MathNotations

\displaystyle \sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}

The remarkable identity above could be the subject of many math blog posts but we will look at a variation, one that is accessible to precalculus and calculus students. With the AP Calculus BC Exam looming, the following investigation can be used to introduce or to review the topic.

I'm not sure if I have ever made it really clear on this blog that I routinely used these kinds of investigations in the classroom. For those who wonder how I could possibly have completed the required coursework for the AP Calculus BC syllabus or who might question my sanity, a couple of points here:

(1) Of course I didn't do this every day. I might have done an extensive investigation once per unit.
(2) Imagine my surprise when I first saw the Finney, Demana, Waits and Kennedy text, a book that has these kinds of explorations in every chapter! I thought they had found my old lesson plans.
(3) Most of the extensive investigations were assigned for work outside the classroom. In fact, for a while, the first investigation of the year was posted on my web site and emailed to students at the end of August before they arrived in school (I met them in June before they left for the summer or I got their phone numbers from guidance and called each of them to tell them to look for the assignment online, and to download and print it.)
(4) Even if I didn't prepare an exploration every day, most every lesson plan which introduced a new topic included a series of leading questions like these. My intent was always to have them think more deeply about a topic, i.e., to understand

• the historical origins of the topic
  
• how it was connected to their prior learning
• its usefulness and application
• why a method or theorem works (derivation, justification)

Developing these lessons initially was labor-intensive but a work of love. Perhaps no more laborious than what another of my colleagues did for his students: developing a PowerPoint presentation for every lesson for the entire year. As time-consuming as all of that sounds, once you've done a few of these, they start to flow naturally and the following year you only have to revise!

A Series Investigation

Consider the following finite series:

\frac{1}{3}+\frac{1}{8}+\frac{1}{15}+...+\frac{1}{n^2-1}+...+\frac{1}{99}

(a) Write the series using summation notation.

(b) Verify the following identity for n> 1:

\frac{1}{n^2-1} = \frac{1}{2}(\frac{1}{n-1}-\frac{1}{n+1})

(c) Use the identity in (b) to show that the value of the series above is

(d) Using a method similar to (c) verify the following for n, even:

\frac{1}{3}+\frac{1}{8}+\frac{1}{15}+...+\frac{1}{n^2-1}=\frac{1}{2}((1-\frac{1}{n+1})+(\frac{1}{2}-\frac{1}{n}))

Note: If n = 2, the right side would be accurate however the left side would consist of only one term. I could have used summation notation for the left side but I didn't want to give away the answer to part (a).

(e) If n is odd, show that the series on the left of part (d) can be written:

\frac{1}{2}((1-\frac{1}{n})+(\frac{1}{2}-\frac{1}{n+1}))

(f) Show that the expression on the right side of the equation in (d) and the expression in (e) are algebraically equivalent.

(g) Use the expressions from (d) and (e) to show that the sum of the following infinite series is 3/4:

\frac{1}{3}+\frac{1}{8}+\frac{1}{15}+...+\frac{1}{n^2-1}+...


(h) There are many ways (p-series, integral test, etc.) to prove that the series \displaystyle \sum_{n=1}^\infty\frac{1}{n^2}

converges. However, for this exploration, we will use the convergence of the series in (g) to do this:
Demonstrate that this series converges using both the Comparison Test and the Limit Comparison Test by using the series in (g).

Notes:

• More commonly, the convergence of the series in (g) is demonstrated by comparing it to the p-series. We're doing the reverse here.
• Another important aspect for precalculus and calculus students is to have them compare the partial sums to the sum of the infinite series. Thus, it's worth taking the time to have them see how close the sum is to 0.75 when adding the first 100 terms, the first 1000 terms etc. Also, indicate that the difference can be thought of as the "error" in the approximation. All of this is needed for further study and it deepens their understanding of infinite series.
  
• As indicated above, this investigation may be too time-consuming for a regular period of 40-45 minutes. I would recommend doing parts (a)-(c) (or (d)) in class and assigning the rest for homework to be collected after 2-3 days.
• Teachers of precalculus can use parts of this investigation when developing the concepts of series. Much of the groundwork for infinite series can be laid before students get to calculus!
  

Posted by Dave Marain at 5:55 PM 1 comments

Labels: calculus, investigations, series

With existing technologies and all the help available on the web I am incredulous that students still feel lost at sea the night before a major exam or just getting through an assignment. I've asked so many students what they do when they're frustrated by some math problem at 10 PM: "So who do you call? CalcBusters!"Seriously, they often just look at me as if I have two heads. Well, what are their options?

(a) Call their teacher/professor? Uh, not likely...

(b) Email their teacher/professor? Assuming one has this email address, what are the odds you will receive an immediate reply which will illuminate everything...

(c) Go into a Calc help chat room enabled by your teacher or one set up by some student. A good option if anyone is actually online at that moment. Your teacher or another student can establish guidelines for this so that students know a help session will always be available from 9 PM to 11 PM for example. See (e) below for a similar idea.

