Football and the hypotenuse

An amusing video from Training Camp 2013 for illustrating that each side of a triangle (including right triangles) is shorter than the sum of the other two sides. The speaker is Jason Garrett, head coach of the Dallas Cowboys.

Engaging students: Introducing the two-column, statement-reason paradigm of geometric proofs

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Derek Skipworth. His topic, from Geometry: introducing the two-column, statement-reason paradigm of geometric proofs.

A. Applications – How could you as a teacher create an activity or project that involves your topic?

I would have the students get in groups and come up with 5-8 statements on one sheet of paper, numbering each one. This could be a statement about the weather, something that happened the day before, anything. My example would be “I wore a long sleeve shirt today”. After coming up with these statements, I would then have the students create reasons behind these statements on a separate sheet. For each statement, the students would have to ask “why…”. For my example, it might be that it was laundry day and it was my only clean shirt, or that it was cold outside. Upon generating all reasons behind each statement, I would then introduce the proof model.

B. Curriculum – How can this topic be used in your students’ future courses in mathematics or science?

The two-column, statement-reason paradigm is a system that can actually be used in all subjects. The idea behind it, giving a statement on the right and a reason on the right, can be applied to almost everything. For problem solving, you can work through an entire problem step by step and explain why you think that is the correct process. In a class such as Calculus, this could be used to help them memorize derivatives by doing the problem on the left and listing what “tool” they used for each step of the process. Even for something like social studies, this process could be adapted into a tool similar to the Cornell Notes (http://coe.jmu.edu/LearningToolbox/cornellnotes.html). In this process, you use the two-column approach. On the left, you list your main ideas, while the right column “explains” what you know about the idea.

E. Technology – How can Technology be used to effectively engage students with this topic?

I had issues tying this topic to a third question. While it’s a good topic, its more of a process than an actual concept. This would actually qualify as an engage activity, slightly different to the one mentioned above, but I would see it working better as a take-home assignment than an in-class one. The assignment would be to use YouTube and pull up a video that piques their interest. Obviously, it needs to be school-appropriate. This could be their favorite music video, a funny video of cats, whatever it is would work. They would write the name of the video at the top and provide a link to the video if possible. Then, they would take the paper and fold it in half, hot-dog style. On the left, they list the names of videos on the suggested pane to the right, in order as they appear. On the right, they would add comments about how that video was related to the video they chose and why it was in that order. The idea is that this should take a bit of thinking since often times the videos appear to be randomly added to that queue. This would reinforce the model while hopefully developing a better idea of how a website they are familiar with operates. Though this could be done with any search engine as well, I feel those are just too similar to offer any “investigative” work for the students.

Seen above, one would likely suspect that a Florence and the Machine video would pull up various other Florence videos in the top 9; however, the snapshot shows that this is not the case. We see that most results have nothing to do with Florence. We see that there are a few matches based on the KEXP live performance. When listening to some others, it might be reason to believe they were in the queue not only based on a performance, but also because the genres are very similar. That would be my main conclusion about Gotye’s music video being included in the list: it’s a very popular song and is in the same genre as Florence.

Engaging students: Congruence

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Kelsie Teague. Her topic, from Geometry: congruence.

Application.

Many students in high school go to the county or state fair yearly. I would start off by giving students a picture of a ferris wheel and having them find as many triangles in that ferris wheel that have what seem to be the same sides and angles and see how many different answers I get. After defining congruence, I would continue to ask the students if they thought this ferris wheel could be constructed without the idea of congruence. If the shapes in this ferris wheel were different sizes would it still work properly? I would then use this as a basis of what people need the idea of congruence to do their job.

Culture.

Congruence shows up in art work all over the place. It can show up in photography with taking picture of identical twins. Those twins are congruent but they are not the same person therefore they are not equal. I would post some pictures of art work and talk about the differences and have the student explain to me what they see. The bottom piece is made using the exact same shape and the idea of congruence. I would show my students some pictures and how the lesson for that day can be related to art work in real life.

