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Parametrizing a Circle - Problem 1 2,876 views

Teacher/Instructor 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.

Let’s do a little example where you parameterize a circle. Here I’m asked to parameterize the semi circle that’s given here counter-clockwise. Now you can see that the semi circle has radius 12, and previously we learnt how to parameterize a circle counter-clockwise. And the equations are going to be x equals the radius, which in this case is 12, cosine theta, and y equals the radius sine theta.

Only, we don’t want the whole circle here. So we don’t what to use theta between 0 and 2pi. So I’m going to experiment a little bit and see what values of theta I need for this parametric equations to work. So let me make a table. Theta x equals 12 cosine theta, y equals 12 sine theta and I’ll start with 0. Now the cosine of 0 is 1, so I’m going to get 12 and the sine of 0 is 0; (12,0) that’s this point. So I get this point when theta equals 0.

What about pi over 2? Well cosine of pi over 2 is 0, sine of pi over 2 is 1, so I get (0,12). That’s this point so I get this point when theta is pi over 2. So I’m going to guess that I need negative pi over 2 to get this point down here.

And the cosine of negative pi over 2, cosine is an even function so it’s going to have the same value at opposite inputs. It’s also going to have 0. And sine is an odd function so it’s going to have an opposite value of what it had at pi over 2. It was 1, now is got -1. So I’ll get -12. (0,-12) I get when theta equals negative pi over 2. So if I let theta start at negative pi over 2 and go through to pi over 2, I’ll get my semicircle. That means I need theta to be between negative pi over 2 and pi over 2.