Summer Fun

Posted on June 12, 2018 by Lin McMullin

Every Spring I have a lot of fun proofreading Audrey Weeks’ new Calculus in Motion illustrations for the most recent AP Calculus Exam questions. These illustrations run on Geometers’ Sketchpad. In addition to the exam questions Calculus in Motion (and its companion Algebra in Motion) include separate animations illustrating most of the concepts in calculus and algebra. This is a great resource for your classes.

The proofreading and the cross-country conversations with Audrey give me a chance to learn more about the questions.

This year, I really got into 2018 AB 6, the differential equation question. I wrote an exploration (or as the kids would say “worksheet”) on a function very similar to the differential equation in that question. The exploration, which is rather long, includes these topics:

Finding the general solution of the differential equation by separating the variables
Checking the solution by substitution
Using a graphing utility to explore the solutions for all values of the constant of integration, C
Finding the solutions’ horizontal and vertical asymptotes
Finding several particular solutions
Finding the domains of the particular solutions
Finding the extreme value of all solutions in terms of C
Finding the second derivative (implicit differentiation)
Considering concavity
Investigating a special case or two

I also hope that in working through this exploration students will learn not so much about this particular function, but how to use the tools of algebra, calculus, and technology to fully investigate any function and to find all its foibles.

Students will need to have studied calculus through differential equations before they start the exploration. I will repost it next January for them.

The exploration is here for you to try. Try it before you look at the solutions. It will give you something to do over the summer – well not all summer, only an hour or so.

As always, I appreciate your feedback and comments. Please share them with me using the reply box below.

There will be only occasional, very occasional, posts over the Summer. More regular posting will begin again in August. Enjoy the Explorations, and, more important, enjoy the Summer!

Extreme Values and Linear Motion

Posted on October 24, 2017 by Lin McMullin

Two more applications of differentiation are finding extreme values and the analysis of linear motion.

Extreme Values

The Marble and the Vase

Extremes without Calculus

A Standard Problem

Far Out!

Linear Motion – Motion on a Line

Type 2 Problems

Motion Problems: Same Thing, Different Context

The Ubiquitous Particle Motion Problem – a PowerPoint Presentation and its Handout

Brian Leonard’s Particle Motion Game Velocity Game and answers Velocity game Answers

Matching Motion – an activity

Speed

Good Question 9

Posted on February 14, 2016 by Lin McMullin

This is a good question that leads to other good questions, both mathematical and philosophical. A few days ago this question was posted on a private Facebook page for AP Calculus Readers. The problem and illustration were photographed from an un-cited textbook.

Player 1 runs to first base [from home plate] at a speed of 20 ft/s while player 2 runs from second base to third base a speed of 15 ft/s. Let s be the distance between the two players. How fast is s changing when player 1 is 30 feet from home plate and Player 2 is 60 feet from second base. [A figure was given showing that the distance between the bases is 90 feet.]

Some commenters indicated some possible inconsistencies in the question, such as assuming the Player 2 is on second base when Player 1 leaves home plate. In this case the numbers don’t make sense. So, someone suggested this must be a hit-and-run situation. To which someone else replied that with a lead of that much it’s really a stolen base situation. So, the first thing to be learned here is that even writing a simple problem like this you need to take some of the real aspects into consideration. But this doesn’t change the mathematical aspects of the problem.

One of the things I noticed before I attempted to work out the solution was that Player 2 is the same distance from third base as Player 1 is from home plate. I verbalized this as “players are directly across the field from each other.” I filed this away since it didn’t seem to matter much. Wrong!

Then I worked on the problem two ways. These are shown in the appendix at the end of this post. I discovered (twice) that s’ = 0; at the moment suggested in the question the distance is not changing.

Then it hit me. Doh! – I didn’t have to do all that. So, I posted this solution (which I now notice someone beat me to):

At the time described, the players are directly across the field from each other (90 feet apart). This is the closest they come. The distance between them has been decreasing and now starts to increase. So, at this instant s is not changing (s‘ = 0).

