or like me are "un-

14

retired" college readers with more than six years of reading experience. I have been a Form A reader for the five years that the Form A room has existed. The actual reading of the Form A and Form B Exams is supposed to be a miniature version of the reading as a whole ­ here is how the reading works. The Form A and Form B readers spend the first two or three days in the flow of the operational AB Exam ­ reading the first two or three questions in our assigned part of the operational flow. We thus are familiar with at least two of the common problems from the AB/BC operational exam. We are taken out of the operational flow mid-week to read the other exams, and then we end the week once again reading the operational AB Exam. This year we were moved from the cavernous reading room to small conference rooms to do our non-operational work. This was especially important for the Form A Exams, since those exams must be kept secure. The reading now follows its own flow. The reading room is given standards and notes on anticipated student errors: reversals, degree mode, and the like. The notes and standards for the Form A Exam are returned to ETS after the reading for disposal. These materials are prepared by the Exam Leader (the Form A Exam Leader for the past several years has been our own Stephen Davis of Davidson College) with the input of the other Exam Leaders, the operational Question Leaders, and the Chief Reader. The briefings on the standards are much briefer than those of the operational exam and are conducted by the Exam Leader or the Assistant Chief Reader. Due to the small number of exams, no samples are used in the briefing. Our goal is to provide as consistent a reading as the operational exam, so we look very carefully for responses that bring up issues of interpretation of the standard. When such an issue arises, the Table Leaders and the Exam Leader determine how we will proceed ­ often the entire room will discuss how we will handle the issue and come to a consensus, which is ratified by the Exam Leader. The physical act of reading the Form A and Form B Exams also differs from the operational exam. The scores are not recorded on bubble sheets; instead, they are recorded the old-fashioned way ­ by placing the score on the back page of the exam itself. Readers record their ID number on each exam so one reader will not read more than one question on an exam. Since there are no bubble sheets, exams need not be kept in order within a packet, or even within the same packet. That is the ideal process for reading the Form A and Form B Exams. This year was actually the first year that the process went exactly as described. Late arrival of books is a standard problem; in fact, the Form A rooms were created in part because the short amount of time between the administration of the exam and the reading meant it was likely that very few books would make it to the reading site in time to be read prior to the reading. When a shipment of books arrives (or is discovered) after we have begun reading the alternate exam, the number of books is generally too small to restart the flow. Instead, the readers in the room are divided into subgroups that handle each question that the room has already read simultaneously. Books are passed around the room for the readers to read their questions. Since the books and packets need not be kept in order, the action can be a bit antic as books are spun around the room. It is common to read several different problems in the course of one day -- quite a different experience from reading the operational exam. My five years reading Form A Exams have given me a broader respect for the reading process. Being a minor participant in the standard-setting process has shown me the care with which the standards are developed and tested. Being surrounded by extremely able readers has helped me to become a much better reader, both in efficiency and accuracy. There is much more variety during the reading: at this year's reading my room read 10 distinct questions ­ last year my room read 13 questions. Watching a miniature reading flow in one room (and how easily it can get out of sync) has increased my appreciation for how the flow of the operational exam is managed. Lastly, my experience with reading Form A Exams has convinced me that the level of consistency that the reading achieves is remarkable. The Form A and Form B Exams are indeed read to the same degree of fairness and consistency as the operational exam, and the students who take these alternate exams are at no appreciable advantage or disadvantage.

15

2007 Unofficial Means Doug Kuhlmann ­ Phillips Academy, Andover, MA

AB 1) 2) 3) 4) 5) 6) BC 1) 2) 3) 4) 5) 6) 5.43 4.38 2.83 AB part 3.98 (out of 5) BC part 1.80 (out of 4) 4.60 2.28 5.84 4.59 3.47 AB part 4.26 BC part 2.68 5.16 3.84 Average 4.33 3.02 0.95 ( this is not a typo) 2.91 2.47 3.49 Average (without including zeros) 5.12 3.54 2.36 4.34 3.82 4.20

Common questions were AB1/BC1, AB2/BC2 and AB5/BC5.

2007 AP Reading Events Calculus Development Committee Night Norma Royster ­ Central Cabarrus High School, Concord, NC

Committee members for 2007 will be: Janet Berry, Stephen Davis (new chair), Ruth Dover, Guy Mauldin, Monique Morton, James Epperson, and Tara Smith. Ex Officio members: Caren Defendorfer, Mike Boardman Responsibilities: · Review AP course description ­ New one will be out this week ­ no new changes. This seems to be a transition year and will cover just 2007. Future ones will be valid for a 2 or 3-year period. · Review exam policies · Consult and advise College Board · Conduct college surveys · Write and review MC and FR questions · Write approximately 27 FR questions for operational, Form B, and Alternate AB and BC exams No audit questions were entertained and they were to be deferred for the Open Forum the next night. The question construction was discussed and not all topics are replicated across all three tests. Answers to questions were: · Most FR and MC questions are written in-house · Rubrics are suggested prior to administration of tests · FR questions are not field-tested · College Board decides which test will be released ­ the 2008 will be the next one released · Infinite Series are usually tested because it is so particular to the BC curriculum · Problems are constructed so that they become memorable · Conceptual questions are valued over mechanics-driven questions

16

· · · · ·

Topics that have not been tested on the FR section in recent years will be viable and likely to be on future test questions Universities collect much data on students and will be looking for alternatives for students coming in with AP credit Problems that ask students to look at more than one idea will be typical on future tests TI-89 usage still brings up equity issues since they may be an advantage on the AP tests At least one question on the FR section will always contain a non-calculator and calculator active portion

New Table Leader Experience Charlene Garrett ­ Thompson High School, Alabaster, AL