(d) Call or email a friend (remember you will then have only 2 lifelines left!). Of course this presumes you have a friend who understands it better than you and can communicate a solution over the phone or via email. Remember what time it is...

(e) Go to a Calculus forum/discussion group in which you can post your question and someone with the knowledge will reply in short order. This is a viable option as there are now many such help groups out there and I will review these and provide links in another post.

OR...

Go to YouTube and find a free video tutorial demonstrating a similar problem in detail.
For example, suppose you're floundering with "integrating by partial fractions." You can just Google "YouTube partial fractions calc video" or something like that and, presto, you are transported here. Ok, the video covers a more sophisticated problem involving a rationalizing substitution as well as partial fractions, but you immediately see dozens of related videos in the sidebar. There are many excellent free videos online from many talented teachers/professors but on this post I will feature one of the best.

If you're interested in seeing an exceptionally clear presentation of calculus or other math topics you cannot do better than Patrick's videos which he offers at no cost. Patrick does not know that I am writing this review so rest assured I am not getting any commission here! The link above is one of his lessons.

He tries to limit his lessons to 10 minutes to make the file size manageable. His writing on the whiteboard is crystal clear, his organizational skills are exemplary and his speaking voice is soooo calming. Further, his explanations are mathematically precise and include just enough rigor to make the purists out there happy without sacrificing clarity. Compare his videos to the amateurish attempts I have posted on this blog - uh, there is no comparison.

Patrick also has a website where you can find all of his free videos. Look here first on YouTube or go directly to his site.
Patrick's background (from his site)
About me: I have been teaching mathematics for over 8 years at the college/university level and tutoring for over 15 years. Currently I teach part time at Austin Community College, but have also taught at Vanderbilt University (a top 20 ranked university) and at the University of Louisville.

He runs a tutoring service in Austin and I'm quite sure he is doing well considering the quality of what he is offering for free. And his videos are not exclusively calculus. Enjoy!

Posted by Dave Marain at 7:48 AM 0 comments

Labels: calculus, video lesson, video review

Remember when I originally posted this problem back in January? Look here .

Here is the original problem:

A hole is drilled (bored) completely through a solid sphere, symmetrically through its center. If the resulting "hole" is 6 inches in height (or depth), show that the remaining volume must be 36π inches cubed.

OBJECTIVE: Motivation, explanation and application of method of cylindrical shells for finding volume of solid of revolution

TOTAL LENGTH: about 45 min

Please Note: These videos are not intended for students who want quick simple explanations for standard homework or typical exam items. This problem is above that level and the explanations are lengthy and very detailed!


Well, this 'video' is fragmented into 7 parts, the transitions are amateurish, it was composed over a few days (therefore different outfits!), cheap props and the quality is well, you know...

In spite of all the negatives, I'm hoping someone will find this helpful. Remember I'm doing this to cover a broad audience -- the Calc I/II student who wants understanding and clarity (not skipping steps!) to the AP student/Math-Sci-Engineering major who wants some theory and rigor. I'm also demonstrating some aspects of pedagogy here for the new calculus instructor who may have to prepare a similar lesson.

As mentioned in the video, there are many wonderful websites and videos which will provide better graphics, animation and quality. A couple of links are provided below. However, my purpose here to provide a highly detailed development of a classic calculus problem which reviews the method of cylindrical shells for volumes of solids of revolution.

Finally, my original intent was to find the volume that was removed by at least two methods and to generalize to a hole of depth h, but this is way too long as it is! Of course, I don't expect many views or comments but it will be out there for anyone who might have use of this for as long as this blog exists! I'm really hoping comments will look past the low-tech aspect and address the content and pedagogy.

Instructors
Please feel free to share this with your students or for whatever purpose you may have.
As stated above, the total length of all parts is about 45 minutes, the length of a typical hs class period, so it wouldn't make sense for the classroom. You might want to recommend students view this after learning the basic idea of the 'shell' method as reinforcement or after assigning this problem for hw or extra credit. These days students are savvy enough to locate, on the web, solutions and videos to most any problem we assign, so be careful! (You already knew that!)

Some Recommended Links
Volumes of Revolution - Cylindrical Shells
As mentioned in the video, patrickJMT is as good as it gets for clear, simple and mathematically accurate explanations.

Volumes - Cylindrical Shell Method
Wonderful explanations and excellent graphics and animation of the shell method (in Flash) from one of the best calculus sites on the web - utk (U Tenn Knoxville)

There are many other outstanding sites - I apologize in advance for omissions here. Just keep searching until you find the one that works for you!

As always, I am responsible for any errors - don't hesitate to point them out! At least we made it before XMAS 2008 ended!

The videos below are connected, so you might want to watch them in sequence.
However, the actual solution to the problem starts in the 5th segment below.
Read the descriptions of the segments to guide you in deciding where to begin. If you do not want a lengthy introduction, and already know the shell method, skip down to the 5th clip.


