7.3 Polar Coordinates

[STARTS AT 11:08]

Professor: Example three-- graph the curve defined by the function r equals 4 sine theta. Identify the curve. And then rewrite the equation in rectangular coordinates. 

So to graph this, I'm just going to use this table of points. This is just taking various values of theta and evaluating, the same way you would with x and y for our independent, dependent. Here, our independent is theta and our dependent is r. So we just do the exact same thing. So if we evaluate it at 0, we get 0. At pi, we get 0, et cetera. We get lots of points here. 

So I'm going to go and plot these points in blue. So (0,0). A theta of pi over six, my radius is 2. There's a point right there. At pi over 4, I'm at 2.8. So looking more something like that. 3.4 at pi/3 for my radius. Pi/2, I'm at four. 2 pi over 3, 3.4, something like that. 

3 pi over 4, it is 2.8. That doesn't seem right. 2.8, little less. And 2 at 5 pi over six. OK, pi is 0. So we're right back there again. 7 pi over 6 is negative 2. If you notice, that's actually going to end me up right here. 

So these negative values for our radius are all going to end up instead of here, and here, they're going to end up here, and here, respectively. So it appears that this actually might be forming what looks like a circle, or at least some strange version of a circle. So I want to go ahead and claim that that's a circle. 

Now, rewrite the function in rectangular coordinates. I'm going to use some space over here. 

So I've got the function r equals 4 sine theta. Now there are a few identities I want to hold on to-- y equals r sine theta, x equals r cosine theta, and r squared equals x squared plus y squared. 

So to begin with, I'd really like to make this equation in the form of r squared. That would be great. So I'm going to multiply the whole equation by r so that I get r squared equals 4r sine theta. 

If you notice, I now have r sine theta, which I can say is y. So this is r squared equals 4y. Well r squared is x squared plus y squared equals 4y. Here's my equation. Now some more algebra. I can subtract the 4y to both sides. That-- back-- equals zero. And what I'm going to do is complete the square. 

I'm going to complete the square here. And that would be a perfect square if I had a value 4 there. And so this is x squared plus y minus 2 squared equals 4, which is a circle with a radius-- with a radius of 2 centered at the point (0,2). Which, if you go back to that graph, I think you might be able to agree that the center of that circle is (0,2). 

[ENDS AT 15:30]