"So--what's it like being a table leader instead of a reader?" you ask. Well, in some ways it is the same and in other ways it is somewhat different. Here is how it is different. First of all, I was in Louisville (or wherever) two days longer. There was not a welcome table at the hotel and I had to figure out where to register by going up to people in the elevator of the hotel and asking, "Where did you get that?" while pointing to their nametag. I knew enough about the reading to realize that I could do nothing without a nametag! On the first day for table leaders, we were given a list and asked to find and sit with our table partner. During that day and the next, we went over the common problems and then split into AB and BC groups to go over the other three problems. During this time, we were briefed on the problems just like during the reading, except perhaps longer. We also looked through books looking for whatever the QL asked us to find - perhaps a paper that was a 9, one that had an alternate approach, whether the problem read well, or whether there were unexpected things coming up. There were some things about the standard that had not been decided at that time. I was a little surprised that we did not really read any books; I had assumed that was how the TL's seemed to be so familiar about the standards each year. Once the reading began, we had table leader meetings each morning to find out announcements, get certain handouts, find out what percentage of the books had been read, and the current mean on the problems. It was nice to have a little information about those kinds of things. Some days we were asked to volunteer to sell t-shirts or to hand out samples and standards. At the briefings, the TL's had to pay as much attention as everyone else because the official standard was being discussed at that time. The first two days of the actual reading were the hardest for me. I was trying to read books and immediately people were asking me about the question. I was thinking, "I'm still trying to figure it out myself!" but with the help of my eight previous years of reading, my table partner, and the occasional visit to the QL, I was able to answer the questions. Once my first folder was read, the backreading began. Backreading really helps you to understand the problem and to gain some confidence because most of the time, what you think about the problem is confirmed by the reader's score. I developed my own way to question the reader when I did not agree and was convinced that the majority of us are just here to do a good job and treat all of the students fairly. The people in my room were so nice about the whole situation and wanted me to tell them when they were wrong (although they did not want to make a mistake!). By Day 6 of the reading, I was preferring backreading to reading to some extent. Being a TL also gave me an opportunity to get to know the readers in my room a little better. I tend to not be a very outgoing person, so as a reader, I came in and read the books and talked to my partner, the TL's and maybe a few others, but I frequently did not even know the names of many of the people in my room. As table leader, I was forced to talk to people more. I also made an effort to speak to them during breaks, at meals, and when I saw them around during the evenings. In what ways was this experience the same? I worked all day long grading calculus questions, trying to apply the standards in order to be fair and consistent with every paper. I met people from many states and a few countries and got to discuss books and practices and students with teachers who understand and may be able to help. Being a table leader may not be for everyone, but I think it may have worked for me.

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Professional Night: The AP® Multiple-Choice Section: From Conception to Delivery Craig L. Wright ­ [email protected], Mathematics Group, Assessment Development, Educational Testing Service, Princeton, NJ

The focus of the AP Reading is the grading of the free-response problems of the Calculus AB and Calculus BC exams. The free-response section (Section II) contributes 50% of the points that determine a student's AP grade. The other 50% of the points comes from the multiple-choice section of the exams (Section I). The format of Section I is as follows: Part A: 28 questions, 55 minutes. No calculator is allowed. Part B: 17 questions, 50 minutes. A graphing calculator is required. Only some questions require the use of the calculator. Why multiple-choice questions? · They are useful in measuring a student's level of calculus competence in a variety of contexts. · They allow for coverage of the topic outline as presented in the AP Calculus Course Description. The May 2008 Course Description is now available for free download at AP Central® (http://apcentral.collegeboard.com). They are highly reliable in providing statistical information on student performance. They allow for a selection of questions at various levels of difficulty (easy, medium, hard). They allow comparison of the ability level of the current students with those from a previous year. A subset of the exam questions has been used on a previous AP exam. They complement the free-response section where students have the opportunity to demonstrate their problem-solving ability and mathematical communication skills. If a topic is not included in the free-response section, it is very likely covered in the multiple-choice section.

· · · ·

Where do the questions come from, and how are the questions reviewed? (the "conception" part) · The multiple-choice section is constructed using a set of exam specifications or "blueprint" for the exam. These specifications are set by the AP Calculus Development Committee and are based on the topic outline in the Course Description. They include a breakdown by content, function type (e.g., algebraic, exponential, trigonometric), representation (analytical, graphical, numerical, verbal), and calculator activity. The specifications allow different exams to have similar characteristics though they have different questions. · Development Committee members and external "item writers" receive targeted multiple-choice assignments from ETS mathematics staff to contribute to a pool of approved questions for use in the exams. Guidelines are used in writing the questions. Questions should be clear and concise, mathematically accurate, and unambiguous for students. The content should be appropriate for testing calculus and should generally focus on one concept. The question should have a single best answer, and the incorrect answers (what we call the distracters) should reflect common mistakes and misconceptions - choices that are very attractive to students! Given that each question is a part of a collection of questions, the amount of time that the question takes to answer needs to be considered. If the question requires the use of the graphing calculator, care should be taken to ensure that the question is fair and appropriate regardless of the calculator used. Questions should vary in difficulty. Questions receive a preliminary review by ETS mathematics staff before being reviewed by the Development Committee. They receive additional reviews at ETS for content, style, and fairness. The questions are formatted to appear as they would on an exam; artwork is incorporated as needed. The questions are added to a pool of approved questions. Many questions are pretested in calculus classes before they are used on an exam in order to determine their level of difficulty and how they perform statistically.