The Philosophical Question

Then the original poster asked for someone “to post [actual] work done in calculus” and “to see some related rates.” So, I posted some “calculus” and got to thinking – the philosophical question – isn’t my first answer calculus?

I think it is. It makes use of an important calculus concept, namely that as things change, at the minimum place, the derivative is zero. Furthermore, the justification (that the distance changes from decreasing to increasing at the minimum implies the derivative is zero) is included. * Why do you need variables?

Also, this solution is approached as an extreme value (max/min) problem rather than a related rate problem. This shows a nice connection between the two types of problems.

The Related (but not related rate) Good Question

So here is another calculus question with none of the numbers we’ve grown to expect:

Two cars travel on parallel roads. The roads are w feet apart. At what rate does the distance between the cars change when the cars are w feet apart?

Notice:

That the cars could be travelling in the same or opposite directions.
Their speeds are not given.
You don’t know when or where they started; only that at some time they are opposite each other (w feet apart).
In fact, they could start opposite each other and travel in the same direction at the same speed, remaining always w feet apart.
One car could be standing still and the other just passes it.

But you can still answer the question.

(*Continuity and differentiability are given (or at least implied) in the original statement of the problem.)

Appendix

My first attempt was to set up a coordinate system with the origin at third base as shown below.

Then, taking the time indicated in the problem as t = 0, the position of Player 1 is (90, 30 + 20t) and the position of player 2 is (0, 30 – 15t). Then the distance between them is

$s=\sqrt{{{90}^{2}}+{{\left( 30-15t-\left( 30+20t \right) \right)}^{2}}}=\sqrt{{{90}^{2}}+{{\left( -35t \right)}^{2}}}$

and then

$\displaystyle {s}'\left( t \right)=\frac{2\left( -35t \right)\left( -35 \right)}{2s}\text{ and }{s}'\left( 0 \right)=0$

This is correct, but for some reason I was suspicious probably because zeros can hide things. So I re-stated this time taking t = 0 to be one second before the situation described in the problem. Now player 1’s position is (90, 10+20t) and player 2’s position is (0, 45-15t).

$s\left( t \right)=\sqrt{{{90}^{2}}+{{\left( 45-15t-\left( 10+20t \right) \right)}^{2}}}=\sqrt{{{90}^{2}}+{{\left( 30-35t \right)}^{2}}}$

$\displaystyle {s}'\left( t \right)=\frac{2\left( 35-35t \right)\left( -35 \right)}{2s}\text{ and }{s}'\left( 1 \right)=0$

____________________________________________

Curves with Extrema?

Posted on October 19, 2015 by Lin McMullin

We spend a lot of time in calculus studying curves. We look for maximums, minimums, asymptotes, end behavior, and on and on, but what about in “real life”?

For some time, I’ve been trying to find a real situation determined or modeled by a non-trigonometric curve with more than one extreme value. I’ve not been very successful. I knew of only one and discovered a second in writing this post. Here is an example that illustrates what I mean.

Example 1: This is a very common calculus example. Squares are cut from the corners of a cardboard sheet that measures 20 inches by 40 inches. The remaining sides are folded up to make a box. How large should the squares be to make a box of the largest possible volume?

If we let x = the length of the side of the square, then the volume of the box is given by V = x (20 –2 x)(40 – 2x)

The graph of the volume is shown in Figure 1 and sure enough we have a polynomial curve that has two extreme values. But wait. Do we really have two extreme values? The domain of the equation appears to be all real numbers, but in fact it is 0 < x < 10, since x cannot be negative, and if x > 10, then the (20 –2 x) side is negative and that won’t work either. Figure 2 shows the true situation. There is only one extreme value.

: Figure 1

: Figure 2

Example 2: This is also an optimization problem, but a bit more difficult. A sector is cut from a circular paper disk of radius 1. The remaining part of the disk is formed into a cone. How long should the curved part of the sector be so that the cone has the maximum volume? You might want to try this before you read further.

Let x be the length of curved part of the sector See Figure 3.

