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# Parametrizing a Circle - Problem 3

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

So we’ve been talking about parameterizing circles, and up till now we have the parameterization x equals r cosine theta and y equals r sine theta. So the question is what would happen if we switched the sine and the cosine? That’s the parameterization I have here, x equals to r sine theta and y equals r cosine theta. I want to see what that does, what effect it has.

So first let me observe that x over r equals sine theta. And y over r is cosine theta. So I can still make use of the fact that sine squared theta plus cosine squared theta equals 1. Now sine squared theta is x² over r².

And cosine squared theta is y² over r². So you can see that when I multiply both sides by r², I’m still going to get a circle; x² plus y², equals r². A circle centered at the origin with radius r. Just like our other parameterization, so what’s the difference? Let me calculate a few values of this parameterization. Theta x is r sine theta y is r cosine theta. So just a few starting with 0. Sine of 0 is 0 and cosine of 0 is 1. So for y I’ll get r.

Pi over 2; Sine of pi over 2 is 1, so I get r, and cosine of pi over 2 is 0. How about pi? Sine of pi is 0, cosine of pi is -1 so I get -r. 3Pi over 2; sine of 3Pi over 2 is -1 so I get negative r, and cosine of 3Pi over 2 is 0.

Finally 2Pi, remember sine and cosine of 2Pi are the same as at 0. So I’m going to get 0,r again. Now let’s take a look how this traces through the circle x² plus y² equals r². We started at 0,r when theta equaled 0. And then we got our 0 at pi over 2, that’s this point. Then we got 0,-r at theta equals pi, and then at 3 pi over 2, we got this point. And then we are back here at 2 pi. We are tracing through the circle starting at the top clockwise.

The big difference between this parameterization, and the one we’ve been studying is that this traces through the circle clockwise and it starts at the top. So let’s use that in our problem. Parameterize the semicircle in the clockwise direction. In a clockwise direction we want to use x equals r sine theta. In this case r the radius, is 12. Y equals 12 cosine theta and we want theta to go between 0 and 2Pi or wait a second do we?

We don’t want the whole circle, we want just this half of it. And as we just saw, we get this top point when theta equals 0. This point when theta equals pi over 2, and this point when theta equals pi. We need to stop at pi, so let me change that. We need theta to go from 0 to pi. And so this is my parameterization for the semi circle starting at the top and going clockwise.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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