·

·

18

How are the exams constructed? · Draft Calculus AB and Calculus BC exams are assembled by ETS mathematics staff, based on the exam specifications mentioned above. Questions are selected from the approved pool of questions. · The exams include a set of questions from a previously administered exam. This set of questions is a "mini exam" within the exam, and it allows comparisons to be made between different groups of students. The draft exams are reviewed by the Development Committee. They make revisions to individual questions as needed and to the mix of questions as a whole. (During this process, the Development Committee is also reviewing and finalizing the free-response sections of the exams. They can see the "big picture" to see the complete coverage of calculus topics.) The exams are revised by ETS mathematics staff, and the Development Committee reviews the exams again. The exams also receive an independent review by someone who has not seen the exams before. The final layout version of the exams - what students will see on the exam day - is sent to the Development Committee Chair and the AP Calculus Chief Reader for final "sign off." After some additional checks by ETS mathematics staff and other quality control steps at ETS, the exams are sent to be printed.

·

·

What happens when the exams are administered? (the "delivery" part) · Students take the Calculus AB or Calculus BC Exam on its scheduled administration day in early May. · · The multiple-choice answer sheets are machine scored after they are received at ETS. Data on student performance are shared with the Chief Reader at the AP grade setting session. This information together with student performance on the free-response questions is used to make decisions about the number of points needed for the various AP grades (1 - 5). This grade setting takes place at the end of the AP Reading in June. AP grades are available to students by phone on July 1; grade reports are mailed to students to arrive mid to late July. Final analysis of student performance on the exam is shared with the Development Committee to influence future exam development and question writing.

·

Additional information is available in the 2003 AP Calculus AB and Calculus BC Released Exams, which is available for purchase at AP Central. Collections of multiple-choice questions from released AP Exams are also available for purchase at AP Central.

Reader Perspective of Professional Night: The AP Calculus Multiple Choice Section Linda Hollar ­ Watauga High School, Boone, NC

The multiple-choice section is used to cover the AP curriculum, to allow comparison of students to those of another year, and to allow for a wide level of difficulty. The multiple-choice section is found to be highly reliable. The blueprint for a multiple-choice section is set by the test development committee. It is based on the current course description, and breaks down the questions by content, function type and representation, and calculator activity. Writing the multiple-choice section begins with each of the test development committee members being assigned to write particular items. There are also questions written by external item writers, people not on the test development committee. The next step is reviewing the questions submitted, a process done by both the committee and Education Testing Service for content and fairness. The approved questions are then added to a pool of previously submitted questions. These questions are pre-tested in college classes and edited again for grammar, fairness, and sensitivity to particular groups. A list of guidelines followed by the question writers was given. Questions should contain original ideas if possible, focus on a single concept, be clear and concise ­ not too heavy on "reading load," unambiguous, have a single best answer, and not take too much time. The distractors in each question should reflect

19

common errors, but should not use intermediate answers. Writers are also instructed that the correct answer should not be too obvious. Other concerns are that questions on the calculator section should not be easier when done using TI-89 than done using a TI-83 for example, questions should not need any outside "special knowledge," and questions should contain a variety of functions types and representations. Writers are to avoid questions where "button pushing" is the only knowledge needed. Also, notation should be consistent throughout the problem, not 2 < x < 5 in one part and (2,5) in another part for interval notation, for example. How is the multiple choice section put together? A draft is constructed using statistics from field tests that help indicate difficulty level. Included within the draft is the "mini" exam that was also included in previous year's multiple-choice sections. These questions are called the equating set and are used each year, starting over every 5th year after a multiple-choice test is released. The next released exam will be the 2008 multiple-choice, and it will be released after the 2009 test has been given. The draft is reviewed by the test development committee, first by mail. The test is revised and reviewed again during face-to-face meetings. The meetings occur 4 times per year. The July 2007 meeting will finalize the 2008 free response section, and the October 2007 meeting will finalize the 2008 multiple-choice section. To assure quality even further there is also an independent review by someone who has not seen the test at all. The test layout goes to the test development committee chair and to the chief reader for a final sign-off before going to the printer in November. The multiple-choice section is graded and awaits the grade setting session, this year on June 20, after the free response questions are scored. The final analysis is shared with the test development committee, which will then influence future reviews of exams. Two final tidbits: Problem number 76, which begins the calculator section on the multiple-choice test, is numbered as such so that students will be on a different page of the answer sheet, and not confuse answers with numbers 1 - 28. This year for the first time, the multiple choice questions on AB and BC Calculus, as well as US History were scrambled so that seating for large numbers of students taking the exams could be more easily managed.

Presentation: Has Calculus Become a High School Course David Bressoud ­ Macalester College, St. Paul, MN

On the following page are two graphs from the presentation. The entire PowerPoint can be found at www.macalester.edu/~bressoud/talks

20

AP Calculus currently growing at >14,000/year (about 6%)

Estimated # of students taking Calculus in high school (NAEP, 2005): ~ 500,000 Estimated # of students taking Calculus I in college: ~ 500,000 (includes Business Calc)

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The Water-Skier Problem Dan Teague ­ North Carolina School of Science and Mathematics, Durham, NC

What is the path a water-skier takes as she is being pulled behind a boat? To determine this path, students must be able to · set up a differential equation by finding the slope of a line given two points · use the Pythagorean Theorem · integrate using substitution · divide polynomials · integrate using partial fractions All of these techniques are a part of the BC curriculum, so this problem is a nice review of many topics. Another interesting feature of the problem is that the TI-89 cannot find the solution. So, if you need a problem to show why we still need to be able to do integrals by hand (or by brain), this is a good example. Consider the situation in which a boat and water-skier are alongside a dock, connected by a tightly stretched rope. At time t = 0, the boat moves with a constant velocity v away from the dock in a direction perpendicular to the dock. Initially the skier is at (r, 0) and the boat at (0, 0). The rope has length r and the boat is traveling up the y-axis with velocity v. So, at time t, the boat is at (0, vt) and the skier at (x(t), y(t)). We want to determine the path that the skier follows if i) ii) the rope is always tangent to the skier's trajectory. the rope has a constant length r.

