# User:Wyattsc

Hello all, these are the things i am looking to accomplish

1. I am looking to talk about the relationship of roulettes to the game spirograph. When I read through the roulette page, I suddenly made the connection that spirograph is a real-life and common example of a roulette. I was also surprised that spirograph did not appear under "Interesting Application of the Concept" so I hope to add spirograph to that.

2. I am looking to connect the involute roulette to the golden spiral. Immediately when i saw the gif of the involute spiral i thought it looked awfully like a golden spiral. Upon further research I saw some websites that said it was a golden spiral and others that said it "resembled an Archimedean spiral" but was not one. I would like to do some of the mathematics behind this and graph the involute of a circle and find if it is a golden spiral or an archimedean spira, or neither. Also, if it is a golden spiral i would like to superimpose a golden rectangle over it making a cool math image.

Call For Help

I am a 9th grade student taking geometry right now and have little to no understanding of polar coordinates and trig. I was wondering if anyone has a way to graph the involute of a circle in order to do what i am trying to accomplish in #2. If you have any help it would be great if you could contact me at wyattsaintclair@gmail.com

thanks!

Update 2/27/12

I have decided to use khanacademy.org to teach myself some concepts of trig and pre calculus. With this i hope to graph the involute of a circle.

Once you feel like you're getting a bit of a handle on polar coordinates, a good way to both practice using that graphing structure and draw some connections to your topic would be to trace a simple spirograph pattern onto x,y axes on a piece of graph paper. Put the center of the pattern at (0,0) and see if you can work out a function over $(r,\theta)$ that generates the same pattern.

Other than that, keep it up! If you come across any questions as you go through the Khan Academy materials, just post the question up on Masterman Students Work.

-Diana (18:33, 3/4/12)

Update 3/7/12

i have a pretty good understanding now of most of the concepts and i believe that I am ready to take the equation for the involute of a circle, and then connect and compare it to a golden or archimedian spiral. Also, i have decided that my image will be a picture of a tetherball (that i will take using a model) from above, hopefully with a graph of the involute of a circle based on the that tetherball's pole width laid over top. So if anyone has a way to graph the involute of a circle, and maybe archimedian and golden spiral (though i will attempt to find these out myself) if you could contact me at wyattsaintclair@gmail.com

Email sent to Wyatt, 23:32, 3/11/12:

Hi Wyatt,
The involute of a circle is actually something you can derive pretty easily for yourself, because it's essentially a matter of measuring right triangles. Check out the file I attached. In that diagram, points B,C,E,G, and I all lie along the involute of the circle with center A. But I've shown them here as vertices of triangles instead to indicate how you can go about finding a formula for the involute.
First, these triangles all have the same base with length equal to the radius of the circle -- in this case, 2. Second, you can find the length of the side perpendicular to the base (tangent to the circle) because it's directly related to the circumference of the circle and the angle between the base and the positive x-axis -- that is, how far you have "unwound" the involute.
Knowing those, you can easily find the length of the hypotenuse and the measure of the angle between the base and the hypotenuse. Then you can define any point on the involute, (r,t), with r equal to the length of that hypotenuse and t the difference between the angle between the base and the hypotenuse and the angle of unwinding (that angle between the base of the triangle and the positive x-axis again).
So in the picture I attached, the triangle ADC has base=$2$ and height=$\frac{4\pi}{4}$=$\pi$. So the point C on the involute lies at distance $\sqrt{2^2 + \pi^2}$ from the origin and at angle $\frac{\pi}{2}-\tan^{-1}\left(\frac{\pi}{2}\right)$. Can you see why that follows from the reasoning above?
I'm going to post this email on your user page as well, where the math might be easier to read.
-Diana

More help needed

Hey Diana,

I have a question about the involute equation. I worked out that $r= sqrt((atan(thetaprime))^2+a^2)$ where a is the circle width. I also found that $thetaprime=arccosign(a/r)+theta$ where a is the circle width.

The problem is that if i just put the second equation inside the first, I will have r on both sides. I was wondering the easiest way to work it out so i have 'r' on only one side.

Thank you, Wyatt

Alright, recall the equation you put together Monday: $\theta=\tan^{-1}\left( \frac{\sqrt{r^2-a^2}}{a}\right) -\cos^{-1}\left( \frac{a}{r}\right)$
This equation was SO close! I went back a re-derived it using roughly the same methods you'd used, and I'm pretty sure what happened was at some point, you threw in a $\theta '$ that didn't need to be there. Anyway, let's say we have the triangle representation of the involute we've been using, where a is the radius, d is the length of the tangent generating out point, r is the distance from the origin to our point (also the hypotenuse of our triangle), $\theta$ is the angle of our point, and $\theta '$ is the angle to the base of the triangle. Then:
$d=\sqrt{r^2-a^2}$, and
$d=a\theta '$
Looking at the internal angle of our triangle, the difference between $\theta '$ and $\theta$, we see:
$\theta '=\theta+\cos^{-1}\left( \frac{a}{r}\right)$
$\Rightarrow\theta=\theta '-\cos^{-1}\left(\frac{a}{r}\right)$ [Eq. 1]
By our previous expressions for d, we have:
$\theta '=\frac{d}{a}$
$\Rightarrow\theta '=\frac{\sqrt{r^2-a^2}}{a}$
So we can plug this in to [Eq. 1], and we get:
$\theta=\frac{\sqrt{r^2-a^2}}{a}-\cos^{-1}\left(\frac{a}{r}\right)$
Very close to what you got! And note that most of the math I used here is stuff that you already derived. I think something just got mixed up somewhere along the way.
-Diana (12:53, 4/20/12)