Technology

http://www.khanacademy.org/math/geometry/congruent-triangles

The above website is a great hands on activity. It lets the students move triangles around to see if they can form triangles that aren’t the same. It also uses previous knowledge to guide them into the idea of congruence. Khanacademy.org also has other activities that can help with previous knowledge and then activities that take the concept of congruence and build on it. The activity I did was really good, it let me drop and stretch triangles to try and make them non congruent. It also gives one where you can’t lengthen the side but you can move it around and try to make a triangle out of it. I think this activity could show students about congruence in a different kind of way.

Engaging students: Solving for unknown parts of triangles and rectangles

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Michelle McKay. Her topic, from Algebra I: solving for unknown parts of triangles and rectangles.

A. How could you as a teacher create an activity or project that involves your topic?

There are several different ideas that immediately come to mind on how to center a lesson around solving for unknown parts of rectangles and triangles. I would like to focus on and describe one. For this particular lesson, the student will start by making a prediction of which side(s) of a shape (triangle or rectangle) has the greatest length. Then, with a partner, they will use rulers and a handout to record the dimensions of both shapes. On the handout, they will work to fill out the chart provided. Then, we will reconvene as a class and talk about the discoveries made. For rectangles, I would ask first about what we found to be consistent for every rectangle. Using what we know, how we could find or solve for the length of one side if we only had certain parts of information? Similarly for triangles, I would begin by asking how each side differed from one another. Did the general shapes of the triangles make a difference? What was special about the right triangles? After these questions, I would introduce Pythagorean’s Theorem and have them solve for the side of triangles without rulers, then follow up with using rulers to verify their information.

D. What interesting things can you say about the people who contributed to the discovery and/or the development of this idea?

Pythagoras of Samos: During Pythagoras’ time, math was considered to be a mixture of both religious and scientific beliefs and was often associated with secret societies and only those of very high social standing. As Pythagoras was one of the more influential mathematicians of his time, most details of his life were kept secret until centuries after his death, leaving very little reliable information to be pieced together in form of a biography. It is generally accepted that he was born on the island of Samos, which is now incorporated into the country of Greece. Little is known about his childhood, but most agree that he was very well educated and was acquainted with geometry before he traveled to Egypt. He was known to be almost sacrosanct and divine to those alive during his time and even a few well after his death. He founded a religious, and simultaneously mathematical, movement called Pythagoreanism, which consisted of two schools of thought: the “learners” and the “listeners”.

D. What are the contributions of various cultures to this topic?

Time Period	Civilization	Contribution
Earliest known references: 23^rd Century B.C.	Babylonians	– Had rules for generating Pythagorean triples. – Comprehended the relationship of a right triangle’s sides. – Discovered the relationship of $\sqrt{2}$ .
500 – 200 B.C.	Chinese	– Gives a statement and geometrical demonstration of the Pythagorean Theorem (possibly before Pythagoras’ time).
570 – 495 B.C.	Greek	– Golden rectangles were very vaguely referenced by Plato. – Euclid wrote a clear definition of what a rectangle is. – Pythagoras discovered a relationship between the sides of right triangles.
Earliest known references: 800 – 600 B.C.	Indian	– Pythagorean Theorem was utilized in forming the proper dimensions for religious altars.

It is very hard to for historians to pinpoint with exact certainty which civilization was the first to discover what we know now as the Pythagorean Theorem. Many of the civilizations listed above existed during the same time period, but were geographically located on opposite ends of the map. Also due to loss of information from translations, damaged or completely destroyed texts, these dates and the authenticity of certain contributions are still debated to this day.

Sources

Engaging students: right-triangle trigonometry

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Kelsie Teague. Her topic, from Precalculus: right-triangle trigonometry.

How has this topic appeared in popular culture?

In the famous T.V. show Numbers they do an episode using trig to find the angle of origin of the blood spatter. In forensic science they use trig every day to determine where the victim was originally injured. They can also use this to find the angle of impact, area/point of convergence, and area of origin. The following power point goes into more detail: http://cmb.physics.wisc.edu/people/gault/Blood%20Splatter%20Trig.pdf

If the blood was dropped by a 90-degree angle, the stain will appear to be an almost perfect circle.

We could get out some long paper and colored water and experiment with the idea of change of angle in the drop of blood and calculate the angles.