These first two video clips provide an overview for what I intend to cover.
Also the key relationship R² - r² = 9 is developed.

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These next two segments motivate and derive the method of cylindrical shells.

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The actual solution to the problem starts below!

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Yes folks I know how drawn out this all was. I will try to improve on these but I will take an hiatus from my busy movie production schedule for awhile!

Happy New Year!

Posted by Dave Marain at 2:18 PM 3 comments

Labels: calculus, hole in sphere problem, method of shells, video lesson, volume

Well, I felt badly about that error I made in the original version from a few days ago (which will now be deleted). I also decided to change some portions, electing to solve for the critical values using algebra rather than by the graphing calculator. Finally, I took a tighter view of the whiteboard so that the writing will appear larger. There will be some glare on the board which I hope will not be too distracting. I hope you will find this more helpful and again I apologize for any confusion caused. If you stored the original video, I would ask you to delete that.

The problem in the video below demonstrates important concepts as well as the standard procedure for solving optimization problems. There is also a brief discussion of a heuristic I have found very useful when teaching these kinds of applications. As always I depend on you to share your thoughts. I keep saying this knowing there might not be too many comments!



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Posted by Dave Marain at 10:04 AM 3 comments

Labels: calculus, maxima/minima problem, optimization, video lesson

As promised, here is a video of a related rate application. This one is fairly straightforward but it does demonstrate the procedure to follow when doing the more sophisticated types.

These videos build on the foundation videos I posted which developed the Leibnitz form of the derivative and the Chain Rule as well as the basics of implicit differentiation. I strongly encourage the student to view these first. If you're more familiar with the topic and want a quick review, then start anywhere!

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As always, I depend on my readers to make suggestions to improve the quality of these videos, both mathematically and technologically. Please indicate any errors I may have made so that I can correct them or at least indicate where they are in the video.

Finally, there are DVD sets of Calculus lessons out there as well as other videos you can find on YouTube or by searching. I am interested in knowing if you feel there is a reason for my producing more of these. I hope you enjoy it and, if it helps just one student feel a bit more comfortable with these topics, then it has been worth the effort!

Posted by Dave Marain at 10:22 PM 0 comments

Labels: calculus, implicit differentiation, related rates, video lesson

I 've been promising this for awhile now so I thought I would at least upload the first 3 video segments. The more complicated implicit differentiation example will follow...

Note: There are usually some minor glitches in these productions, no exception here!
(1) In the first video below I inadvertently referred to one of the variables as 'dependent' rather than independent.
(2) On a few occasions the writing near the bottom of the board gets cut off but it should still be possible to follow along.

The three video segments below are linked. I had to split the original video to make it manageable.

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Posted by Dave Marain at 4:09 PM 0 comments

Labels: calculus, chain rule, implicit differentiation, video lesson

Update: The limit below can easily be derived using L'Hopital's theorem. The purpose of this article is to provide practice in algebra and limit manipulations, limit properties and the definition of the derivative prior to using this theorem.











Important Notes:
(1) The condition n ≠ 0 can be relaxed. At some point, students should be asked to analyze the need for restrictions.
(2) The instructor may well want to avoid giving students the above formula, preferring to have them derive it at least in the positive integer case (see comments below in red under "Developing the Problem").
(3) Note that m and n are not restricted to be positive integers. It is recommended that the instructor begin with this restriction on m and n to allow for an algebraic derivation.
(4) This is not an introductory limit exercise! Please read comments (red, bold) below about starting with concrete numerical values before attempting this generalization.

BACKGROUND/OVERVIEW
This is the time of year when Calculus students quickly move into those wonderful limit problems. Epsilon-delta arguments may not be as popular these days but the mechanics of limits are still the challenge for students. Those who have taught this know that students generally struggle with the algebraic simplifications and procedures. Other than these manipulations students generally feel this topic is easy:

Possible Student Thinking: "You just do some algebra, eliminate the "bad" factor in the denominator and plug in. Easy stuff!"

Naturally, if the assignments contain more theoretical limit problems, they may not feel that way!

On the other hand, the algebra can be a major stumbling block for the more challenging exercises. In this post, I will uncharacteristically deemphasize the theory behind the "cancel and plug in" technique and focus on the algebra at first. Then we will move on to relating the limit to the definition of the derivative and application of some important limit properties, in other words, theory! Of course, if L'Hopital's Theorem were introduced early on (in the chapter on differentiation), that would clearly be the method of choice for students!

DEVELOPING THE PROBLEM (SCAFFOLDING)
If m and n in the limit above are positive integers, students can attempt to factor out "x-a" from the numerator and denominator and substitute x = a into the resulting reduced expression. However, many students struggle with such general factoring formulas (or may not have seen them.) Therefore, synthetic division can be used to generate the other factor. This reviews some nice Algebra 2 but what if m and n are not positive integers? What if they are rational or even irrational? Standard factoring techniques would not apply in general so what to do?

