2019 CED Unit 5 Analytical Applications of Differentiation

Unit 5 covers the application of derivatives to the analysis of functions and graphs. Reasoning and justification of results are also important themes in this unit. (CED – 2019 p. 92 – 107). These topics account for about 15 – 18% of questions on the AB exam and 8 – 11% of the BC questions.

Reasoning and writing justification of results are mentioned and stressed in the introduction to the topic (p. 93) and for most of the individual topics. See Learning Objective FUN-A.4 “Justify conclusions about the behavior of a function based on the behavior of its derivatives,” and likewise in FUN-1.C for the Extreme value theorem, and FUN-4.E for implicitly defined functions. Be sure to include writing justifications as you go through this topic. Use past free-response questions as exercises and also as guide as to what constitutes a good justification. Links in the margins of the CED are also helpful and give hints on writing justifications and what is required to earn credit. See the presentation

Topics 5.1

Topic 5.1 Using the Mean Value Theorem While not specifically named in the CED, Rolle’s Theorem is a lemma for the Mean Value Theorem (MVT). The MVT states that for a function that is continuous on the closed interval and differentiable over the corresponding open interval, there is at least one place in the open interval where the average rate of change equals the instantaneous rate of change (derivative). This is a very important existence theorem that is used to prove other important ideas in calculus. Students often confuse the average rate of change, the mean value, and the average value of a function – See What’s a Mean Old Average Anyway?

Topics 5.2 – 5.9

Topic 5.2 Extreme Value Theorem, Global Verses Local Extrema, and Critical Points An existence theorem for continuous functions on closed intervals

Topic 5.3 Determining Intervals on Which a Function is Increasing or Decreasing Using the first derivative to determine where a function is increasing and decreasing.

Topic 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema Using the first derivative to determine local extreme values of a function

Topic 5.5 Using the Candidates’ Test to Determine Absolute (Global) Extrema The Candidates’ test can be used to find all extreme values of a function on a closed interval

Topic 5.6 Determining Concavity of Functions on Their Domains FUN-4.A.4 defines (at least for AP Calculus) When a function is concave up and down based on the behavior of the first derivative. (Some textbooks may use different equivalent definitions.) Points of inflection are also included under this topic.

Topic 5.7 Using the Second Derivative Test to Determine Extrema Using the Second Derivative Test to determine if a critical point is a maximum or minimum point. If a continuous function has only one critical point on an interval, then it is the absolute (global) maximum or minimum for the function on that interval.

Topic 5.8 Sketching Graphs of Functions and Their Derivatives. First and second derivatives give graphical and numerical information about a function and can be used to locate important points on the graph of the function.

Topic 5.9 Connecting a Function, Its First Derivative, and Its Second Derivative. First and second derivatives give graphical and numerical information about a function and can be used to locate important points on the graph of the function.

Topics 5.10 – 5.11

Optimization is important application of derivatives. Optimization problems as presented in most text books, begin with writing the model or equation that describes the situation to be optimized. This proves difficult for students, and is not “calculus” per se. Therefore, writing the equation has not be asked on AP exams in recent years (since 1983). Questions give the expression to be optimized and students do the “calculus” to find the maximum or minimum values. To save time, my suggestion is to not spend too much time writing the equations; rather concentrate on finding the extreme values.

Topic 5.10 Introduction to Optimization Problems

Topic 5.11 Solving Optimization Problems

Topics 5.12

Topic 5.12 Exploring Behaviors of Implicit Relations Critical points of implicitly defined relations can be found using the technique of implicit differentiation. This is an AB and BC topic. For BC students the techniques are applied later to parametric and vector functions.

Timing

Topic 5.1 is important and may take more than one day. Topics 5.2 – 5.9 flow together and for graphing they are used together; after presenting topics 5.2 – 5.7 spend the time in topics 5.8 and 5.9 spiraling and connecting the previous topics. Topics 5.10 and 5.11 – see note above and spend minimum time here. Topic 5.12 may take 2 days.

