Anita E. Solow
Randolph-Macon Women's College

Editor's Note: Anita Solow is the past chair of the AP Test Development Committee for calculus. She graciously agreed to write this article for the NCAAPMT newsletter.

When I was asked to be a member of the AP Test Development Committee, I had no idea what I was getting into. I spent six years on the committee writing and working more calculus problems than I could have ever imagined. As part of the preparation for each of the committee meetings, the members are given assignments to create problems. In this article, I will attempt to explain how a problem goes from the assignment stage to appearing on an actual AP examination.

A typical assignment is to write a collection of either multiple-choice or free-response questions. Usually, the topic areas (based on the course description) are assigned because the pool of questions will be lacking problems for some topics. When we write a multiple-choice problem, we are also required to create the wrong answers and to consider whether the problem is designed to be done with or without a graphics calculator. Once the problems are submitted, the entire committee reviews them. If any two members of the committee dislike the problem (for any reason), it is discarded from the pool. I would estimate that approximately one-third of the problems are killed at this stage. The rest of the problems are discussed and extensively rewritten (or discarded) by the committee.

The committee tries to be very careful that the wording is mathematically accurate, that there is only one correct answer, and that the problem is not ambiguous in any way. The committee has made a concerted effort in the past few years to write problems that reflect the new course description. For example, there is an increased emphasis on functions defined graphically or numerically and on various applications of the integral.

I must emphasize that by the time a problem appears on the exam, it rarely resembles its original form. In fact, by the time it appears, we often no longer know who wrote it, for, in reality, it has been rewritten so many times that it may resemble the original version in only the germ of the idea. For example, the original version of AB-21B0-2 from the 1998 exam was an examination of the family of functions y = bxe^bx. There was no preliminary analysis of the case of b = 2. And the original version of AB-3 asked questions about the motion of a car where the only data given was the table of velocity values.

One characteristic of multiple-choice problems is that we do not see the student's work; we only see the answer. This means that we have to be very careful that the problem the student does actually tests the idea that we are trying to test. A simple example of this is the following. Suppose we wish to test the definition of the derivative by asking the student to evaluate
  for a particular function f(x) for x = a. The answer, of course, is the value f '(a) . But if we ask this question when a student has a calculator in hand, the student can arrive at the correct numerical answer by exploring the value of the limit by plugging in small values of h or by graphing the quotient. In any case, the student is not using any knowledge of the definition of the derivative, the idea that we were trying to test. Of course, even if we place this problem on Part A, the student can get the correct answer by using L'Hopital's rule, again circumventing the definition of the derivative.

One impact of technology is that we can no longer ask the student some questions on Part B of the exam. For example, we cannot ask the student to use calculus to find the maximum value of a function on Part B. Since we cannot control how the student is arriving at his/her answer-using critical points, using the max function on the calculator, or graphing and zooming-we would not know what exactly we have tested. We can, of course, ask the question on Part A or on the free-response section of the exam. We could also ask for the critical point of a function on Part B, because in order to do the problem, the student must use calculus to translate critical point to either some property of a graph or to the derivative of the function.

To be honest, the committee has found it challenging to deal with the use of the calculator in the testing environment. On the free-response section, the student is required to show all work, thereby making it a bit easier to deal with calculator use. So, for example, if the student is asked to find the maximum value of a function on the free-response section, it is not satisfactory to use the max button or to graph and zoom.

The calculators with symbolic capability pose additional problems for the free response portion of the exam. One of the major concerns of the committee is that there be a level playing field for all students using any accepted calculator for the exam. Although the committee made the decision to allow symbolic calculators on the test, it is dedicated to maintaining the equity of the examination. Last year, when we became aware of the new symbolic calculators, we rewrote several questions on the 1999 exam to neutralize the effect of the calculators. Unfortunately, since that exam has not been given yet, I cannot give specific details about how we changed the problems; the folks at ETS are sort of fussy about things like this.

Starting with the May 2000 examinations, the free-response section will be split into two parts, one with a calculator and one without. This split is being done both to maintain a level playing field for the exam and to enable the committee to ask questions, e.g., the solution of separable differential equations, that we all agree are important to ask on a calculus test but which cannot be asked if the student uses a calculator. One of the major challenges for the committee is to respond to the changes in technology so that the examinations remain fair and accurate tests of calculus. And the technology does keeps changing.

THERE ARE NO CHANGES IN THE 1999 EXAMINATION FORMAT. The 1999 format will be the same as the 1998 format.