I certainly am not suggesting that the instructor begin with the general problem. In fact, I would 'concretize' the problem using a few special cases:
n=2,m=1
n=3,m=1
n=3,m=2
n=1,m=2
n=2,m=2 (This special case is worthwhile as it reviews basic definitions and limit properties).
n=4,m=5 (requires more sophisticated factoring or synthetic)

Based on these exercises, the instructor may ask students if they can develop a general formula for any positive integer exponents. this is in lieu of giving them the formula at the beginning.

FOLLOW-UP: THE GENERAL CASE
After the definition of the derivative is given, students can attempt the more general version. This is a fairly sophisticated limit manipulation but one worth assigning. I may outline the method in an addendum to this post or in the comments or wait for one of our astute readers to contribute! As a hint, the technique I used is related to the derivation of L'Hopital's Theorem!

Posted by Dave Marain at 4:50 PM 7 comments

Labels: calculus, generalization, limits

From Savage Research, Humor...

Math Knowledge

Two mathematicians were having dinner in a restaurant, arguing about the average mathematical knowledge of the American public. One mathematician claimed that this average was woefully inadequate, the other maintained that it was surprisingly high.

"I'll tell you what," said the cynic. "Ask that waitress a simple math question. If she gets it right, I'll pick up dinner. If not, you do." He then excused himself to visit the men's room, and the other called the waitress over.

"When my friend comes back," he told her, "I'm going to ask you a question, and I want you to respond `one-third x cubed.' There's twenty bucks in it for you." She agreed.

The cynic returned from the bathroom and called the waitress over. "The food was wonderful, thank you," the mathematician started. "Incidentally, do you know what the integral of x squared is?"

The waitress looked pensive; almost pained. She looked around the room, at her feet, made gurgling noises, and finally said, "Um, one-third x cubed?"

So the cynic paid the check. The waitress wheeled around, walked a few paces away, looked back at the two men, and muttered under her breath, "...plus a constant."


I'm sorry, but that did make me smile! Reminds me of when I was teaching calc, I would tell my students that if they forgot the +C in an indefinite integral, their grade would be C+! Actually, I wasn't kidding...

BTW, there are many more of these at the above web site. Many are one-liners with that twisted sense of humor characteristic of Steven Wright or Jackie Vernon. I will not apologize for laughing!

Here are a few more...

1) Save the Whales -- collect the whole set.

2) If you believe in telekinesis, raise my hand...

3) The early bird may catch the worm, but it's the 2nd mouse that gets the cheese.

Ok, enuf' already (for now)...

Posted by Dave Marain at 2:32 PM 3 comments

Labels: calculus, math humor

Important Note: It took forever but I finally posted the detailed video explanation of this problem here.


Please don't gag on my feeble attempt at humor in the title (my wife actually had bought a sign with that quote -- it's hanging on the dining room wall).

There are a couple of classic volume problems in calculus which have always been my favorites:

• The Volume of the Torus Problem (using 2 methods: cylindrical shells and by disks)

• The Hole in the Sphere Problem (also by 2 methods)

I always assigned one or both of these to my BC Calculus classes, most often as Extra Credit problems. Because of the extra points they could earn, most students tried these and submitted solutions. My feeling is that if a student could do both of these by both methods, they really understood disk and shell!

In this post we will focus on the 2nd problem as it always seems to generate curiosity and interest. I'm guessing that most of you know the puzzle version of this question that was answered by Marilyn vos Savant in her Ask Marilyn column over a decade ago. It's just possible that some calculus student in some second semester class is feeling some anxiety over this problem!

Here's one version of that famous conundrum. There are many approaches here, even the clever mathematical approach of assuming that the problem is well-defined and therefore independent of the radii involved (I expect at least one of our readers to do it that way!).

A hole is drilled (bored) completely through a solid sphere, symmetrically through its center. If the resulting hole is 6 inches in height (or depth), show that the remaining volume must be 36π inches cubed.

That's right, the answer is independent of the radius of the sphere and the diameter of the hole! The total volume of the sphere and the volume removed however do depend on the radii. Note that the volume removed is a cylinder with two spherical caps.


The original problem was worded ambiguously in Marilyn's column and then clarified somewhat. My version is not perfect but hopefully you'll get the 'picture', although a real picture would be far better. I will probably do a video presentation of the solution and a discussion of the problem because the diagram and the math expressions are cumbersome and it's not worth the time to play with Draw programs or LaTeX right now. I plan on presenting in detail the disk-washer and cylindrical shells method using a general depth of h inches for the hole.

For now, have fun playing with this. This is a well-known problem and therefore searchable on the web but try it yourself first. Try to use calculus to set up the integral and if you're brave you'll evaluate those integrals without Mathematica or the TI-89! Can you see why the answer for the volume remaining depends only on the depth of the hole?

Posted by Dave Marain at 2:09 PM 8 comments

Labels: calculus, paradox, puzzle, volume

As a result of the numerous views of a calculus problem I published in November, I decided to present the following video mini-lesson. As before, I had to break it up into parts to control the file size for uploading. I hope this has some value for those who were looking for a more detailed discussion of this question. Much of this is highly appropriate for precalculus students.

