Polar Equations

By

Brandon Samples

I wanted to begin by exploring the polar equation

r = sin(n theta)

for various values of n.

Can you guess which value of n corresponds to which graph?

Hint: n varies from 1 to 4.

Let's think about n = 1 first, i.e., r = sin(theta).

First, the length is given by sin (theta), so we expect it to trace out a circle at the origin since it goes from 0 to 1 and back to 0. Then as theta goes from Pi to 2Pi, the value of r is negative, but the angle is in the third and fourth quadrant, so it will trace the same circle again. Is this what you expected?

Okay, so now let's look at the graph r = sin(2theta).

Obviously, the value for sine is reached twice as fast by increasing the period, so we should expect that the graph has more petals. Does a function of r = sin(ktheta) yield a graph with 2k petals? Let's try r= sin(4theta).

Hmm, seems to be working. Let's try r= sin(6theta).

So we conjecture that the number of petals for the graph r = sin(ktheta) is equal to 2k for k even and positive.

Now, let's return to odd k and look at r = sin(3theta).

So it appears that the number of petals for k odd is actually equal to k. Let's test this theory with a few more examples. First, r = sin(5theta).

How about r = sin(7theta)?

Okay, so we conjecture that the number of petals for r=sin(ktheta) is equal to k for k odd and positive.