First, draw a diagram. The rope always points towards the boat and is always tangent to the curve. We can use this information to write a differential equation for the slope of this tangent line. When the water-skier is dy vt ! y = at the point (x, y), the boat is at (0, vt), so . dx 0 ! x This equation involves the variables t, x, and y and the parameter v. Before we can solve the equations, we need to eliminate t. From the diagram students will notice that there is a right triangle formed with vertices (x, y), (0, vt), and (0, vt - y). So x 2 + (vt ! y ) = r 2 .

2

dy vt ! y 2 = and x 2 + (vt ! y ) = r 2 . Our present goal is to dx 0 ! x eliminate t, but we see that we have the expression vt ! y in both equations. Solving the second equation for vt ! y and substituting, we have

So, we have two pieces of information,

dy r2 ! x2 . =! dx x By solving this equation we can find the path. Note: This is an equation the TI-89 has difficulty solving. Try it and graph the resulting function. This equation must be solved by using calculus.

vt ! y = r 2 ! x 2 and so

The integral

"

!

r2 ! x2 dx can be evaluated using a u-substitution. Let u = r 2 ! x 2 , so u 2 = r 2 ! x 2 x

and 2udu = !2xdx with x = r 2 ! u 2 .

22

So, we have

r2 ! u2 r2 ! u2 integral has the degree of the numerator the same as the degree of the denominator. We need to divide to reduce the order of the integrand. By division we find that u2 r2 du = " !1 + 2 du . " r2 ! u2 r ! u2 A B + This second integral requires integration by partial fractions with and eventually r!u r+u !r" 1 !r" 1 ( #1 + $ 2 % r # u + $ 2 % r + u du . Finally, an integral we can evaluate. Solving, we find that & ' & ' r" 1 ! !r" 1 !r" !r" ! r " !r+u" ( #1 + $ 2 % r # u + $ 2 % r + u du = #u # $ 2 % ln (r # u ) + $ 2 % ln (r + u ) + C = #u + $ 2 % ln $ r # u % + C . & ' & ' & ' & ' & ' & '

2 2 ! r " !r+ r #x Rewriting in terms of x and y, we have y = # r 2 # x 2 + $ % ln $ & 2 ' $ r # r2 # x2 &

! r 2 ! x 2 dx = " " x

!u

u du

="

u2 du . Can we integrate r2 ! u2

u2 " r 2 ! u 2 du ? This

" % + C as the path taken by the % ' water-skier. At the start, we know that x = r and y = 0 , so C = 0 . Finally, we have a solution to the " % is a solution to the % ' equation. The path of the water-skier is graphed below along with the "solutions" found by the TI-89 and Maple.

water-skier problem, if

2 2 dy ! r 2 ! x 2 ! r " !r+ r #x , then y = # r 2 # x 2 + $ % ln $ = dx x & 2 ' $ r # r2 # x2 &

Result from TI-89:

600

Result from Maple:

500

" # # 2 2 h ( x) := $! r ! x

(

)

1% 2

& &+ '

r

2 1

( atanh

(r2)2

Result from Calculus:

2 " r # # 1 # 2 # r2 ( r2 ! x2 $

() (

)

% & 1& 2& & '

400 f( x) g( x) h( x) 200 300

" r + r2 ! x2 % & g ( x) := ! r ! x + ( ln# 2 # 2 2& $r! r !x '

2 2

r

100

10

0

10

20 x

30

40

50

60

Path of Water Skier

23

The Big Bang and the Age of the Universe Dan Teague ­ North Carolina School of Science and Mathematics, Durham, NC

In 1929, Edwin Hubble discovered the cosmos. His 1929 groundbreaking paper, A Relation Between Distance and Radial Velocity Among Extra-Galactic Nebulae, published in the Proceedings of the National Academy of Science http://antwrp.gsfc.nasa.gov/diamond_jubilee/1996/hub_1929.html was the first step in what was ultimately to be known as the Big Bang Theory. In his paper, Hubble presented data for 24 nebulae illustrating the relationship between the recession velocity, in kilometers per second, and the distance from the earth, measured in mega parsecs (one mega parsec is about 30.9 million trillion kilometers). The recession velocity was measured very accurately by the red shift in the spectrum of light while the distance was measured somewhat inaccurately by comparing mean luminosities of the nebulae to luminosities of known star types. The data are given in Table 1 and the scatter plot in Figure 1 illustrates the relationship. Velocity Distance Velocity Distance 170 0.032 650 0.900 290 0.034 150 0.900 -130 0.214 500 0.900 -70 0.263 920 1.000 -185 0.275 450 1.100 -220 0.275 500 1.100 200 0.450 500 1.400 290 0.500 960 1.700 270 0.500 500 2.000 200 0.630 850 2.000 300 0.800 800 2.000 -30 0.900 1090 2.000

Table 1: Recession Velocity and Distance for Hubble's Original 24 Nebulae

Figure 1: Scatter plot of Hubble's Original Data We can use these data to estimate the age of the universe. First, we must develop a mathematical model that describes the positions of the earth and the observed nebulae assuming the Big Bang Theory is correct. According to the Big Bang Theory, both the earth (E) and the nebula (N) have been moving directly away from the site of the Big Band (BB) at velocity v for time T, where T is the age of the universe. As these objects move away from BB, they also move away from each other. This is called recession. The rate at which they recede from each other is R, where R is the component of v in the direction of the segment connecting E and N. The distance between earth and the nebula at time T is D, as shown in Figure 2.