Beginning with the May 2000 Calculus AB and Calculus BC Examinations, Section II will consist of two parts. There will be no change to the format for Section I.

Part A of Section II will consist of three(3) free-response questions and will have a time limit of forty-five minutes. It will contain some questions or parts of questions for which a graphing calculator is required.

Part B of Section II will consist of three(3) free-response questions and will have a time limit of forty-five minutes. The use of a calculator will not be permitted to solve these problems. During the timed portion for Part B, students will be permitted to continue to work on questions in Part A, but they will not be permitted to use a calculator during this time.

The change in the format is an effort to respond to heightened concerns with equity as more students may use graphing calculators with computer algebra system(CAS) features. The AP Calculus Examinations are designed to accurately assess student mastery of both concepts and techniques of calculus. The two-part format for the free-response section provides greater flexibility in the types of questions that can be asked while ensuring fairness to all students taking the examination, regardless of the graphing calculator used.

Specific details of the new format will be available in the spring of this year when the Acorn Book "May 2000, May 2001 Course Description" becomes available. Details can also be found in the "What's New"? section at the following Internet address: http://www.collegeboard.org/ap/calculus/. The distribution of the grades for the 1998 examination candidates and the members of the current Test Development Committee are also listed at this address.

GENERAL INSTRUCTIONS FOR SECTION II OF THE AP EXAMINATIONS

It is helpful for students to be familiar with the instructions for the free-response section of the advanced placement examination before going in to take the examination. The general instructions for section II of the 1998 examinations are as follows.

You may wish to look over the problems before starting to work on them, since it is not expected that everyone will be able to complete all parts of all problems. AU problems are given equal weight, but the parts of a particular problem are not necessarily given equal weight. The problems are printed in the booklet and in the green insert; it may be easier for you to first look over all problems in the insert. When you are told to begin, open your booklet, carefully tear out the green insert, and start to work.

A GRAPHING CALCULATOR IS REQUIRED FOR SOME PROBLEMS OR PARTS OF PROBLEMS ON THIS SECTION OF THE EXAMINATION.

• You should write all work for each part of each problem in the space provided for that part in the booklet. Be sure to write clearly and legibly. If you make an error, you may save time by crossing it out rather than trying to erase it. Erased or crossed-out work will not be graded.

• Show all your work. You will be graded on the correctness and completeness of your methods as well as the accuracy of your final answers. Correct answers without supporting work may not receive credit.

• Justifications require that you give mathematical (non-calculator) reasons and that you clearly identify functions, graphs, tables, or other objects you use.

• You are permitted to use your calculator to solve an equation, find the derivative of a function at a point, or calculate the value of a definite integral. However, you must clearly indicate the setup of your problem, namely the equation, function, or integral you are using. If you use other built-in features or programs, you must show the mathematical steps necessary to produce your results.

• Your work must be expressed in standard mathematical notation rather than calculator syntax. For example, may not be written as fnInt(x², x, 1, 5).

• Unless otherwise specified, answers (numeric or algebraic) need not be simplified. If your answer is given as a decimal approximation, it should be correct to three places after the decimal point.

• Unless otherwise specified, the domain of a functions assumed to be the set of all real numbers x for which f(x) is a real number.

A MAXIMUM PROBLEM? A MINIMUM PROBLEM?
Earl Mitchelle

The figure shown below consists of the rectangle ABCD, the isosceles triangle with base AB and a height six times the base, the square with side DC, and the isosceles right triangle with hypotenuse AD.

If the area of the rectangle ABCD remains constant, determine the maximum or minimum possible area of the entire figure relative to the area of the rectangle, Identify whether the area found is the maximum area or minimum area and justify the choice. 

Solution

When which means the area found is a minimum. The minimum area of the figure is three times the area of the rectangle ABCD.

NOTES FROM THE 1998 ANNUAL MEMBERSHIP MEETING
Carolyn Walmaley, NCAAPMT President

The annual meeting of NCAAPMT was held in Greensboro October 30, 1998, during the NCCTM Conference. Thanks to over forty-five members in attendance and the distinguished presenters for making this a successful event.

Two special people were recognized for their contributions to NCAAPMT. A charter board member, Bernice Kenan of Greensboro, a master teacher of AP Calculus at Ragsdale High School, is retiring from the Board of Directors. Many thanks to Bernice for her tireless, insightful and always cheerful support to me and our organization. In 1992, Charles Bodine, a teacher at Charlotte Country Day School, took a little seed money from an organization in Massachusetts and a lot of enthusiasm and energy to form NCAAPMT. This greatly needed network for advanced placements mathematics teachers has grown to over 300 members from all over the country. As he retires from the Board of Directors, we extend sincere gratitude to Charlie and share in his pleasure of how his vision has developed.