The radius of the disk becomes the slant height of the cone. The circumference of the disk is $2\pi \left( 1 \right)$ and so the circumference of the base of the cone is $C=2\pi \left( 1 \right)-x$ and its radius is $\displaystyle r=\frac{2\pi -x}{2\pi }=1-\frac{x}{2\pi }$ . The height, h of the cone is $\displaystyle h=\sqrt{1-{{\left( \frac{x}{2\pi } \right)}^{2}}}$ . See figure 4.

: Figure 3

: Figure 4

The volume of the cone is . $\displaystyle V=\frac{\pi }{3}{{\left( 1-\frac{x}{2\pi } \right)}^{2}}\sqrt{{{1}^{2}}-{{\left( 1-\frac{x}{2\pi } \right)}^{2}}}$

The graph of this equation is shown in Figure 5, and has two maximum values and a minimum. The domain appears to be $0<x<4\pi$ . But if $x>2\pi$ the piece you cut out will be larger than the original disk (and the expression under the radical will be negative). So our domain will be $0<x<2\pi$ (the endpoints correspond to not cutting any sector or cutting away the entire disk. The graph is shown in Figure 6 with, alas, only one extreme value.

: Figure 5

: Figure 6

(For the original expression the minimums are at $x=0,\ 2\pi ,\text{ and }4\pi$ and the maximums are at $x\approx 1.153\text{ and }x\approx 11.413$ . A CAS will help with these calculations or just use a graphing calculator.)

Extension 1: Find the value of x that will make the largest volume when the piece cut out is formed into a cone. Compare the two graphs and explain their congruence. See Figure 7.
Extension 2: Here I finally found what I was after. – a situation with more than one extreme value. Find the value of x that will make the largest total volume formed when the volume of the original cone and the cone formed by the piece cut out. Compare the first two graphs and the graph of this volume. See figure 8 – the magenta graph.

: Figure 7

: Figure 8

The Mae West Curve

There is at least one real situation that is modeled by a function with several extreme values. Spud’s blog gives the following explanation and illustration.

“When a Uranium (or Plutonium) atom fission, or splits, you end up with two much lighter atoms, called fission products, or daughter nuclides. The U-235 nucleus can split into a myriad of combinations, but some combinations are more likely than others.

“[Figure 9 below] shows the percentage of fission products by [atomic] mass[, A]. [This is] called the Mae West curve. … Note that the more likely fission products have two peaks at a mass of about 95 and 135.”

Thus we have a real life illustration of a model that has three extreme values in its domain.

The model graphed in Figure 9 is known as the “Mae West Curve,” named after Mae West (1893 – 1980) and actress, playwright and screenwriter.

: Figure 9: The “Mae West” Curve

If you know of any other real situations with more than one extreme, please let us know. Use the “comment” button below.

Good Question 7 – 2009 AB 3

Posted on October 5, 2015 by Lin McMullin

Another in my occasional series on Good Questions to teach from. This is the Mighty Cable Company question from the 2009 AB Calculus exam, number 3

This question presented students with a different situation than had been seen before. It is a pretty standard “in-out” question, except that what was going in and out was money. Students were told that the Mighty Cable Company sold its cable for $120 per meter. They were also told that the cost of the cable varied with its distance from the starting end of the cable. Specifically, the cost of producing a portion of the cable x meters from the end is $6\sqrt{x}$ dollars per meter. Profit was defined as the difference between the money the company received for selling the cable minus the cost of producing the cable.

Students had a great deal of trouble answering this question. (The mean was 1.92 out of a possible 9 points. Fully, 36.9% of students earned no point; only 0.02% earned all 9 points.) This was probably because they had difficulty in interpreting the question and translating it into the proper mathematical terms and symbols. Since economic problems are not often seen on AP Calculus exams, students needed to be able to use the clues in the stem:

The $120 per meter is a rate. This should be deduced from the units: dollars per meter.
The cost of producing the portion of cable x meters from one end cable is also a rate for the same reason. In economics this is called the marginal cost; the students did not need to know this term.
The profit is an amount that is a function of x, the length of the cable.