Angle of Impact =Sin (theta)= Width of drop/Length of drop.

How can technology (YouTube, Khan Academy [khanacademy.org], Vi Hart, Geometers Sketchpad, graphing calculators, etc.) be used to effectively engage students with this topic?

YouTube is a great website for engagement because you can find many videos to start the lesson off with some previous knowledge that they will be using for that days lesson. The following video would be a good way to engage the students when talking about right triangle trig.

It’s to a song that they probably have already heard and it’s teaching them something they already know. Since the students already have knowledge of this, the video isn’t teaching them the topic but refreshing their memory in a entertaining fashion.

When looking for a good video, I ran across many that would work for this lesson, but this one seemed like it would grab the student’s attention more and keep their attention.

The above video is also a good one, and it shows the lyrics in the description so you can make sure what they are saying is mathematically correct so it doesn’t give the students any misconceptions.

How could you as a teacher create an activity or project that involves your topic?

If I was teaching at a school that was close to a hill or a mountain outside I could take my students outside and have them figure out how far up the mountain they would have to walk to get to the top. We could use a tap measure to measure how high they had the protractor in the air and then we could look up the height and distance away of the mountain. They then could use the protractor to find the angle between themselves and the top of the mountain. We could then use this information inside the classroom to solve how far to travel up the mountain.

Similar to the above picture except they will know the height of the mountain. This would show the length of the hypotenuse of the right triangle. They will have to subtract the height they have the protractor at from the height of the mountain to be accurate since the height of the mountain is from the ground up.

Area of a circle (Part 4)

Math majors are completely comfortable with the formula $A = \pi r^2$ for the area of a circle. However, they often tell me that they don’t remember a proof or justification for why this formula is true. And they certainly don’t remember a justification that would be appropriate for showing geometry students.

In the previous three posts, I discussed various ways that calculus can be used to show that $A = \pi r^2$ . Still, most future high school teachers would like to know a justification for why $A = \pi r^2$ . After all, the definition of $\pi$ is

$\pi = \displaystyle \frac{ \hbox{Circumference} }{ \hbox{ Diameter}}$ , or $C = 2\pi r$

So there ought to be a reasonable explanation for why $\pi$ reappears in the formula for the area of a circle. Furthermore, this explanation should within the grasp of geometry students — so that the explanation should not explicitly use calculus. Even better, they’d prefer a hands-on classroom activity so that students could discover the formula for themselves.

The video below shows a completely geometric justification for why $A = \pi r^2$ that meets the above criteria. I have a couple of small quibbles with the narrated text — I’d prefer to say that the each rearrangement of pieces is approximately a parallelogram (as opposed to a rectangle), and that figures get closer and closer to a real parallelogram with area $A = \pi r^2$ .

In other words, I would avoid saying that we ultimately divide the circle into infinitely many wedges of infinitesimal width to get a perfect rectangle, as this promotes a misconception concerning the definition of a limit that they shouldn’t carry into a future calculus course.

However, the graphics are excellent in this video. In my mind, that more than counterbalances the preferred way that I would describe the process of taking a limit to students.

Pedagogically, I would recommend a hands-on activity along these lines. Let the students use a protractor to draw a 5- or 6-inch circle on a piece of paper. Then have them mark $18$ points on the circumference of the circle at every $20^o$ , and then draw the lines connecting these points and the center of the circle. Then have the students cut out these wedges and physically rearrange them as in the video. They should discover for themselves that the wedges approximately form a parallelogram, and they know how to find the area of a parallelogram.

After they do this activity, then I would show the above video to geometry students.

If anyone knows a video that (1) is as visually appealing as the one above and (2) correctly states the principle of limit for geometry students, please let me know.

Area of a circle (Part 3)

$A = \displaystyle \iint_R 1 \, dx \, dy$

This double integral may be computed by converting to polar coordinates. The distance from the origin varies from $r=0$ to $r=a$ , while the angle varies from $\theta = 0$ to $\theta = 2\pi$ . Using the conversion $dx \, dy = r \, dr \, d\theta$ , we see that

$A = \displaystyle \int_0^{2 \pi} \int_0^a r \, dr \, d \theta$

$A = \displaystyle \int_0^{2\pi} \left[ \frac{r^2}{2} \right]_0^a \, d\theta$

$A = \displaystyle \int_0^{2\pi} \frac{a^2}{2} \, d\theta$

$A = \displaystyle 2 \pi \cdot \frac{a^2}{2}$

$A = \displaystyle \pi a^2$

We note that the above proof uses the fact that calculus with trigonometric functions must be done with radians and not degrees. In other words, we had to change the range of integration to $[0,2\pi]$ and not $[0^o, 360^o]$ .

Area of a circle (Part 2)

A circle centered at the origin with radius $r$ may be viewed as the region between $f(x) = -\sqrt{r^2 - x^2}$ and $g(x) = \sqrt{r^2 - x^2}$ . These two functions intersect at $x = r$ and $x = -r$ . Therefore, the area of the circle is the integral of the difference of the two functions:

$A = \displaystyle \int_{-r}^r \left[g(x) - f(x) \right] \, dx= \displaystyle \int_{-r}^r 2 \sqrt{r^2 - x^2} \, dx$

This may be evaluated by using the trigonometric substitution $x = r \sin \theta$ and changing the range of integration to $\theta = -\pi/2$ to $\theta = \pi/2$ . Since $dx = r \cos \theta \, d\theta$ , we find

$A = \displaystyle \int_{-\pi/2}^{\pi/2} 2 \sqrt{r^2 - r^2 \sin^2 \theta} \, r \cos \theta d\theta$

$A = \displaystyle \int_{-\pi/2}^{\pi/2} 2 r^2 \cos^2 \theta d\theta$

$A = \displaystyle r^2 \int_{-\pi/2}^{\pi/2} (1 + \cos 2\theta) d\theta$

$A = \displaystyle r^2 \left[ \theta + \frac{1}{2} \sin 2\theta \right]_{-\pi/2}^{\pi/2}$

$A = \displaystyle r^2 \left[ \left( \displaystyle \frac{\pi}{2} + \frac{1}{2} \sin \pi \right) - \left( - \frac{\pi}{2} + \frac{1}{2} \sin (-\pi) \right) \right]$

$A = \pi r^2$

Area of a circle (Part 1)

In this series of posts, I’ll discuss several ways that the area of a circle can be found using calculus. I’ll also discuss a straightforward classroom activity by which students can discover for themselves why $A = \pi r^2$ .In the first few weeks after a calculus class, after students are introduced to the concept of limits, the derivative is introduced for the first time… often as the slope of a tangent line to the curve. Here it is: if $y = f(x)$, then

$\displaystyle \frac{dy}{dx} = y' = f'(x) = \lim_{h \to 0} \displaystyle \frac{f(x+h) - f(x)}{h}$

From this definition, the first few rules of differentiation are derived in approximately the following order:

1. If $f(x) = c$ , a constant, then $\displaystyle \frac{d}{dx} (c) = 0$ .

2. If $f(x)$ and $g(x)$ are both differentiable, then $(f+g)'(x) = f'(x) + g'(x)$ .

3. If $f(x)$ is differentiable and $c$ is a constant, then $(cf)'(x) = c f'(x)$ .

4. If $f(x) = x^n$ , where $n$ is a nonnegative integer, then $f'(x) = n x^{n-1}$ . This may be proved by at least two different techniques:

The binomial expansion $(x+h)^n = x^n + n x^{n-1} h + \displaystyle {n \choose 2} x^{n-2} h^2 + \dots + h^n$
The Product Rule (derived later) and mathematical induction

5. If $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ is a polynomial, then $f'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + a_1$ . In other words, taking the derivative of a polynomial is easy.

After doing a few examples to help these concepts sink in, I’ll show the following two examples with about 3-4 minutes left in class.

Example 1. Let $A(r) = \pi r^2$ . Notice I’ve changed the variable from $x$ to $r$ , but that’s OK. Does this remind you of anything? (Students answer: the area of a circle.) What’s the derivative? Remember, $\pi$ is just a constant. So $A'(r) = \pi \cdot 2r = 2\pi r$ . Does this remind you of anything? (Students answer: Whoa… the circumference of a circle.)

Example 2. Now let’s try $V(r) = \displaystyle \frac{4}{3} \pi r^3$ . Does this remind you of anything? (Students answer: the volume of a sphere.) What’s the derivative? Again, $\displaystyle \frac{4}{3} \pi$ is just a constant. So $V'(r) = \displaystyle \frac{4}{3} \pi \cdot 3r^2 = 4\pi r^2$ . Does this remind you of anything? (Students answer: Whoa… the surface area of a sphere.)

Hmmm. That’s interesting. The derivative of the area of a circle is the circumference of the circle, and the derivative of the area of a sphere is the surface area of the sphere. I wonder why this works. Any ideas? (Students: stunned silence.)

This is what’s known on television as a cliff-hanger, and I’ll give you the answer at the start of class tomorrow. (Students groan, as they really want to know the answer immediately.)

In the spirit of a cliff-hanger, I offer the following thought bubble before presenting the answer.

By definition, if $A(r) = \pi r^2$ , then

$A'(r) = \displaystyle \lim_{h \to 0} \frac{ A(r+h) - A(r) }{h} = 2\pi r$

The numerator may be viewed as the area of the ring between concentric circles with radii $r$ and $r+h$ . In other words, imagine starting with a solid red disk of radius $r +h$ and then removing a solid white disk of radius $r$ . The picture would look something like this:

Notice that the ring has a thickness of $r+h -r = h$ . If this ring were to be “unpeeled” and flattened, it would approximately resemble a rectangle. The height of the rectangle would be $h$ , while the length of the rectangle would be the circumference of the circle. So

$A(r + h) - A(r) \approx 2 \pi r h$

and we can conclude that

$A'(r) = \displaystyle \lim_{h \to 0} \frac{ 2 \pi r h}{h} = 2\pi r$

By the same reasoning, the derivative of the volume of a sphere ought to be the surface area of the sphere.

Pedagogically, I find that the above discussion helps reinforce the definition of a derivative at a time when students are most willing to forget about the formal definition in favor of the various rules of differentiation.

In the above work, we started with the formula for the area of the circle and then confirmed that its derivative matched the expected result. However, the above logic can be used to derive the formula for the area of a circle from the formula $C(r) = 2\pi r$ for the circumference. We begin with the observation that $A'(r) = C(r)$ , as above. Therefore, by the Fundamental Theorem of Calculus,

$A(r) - A(0) = \displaystyle \int_0^r C(t) \, dt$

$A(r) - A(0) = \displaystyle \int_0^r 2\pi t \, dt$

$A(r) - A(0) = \displaystyle \left[ \pi t^2 \right]_0^r$

$A(r) - A(0) = \pi r^2$

Since the area of a circle with radius $0$ is $0$ , we conclude that $A(r) = \pi r^2$ .

Pedagogically, I don’t particularly recommend this approach, as I think students would find this explanation more confusing than the first approach. However, I can see that this could be useful for reinforcing the statement of the Fundamental Theorem of Calculus.

By the way, the above reasoning works for a square or cube also, but with a little twist. For a square of side length $s$ , the area is $A(s) = s^2$ and the perimeter is $P(s) = 4s$ , which isn’t the derivative of $A(s)$ . The reason this didn’t work is because the side length $s$ of a square corresponds to the diameter of a circle, not the radius of a circle.

But, if we let $x$ denote half the side length of a square, then the above logic works out since

$A(x) = s^2 = (2x)^2 = 4x^2$

and

$P(x) = 4s = 4(2x) = 8x$

Written in terms of the half-sidelength $x$ , we see that $A'(x) = P(x)$ .

Collaborative Mathematics: Challenge 07

My colleague Jason Ermer has posted his 7th challenge video, shown below. It’s both an experiment and an exercise in probability.

Video responses can be posted to his website, http://www.collaborativemathematics.org. In the words of his website, this is a unique forum for connecting a worldwide community of mathematical problem-solvers, and I think these unorthodox but simply stated problems are a fun way for engaging students with the mathematical curriculum.