The suggested time for Unit 5 is 15 – 16 classes for AB and 10 – 11 for BC of 40 – 50-minute class periods, this includes time for testing etc.

Finally, were I still teaching, I would teach this unit before Unit 4. The linear motion topic (in Unit 4) are a special case of the graphing ideas in Unit 5, so it seems reasonable to teach this unit first. See Motion Problems: Same thing, Different Context

Previous posts on these topics include:

Then There Is This – Existence Theorems

What’s a Mean Old Average Anyway

Did He, or Didn’t He?   History: how to find extreme values without calculus

Mean Value Theorem

Fermat’s Penultimate Theorem

Rolle’s theorem

The Mean Value Theorem I

The Mean Value Theorem II

Graphing

Concepts Related to Graphs

The Shapes of a Graph

Joining the Pieces of a Graph

Extreme Values

Extremes without Calculus

Concavity

Reading the Derivative’s Graph

Real “Real-life” Graph Reading

Far Out! An exploration

Open or Closed  Should intervals of increasing, decreasing, or concavity be open or closed?

Others

Lin McMullin’s Theorem and More Gold  The Golden Ratio in polynomials

Soda Cans  Optimization video

Optimization – Reflections

Curves with Extrema?

Good Question 10 – The Cone Problem

Here are links to the full list of posts discussing the ten units in the 2019 Course and Exam Description.

2019 CED – Unit 1: Limits and Continuity

2019 CED – Unit 2: Differentiation: Definition and Fundamental Properties.

2019 CED – Unit 3: Differentiation: Composite , Implicit, and Inverse Functions

2019 CED – Unit 4 Contextual Applications of the Derivative  Consider teaching Unit 5 before Unit 4

2019 – CED Unit 5 Analytical Applications of Differentiation  Consider teaching Unit 5 before Unit 4

2019 – CED Unit 6 Integration and Accumulation of Change

2019 – CED Unit 7 Differential Equations  Consider teaching after Unit 8

2019 – CED Unit 8 Applications of Integration   Consider teaching after Unit 6, before Unit 7

2019 – CED Unit 9 Parametric Equations, Polar Coordinates, and Vector-Values Functions

2019 CED Unit 10 Infinite Sequences and Series

Did He, or Didn’t He?

Pierre de Fermat (1607 – 1655)

Since it’s soon time to start derivatives, today we look at one way people found maximums and minimums before calculus was invented. Pierre de Fermat (1607 – 1655) was a French lawyer. His hobby, so to speak, was mathematics. He is considered one of the people who built the foundations of calculus. Along with Descartes, he did a lot of the early work on analytic geometry and did much work leading to calculus, which today we would consider calculus. Neither he nor other mathematicians of his time had the tool of limits to aid him. Derivatives were not yet invented. Nevertheless, he came awfully close to both.

Fermat was working on optimization and that required him to find the maximum or minimum values of functions. Here is how he found extreme values of polynomial functions. Even though he didn’t know limits and derivatives, maybe he used them.

Using Fermat’s method, and some modern notation and terminology, we will demonstrate has method by finding the extreme values of the function:

(1)          $\displaystyle f\left( x \right)={{x}^{4}}-3{{x}^{2}}+2x$

Near an extreme value Fermat reasoned that a horizontal line would intersect the graph in two points, say where x = a and x = b. (There is a third intersection on this function, since there are three extreme points. The method finds all three.) He thought of this line moving up or down and the two points coming close to the extreme value. Since these two points are on the same horizontal line f(a) = f(b):

(2)          ${{a}^{4}}-3{{a}^{2}}+2a={{b}^{4}}-3{{b}^{2}}+2b$

After moving everything to the left side and grouping terms by their powers and factoring we have:

(3)          $\displaystyle \left( {{{a}^{4}}-{{b}^{4}}} \right)-3\left( {{{a}^{2}}-{{b}^{2}}} \right)+2\left( {a-b} \right)=0$

(4)          $\displaystyle \left( {a-b} \right)\left( {\left( {a+b} \right)\left( {{{a}^{2}}+{{b}^{2}}} \right)-3\left( {a+b} \right)+2} \right)=0$

Near where a = b, (a – b) is not zero, so we may divide by it.