Part (a): Students were required to find the profit from the sale of a 25-meter cable. This is an amount. As always, when asked for an amount, integrate a rate. In this case integrate the difference between the rate at which the cable sells and the cost of producing it.

$\displaystyle P(25)=\int_{0}^{25}{\left( 120-6\sqrt{x} \right)dx}=\$2500$

$\displaystyle P(25)=120(25)-\int_{0}^{25}{6\sqrt{x}\ dx}=\$2500$

Part (b): Students were asked to explain the meaning of $\displaystyle \int_{25}^{30}{6\sqrt{x}dx}$ in the context of the problem. Since the answer is probably not immediately obvious, here is the reasoning involved.

This is the integral of a rate and therefore, gives the amount (of money) needed to manufacture the cable. This can be found by a unit analysis of the integrand: $\displaystyle \frac{\text{dollars}}{\text{meter}}\cdot \text{meters}=\text{dollars}$ .

Let C be the cost of production, so $\displaystyle \frac{dC}{dx}=6\sqrt{x}$ , and therefore, $\displaystyle \int_{25}^{30}{6\sqrt{x}dx}=C\left( 30 \right)-C\left( 25 \right)$ by the Fundamental Theorem of Calculus (FTC).

Therefore, $\displaystyle \int_{25}^{30}{6\sqrt{x}dx}$ is the difference in dollars between the cost of producing a cable 30 meters long, C(30), and the cost of producing a cable 25 meters long, C(25). (Another acceptable is that the integral is the cost in dollars of producing the last 5 meters of a 30 meter cable.)

Part (c): Students were asked to write an expression involving an integral that represents the profit on the sale of a cable k meters long.

Part (a) serves as a hint for this part of the question. Here the students should write the same expression as they wrote in (a) with the 25 replaced by k.

$\displaystyle P(k)=\int_{0}^{k}{\left( 120-6\sqrt{x} \right)dx}$

$\displaystyle P(k)=120k-\int_{0}^{k}{6\sqrt{x}\ dx}$

Part d: Students were required to find the maximum profit that can be earned by the sale of one cable and to justify their answer. Here they need to find when the rate of change of the profit (the marginal profit) changes from positive to negative.

Using the FTC to differentiate either of the answers in part (c) or by starting fresh from the given information:

$\displaystyle \frac{dP}{dx}=120-6\sqrt{x}$

$\displaystyle \frac{dP}{dx}=0$ when x = 400 and P(400)= $16,000.

Justification: The maximum profit on the sale of one cable is $16,000 for a cable 400 meters long. For $0<x<400,{P} '(x)>0$ and for $x>400,{P} '(x)<0$ therefore, the maximum profit occurs at x = 400. (The First Derivative Test).

Once students are familiar with in-out questions, this is a good question to challenge them with. The actual calculus is not that difficult or unusual but concentrating on the translation of the unfamiliar context into symbols and calculus ideas is different. Show them how to read the hints in the problem such as the units.

Steel cable or steel wire rope as it is called also has some interesting geometry in its construction. You can find many good illustrations of this, such as the ones below, by Googling “steel wire rope.”

This slideshow requires JavaScript.

Good Question 5: 1998 AB2/BC2

Posted on July 29, 2015 by Lin McMullin

Continuing my occasional series of some of my favorite teaching questions, today we look at the 1998 AP Calculus exam question 2. This question appeared on both the AB and BC exams. I use this problem to illustrate two very different questions that come up almost every time I lead a workshop or an AP Summer Institute. The first is if a limit is infinite, should you say “infinite” or “does not exist (DNE)”? The second is if the student solves the problems correctly, but by some other method, maybe even one not using the calculus, do they still earn full credit? In addition to discussing these two questions I’ll have a few suggestions for how to use this kind of question for teaching (maybe in other than a calculus class).

The question had the student examine the function $f\left( x \right)=2x{{e}^{2x}}$ and, although it is easy enough to answer without, students were allowed to use their graphing calculator. A reasonable student probably looked at a graph of the function.