Note: Before playing the videos below, a correction and comments:
(1) In error, I referred to the cross-section of the cone as an isosceles right triangle. Make that isosceles only!
(2) The video and audio quality is far from perfect. Bear with me on this!
(3) I didn't discuss the case where the height of the cone is less than or equal to the radius. This will not produce maximum volume but should have been noted. I will have more to say about this later.
(4) There is so much more to discuss about this question, in particular, the result that the cone of maximum volume has height equal to (4/3)R or that the center of the sphere divides the altitude into a 3:1 ratio. These may be discussed in upcoming videos. In particular, as suggested in the videos below, there will be a treatment of the 2-dimensional analogue of this problem, namely, the isosceles triangle in the circle problem.
(5) These video 'mini' lessons are designed for the university or secondary calculus student (probably comes too late for the college final exam) or for anyone wanting a refresher. Beyond my personal style of presentation, there are pedagogical issues (instructional tips) that arise in the videos that might be of interest to someone teaching calculus for the first time.

If you're getting bored of watching the same chalkboard and my same drab outfit, well, it is a low-budget video! I hope you will let me know if this proves helpful and if you'd like me to continue these. As mentioned previously, I will also be employing other technologies for demonstration purposes.

Happy Holidays!


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Posted by Dave Marain at 4:14 PM 19 comments

Labels: calculus, cone in sphere, maxima/minima problem, precalculus, video lesson

It's been awhile but something this good is always worth waiting for!
TC has sent me some fascinating challenge problems for our readers. If you are now sick of watching amateur videos on the Arithmetic and Geometric Mean Inequality, it's time to raise the bar. The following involves a well-known generalization of these means but the results are worth your efforts, particularly parts (c) and (d) below.

If a and b are positive, we can define their generalized mean to be:

GNM = ((a^k + b^k)/2)^(1/k)

This would look far prettier in LaTeX but I'm hoping it's readable. In words, we're looking at:
The kth root of the arithmetic mean of the kth powers of a and b.

(a) What is another name for the result when k = 1? (we're starting off easy here!)
(b) What is another name for the result when k = -1? (slightly harder algebraically)
(c) Ok, now for the real challenge for you Calculus lovers:
What is the limit of GNM as k-->0? The result is totally cool!
(d) TC's Super Bonus: Show that the limit of GNM as k-->∞ is the maximum of a and b.

Note: These have been slightly edited from tc's original problems, but they are essentially the same. Solutions may be posted in a couple of days although the notations will be hard to render. I might just have to do another video or wait for that special technology I mentioned earlier! We're hoping some of you will tackle the harder ones and comment!

Posted by Dave Marain at 5:49 AM 14 comments

Labels: calculus, generalized means, limits, totally clueless challenge

Update: View the series of videos here explaining the procedure for solving the cone in the sphere problem below as well as related questions.

Many Algebra 2 and Precalculus textbooks have begun to include those challenging 3-dimensional geometry questions involving 2 or more variables and/or constants. However, we know from the difficulty that calculus students continue to have with these, that we need to do more before students do their first optimization problems in calculus. You know the kind: Determine the radius of the __________ of maximum volume that can be inscribed in a _________ of radius R. These problems have fallen out of favor somewhat with the AP Development Committee, perhaps because they lack that real-world flavor or perhaps because they had become predictable or perhaps too hard. I would argue they have been part of the rites of passage for calc students for many generations for a reason - they blended the spatial reasoning of geometry with the need to identify variable relationships and reduce the number of conditions down to one function of one variable if possible. In other words, they help to develop mathematical sophistication. I 'cut my teeth' on these -- did you? Any calculus teachers reaching this topic yet in AP Calc?

Anyway here's an activity for you Algebra 2 or Precalculus students to prepare them for these challenges. As usual we proceed from the concrete (i.e., given numerical dimensions) to the abstract. Rather than attempt to draw the diagram, which is fairly challenging for me given the tools I have, I will describe the problem verbally. Good luck!

STUDENT ACTIVITY

(1) A right circular cone of height 32 is inscribed in a sphere of diameter 40.
Note: Students need to learn how to make a diagram of this problem situation.

(a) Determine the radius of the cone.
(b) Determine the volume of the cone. [Imagine asking students to memorize the formula!]
(c) Keep the diameter of the sphere at 40. This time, determine both the radius and volume of the inscribed cone whose height is 80/3. The numbers are messy but try to work in exact form (fractions, radicals) before rushing to the calculator to convert everything to decimals. Oh well, we all know what will happen here!
(d) Try another value for the height of the cone, keeping the diameter of the sphere at 40. See if you can produce a volume greater than in (c). Any conjectures?