Figure 2: Schematic Drawing of Recession of Earth and Nebula

24

1) If the Big Bang Theory is correct, then the data in Table 1 should approximate the line D = R ! T . If so, then the slope of this line is an estimate of T, the age of the universe measured in mega parsecs/(kilometer/second) or mega parsecs-seconds/kilometer. We need to fit a line with no intercept, so we cannot just plug the data in our calculator. We need a line of the form D = R ! T , not D = R ! T + b . a) First, use your calculator to fit a line D = R ! T + b . Use the formula for the standard error of the intercept, s b = s e

(x ) 1 , to determine if the intercept is significant. If it is, then this data + n " (x i ! x )2

2

contradicts the Big Bang Theory. b) Now, let us fit a line without an intercept. This is a job for CALCULUS! To find the least square line without an intercept, define S =

# (D ! T " R )

i i i =1

24

2

. Find the value of T that minimizes S in terms of Di and

Ri .

2) Use the data to compute the least squares estimate of T. How well does it appear to fit the data? If one mega parsec-sec/km is about 979.8 billion years, how old is the universe according to Hubble's original data? The accepted age of the universe is around 13 billion years. How well did Hubble do? (Of course, the universe is older now, since this data was collected in 1929.) SOLUTIONS: 1) If the Big Bang Theory is correct, then the data in Table 1 should approximate the line D = R ! T . If so, then the slope of this line is an estimate of T, the age of the universe measured in mega parsecs/(kilometer/second) or mega parsecs-seconds/kilometer. We need to fit a line with no intercept, so we cannot just plug the data in our calculator. We need a line of the form D = R ! T , not D = R ! T + b . a) If we use our calculator to fit a linear model, we find that D = 0.001373R + 0.3991 is our least squares regression model. Is the intercept 0.3991 significantly different from 0? If so, this data contradicts the Big Bang Theory. To answer this question, we compute a 95% confidence interval for the intercept and see if it contains 0. All 95% confidence intervals are created by using the equation point estimate ± t * (standard error ) . df The standard error for this estimate is s b = s e

(x ) 1 . + n " (x i ! x )2

2

Since there are 24 data points, there are 22 degrees of freedom, so t 22 = 2.074 . The standard error of the

*

residuals reported by the calculator is s e = 0.4050 . The mean recession velocity is 373.125 km/sec. With this information, we can compute s b .

(373.125) 1 s b = 0.4050 + = 0.1185 . 24 3170090.625

2

Our confidence interval, then, is 0.3991 ± (2.074 )(0.1185 ) or (0.153, 0.645). Based on this data, the true intercept is somewhere between 0.153 and 0.645. Zero is not a plausible value. This data does not support the Big Bang Theory!

25

b) To find the least squares line without an intercept, define S = that minimizes S in terms of Di and R i . If S =

# (D ! T " R )

i i i =1

24

2

. Find the value of T

# (D ! T " R )

i i i =1 24

24

2

, then

dS = dt

# !2R (D ! T " R ) = 0

i i i i =1

24

will find the value of T that minimizes S

24

(and is therefore our estimate of the age of the universe). Solving for T, we find

# !R D + #

i i i =1 i =1

24

T " R i2

= 0 so,

!R D

i i =1

24

i

=T

!

i =1

24

R i2

, and T =

!R D

i i =1 24

i

!R

i =1

.

i

2) Use the data to compute the least squares estimate of T. How well does it appear to fit the data? If one mega parsec/km is about 979.8 billion years, how old is the universe? With this data, we have so T =

!

i =1

24

R i Di = 12513.695 ,

!R

i =1

24

2 i

= 6511425S ,

12513.695 = 0.0019218 mega parsecs/kilometer. This is 1.883 billion years. 6511425

You may have noticed that the line appears to lie too low in the data. A model with a non-zero intercept appears to fit better. This is a result of the poor measurements of distance made in 1929. If the model truly has a non-zero intercept, then the Big Bang Theory could not be correct. It turns out that Hubble's original data, which gave rise to the Big Bang Theory, does not actually support the Big Bang Theory. By such steps and missteps, science progresses. Reference: Ramsey, Fred L. and Daniel W. Schafer, The Statistical Sleuth, Duxbury Press, 1997.