(5)          $\displaystyle \left( {a+b} \right)\left( {{{a}^{2}}+{{b}^{2}}} \right)-3\left( {a+b} \right)+2=0$

Again, near the extreme value $\displaystyle a\approx b$, so, substituting  $\displaystyle a=b$ we have

(6)          $\displaystyle 2a\left( {2{{a}^{2}}} \right)-3\left( {2a} \right)+2=4{{a}^{3}}-6a+2\approx 0$

Now all that remains is to solve this equation (factoring by synthetic division).

(7)          $\displaystyle 4{{a}^{3}}-6a+2=2\left( {a-1} \right)\left( {2{{a}^{2}}+2a-1} \right)=0$

(8)          $\displaystyle a=1,a=\frac{{-1+\sqrt{3}}}{2},a=\frac{{-1-\sqrt{3}}}{2}$

These are the x-coordinates of the critical points.

Now, let’s look at all the “calculus” Fermat did not do.

The left side of equations (3) and (4) are

(9)          $\displaystyle f\left( a \right)-f\left( b \right)$

And after dividing this by (a – b) he had in (5)

(10)         $\displaystyle \frac{{f\left( a \right)-f\left( b \right)}}{{a-b}}$

Then by letting a be approximately equal to b, which sounds a lot like finding a limit, he had in modern terms

(11)         $\displaystyle \underset{{a\to b}}{\mathop{{\lim }}}\,\frac{{f\left( a \right)-f\left( b \right)}}{{a-b}}={f}'\left( a \right)$

Expression (6) is $\displaystyle {f}'\left( a \right)$ – Fermat unknowingly is using the derivative! After which he set it equal to 0 and solved to find the x-coordinates of the critical points.

So, did he or didn’t he use calculus? You decide.

Optimization – Reflections

First, a new resource has been added to the resource page. listed by major topics. These were researched by by Kalpana Kanwar a teacher at Wisconsin Heights High School. Thank you Kalpana! They include precalculus topics that were tested on the exams before 1998. These may be good for your precalculus classes. (Remember that the course description underwent major changes in 1998 and some topics were dropped at that time. These include the “A” topics (precalculus), Newton’s Method, work, volume by cylindrical shells among others. Be careful, when assigning old questions; they’re good, but they may no longer be tested.)

Optimizations problems are situations in which some item is to be made as large or small as possible. Often this is the minimum cost of producing something, or to maximize profit, or to make the largest area or volume with the least material.

While these problems are found in most of the textbooks, they almost never appear on the AP Calculus Exams. The reason for this is that the first step is to write the equation that models the situation. This step does not involve any “calculus.” If a student cannot do this or does it incorrectly, then there is no way to earn the calculus points that follow. On the exams, students are given an expression and asked to find its maximum or minimum value.

Nevertheless, the problems can be interesting and are useful in a practical sense. Reflection is one of my favorites: show that the angle of incidence equals the angle of reflection. In the figure below, light travels from a point A to point D on a reflecting surface CE and then to point B by the shortest total distance. Show that this implies that the angle α between AD and the normal to the surface is equal to the angle β between the normal and DB. The angle α is called the angle of incidence and the angle β is called the angle of reflection.

Using the lengths marked in the drawing,  $\overline{{AC}},\overline{{PD}}$ and $\overline{{BE}}$ are all perpendicular to  $\overline{{CE}}$ the total distance is AD + DB. Therefore:

$AD+DB=\sqrt{{{{a}^{2}}+{{x}^{2}}}}+\sqrt{{{{b}^{2}}+{{{\left( {CE-x} \right)}}^{2}}}}$

To find the minimum distance find the derivative of AD + DB and set it equal to zero.