(2) We could throw in an intermediate step by using a parameter R to denote the radius of the sphere, and use numerical values for different possible heights of the cone, but I'll leave that to the instructor. Instead, we'll jump to the abstract generalization:

A right circular cone of height h is inscribed in a sphere of radius R.
(a) Express the radius, r, of the cone in terms of R and h.
(b) Express the volume, V, of the cone as function of h alone (R is a constant here).
(c) Use your expression for r and your function for V to verify your results in (1).
(d) Calculus Students: You know what the question will be! Oh, alright:
Determine the dimensions and volume of the right circular cone of maximum volume that can be inscribed in a sphere of radius R. Anything strike you as interesting in this result?

Posted by Dave Marain at 7:36 AM 0 comments

Labels: calculus, cone, functional relationships, geometry, optimization, precalculus, sphere

As you may have read in an earlier comment, I've invited one of MathNotations' most dedicated and talented contributors to go beyond commenting and share some of his creative ideas and insights by being an occasional guest blogger - he has graciously accepted.

For his inaugural offering, tc is challenging you and/or your students to solve a classic calculus problem using non-calculus methods. I have made a few minor edits, but the activity is essentially what tc sent to me.
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I give you tc's Total Challenge I:

One of my math professors in college used to say there were three ways
of tackling any problem: the right way, the wrong way and the Navy way
(correct, but extremely roundabout).

In this exercise, we will look at three ways (not necessarily the ones
named above) of doing the following problem:

Determine the rectangle of maximum area that can be inscribed
in a circle of given radius r.

Let the inscribed rectangle have sides a and b. The diagonal of the rectangle passes through the center of the circle (this can be shown, but you can assume it is true).

(1) Express r in terms of a and b.

(2) Express the area in terms of a and r.

(3) Instead of maximizing the area, we can maximize the square of the area.
(a) Express the square of the area as a quadratic in a² (you may want to substitute c for a²).
(b) By completing the square, determine the value of a for which the area is a maximum.
(c) Determine the value of b and the maximum area.
(d) What conclusion can you draw about the rectangle of maximum area?
(This is the first way, which I call the Algebra way)

(4) Divide the rectangle into 2 congruent triangles, using a diagonal. Draw a half
diagonal that intersects this diagonal.
(a) Write an inequality for the area of one of these triangles in terms of r alone. The inequality should be of the form Area ≤ _______.
(b) If you can achieve equality, then you have maximized the area of the rectangle! Find out when this occurs, and if it does, find the lengths of a and b. (The Geometry way).

(5) Method 3 - the Calculus way of course.
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Additional comments from DM:
(i) thanks, tc!
(ii) tc's geometric approach in (4) also suggests a connection to the famous AM-GM Inequality. Visit this link and see if you can make the connection. This is not obvious.
Hint: Apply the AM-GM to a² and b².

Posted by Dave Marain at 10:18 PM 2 comments

Labels: calculus, inscribed rectangle, maxima/minima problem, tc'sTotal Challenges

With the AP Exams this week and the year coming to an end, the timing of this post is way off but I wanted to share some thoughts about using u-substitutionin integration (and differentiation). Every text promotes it, I've been teaching it for years, but, like so much else in mathematics, u-substitution is really a form of 'information-hiding' that may sacrifice understanding for the sake of brevity. While I believe using u in place of f(x) is an important technique for some calculus algorithms, experience has shown me that keeping f(x) in the formulas works better during the learning phase. In this post I will show how using f(x) in the formulas for integrals may help students deal with the variety of functions they encounter when integrating using the 'Inverse Chain Rule' (I coined that phrase to help students recognize that the f'(x) factor disappears when anti-differentiating (integrating) and appears when differentiating. If your first reaction is that these formulas are more complicated, I think you might need to actually experience the difference in students' reactions. There's no substitute for that!!

If you've already tried both approaches, let me know which form worked better for a majority of your students. My impression is that many students are at first overwhelmed by the varieties of different forms that appear in the typical set of exercises following this lesson. We can say that they're basically all the same, but it doesn't appear that way to the novice!

Outline of Lesson:
The topic was introduced by reviewing several differentiation problems using the General Power Rule for d/dx(f(x)), using the Chain Rule. Each derivative was immediately followed up with the corresponding antiderivative problem. The idea that both (f(x))ⁿ and f'(x) must be present in the integrand was emphatically stressed. I repeatedly asked the question, "What is the f(x) here?" then, "What would f'(x) be?" Also, "Is the exact f'(x) in the integrand?" If not, we discussed the method of supplying a missing numerical factor and its reciprocal, using properties of antiderivatives, etc. They seemed to get this, struggling a bit with the idea of using differentiation in an integration problem, a natural source of confusion at first. We made sense of all this for the first few examples by differentiating our answer to check it. As with all new ideas, some caught on quickly, others had not yet internalized it after 3-4 examples because I too quickly moved into variations on the basic form before they had really processed it.

The following are the 'forms' that helped them adjust to the variety of trigonometric and other types of examples. This seemed to eliminate much confusion...

