Category Archives: Derivative Applications

Is this going to be on the exam?

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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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Good Question 10 – The Cone Problem

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Today’s good question is an optimization problem, but its real point is choosing how to do the computation. As such it relates to MPAC 3a and 3b: “Students can  … select appropriate mathematical strategies [and] sequence algebraic/computational processes logically.” The algebra required to solve this questions can be quite daunting, unless you get clever. Here’s the question.

A sector of arc length x is removed from a circle of radius 10 cm. The remaining part of the circle is formed into a cone of radius r and height h,

  1. Find the value of x so that the cone has the maximum possible volume.
  2. The sector that was removed is also formed into a cone. Find the value of x that makes this cone have it maximum possible volume. (Hint: This is an easy problem.)
  3. In the context of the problem, the expression for the volume of the cone in part a. has a domain of 0\le x\le 20\pi . Why? Ignore the physical situation and determine the domain of the expression for the volume from a. Graph the function. Discuss.

Solutions:

Part a: As usual, we start by assigning some variables.

cone-1

Let r be the radius of the base of the cone and let h be its height. The circumference of the cone is 2\pi r=20\pi -x, so r=10-\frac{x}{2\pi } and h=\sqrt{{{10}^{2}}-{{r}^{2}}}. The volume of the cone is

\displaystyle V=\frac{\pi }{3}{{r}^{2}}h=\frac{\pi }{3}{{r}^{2}}\sqrt{{{10}^{2}}-{{r}^{2}}}=\frac{\pi }{3}{{\left( 10-\frac{x}{2\pi } \right)}^{2}}\sqrt{{{10}^{2}}-{{\left( 10-\frac{x}{2\pi } \right)}^{2}}}

To find the maximum, the next step is to differentiate the volume. The expression on the right above looks way complicated and its derivative will be even worse. Simplifying it is also a lot of trouble, and, in fact, does not make things easier.* Here’s where we can be clever and avoid a lot of algebra. Let’s just work from \displaystyle V=\frac{\pi }{3}{{r}^{2}}\sqrt{{{10}^{2}}-{{r}^{2}}}

To find the maximum differentiate the volume with respect to x using the chain rule.

\displaystyle \frac{dV}{dx}=\frac{dV}{dr}\cdot \frac{dr}{dx}=\frac{\pi }{3}\left( {{r}^{2}}\frac{-2r}{2\sqrt{{{10}^{2}}-{{r}^{2}}}}+2r\sqrt{{{10}^{2}}-{{r}^{2}}} \right)\left( -\frac{1}{2\pi } \right)

Setting this equal to zero and simplifying (multiply by -6\sqrt{{{10}^{2}}-{{r}^{2}}}) gives

-{{r}^{3}}+2r\left( 100-{{r}^{2}} \right)=200r-3{{r}^{3}}=0

\displaystyle r=0,r=\sqrt{\frac{200}{3}}=\frac{10\sqrt{6}}{3}

The minimum is obviously r = 0, so the maximum occurs when  \displaystyle r=10-\frac{x}{2\pi }=\frac{10\sqrt{6}}{3}. Then, solving for x gives

\displaystyle x=2\pi \left( 10-\frac{10\sqrt{6}}{3} \right)\approx 11.52986

Aside: We often see questions saying, if y = f(u) and u = g(x), find dy/dx. Here we have put that idea to practical use to save doing a longer computation.

Part b: The arc of the piece cut out is the circumference, x, of a cone with a radius of \displaystyle {{r}_{1}}=\frac{x}{2\pi } and a height of \displaystyle {{h}_{1}}=\sqrt{{{10}^{2}}-{{r}_{1}}^{2}}. Its volume is

\displaystyle V=\frac{\pi }{3}{{r}_{1}}^{2}\sqrt{{{10}^{2}}-{{r}_{1}}^{2}}

This is the same as the expression we used in part a. and can be handled the same way, except that here \displaystyle \frac{d{{r}_{1}}}{dx}=+\frac{1}{2\pi }. The computation and result will be the same. The result will be the same. The maximum occurs at

\displaystyle x=2\pi \left( 10-\frac{10\sqrt{6}}{3} \right)\approx 11.52986

This should not be a surprise.  The piece cut out and the piece that remains are otherwise indistinguishable, so the maximum volume should be the same for both.

Part c: From part a we have \displaystyle V=\frac{\pi }{3}{{r}^{2}}\sqrt{{{10}^{2}}-{{r}^{2}}}=\frac{\pi }{3}{{\left( 10-\frac{x}{2\pi } \right)}^{2}}\sqrt{{{10}^{2}}-{{\left( 10-\frac{x}{2\pi } \right)}^{2}}}. To graph there is no need to simplify the expression in x:

Tthe x scale marks are at multiples of $latex 5\pi $

The x-scale marks are at multiples of 5\pi

The domain is determined by the expression under the radical so

-10\le r\le 10

-10\le 10-\frac{x}{2\pi }\le 10

0\le x\le 40\pi

This is the “natural domain” of the function without regard to the physical situation given in the original problem. I cannot think of a reason for the difference.

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*Fully simplified in terms of x the volume is \displaystyle V=\frac{1}{24{{\pi }^{2}}}{{\left( 20\pi -x \right)}^{2}}\sqrt{40\pi x-{{x}^{2}}}. This isn’t really easier to differentiate and solve.

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Good Question 9

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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.]

baserunners 2-8-16

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.