$f\left( x \right)=2x{{e}^{2x}}$

Part a: First the question asks the student to explore the end behavior of the function by finding two limits: $\underset{x\to -\,\infty }{\mathop{\lim }}\,f\left( x \right)$ and $\underset{x\to \infty }{\mathop{\lim }}\,f\left( x \right)$ . The students should not depend on the graph here. As $x\to -\infty$ , ${{e}^{2x}}$ approaches zero and since the exponential function dominates the polynomial, $\underset{x\to -\,\infty }{\mathop{\lim }}\,f\left( x \right)=0$ . In passing note that for x < 0 the function is negative and approaches zero from below. No work or explanation was required, but when teaching things like this be sure students know and can explain their answer without reference to their calculator graph. For the second limit, since both factors increase without bound $\underset{x\to \infty }{\mathop{\lim }}\,f\left( x \right)=\infty$ If the student wrote $\underset{x\to \infty }{\mathop{\lim }}\,f\left( x \right)=\text{DNE}$ , he received full credit.

Infinity is not a number, so there really is no limit in the second case; the limit DNE. But there are other ways a limit may not exist such as a jump discontinuity or an oscillating discontinuity. DNE covers these as well as infinite limits. Saying a limit is infinite tells us more about the limit than DNE. It tells us that the function increases without bound; that eventually it becomes greater than any number.

Both answers are correct.

But we’re not done with this yet. We will come back to it before the question is done.

Part b: Students were asked to find and justify the minimum value of the function. Using the first derivative test, students proceeded by finding where the derivative is zero..

${f}'\left( x \right)=\left( 2x \right)\left( 2{{e}^{2x}} \right)+2{{e}^{2x}}=2{{e}^{2x}}\left( 2x+1 \right)=0$

$x=-\frac{1}{2}$

$f\left( -\tfrac{1}{2} \right)=2\left( -\tfrac{1}{2} \right){{e}^{2\left( -\tfrac{1}{2} \right)}}=-\frac{1}{e}\approx 0.368\text{ or }0.367$

Justification: If $x<-\tfrac{1}{2},\ {f}'\left( x \right)<0$ and if $x>-\tfrac{1}{2},\ {f}'\left( x \right)>0$ , therefore the absolute minimum is $-\frac{1}{e}$ and occurs at $x=-\frac{1}{2}$ .

All pretty straightforward

Part c: This part asked for the range of the function. Here the student must show that if he wrote DNE in part a, he knows that in fact the function grows without bound.

Putting together the answers from part a and part c, the range is $f\left( x \right)\ge -\frac{1}{e}$ , which may also be written as $\left[ -1/e,\infty\right)$ . (The decimals could also be used here.)

Part d: asked students to consider functions given by $y=bx{{e}^{bx}}$ where b was a non-zero number. The question required students to show that the absolute minimum value of all these functions was the same.

Most students did what was expected and preceded as in part b. The work is exactly the same as above except that all of the 2s become bs. The absolute minimum occurs at $x=-\frac{1}{b}$ and $y\left( -\tfrac{1}{b} \right)=-\frac{1}{e}$ .

BUT ….

Other students found a way completely without “calculus.” Can you find do that?

They realized that the given function as a horizontal expansion or compression, possibly including a reflection over the y-axis, of and therefore the range is the same for all these functions and so the minimum value must be the same. This received full credit. The rule of thumb is “don’t take off for good mathematics.”

Pretty cool!

The graphs of several cases are shown below

$y=bx{{e}^{bx}}$
b = -5 in blue, b = -1 in red, b = 2 in green, and b = 4 in magenta.

Teaching Suggestions

I can see using this in a pre-calculus class. The calculus (finding the minimum for b = 2 or in general) is straightforward. In a pre-calculus setting as an example of transformations it may be more useful. You could give students 6, or 8, or 10 examples with different values of b, both positive and negative.