Some ∫ form(ula)s:

1. (General Power Rule): ∫f'(x)(f(x))ⁿdx = (f(x))ⁿ⁺¹ /(n+1) + C,
n ≠ -1

2. ∫cos(x)sinⁿ(x)dx = (sinⁿ⁺¹(x))/(n+1) + C, n≠ -1

3. ∫f'(x)sin(f(x))dx = -cos(f(x)) + C
OR
4. ∫f'(x)cos(f(x))dx = sin(f(x)) + C
Note: There is a critical but subtle difference between #2 and #3, even though they are both applications of the Chain Rule.

5. ∫f'(x)sec²(f(x))dx = tan(f(x)) + C
6. ∫f'(x)sec(f(x))tan(f(x))dx = sec(f(x)) + C

7. ∫f'(x)e^f(x)dx = e^f(x) + C

8. ∫f'(x)/f(x)dx = ln|f(x)| + C vs. ∫du/u = ln|u| + C

There are many others, but you get the idea.
If you still believe that the u-forms are so much simpler, that's fine, since I like the u-method as well and it is needed later on. I think all experienced math teachers latch on to their favorite methods or models because they have worked effectively with many students. However, I do not believe that any method will be effective for all due to learning style issues. There may be some students who will simply feel more comfortable and perform better if they use Method A vs. Method B.

I should have included some of the actual examples used, but...

Posted by Dave Marain at 12:58 AM 5 comments

Labels: calculus, chain rule, differentiation, integration, u-substitution

Update: I've added another 'paradox' in the comments. With the AP Calculus (BC) Exam looming, AP teachers may want to share this with their students for review.

[The following AP level question is designed for upper level students.]

Why do mathematicians have to be so rigid, um, I mean, rigorous?

Here's an AP Calculus problem brought to my attention this morning by one of our outstanding Calculus teachers, Mr. D. He found this in an AP Review book.

[Rather than play with the symbols, I'll 'write it out']:

The definite integral of sec²(x) from x = 0 to x = 3pi/4 is?
It was multiple choice and the answer given was -1.

I shared this problem with my AP group later on in the morning and I asked them why that answer makes no sense. R.J. immediately replied, "The answer can't be negative since sec²(x) is never negative."
Of course, but let's work it out!

[I intentionally did it incorrectly at first]:
By the Fundamental Theorem, the integral equals tan(3pi/4) - tan(0) = -1!

What's going on here! I was gratified that one of my students recognized that the function sec²(x) has an infinite discontinuity at x = pi/2, so the original integral is improper. When we integrate from 0 to pi/2, then from pi/2 to 3pi/4, and apply the rigorous limit definition of an improper integral, we see that the integral diverges! If anything, the 'area' is infinite or unbounded.

Using these kinds of 'paradoxical' examples and asking students to 'FIND THE ERROR' is a wonderful device many educators use to deepen student understanding of mathematics and demonstrate the need to be rigorous!

Now why isn't the definite integral of 1/x from -1 to 1 equal to zero, since the region in the first quadrant 'clearly cancels' the part in the 3rd quadrant?? Hmmm... I'll bet some of you could explain this and find many other such 'paradoxes'!

Posted by Dave Marain at 11:43 AM 4 comments

Labels: calculus, improper integral, paradox

MAA members will likely recognize the following challenge that appeared on the outside of the envelope in the mailing to members or prospective members. I plan on giving this to my AP Calculus students as review for the exam or afterwards. As usual I will modify it for the student, place it in the context of an activity, broken into several parts with some hints. The original problem comes with a helpful diagram, however, unless I scan it, it would be difficult to reproduce. The problem involves a property of a point on an ellipse and requires basic understanding of the parametric form of this curve and some basic calculus and trig. The last part of the activity suggests a possible significance of this property but I'll leave the details to our astute readers.

Consider a standard ellipse, center at (0,0), with major axis of length 2a on the x-axis and minor axis of length 2b.
Let P(x,y) be a generic point on this ellipse with the restriction that P is not one of the endpoints of the major or minor axes. Consider the tangent and normal lines at P. Let P denote the point of intersection of the normal line with the x-axis and Q, the point of intersection of the tangent line with the x-axis.
Prove that (OP)(OQ) = a²-b², where OP represents the distance between the origin and P and similarly for OQ.
Here is an outline with several parts for the student:
(a) Show that x = acos(t), y = bsin(t), 0<=t<2pi,>2-b²)/a)cos(t).
(f) Use (d) and (e) to derive the desired result: (OP)(OQ) = a²-b²
(g) Explain why we did not allow P to be an endpoint of the major or minor axes.
(h) What does the expression a²-b² have to do with the foci of the ellipse? For EXTRA CREDIT, investigate this 'focal' property further.

Posted by Dave Marain at 8:40 AM 0 comments

Labels: calculus, ellipse, MAA, parametric, tangents, trigonometry

To continue our discussion of infinite series, I usually show students the famous proof that the harmonic series 1+1/2+1/3+1/4+... diverges. This series is paradoxical to students because, in their minds,there is convergence, since the terms themselves approach zero. With some exploration they can begin to appreciate that convergence of the sum of the terms depends on how fast the terms approach zero! Most of the content of the student investigation below can be found in MathWorld or Wikipedia but my intent, as it almost always is on this blog, is to produce a classroom experience for students and an activity for teachers to use, not just an expository piece of writing.