26

Items of Interest Wolfram Research just announced Mathematica version 6 ­ a significant upgrade. They have completely rewritten graphics, adding sliders and lots of other good stuff. Even if you do not have or don't plan to have Mathematica at your school, they have also created a Demonstrations Project which currently has over 1000 ready-to-go demos on just about everything. (There is a search engine, too.) There is a free Mathematica Player program to download, and then you may view any of the demos you wish ­ whether or not you own a copy of Mathematica. http://demonstrations.wolfram.com/ There is a wide variety of demos for math, science, and other fields, including many for calculus and precalculus. See www.wolfram.com for more info. Ruth Dover, IL Math & Science Academy Calculus: The Musical Some of you may have heard of it, but I thought I would send a link to the website. It has some great songs that you can download, and the lyrics are on the site too. The website is www.calculusthemusical.com Meghan Maher, Notre Dame Academy, Park Hills, Kentucky New textbooks for AP Advanced Placement Calculus Textbook It is suited primarily for teaching the AB course. Here is the webpage. We have done extremely well and the students are really pleased. For example, this year, of 26 students who wrote this year, 20 received 5's and 6 scored 4's. Now they are good students, to be sure. But they also say that the explanations are clear and concise. The problems mirror those on the AB exam in type and in level of difficulty. However, we have also added a lot of breadth in areas not emphasized by the AP (e.g. related rates, optimization problems). Feel free to ask for a sample chapter on any topic. I can send it to you electronically. www.trafford.com/05-0775 Jack Koenka, Havergal College, Toronto, ON, Canada Calculus, by John Rogowski There is a new textbook that was just published by W.H. Freeman: Calculus, by Jon Rogawski. This textbook presents calculus with solid mathematical precision but with an everyday sensibility, that puts the main concepts in clear terms. It is rigorous without being inaccessible and clear without being too informal - it has the perfect balance for instructors and their students. Throughout the text, student difficulties are anticipated and addressed. There is a wide range of applications provided throughout. Caution notes warn students of common pitfalls they can encounter in understanding the material. Reminders to previously learned materials are provided. This is a student-friendly text. Conceptual Insights encourage the student to develop a conceptual understanding of calculus by explaining important ideas clearly but informally. Graphical Insights enhance the students' visual understanding by making the crucial connection between graphical properties and the underlying concept. The text also includes summaries, and reminder and caution notes. There will also be an AP toolkit available in the Fall that was written by long time AP Calculus High School teacher Lin McMullin. Check out the companion website at http://www.whfreeman.com/rogawskipreview/ There are 4 different versions available: 1. Combined Early Transcendental 2. Combined Late Transcendental 3. Single Variable Early Transcendental 4. Single Variable Late Transcendental. Danielle Lewandowski, High School Marketing Assistant, Bedford/St. Martins, W.H. Freeman, & Worth Publishers, Toll Free: 866-843-3715 X1753, [email protected] There is a second printing of the text being made available this summer. For this printing, there will be corrections, as well as a specific high school version. This means that the binding will be stronger and it will have the high school stamp imprint on the front cover. Testimonial After a lot of discussion, we have decided to adopt Jon Rogawski's new calculus textbook, Calculus of a Single Variable Early Transcendentals that will be available this summer. We had to adopt a new book this year and after teaching out of several books and reviewing several more, we chose this one. The clincher for us was the student's response to this book. They really liked how it was organized, how the problems were set up in each section and the general tone of the book. They felt that it was a good book for the AP Calculus course. Theresa Reardon Offerman, Mathematics Department, Upper School Technology Coordinator, Mounds Park Academy

27

Movie: The Great Debate: e or pi This DVD is not yet listed on MAA's online catalog. To get it call 1-800-331-1622. The catalog code is PIE ISBN:978-0-88385-900-1. The price for non-members is $24.95. The price for members is $19.95. The MAA has released the DVD recording of the Great Pi/e Debate filmed at Williams College. This debate between Colin Adams and Tom Garrity settles once and for all the burning question "Which is the better number, e or pi?" Ed Burger "moderates" this debate. Some have commented on the absolutely ridiculous nature of the arguments. Others have remarked that this 40-minute video is hilarious. www.maa.org Movie: Flatland Contact [email protected] for information about ordering this one. It is not yet released but will be delivered in the fall and can be pre-ordered. It is based on the book. Cost is $34.95, which includes shipping. eBay item including several useful programs for AB and BC Calculus written for several different TI calculators: http://cgi.ebay.com/TI-Programs-I-wrote-for-my-Students-in-Calculus-III_W0QQitemZ9722773515QQcategoryZ41391QQcmdZViewItem Jorge Garcia, mailto:[email protected], CALCPAGE: The Calculus and Computer Science Archive http://calcpage.tripod.com This site contains poems and songs (performances) about Mathematics. You might want to start with the Twelve Days of Factoring. The URL is http://www.mscc.cc.tn.us/webs/vyoung/songs/Main_Pages/Tables.htm, Lin McMullin Last year I created a song so my students will NEVER have a problem between the two averages. It helped them OWN the Mean Value Theorem - It is their favorite song by far - to the tune of "Those Were the Days My Friend". Jan Komancheck, Escondido High School, Escondido, California Mean Value Theorem (to the tune of "Those Were the Days") To find the slope of lines in Math-e-matics We often use the phrase rise over run. Divide the change in y over the x's Care-fully, remember how it's done? CHORUS: If f 's continuous and differenti'ble On a closed in-terval from a to b. Slope of the secant line equals f prime sometime, This the Mean Val-ue Theorem guarantees. Cal-culate the slope, first, of the se-cant Also called the average rate of change. Set f prime at c equal to this number, Then find all the values c can be. Sing CHORUS When your points have got the same y value Zero is the slope you will agree. Set f prime at c equal to that zero Max or min's are possibilities. Sing CHORUS

28

When you need to find the average value When integrating f from a to b. You take the integral and just di-vide it By the length of b minus the a. Sing CHORUS Teaching Derivative of Inverse Functions: 1 - briefly review relation between graphs of a function and its inverse: http://mph.net:8080/academic/math/APCalcAB/Notes/ch3/inverfun.htm 2 - Compare the graphs of the lines tangent to the function and its inverse: http://mph.net:8080/academic/math/APCalcAB/Notes/ch3/Inverse%20Function%20Theorem.htm This page links to a Geometer's Sketchpad demo that the tangents are also reflections through y=x, giving evidence that dx/dy = 1/(dy/dx). http://www.mph.net/coelsner/calcapps/invfunthm_b.htm (Move the red x back and forth along the axis.) So the inverse function theorem looks like a 6th grade property of fractions. Wonder of Leibniz's symbolism! My students have then been ready to appreciate a proof. Chuck Oelsner Projects idea: Try this link. http://www.math.duke.edu/education/webfeatsII/gdrive/Team%20A/OurPage.htm Then have the students do their own object. Ted Gott Calculus site: http://www.gummy-stuff.org/Calculus/Essays-on-Calculus.htm Peter Ponzo