First ask students to investigate the end behavior by finding the limits as x approaches positive and negative infinity. The results will be similar. Have them write a summary considering two cases: b > 0 and b < 0.
Graphing calculators have built-in operations that will find the x-coordinates or both coordinates of the minimum point of a function. Since we’re concerned with the transformation and not the calculus, let students use their graphing calculators to find the coordinates of the minimum point of each graph (as decimals). See if they can determine the x-coordinate in terms of b. They should also notice that y-coordinates will all be the same (about -0.367880).
Finally, set the class to proving using their knowledge of transformation that the minimums are really all the same.

Soda Cans

Posted on May 13, 2015 by Lin McMullin

A typical calculus optimization question asks you to find the dimensions of a cylindrical soda can with a fixed volume that has a minimum surface area (and therefore is cheaper to manufacture).

Let r be the radius of the cylinder and h be its height. The volume, V, is constant and $V=\pi {{r}^{2}}h$ . The surface area including the top and bottom is given by

$S=2\pi rh+2\pi {{r}^{2}}$

Since $\displaystyle h=\frac{V}{\pi {{r}^{2}}}$ , the surface area, S, can be expressed as

$S=2V{{r}^{-1}}+2\pi {{r}^{2}}$

To find the value of r that will give the smallest surface area we find the derivative, set it equal to zero and solve for r:

$\displaystyle \frac{dS}{dr}=-2V{{r}^{-2}}+4\pi r$

This will equal zero when $\displaystyle r=\sqrt[3]{\frac{V}{2\pi }}$ and substituting into the expression above $\displaystyle h=\sqrt[3]{\frac{4V}{\pi }}$ .

Then $\displaystyle \frac{h}{r}=\sqrt[3]{\frac{\frac{4V}{\pi }}{\frac{V}{2\pi }}}=2$ , so $h=2r$ . In the optimum can the height is equal to the diameter.

The thing is that very few cans, especially beverage cans are anywhere near this “square “ shape. The closest I could find in my pantry was a tomato sauce can holding 8 oz. or 277 mL. The inside dimensions are about 65 cm. by 75cm. Compare this to the 12 oz. soda can holding 355 mL. The usual reason given for this departure from the mathematically best shape is the taller can is easier to hold especially for children.

What got me interested in this was the video below. While there is no overt calculus mentioned, there is a lot of math. There are also STEM considerations, specifically engineering. As you watch look for the math and engineering ideas that are mentioned and discuss them with your class. Here are a few:

Geometry: Why a cylinder? Why not a sphere or a cube?
Engineering: When cutting circles out of rectangular sheets of aluminum there is a lot of unused metal. Why is all this waste not a problem? This goes to materials engineering; steel is more difficult to recycle than aluminum.
Math: Efficient packing is also a consideration. Check the calculations in the video as to the most efficient way (least empty space) to pack containers. Why do they not use the most efficient?
Geometry: The (spherical) dome is a very strong shape. In what other places are domes used? Why?
Engineering: How does pressurizing the cans make them stronger?
Geometry and Engineering: The elongated ridges on the sides of non-pressurized steel cans strengthen the sides. How are these ridges similar to the dome or circular arch?
Physics: Look for a discussion of first- and second-class leavers.
Engineering: What other advantages are there to using the very thin aluminum can.

At the end of the video 6 other videos are mentioned. These are also interesting and show the same process in cartoon form and in video of the machines making cans. The links to these are here:

Rexam: http://www.youtube.com/watch?v=7dK1VV…
How It’s Made: http://www.youtube.com/watch?v=V7Y0zA…
Anim1: https://www.youtube.com/watch?v=WU_iS…
Anim2:https://www.youtube.com/watch?v=hcsDx…
Drawing: https://www.youtube.com/watch?v=DF4v-…
Redrawing: http://www.youtube.com/watch?v=iUAijp…

Teaching Calculus

The pleasure lies not in discovering the truth, but in searching for it.

Tag Archives: Extreme Value

Summer Fun

Extreme Values and Linear Motion

Curves with Extrema?

Good Question 7 – 2009 AB 3

Good Question 5: 1998 AB2/BC2

Soda Cans

Share this:

Share this:

Share this:

Share this:

Share this:

Share this:

Share this: