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The Distance Formula in Polar Coordinates - Problem 3 3,912 views
Here is an interesting problem you can do with the distance formula and polar coordinates. We can find a polar equation of any circle. Let's find the polar equation of a circle with center at 4, 2 pi over 3 and radius 3. I've got it drawn on the polar coordinate system here.
Here is my center; 4, 2 pi over 3 and my radius 3. Now how do we come up with an equation of a circle? Actually how do we do it originally with rectangular coordinates? We did something like this. We drew a picture. We picked a point on the edge of the circle, and let's call that point P. It has coordinates r theta. We want an equation that involves r and theta, the coordinates of this point. We use the distance formula. That what we're about to do.
Now, the distance that we want to use the formula on, is the distance between this point P and the center A. Remember in a circle, every point on the circle is the same distance from point A, and in this case it's 3. So this distance is 3.
So I want to use the distance formula to show this relationship. My two points are going to be P, and A. The distance between them is 3. So let me write that down. Distance squared is r1² plus r2² minus 2r1r2, cosine theta 2 minus theta 1. So I have to decide, which points are 1 theta 1 and which points are 2 theta 2. So how about point P, which is r theta be r1, theta 1, then the other point will be r2 theta 2.
So the distance between them squared, I already know is 9. The distance is 3, so squared is 9. Well, r1 is just r, so that's just r², and the other point 4, 2 pi over 3, that's going to be my r2, theta 2. So r2² is 16, minus 2r1 which is r, r2 which is 4, times the cosine of the angle between them.
Now the angle between them is theta minus 2 pi over 3. Now if you're paying close attention you may realize that I just did theta 1 minus theta 2. Remember that the order of subtraction doesn't matter, because cosine is an even function. Cosine of 2 minus theta 1 is the same as cosine of theta 1 minus theta 2. So this is okay.
And then, we're just simplifying a little bit, r² plus 16 minus 8r cosine theta minus 2 pi over 3. Let me just subtract 9 from both sides, and I'll call it done.
My equation is r² minus 8r, let me put the constant at the end, minus 8r cosine theta minus 2 pi over 3. Then I'm going to have plus 7 at the end. That's my final equation. For this circle, radius 3, centered at the point 4, 2 pi over 3.