The College Board Information The "Acorn Book" is the Course Description. Available free at AP Central. It is the first item on this page http://apcentral.collegeboard.com/apc/public/courses/teachers_corner/2178.html There is a new 2008 one. The new AP Calculus Teacher's Guide is now available at AP Central. Perhaps its greatest feature is that you can download your own copy free from the Calculus AB or BC home page. The Teacher's Guide contains a wealth of information that both new and experienced Calculus teachers should find useful, including: sample syllabi (that cleared the audit); excerpts from the course description; suggestions for covering the major content areas; tips from experienced teachers; journal questions, and more. Mark Howell, Gonzaga High School, Washington, DC As your school prepares to participate in the AP Course Audit, please encourage your local AP colleagues to visit the newest resources online for teachers and administrators, such as: The AP Course Audit Manual District-designed AP Course Audit Activities A Syllabus Self-Evaluation Checklist Sample Evidence Tables for Syllabus Creation Sample Annotated Syllabi These resources and more can be found at: http://apcentral.collegeboard.com/apc/public/courses/teachers_corner/150356.html The College Board will be hosting free Online Presentations on the AP Course Audit Process on November 30, and December 14, 2006. Pre-registration is required. Visit: http://survey.collegeboard.com/ViewsFlash/servlet/viewsflash?cmd=showform&pollid=AP!PreRegCourse. For complete information about the AP Course Audit, please visit http://apcentral.collegeboard.com/apc/public/courses/teachers_corner/46361.html. Questions may be directed to: http://survey.collegeboard.com/ViewsFlash/servlet/viewsflash?cmd=showform&pollid=Audit_Survey!audi tSurvey John F. Mahoney, Benjamin Banneker Academic HS

29

2007 AB EXAMINATION FREE-RESPONSE QUESTION FREQUENCY Prepared by Trish Morris FUNCTIONS '07 Zeros Asymptotes Symmetry Domain Odd/Even Range Inverse Limits Linear Equation Continuity X X X X X X X X `06 `05 `04 `03 '02 '01 `00 '99 '98

DIFFERENTIAL CALCULUS `07 Analyze a Function Given as a Table of Values Tangent Line Equation Differentiation & Evaluation Increasing & Decreasing Functions Critical Numbers, Maximum & Minimum Points ­ Relative & Absolute Concavity Inflection Points Average Rate of Change X X X X X X X `06 X `05 X `04 `03 X '02 '01 '00 '99 X '98 X

X

X

X X X

X X X X

X X X X

X

X X

X

X

X

X

X X X

X X X X

X

X

X X X X

30

DIFFERENTIAL CALCULUS `07 Critical Numbers, Maximum & Minimum Points, Increasing & Decreasing, Concavity, Inflection Points from the graph of f !(x) Critical Numbers, Maximum & Minimum Points, Increasing & Decreasing, Concavity, Inflection Points from a table of values of f(x), f !(x), & f !!(x) Curve sketching or analyzing data using information from a table of values of f(x), f !(x), & f !!(x) or from the graph of f !(x) Mean Value Theorem for Derivatives Implicit Differentiation Linear Approximation Related Rates X X X X X X `06 X `05 `04 X `03 X '02 X '01 '00 X '99 '98

X

X

X

X

X

X

X X X X

X

X X X

INTEGRAL CALCULUS `07 Area and/or Interpretation Riemann Sums using Left, Right, Midpoint Evaluation Points Properties of Integrals Trapezoidal Rule/Approximation Fundamental Theorem of Calculus Mean (Average) Value Volumes of Solids: Disks and Washers Volumes with known Cross Sections X X X X X X X X X X X X X X X X X X X X `06 X X `05 X `04 X `03 X X `02 X `01 X `00 X `99 X X `98 X X

X

X

X X X X X X X

X

X X X X X

31

INTEGRAL CALCULUS `07 Solving Differential Equations: Separation of Variables Drawing Slope Field from Differential Equation Rectilinear Motion: Equation(s) for Position, Velocity, & Acceleration; Direction of Motion; Total Distance Rectilinear Motion: Position, Velocity, & Acceleration; Total Distance from the Graph of Velocity Rectilinear Motion Analysis from a Graph Definite Integral as an Accumulator Accumulation of the Derivative with Initial Condition X X X X X X X X X X X X `06 X `05 X `04 `03 X `02 `01 X `00 X `99 X `98 X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

CALCULATOR `07 Draw a Graph in a Given Window Find the Zeros of a Function Find the Intersection Points of Two Graphs Evaluate a Definite Integral Evaluate a Derivative X X X X X X `06 `05 X `04 `03 '02 X X X X '01 '00 '99 '98 X

X

X X

X X

X X

X X X

X

X X

X

32

2007 BC EXAMINATION FREE-RESPONSE QUESTION FREQUENCY FUNCTIONS `07 Domain & Range Odd/Even Intercepts LIMITS & CONTINUITY `07 Finite Limits Limits at Infinity Infinite Limits Definition of Continuity DIFFERENTIAL CALCULUS `07 Definition of the Derivative Analyze a Function Given as a Table of Values Tangent/Normal Line Equation Derivative of Composition Functions (Chain Rule) Increasing & Decreasing Functions Critical Numbers Concavity Inflection Points Average Rate of Change Extreme Values X X X X X X X X X X X X X X X X X X X X X X X `06 `05 `04 `03 '02 '01 '00 '99 '98 X X `06 `05 `04 `03 `02 `01 X X X `00 `99 `98 `06 `05 `04 `03 '02 '01 `00 X '99 '98 X