$\displaystyle \frac{{2x}}{{2\sqrt{{{{a}^{2}}+{{x}^{2}}}}}}+\frac{{-2\left( {CE-x} \right)}}{{2\sqrt{{{{b}^{2}}+{{{\left( {CE-x} \right)}}^{2}}}}}}=0$

Then

$\displaystyle \frac{x}{{\sqrt{{{{a}^{2}}+{{x}^{2}}}}}}=\frac{{\left( {CE-x} \right)}}{{\sqrt{{{{b}^{2}}+{{{\left( {CE-x} \right)}}^{2}}}}}}$

Now, we need to be clever:

$\displaystyle \frac{x}{{\sqrt{{{{a}^{2}}+{{x}^{2}}}}}}=\cos \left( {ADC} \right)=\cos \left( \beta \right)$ and

$\displaystyle \frac{{\left( {CE-x} \right)}}{{\sqrt{{{{b}^{2}}+{{{\left( {CE-x} \right)}}^{2}}}}}}=\cos \left( {BDE} \right)=\cos \left( \alpha \right)$

And therefore, $\displaystyle \alpha =\beta$ QED.

See the illustration of this in Desmos here and see an easier way to do this problem.

The conic sections all have interesting reflection properties that are quite useful.

The Ellipse: A light ray leaving one focus of an ellipse is reflected by the ellipse through the other focus of the ellipse. The angle of incidence and the angle of reflection are between the segments to the foci and the normal to the ellipse.

The computation is done using a Computer Algebra System (CAS) and is shown below, the line-byline explanation follows:

• The first line starts with the ellipse $\frac{{{{x}^{2}}}}{{{{a}^{2}}}}+\frac{{{{y}^{2}}}}{{{{b}^{2}}}}=1$ with a > b > 0. Solving for y there are two equations, the second one is for the upper half that we will use below. The other for the lower half.
• The second line finds the derivative of y and the third line, m2 is the slope of the normal, the opposite of the reciprocal of the derivative.
• The fourth and fifth lines are the slopes, m1 and m3, are the slopes from the point on the ellipse to the foci. The “such that” bar, |, indicates that what follows it is substituted into the expression.
• The next two lines compute the inverse tangent of angle rotated counterclockwise between the segments to the foci and the normal. This uses the formula from analytic geometry: ${{\tan }^{{-1}}}\left( {\frac{{{{m}_{2}}-{{m}_{1}}}}{{1+{{m}_{1}}{{m}_{2}}}}} \right)$
• The last line shows that the expression from the two lines above it are equal, indicated by the “true” on the right.

An illustration using Desmos is here. Ellipses are used as reflectors in medical and dental equipment so that a relatively dim light source can be concentrated at the place where the doctor or dentist is working without “blinding” everyone in the room. There are also ceilings that reflect sound from one focus to the other without anyone elsewhere in the room hearing. These are only a few of their uses.

The hyperbola: A light ray from one focus of a hyperbola is reflected as though it came from the other focus. This is true whether the reflection is from the side nearer the focus or farther from the focus (reflection from the convex or the concave side.

The computation is like the ellipse computation with only a few sign changes. I will not reproduce it here. If you want to try use $\displaystyle \frac{{{{x}^{2}}}}{{{{a}^{2}}}}-\frac{{{{y}^{2}}}}{{{{b}^{2}}}}=1$ and  $c=\sqrt{{{{a}^{2}}+{{b}^{2}}}}$.

There is a Desmos illustration here. Use the p-slider to move the point. The left side shows the reflection from the “outside” surface; the right side shows the reflection from the “inside” surface.

Hyperbolas are used in telescopes and magnifying mirrors to enlarge the image.

The Parabola: A light ray from the focus is reflected parallel to the axis of symmetry of the parabola Or you can go the other way: light traveling parallel to the axis is reflected to the focus.

If you try to prove this on use  $x=a{{y}^{2}}$ to avoid working with an undefined slope. The focus is at  $\left( {\frac{a}{4},0} \right)$.

There is a Desmos illustration here

Parabolic reflectors are used in various kinds of spotlight and telescopes and for radar dishes. They are also used for satellite dishes for cable TV; you may have one at home.

Is this going to be on the exam?

Recently 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

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.

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 ug(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:

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.

________________

*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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Curves with Extrema?

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. 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  (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

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.

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

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 $00$ 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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