Ben Klein, Stephen Davis, Brenda McSwain and Jeff Lucia, AP readers, reviewed grading standards for several free response questions from the 1998 AB and BC examinations. Debbie Britt, AP reader and ETS consultant, discussed some of the trends and issues related to the constantly changing examinations. The credentials of these presenters are impressive. More impressive is their willingness to give their time to make our meeting so valuable.

SCAAPMT and NCAAPMT plan another joint meeting at the Carolina Conference to be held in Greensboro in the fall of 1999.

Finally, I would like to personally thank Jane Barnett, Jeff Lucia and Earl Mitchelle. Jane, Past President, has given NCAAPMT great leadership and me a lot of support. Jeff is our Treasurer and coordinator of our constantly changing and increasing membership. Earl, Secretary and editor of the newsletter, works all year to make our publication a quality product.

THINGS I HAVE LEARNED AT THE AP READING
Dan Kennedy, The Baylor School

Editor's Note: The following is a reprint of a talk given by Dan Kennedy at the grading of the 1998 AP examinations at Clemson University. Being an AP reader is an extremely valuable experience for those who are selected to participate in the grading process. Readers learn information which increases their understanding of calculus and the AP examinations. They also make new professional contacts which expand their support systems. In this talk, Mr. Kennedy talks about the things that he has learned during his participation in the AP grading process.

It almost goes without saying, at least in this group, that the benefits of the Advanced Placement reading are many and varied. There is, for example, the practical benefit of getting 145,000 exams graded in a smooth and professional manner. For us, there are also the social benefits of seeing old friends and making new acquaintances with teaching colleagues from across the country. There are the nutritional benefits of storing up the valuable reserves of fat and cholesterol that your body will need to make it through the long, hot summer. But one of the most important professional benefits for us is the opportunity for calculus teachers to learn more about calculus, and that is the aspect of the reading that I would like to talk about this evening.

Only the most naive of God's creatures - for example, our students - would assume that their teachers know everything about the subjects that they teach. We who teach calculus have learned to be particularly humble in that respect, encountering on a fairly regular basis questions that at least cause us to tell our students, "Say, that's a good question, but let's not take class time on that right now; I'll explain it to you tomorrow." This buys us enough time, hopefully, to solve the problem in the privacy of our own rooms, where we can consult the solution manual.

We have all learned a lot of calculus that way. But those of us who have read AP Calculus over the years have also learned that the simplest of problems can have remarkable subtleties buried just beneath the surface. Such subtleties might go undetected under normal circumstances, but under the abnormal circumstances of 100,000 panicky students thrashing around with unfamiliar mathematical tools, they are soon exposed and demand our scrutiny. It is in these unexpected moments that I have learned some fascinating calculus facts over the years, and I would like to share a few of them this evening with you. If all goes well, I am hoping that at least a few of these examples will give you the same thrill of discovery that they gave me.

RADICAL LIES THAT WE TELL STUDENTS IN ALGEBRA

AB1 in 1988 was a pretty harmless looking problem, but imagine your best algebra student looking at the radical and thinking, "Uh-oh. Better simplify  that before I go on." We have almost programmed our students to think that way. So, the student writes: (Notice the absolute value. I said that we were imagining your best algebra student.)

Now, how does the student answer part (a)? If she uses the simplified expression, she sets  x² - 16 > 0 and arrives at a domain of , losing the domain point {0} (and, of course, a point from her AB1 score). If she uses the unsimplified expression, she sets x⁴ - 16x² > 0 and arrives at the correct domain of . (Actually, many AB students that year divided by x² and arrived at the wrong domain anyway, but your best algebra student would surely not have done that.)

There are many examples of expressions that pick up domain values when simplified, such as but this problem provides a rare example of an expression that loses a domain value when simplified. The dozens of AB students who lost that point in 1988 were probably not as charmed by this subtlety as we were, but it made for some nice conversation among teachers that year. We all resolved to be a little more careful to think about equations like as identities. Identities have domains of validity, and this one is invalid at x = 0.

PROVING DIFFERENTIABILITY AT A POINT

There have been several "split-function" questions over the years, such as BC4 in 1992.

TO solve part (a), the typical good student will paste together the two sides of the function at x = 1 to make it continuous:

2(1) - 1² = 1² + k(1) + p k + p = 0

then paste together the derivatives of the two sides at x = 1 to make it differentiable:

2 - 2(1) = 2(1) + k k = -2

The two conditions taken together imply that k = -2 and p = 2.

This is all well and good, but it leads many students (and once led me) to conclude that this is how one shows that a function is differentiable at a point: establish continuity and establish that the derivative is the same coming in from the left as it is coming in from the right. Luckily for everyone doing 1992 BC4 above, that approach will successfully establish differentiability if it works. But it might now work, as was shown in 1982 on BC7.

We solve the problem as follows:

Therefore, f' is not continuous at x = 0.

The implication of this problem is subtle, but profound. Notice that f'(x) does exist at x = 0, but that one could not discover that by looking at f'(x) to the left and right of x = 0.

This same problem also taught me the true meaning on right and left hand derivatives. Notice that f is differentiable at 0, so both its right and left hand derivatives exist there.  They are the two limits , both of which equal 0. They are not the same as , both of which fail to exist.

Students could have made perfect scores on these problems and never realized half of what I just said, but because I graded their papers at the reading those same problems afforded me the chance to learn those nuggets of calculus from my colleagues.

I later learned other interesting things about this particular split function, which is shown in the graph on the top below. As we have just seen, this function is both continuous and differentiable at x = 0. The graph of is shown on the bottom. This function is continuous but not differentiable at x = 0. You can almost see why, too: You could "zoom in" endlessly on the bottom graph at the origin and the graph would get no straighter than it is now; whereas if you were to zoom in on the top graph the curve would get flatter and flatter at the origin. If you find that your eye can not quite distinguish why both fail to exist in the top graph, welcome to the club. This is why there will always be the need for algebraic limits in first-year calculus!

THE FIRST DERIVATIVE TEST FOR POINTS OF INFLECTION

There have been so many questions on AP exams over the years asking students to justify points of inflection that it is impossible to say when the issue first arose, so I will put it in the context of the following hypothetical question:

The expected response is because there is a turning point of the graph of y=f'(x) at x=1 (or some equivalent statement about f" changing sign), but what would you do with a student who says this:

There is a point of inflection at x=1 because f'(1) = 0 while f'(x) is positive on either side of x = 1.

This student has noted the "shelf point" at x = 1 and is arguing that it must be a point of inflection, as in the familiar graph of y = x³. There is no mention of the sign of the second derivative; indeed, the whole argument hinges on the sign of the first derivative. Does this student get the justification point?

Since this hypothetical response actually occurred one year, the table leaders had to decide whether or not this was a valid theorem. The smart money being on "no" from the outset, most of the attention was focused on finding a counterexample. My notes do not record who came up with one first, but later conversations have variously attributed it to Robert Ellis, Tom Tucker, or Bruce Peterson. (Bruce, chair of the Committee in those days, actually published a paper on the topic.) This is the function that I found in my notes from the reading:

We can show that f'(0)=0 and that f'(x) is positive on both sides of x=0, and yet the graph of f does not have a point of inflection at the origin. The salient feature of the function is that its graph changes concavity infinitely often on every neighborhood of zero. Here is a sketch of the proof:

1. (Proof that f'(0)=0):

2. (Proof that f'(x) > 0 on some neighborhood of 0):

So for small x on either side of 0, f'(x) has the same sign as 3x², namely positive. 

3. (Proof that f" changes sign infinitely often on any neighborhood of x = 0):

So for small x on either side of 0, f'(x) has the same sign as 18x infinitely often and the same sign as -6x infinitely often. There can be no well-defined change in concavity at x = 0, and so the graph does not have a point of inflection there.

Incidentally, as you might have suspected by now, the student's argument is actually valid if it is known that the function only has a finite number of inflection points near the critical point in question. 

WHEN IS A MACLAURIN SERIES NOT A MACLAURIN SERIES?

BC2 in 1996 was about Maclaurin series. I had no problem recognizing this as a "series manipulation" problem when I took the exam in the privacy of my own room. The series in part (c) was clearly the series for e^x-1 and so my answer in part (d) was:  

When I arrived at the reading I discovered that I had only scored 8 out of 9 on BC2. I am not sure whether the Committee had intended to catch me on a technicality, but I had certainly taken their bait, including the hook, line, and sinker. It was somewhat consoling to note, after all the exams had been graded, that I was not the only sucker in the Committee's boat: Flopping around beside me were about 99% of the BC population. The problem, you see, is that is not even defined at x = 0, let alone infinitely differentiable, and is therefore not eligible to have a Maclaurin series. The correct f should be:  

For additional perspective let us hearken back to 1993 BC5. Notice that in part (b) the clever avoidance of the names "Taylor" and "Maclaurin." This allows the student to construct a power series of the form a_o + a₁x + a₂x²+ ^... +a_nxⁿ+^... without being concerned about whether or not g(0), g'(0), g"(0), etc., exist -- which they do not. Since learning my lesson in 1996 I have enjoyed looking through various textbooks to see how many include exercises in their "series manipulation" sections that look like this: Use the Maclaurin series for cos x to construct a Maclaurin series for [(cos x - 1)/x] Each one I find means that there is one more sucker in the boat. 