X

X

X X

X

X X

X X X X

X X X X X X

X X

X

33

DIFFERENTIAL CALCULUS `07 Higher Order Derivatives Optimization Problems: Max/Min Relative and Absolute Curve Sketching using Information from a Table of Values of f ( x), f !( x), & f !!( x) or from the Graph of f !( x) Mean Value Theorem for Derivatives Implicit Differentiation Linear Approximation Related Rates INTEGRAL CALCULUS `07 Area and/or Interpretation Riemann Sums using Left, Right, & Midpoint Evaluation Points Properties of Integrals Trapezoidal Rule/Approximation Fundamental Theorem of Calculus Mean (Average) Value Volumes of Solids: Disks and Washers Volumes with known Cross Sections Rectilinear Motion: Equations for Position, Velocity, & Acceleration; Direction of Motion; Total Distance X X X X X X X X X X X X X X X X X X X X X X X `06 X X `05 X `04 X `03 X `02 X `01 `00 X `99 X X `98 X X X X `06 X `05 `04 `03 '02 '01 '00 '99 '98

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X X

X

X X X

X X X

34

INTEGRAL CALCULUS `07 Rectilinear Motion: Equations for Position, Velocity, & Acceleration; Direction of Motion; Total Distance Rectilinear Motion Analysis from a Graph Definite Integral as an Accumulator Arc Length, Surface Area, and Work X X `06 `05 X `04 `03 `02 `01 `00 X `99 `98

X

X

X

X

X

X

X

X

X

X

CALCULATOR `07 Draw a Graph in a Given Window Find the Intersection Points of Two Graphs Evaluate a Definite Integral Evaluate a Derivative METHODS of INTEGRATION `07 Integration by Parts Integration by Partial Fractions Improper Integrals DIFFERENTIAL EQUATIONS `07 Solving Differential Equations by Separation of Variables Logistic Differential Equations Slope Fields Euler's Method X X X `06 X `05 `04 `03 X `02 `01 X `00 X `99 `98 X X X X `06 `05 `04 `03 `02 `01 `00 `99 `98 X X X X X X `06 `05 `04 `03 '02 '01 '00 '99 `98 X

X

X X

X

X X X

X

X X

X

X X X X X X X X

35

SEQUENCES and SERIES `07 Tests for Convergence Geometric Series Alternating Series & Error Approximation p-Series Manipulation of a Power Series Power Series: Radius of Convergence Power Series: Interval of Convergence Maclaurin Series Taylor Series Lagrange Error Bound PARAMETRIC, POLAR, and VECTOR FUNCTIONS `07 Parametrically Defined Curves Derivatives of Vector & Parametrically Defined Curves Tangent Lines to Parametrically Defined Curves Arc Length Including Parametrically Defined Curves Velocity, & Acceleration Vectors for Motion on a Curve Polar Coordinate Graphs Area of Polar Curves X X X X X X X X X `06 `05 `04 X `03 X `02 X `01 X `00 `99 X `98 X X X X X X X X X X X X X X X X X X X `06 `05 X X X `04 `03 `02 `01 `00 `99 `98

X

X

X

X X X X X

X X

X

X

X

X

X

X

X

X

X

X

X

X X

X X

36

NCA2PMT 2007 REPORT FROM THE TREASURER/MEMBERSHIP CHAIR Balance as of 6/24/06 3782.92 Deposits (memberships) 1543.00 Deposits (newsletter grants) 355.64 ---------------------------------------------------------------TOTAL $5681.56 Newsletter (August 2006) 738.54 Newsletter (February 2007) 515.48 Balance as of 6/23/07 427.54 ---------------------------------------------------------------TOTAL $5681.56 TOTAL INCOME 1898.64 TOTAL EXPENSE 1254.02 ---------------------------------------------------------------NET GAIN 2006-2007 $644.62 The above gain for 2006-2007 represents our continuing recovery from an operating loss of $622.67 for 2002-2003. The following reasons can be cited: 1. Numerous members pay more than one year ahead of their expiration date. 2. We received a grant for each newsletter through Dr. David Royster from the UNC Charlotte Center for Mathematics, Science and Technology Education (CMSTE). These grants defrayed approximately one-fourth of the cost of each newsletter, a combined $355.64. Dr. Royster has indicated to me that this grant can be ongoing, as long as the CMSTE has funding available. 3. We continue to hold down the cost of printing and mailing our semi-annual newsletter. Current membership is 341, including 68 whose membership expired 3/2007 who will receive a reminder but will not get the summer newsletter unless they renew. This overall number is down from 385 last year at this time, but the number of paid up members is currently 273, as opposed to 280 last year. We continue to send 8 newsletters (professional courtesy exchange) to the officers of SCA2PMT. There are 126 members from NC with 39 Eastern, 47 Central and 40 Western. There are 214 members in other states and one member from Japan. Thirty-seven states and the District of Columbia are represented in the membership. The following states have no members: Alaska, Arkansas, Illinois, Louisiana, Maine, Mississippi, Montana, New Hampshire, North Dakota, Rhode Island, South Dakota, West Virginia, and Wyoming.

Correction to our Sudoku Puzzle in Winter 2007 Newsletter Stephen Davis ­ Davidson College, Davidson, NC

Corrections were needed to the Calculus Sudoku #1 that was printed in the Winter 2007 Newsletter. They are listed below. You can also go to our website and Stephen has corrected it in the electronic version of the newsletter. We apologize for any inconvenience this may have caused. It was a learning experience. · · · · · clue #1: is ambiguous (any nonzero value is correct, but the Sudoku value for this cell is forced anyway and matches the "natural" choice for g(x)) clue #2: "x" in the numerator should be replaced by "2" clue #8: use f(x) = 8 (sin x)6 ­ (cos x)2 [i.e., drop the factor of cos x in the first term] clue #9: only makes sense if the piecewise function is split at ­2; e.g. f(x) = ax + b if x ­2; f(x) = ­4x if x < ­2. clue #10: replace "3 ­ 4x" by "3 ­ x" in the denominator.

37

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