Blog 2-8-16

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

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____________________________________________

Determining the Indeterminate 2

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The other day someone asked me a question about the implicit relation {{x}^{3}}-{{y}^{2}}+{{x}^{2}}=0. They had been asked to find where the tangent line to this relation is vertical. They began by finding the derivative using implicit differentiation:

3{{x}^{2}}-2y\frac{dy}{dx}+2x=0

\displaystyle \frac{dy}{dx}=\frac{3{{x}^{2}}+2x}{2y}

The derivative will be undefined when its denominator is zero. Substituting y = 0 this into the original equation gives {{x}^{3}}-0+{{x}^{2}}=0. This is true when x = –1 or when x = 0. They reasoned that there will be a vertical tangent when x = –1 (correct) and when x = 0 (not so much). They quite wisely looked at the graph.

relation 1

{{x}^{3}}-{{y}^{2}}+{{x}^{2}}=0

The graph appears to run from the first quadrant, through the origin into the third quadrant, up to the second quadrant with a vertical tangent at x = –1, and then through the origin again and down into the fourth quadrant. It looks like a string looped over itself.

What’s going on at the origin? Where is the vertical tangent at the origin?

The short answer is that vertical tangents occur when the denominator of the derivative is zero and the numerator is not zero.  When x = 0 and y = 0 the derivative is an indeterminate form 0/0.

In this kind of situation an indeterminate form does not mean that the expression is infinite, rather it means that some other way must be used to find its value. L’Hôpital’s Rule comes to mind, but the expression you get results in another 0/0 form and is no help (try it!).

My thought was to solve for y and see if that helps:

y=\pm \sqrt{{{x}^{3}}+{{x}^{2}}}

The graph consists of two parts symmetric to the x-axis, in the same way a circle consists of two symmetric parts above and below the x-axis. The figure below shows the top half.

y=+\sqrt{{{x}^{3}}+{{x}^{2}}}

So, the graph does not run from the first quadrant to the third; rather, at the origin it “bounces” up into the second quadrant. The lower half is congruent and is the reflection of this graph in the x-axis.

So, what happens at the origin and why?

The derivative of the top half is \displaystyle \frac{dy}{dx}=\frac{3{{x}^{2}}+2x}{2\sqrt{{{x}^{3}}+{{x}^{2}}}}. Notice that this is the same as the implicit derivative above. Now a little simplifying; okay a lot of simplifying – who says simplifying isn’t that big a deal?

\displaystyle \frac{dy}{dx}=\frac{x\left( 3x+2 \right)}{2\sqrt{{{x}^{2}}\left( x+1 \right)}}=\frac{x\left( 3x+2 \right)}{2\left| x \right|\sqrt{x+1}}=\left\{ \begin{matrix} \frac{3x+2}{2\sqrt{x+1}} & x>0 \\ -\frac{3x+2}{2\sqrt{x+1}} & x<0 \\ \end{matrix} \right.

Now we see what’s happening. As x approaches zero from the right, the derivative approaches +1, and as x approaches zero from the left, the derivative approaches –1.  This agrees with the graph. Since the derivative approaches different values from each side, the derivative does not exist at the origin – this is not the same as being infinite.  (For the lower half, the signs of the derivative are reversed, due to the opposite sign of the denominator.)

The tangent lines at the origin are x = 1 on the right, and x = –1 on the left, hardly vertical.

What have we learned?

  1. Indeterminate forms do not necessarily indicate an infinite value. An indeterminate form must be investigated further to see what you can learn about a function, relation, or graph.
  2. Sometimes simplifying, or at least changing the form of an expression, is helpful and therefore necessary.

Extension: Using a graphing utility that allows sliders (Winplot, GeoGebra, Desmos, etc)  enter A{{x}^{3}}-B{{y}^{2}}+C{{x}^{2}}=0 and explore the effects of the parameters on the graph.

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The Man Who Tried to Redeem the World with Logic

WALTER PITTS (1923-1969): Walter Pitts’ life passed from homeless runaway, to MIT neuroscience pioneer, to withdrawn alcoholic. (Estate of Francis Bello / Science Source)

WALTER PITTS (1923-1969): Walter Pitts’ life passed from homeless runaway, to MIT neuroscience pioneer, to withdrawn alcoholic. (Estate of Francis Bello / Science Source)

I ran across this article that you might find interesting. It is about Walter Pitts one of the twentieth century’s most important mathematicians we, or at least I, have never heard of.  It is from the February 5, 2015 of the science magazine Nautilus

November 2015

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Here is the list of “November Topics,” that is what AP classes usually consider from mid-November into December. There has been a lot of discussion about inverses this month at the AP Calculus Community. While not the most read on this blog, the series on inverses may be helpful in considering all the ins and outs of inverses.

The four featured posts on the first page are the most popular from this month. Speed with 3563 hits this year and 6157 hits since it first appear is one of the most popular overall. “Open or Closed?” is another poplar post.

Thinking ahead into December, the first posts on integration are here and will continue into December. (As I’ve mentioned I try to post a few weeks ahead of where most people are now, so you have some time to read and plan.)

October 13, 2014 Extremes without Calculus

November 2, 2012 Open or Closed?

November 5, 2012 Inverses

November 7, 2012 Writing Inverses

November 9, 2012 The Range of the Inverse

November 12, 2012 The Calculus of Inverses

November 14, 2012 Inverses Graphically and Numerically

November 16, 2012 Motion Problems: Same Thing, Different Context.

November 19, 2012 Speed

April 17, 2013 The Ubiquitous Particle Motion Question  

September 16, 2014 Matching Motion

November 21, 2012 Derivatives of Exponential Functions

November 26, 2012 Integration Itinerary

November 18, 2012 Antidifferentiation

November 30, 2012 The Old Pump

Curves with Extrema?

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mae-west-curve-the-loveliest-distance-between

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.

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.

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.

(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.

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.

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

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Good Question 7 – 2009 AB 3

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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.  Steel Wire Rope 3

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

or

\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}

or

\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.”

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