Is this going to be on the exam?

confused-teacherRecently there was a discussion on the AP Calculus Community bulletin board regarding whether it was necessary or desirable to have students do curve sketching starting with the equation and ending with a graph with all the appropriate features – increasing/decreasing, concavity, extreme values etc., etc. – included. As this is kind of question that has not been asked on the AP Calculus exam, should the teacher have his students do problems like these?

The teacher correctly observed that while all the individual features of a graph are tested, students are rarely, if ever, expected to put it all together. He observed that making up such questions is difficult because getting “nice” numbers is difficult.

Replies ran from No, curve sketching should go the way of log and trig tables, to Yes, because it helps connect f. f ‘ and f ‘’, and to skip the messy ones and concentrate on the connections and why things work the way they do. Most people seemed to settle on that last idea; as I did. As for finding questions with “nice” numbers, look in other textbooks and steal borrow their examples.

But there is another consideration with this and other topics. Folks are always asking why such-and-such a topic is not tested on the AP Calculus exam and why not.

The AP Calculus program is not the arbiter of what students need to know about first-year calculus or what you may include in your course. That said, if you’re teaching an AP course you should do your best to have your students learn everything listed in the 2019 Course and Exam Description book and be aware of how those topics are tested – the style and format of the questions. This does not limit you in what else you may think important and want your students to know. You are free to include other topics as time permits.

Other considerations go into choosing items for the exams. A big consideration is writing questions that can be scored fairly.  Here are some thoughts on this by topic.

Curve Sketching

If a question consisted of just an equation and the directions that the student should draw a graph, how do you score it? How accurate does the graph need to be? Exactly what needs to be included?

An even bigger concern is what do you do if a student makes a small mistake, maybe just miscopies the equation? The problem may have become easier (say, an asymptote goes missing in the miscopied equation and if there is a point or two for dealing with asymptotes – what becomes of those points?) Is it fair to the student to lose points for something his small mistake made it unnecessary for him to consider? Or if the mistake makes the question so difficult it cannot be solved by hand, what happens then? Either way, the student knows what to do, yet cannot show that to the reader.

To overcome problems like these, the questions include several parts usually unrelated to each other, so that a mistake in one part does not make it impossible to earn any subsequent points. All the main ideas related to derivatives and graphing are tested somewhere on the exam, if not in the free-response section, then as a multiple-choice question.

(Where the parts are related, a wrong answer from one part, usually just a number, imported into the next part is considered correct for the second part and the reader then can determine if the student knows the concept and procedure for that part.)

Optimization

A big topic in derivative applications is optimization. Questions on optimization typically present a “real life” situation such as something must be built for the lowest cost or using the least material. The last question of this type was in 1982 (1982 AB 6, BC 3 same question). The question is 3.5 lines long and has no parts – just “find the cost of the least expensive tank.”

The problem here is the same as with curve sketching. The first thing the student must do is write the equation to be optimized. If the student does that incorrectly, there is no way to survive, and no way to grade the problem. While it is fair to not to award points for not writing the correct equation, it is not fair to deduct other points that the student could earn had he written the correct equation.

The main tool for optimizing is to find the extreme value of the function; that is tested on every exam. So here is a topic that you certainly may include the full question in you course, but the concepts will be tested in other ways on the exam.

The epsilon-delta definition of limit

I think the reason that this topic is not tested is slightly different. If the function for which you are trying to “prove” the limit is linear, then \displaystyle \delta =\frac{\varepsilon }{\left| m \right|} where m is the slope of the line – there is nothing to do beside memorize the formula. If the function is not linear, then the algebraic gymnastics necessary are too complicated and differ greatly depending on the function. You would be testing whether the student knew the appropriate “trick.”

Furthermore, in a multiple-choice question, the distractor that gives the smallest value of must be correct (even if a larger value is also correct).

Moreover, finding the epsilon-delta relationship is not what’s important about the definition of limit. Understanding how the existence of such a relationship say “gets closer to” or “approaches” in symbols and guarantees that the limit exists is important.

Volumes using the Shell Method

I have no idea why this topic is not included. It was before 1998. The only reason I can think of is that the method is so unlike anything else in calculus (except radial density), that it was eliminated for that reason.

This is a topic that students should know about. Consider showing it too them when you are doing volumes or after the exam. Their college teachers may like them to know it.

Integration by Parts on the AB exam

Integration by Parts is considered a second semester topic. Since AB is considered a one-semester course, Integration by Parts is tested on the BC exam, but not the AB exam. Even on the BC exam it is no longer covered in much depth: two- or more step integrals, the tabular method, and reduction formulas are not tested.

This is a topic that you can include in AB if you have time or after the exam or expand upon in a BC class.

Newton’s Method, Work, and other applications of integrals and derivatives

There are a great number of applications of integrals and derivatives. Some that were included on the exams previously are no longer listed. And that’s the answer right there: in fairness, you must tell students (and teachers) what applications to include and what will be tested. It is not fair to wing in some new application and expect nearly half a million students to be able to handle it.

Also, remember when looking through older exams, especially those from before 1998, that some of the topics are not on the current course description and will not be tested on the exams.

Solution of differential equations by methods other than separation of variables

Differential equations are a huge and important area of calculus. The beginning courses, AB and BC, try to give students a brief introduction to differential equations. The idea, I think, is like a survey course in English Literature or World History: there is no time to dig deeply, but the is an attempt to show the main parts of the subject.


While the choices are somewhat arbitrary, the College Board regularly consults with college and university mathematics departments about what to include and not include. The relatively minor changes in the new course description are evidence of this continuing collaboration. Any changes are usually announced two years in advance. (The recent addition of density problems unannounced, notwithstanding.) So, find a balance for yourself. Cover (or better yet, uncover) the ideas and concepts in the course description and if there if a topic you particularly like or think will help your students’ understanding of the calculus, by all means include it.


PS: Please scroll down and read Verge Cornelius’ great comment below.


Happy Holiday to everyone. There is no post scheduled for next week; I will resume in the new year. As always, I like to hear from you. If you have anything calculus-wise you would like me to write about, please let me know and I’ll see what I can come up with. You may email me at lnmcmullin@aol.com


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Visualizing Solid Figures 5

To end this series of posts on visualizing solid figures, we will look at a problem that relates to how volumes of solid figures are formed. It has 5 parts which will be presented first. Then the solution will be given.

Consider the region in the first quadrant bounded by the graphs of the parabola y={{x}^{2}} and the line y=9 both for 0\le x\le 3.

half A

This region is revolved around the y-axis to form a solid figure.

half B

  1. Use the disk/washer method to find the volume of this figure.
  2. Use the method of cylindrical shells to find the volume of this figure.
  3. Use the disk/washer method to find a number j , such that when x = j the volume of the figure is one-half that of the original figure.
  4. Use the method of cylindrical shells to find a number k , such that when x = k the volume of the figure is one-half that of the original figure.
  5. The answers to parts a) and b) should be the same, but the answers to parts c) and d) are different. Explain why they are different.

 

Solutions:

1. The volume by the disk/washer method is

\displaystyle V=\int_{0}^{9}{\pi {{x}^{2}}dy}=\int_{0}^{9}{\pi ydy}=\frac{81}{2}\pi \approx 127.235

     2. The volume by the method of cylindrical shells is

\displaystyle \int_{0}^{3}{2\pi x\left( 9-y \right)dx}=\int_{0}^{3}{2\pi x\left( 9-{{x}^{2}} \right)dx}=\frac{81}{2}\pi \approx 127.235

     3. The value is be found by solving the equation for j:

\displaystyle \int_{0}^{{{j}^{2}}}{\pi y\,dy}=\frac{81}{4}\pi , so j\approx 2.52269

     4. The value is be found by solving the equation for k:

\displaystyle \int_{0}^{k}{2\pi x\left( 9-{{x}^{2}} \right)dx}=\frac{81}{4}\pi  and k\approx 1.62359

     5. The reason the values are not the same is this. Think of the revolved parabola as a bowl. If you pour water into the bowl until it is half full, the bowl looks like the figure below. The water is pooled in the bottom of the bowl as you would expect. This is what happens when using the disk/washer method. The washers stack up starting in the bottom of the bowl until it is half full.

half C

On the other hand, the method of cylindrical shells sort of wraps the water in layers (the shells) around the y-axis. Picture the water being sprayed on the y-axes and frozen there. Each new layer (shell) increases the amount and you end up half of the total volume arranged as a cylinder with a rounded (paraboloid shaped) bottom as shown in the figure below. Both bowls contain the same amount of water, arranged differently.

Half D

 

Visualizing Solid Figures 4

Volume by the method of “Cylindrical Shells”

Shells 3

Today I will show you how to visualize the cylindrical shells used in computing the volume using Winplot.

Winplot is a free program. Click here for Winplot and here for Winplot for Macs.

For the example I will use the same situation as in the last post. This was an AP question from 2006 AB1 / BC 1. In part (c) students were asked to find the volume of the solid figure formed when the region between the graphs of  y = ln(x) and y = x – 2 is revolved around the y-axis. This can be done by the washer method, but some students use the method of cylindrical shells. We found that the graphs intersect at x = 0.15859 and x = 3.14619.

Begin as before by graphing the two functions and the vertical segment joining them.

  • In Winplot open a 2D window, click on Equa > 1. Explicit and enter the first equation. Click the “lock interval” box and enter 0.159 for the “low x,” and 3.146 for the “high x,” choose a color and click “ok.”
  • Repeat this for the second equation.
  • Then return to Equa > Segment > (x,y) and enter the endpoints of the vertical segment joining the graphs: x1 = B, y1 = ln(B), x2 = B, and y2 = B – 2. Choose a color and click “ok.”
  • Click Anim > individual > B to open a slider box for B. Type 0.159 and click “set L” and then type 3.146 and click “set R.” Use this slider to move the thin Riemann sum rectangle across the region.

Next draw the 3D graphs:

  • Click One > Revolve Surface. The equations will appear in the drop-down box at the top of the window. Graphs are revolved one at a time. For the first graph click the “y-axis” button to put the correct values for a, b, and c in the boxes. In the “arc start” box type 0.159 and in the “arc stop” box type 3.146, the ends of the interval.  In the “angle stop” box type 2pi@S.  (S for surface.) See the figure below. Click “see surface.”

Solid 4 A

 

  • Repeat this by selecting the second function from the drop-down box. Leave all the values the same and click enter. The two surfaces will be graphed in the same color as the corresponding functions in the 2D set up.
  • Repeat this for the segment, but change the 2p@S in “arc stop” box to 2p@R. (R for Riemann rectangle.) Click “see surface.”
  • You will need to make one adjustment at this point. In the 3D Inventory list choose the segment and click “edit.” In the box change “low t” to 0 and the “high t” to 1. See the next figure.

Solid 4 B

  • In the 3D window, click Anim > Individual and open slider boxes for B, R, and S.
  •                 In the B box enter 0.159 and click “set L” and enter 3.146 and click “set R.” These are the endpoints.
  •                 In the S box make “set L” = -2pi. “Set R” should already be 2pi. Then type 0 and Enter.
  •                 The R box should open with “set L” = 0 and “Set R” = 2pi. no changes are necessary.
  • Move first the B slider, then the R slider, then the B slider again and finally the S slider to explore the situation.
  • Type Ctrl+A to show the axes. Use the 4 arrow keys to move the figure around, and the page up and page down keys to zoom in and out.

That should do it.

In the video above you will see

  • The Riemann rectangle moving in the plane using the B slider.
  • The Riemann rectangle rotated around the y-axis using the R slider.
  • The shell moving through the curves using the B slider again.
  • The two curves rotated part way using the S slider.
  • The shell moving through the solid using the B slider.

The next and last post in  this series will be a question to see if you understand how the washer method and the cylindrical shell method work in a real situation.

 Update:One of my favorite post is Difficult Problems and Why We Like Them from June 10, 2013. In it I mention a sculpture called  Kryptos located at CIA headquarters in Langley, Virginia. The sculpture contains four enciphered messages. Only three of these have been deciphered since the sculpture was erected in 1990. The sculptor has offered a second clue to the fourth message. I’ve added links story and clues in the original post; see if you can decipher the fourth part.

kryptos 2

Challenge Answer

I posted a challenge question on March 28, 2013. Only one person, “April,” replied with an explanation and it was a very good and succinct answer. You may read it here.

The conundrum resulted from trying to calculate when a parabolic bowl was half full. Using the “disk” method there was one answer; using the “shell” method resulted in a different answer. Both are correct; how can that be?

The explanation is this:

When you half fill the bowl in the usual way, by pouring water into it, the water accumulates in the bottom until the bowl is half full, as shown below. This is the result you get using the “disk” method.

Washer 2

But filling the bowl by “shells” is different. Think of spraying the y-axis with paint.  When the paint dries, spray another layer, and then another. Each layer is a “shell.” The paint accumulates as “shells” around the y-axis until the bowl is half full. The paint forms a cylinder with a rounded (parabolic) bottom that fills half the bowl, as shown below.

Shell 2

Geometry Counts

Challenge

A student was asked to find the volume of the bowl-shaped figure generated when the curve y = x2 from x = 0 to x = 2 is revolved around the y-axis. She used the disk method and found the volume to be \displaystyle \int_{0}^{4}{\pi ydy}=8\pi . To check her work she used the method of cylindrical shells and found the same answer: \displaystyle\int_{0}^{2}{2\pi x\left( 4-{{x}^{2}} \right)dx}=8\pi .

The second part of the question asked for the value of x for which the bowl would be half full. So she first solved the equation \displaystyle \int_{0}^{k}{\pi ydy=\frac{1}{2}\cdot 8\pi }  and found k=2\sqrt{2}. This is a y-value so the corresponding x-value is \sqrt{2\sqrt{2}}\approx 1.682. She again checked her work by shells by solving \displaystyle \int_{0}^{k}{2\pi x\left( 4-{{x}^{2}} \right)dx=\frac{1}{2}\cdot 8\pi } and found that k=\sqrt{-2\left( \sqrt{2}-2 \right)}\approx 1.082. (This is the x-value.)

Both computations are correct. Can you explain to her why her answers are different? Use the comment box below to share your explanations. I will post mine in a week or so.

Why You Never Need Cylindrical Shells

Don’t get me wrong; finding volumes of solids of rotation by the method of cylindrical shells is a great method. It’s just that you can always work around it; you don’t ever need to use it. The work-around is often longer and involves more work, but it is interesting mathematically. So here’s an example of how to do it.

The region bounded by the graph of y={{x}^{3}}+x+1, and the lines y = 1 and x = 2, is revolved around the y-axis. What is the volume of the resulting solid?

Supposedly this volume must be found by the shell method. Using the washer method the volume is set up as the integral of the area of the outside circle with constant radius of 2, minus the area of the inside circle of radius x, times the “thickness” of a slice. This “thickness” is in the y-direction and so is dy. The dy also determines the limits of integration.

\displaystyle V=\int_{1}^{11}{\left( \pi {{2}^{2}}-\pi {{x}^{2}} \right)\,dy}

Since the integral contains a dy the usual way is to change the x to a function of y, which in this case involves solving a cubic polynomial. In this case that is very difficult to find x as a function of y and with a different function that may even be impossible. But do we really have to do that? In fact, what is required is to have only one variable and the variable may be x! So we find dy in terms of x and dx and substitute into the expression above. We also change the limits of integration to the corresponding x-values.

dy=\left( 3{{x}^{2}}+1 \right)dx
\displaystyle V=\int_{0}^{2}{\left( \pi {{2}^{2}}-\pi {{x}^{2}} \right)\,\left( 3{{x}^{2}}+1 \right)dx}
\displaystyle V=\pi \int_{0}^{2}{\left( 4+11{{x}^{2}}-3{{x}^{4}}\, \right)dx}

This integral is easy to evaluate and will give the same value, \tfrac{272}{15}\pi, as the shell integral.

To use this idea, the function must either be one-to-one on the interval or the solid must be broken into sections that are one-to-one. This may make the problem longer. Most problems that you want to do by shells are easier by shells. The point is that washers (or disks) may always be used; not that the washer approach is the easiest way.