Consider the following "S-series":
1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + 1/16 + 1/16 +...
(a) Continuing this pattern (of repeating groups of reciprocals of powers of 2), what would the 16th term be?
(b) If S_n represents the sum of the first n terms of this series (where n is a positive integer), what is the value of S₁₆? No calculator!
(c) Develop a formula for S_2ⁿ and verify your formula for S₁₀₂₄. Here, n = 0,1,2,...
(d) What conclusion can you draw about the convergence of the "S-series?" Explain.
(e) Consider the harmonic series (which we will call the "H-series"):
1 + 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/n + ...
Let H_n represent the sum of the first n terms of this series.
Show that H₁₆> 3, H₁₀₂₄> 6 and H₆₅₅₃₆> 9 by comparing the "H-series" to the "S-series" term by term.

(f) Based on the above, what conclusion can you draw regarding the limit of the sequence of partial sums, H_n? What does this imply about the convergence or divergence of the harmonic series? How would you describe the rate at which this series converges or diverges?
(g) Research the harmonic series online. Be prepared to answer the following question:
What does the harmonic series have to do with overtones in music?
(h) Consider generalizations of the harmonic series, such as replacing 1/n by 1/(kn+j). Make two such generalizations and examine convergence in each case.

The possibilities are endless. If two roads diverged in the woods, which one would you take?

Posted by Dave Marain at 12:50 PM 5 comments

Labels: calculus, harmonic series, series

The following may drive away most casual readers but it does describe what I try to do every day. One of my goals in starting this blog was to enable a dialogue for effective instructional strategies. My focus has generally been on middle and secondary school curriculum up to Algebra 2, bordering on Precalculus. Today I am sharing a different experience. I hope some of you will appreciate it beyond its technical aspects. Similar developments can be found in some textbooks and experienced teachers already do most of this but as this scenario is fresh in my mind, I thought I'd re-play it for you...

Although most Advanced Placement Calculus (BC) teachers are completing or have already completed the unit on infinite series, I would like to offer a view that I hope brings a sense of 'shock and awe' to the student of the 21st century who rarely has the time to stop and appreciate the beauty of our subject. To those who have been teaching this for a while, you may not quite feel this. However, I still get goosebumps when I observe student reactions as this unfolds in front of their eyes...

Assume that students already have a basic understanding of infinite series, the infinite geometric series in particular.

Consider the following three infinite geometric series:

1+1/2+1/4+1/8+... = 1/(1-1/2) = 2
1+1/3+1/9+1/27+... = 1/(1-1/3) = 3/2
1-1/4+1/16-1/64+... = 1/(1-(-1/4)) = 4/5

Just a collection of simple geometric series, boys and girls?
Genius is looking at an ordinary collection of objects and seeing something different. Some mathematician or mathematicians (research this and report back with their bios!) may have considered a reverse view of these series. Instead of the goal being a formula for the sum of the series, perhaps the goal was to represent a function in a different way. Step back into history...

Consider the general formula for the sum of all these series: 1/(1-r) provided r is between -1 and 1. Replace r by x, the variable we usually use for functions, and we can write:
1 + x + x² + x³ +... = 1/(1-x) provided x is between -1 and 1.

The 'polynomial' of infinite degree on the left is known as a power series in x. As long as x is between -1 and 1 (the interval of convergence), this 'equation' makes sense and allows us to use algebraic and calculus operations to represent other related functions. Think of how one might have felt when 'discovering' this and I'm just speculating here. The rational function 1/(1-x) is being represented by some kind of polynomial that never ends. Even though x= 1 is not in the interval of convergence, substituting leads to 1/0 = 1 + 1 + 1 + 1 +.... Hmm....
Let's try substitution on this representation.
Replace x by -x²:
(You can show the domain is unchanged)
1/(1 - (-x²)) = 1 + (-x²) + (-x²)² + (-x²)³ + ... OR
1/(1 + x²) = 1 - x² + x⁴ - x⁶ +...

Ok, let's integrate both sides (assuming it's legal to do so):
tan^-1(x) = x - (1/3)x³ + (1/5)x⁵ - (1/7)x⁷ + ... + C
Replacing x by 0, we see that C = 0.
Now, you'll have to accept this for the moment (to be proved later), equality holds for x = 1, even though 1 was not in the original interval of convergence! It is not unusual when integrating a power series to see the domain include one or both endpoints even though the original function excluded them!

Thus, tan^-1(1) = 1 - 1/3 + 1/5 - 1/7 + ...
Anyone recognize the left-hand side?
The bell rings...

Posted by Dave Marain at 7:07 AM 8 comments

Labels: calculus, geometric sequence, power series, series

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