THE DIFFERENTIAL EQUATION WITH TOO MANY SOLUTIONS.

Not only was I chair of the Committee when 1993 BC6 appeared, I actually wrote the thing. It then went through the usual rigorous scrutiny and polishing, where in quite a few competent people solved it multiple times. All this quality control notwithstanding, it was only as the reading itself that someone (the Chief Reader) discovered that there was a lot more involved in pinning f down than anyone had theretofore realized (including the author of the problem).

The solution to the differential equation is pretty straightforward: You separate variables and find that f(x)=tan(x+c). One is then supposed to use the initial condition from part (a) to conclude than tan(c)=0, from which it follows that f(x)=tan(x + kp) = tan x. It looks simple enough, but it is not that simple. Can you see why?

Here is the rub: The initial condition is fine for "pinning down" the function at x=0, but the function we are talking about is the tangent function, which has asymptotes. So we can narrow our solution down to y=tan x on the interval (-p/2, p/2), but outside that interval we could jump to a different, parallel tan curve without affecting f(0) a bit! Fortunately for the integrity of the exam, condition (i) precludes this from happening. Still, nobody expected the BC students to catch this subtlety, let alone prove their way out of it, so the grading standard was written without mentioning it. Those of us who discussed the subtlety at the grading, however, gained some wonderful insights into differential equations. 

METHODS OF SOLUTION THAT SHOULD NOT WORK

Few of us will ever forget the "cola problem" on the 1996 exam.

There were numerous lessons we learned from this problem, including the influence of units of measurements on the constants C and k. The most unexpected lesson, however, came (as usual) from a student solution. Here is the way you expect students to find the average value in part (b):

A few students, however, proceeded quite differently. They found the "special c" value guaranteed by the Mean Value Theorem for derivatives (which I will call p here to avoid confusion with C), then plugged it in to the function S:

Notice that the answers are the same! This is no coincidence arising from the

numbers involved, either; that is why I used all those general constants. It is actually true, for this function, that the average value of the function occurs at the point found by the MVT for derivatives! Is this a general theorem?

Well of course not. Consider the graph of y=sin x on the interval [0, p], for example. The MVT value occurs where the derivative is zero, and the y-coordinate there is the maximum, not the average, value.

It is not even true for monotonic functions in general, but it is true for exponential functions -- so the students got full credit! And that wasn't the only surprise on the 1996 test. Check out AB2.

Here is a reasonable approach to solving part (b):

Again, I left all the letters in the solution so that I could contrast it with another method of solution that was actually used by some students. These creative individuals chose to find k by finding the x in [4,9] at which the average value of the function occurs. (This is the number whose existence is guaranteed by the Mean Value Theorem for Integrals.) Here's the work.

Needless to say, these students got the same answer. In fact, they also got the same credit, since we can see that their algorithm is valid for this function. But is this a general theorem?

Once again, the answer is, "Of course not." Indeed, the same picture of y = sin x on [0, p] that debunked the previous theorem will serve to debunk this one, for virtually the same reason: The area splits evenly where the function value is the maximum, not the average, value of f on the interval.

So, in these two problems did the students who used the methods that should not work actually know what they were doing? In all likelihood, no. Did they really deserve full credit? Yes. Why? Because the Chief Reader said so. But just in case that is not reason enough, notice that these students showed their work and demonstrated what was, like it or not, a valid algorithm for solving the given problem. This is quite a different story from those frequently-seen papers that have correct answers with little or no work. In those cases we take off credit precisely because we cannot tell whether the students are using a valid algorithm or not, We can take off credit for mathematics that is not there, but as Bernie Madison has often pointed out,' We can't take off credit for correct mathematics.

With that philosophical quote from a great modern philosopher ringing in our ears, I will end my brief historical tour of the lessons I have learned from AP Readings Past, For your continued happiness and professional development, I wish that each of you might enjoy similar learning experiences at every reading you attend. God willing, this talk just might be one of them.

Thanks very much for your attention, and good evening one and all.