## Traversetime of r=cos(n*θ)

I have the function in polar coordinates r=cos(n*θ), where r is the radii. I'm supposed to draw the graph of this function, and calculate the area. But to calculate the area, I need to know how fast the function traverses. From the solution in my book, it says if n is even, the function traverses when θ goes from 0 to 2pi, and from 0 to pi if n is odd.

Why is the traverse-time for this function only dependant on if n is odd or even? I do not understand this at all. Can you guys help me develop an intuition for functions in polar coordinates?
 Recognitions: Gold Member Science Advisor Staff Emeritus You will cover one lobe as r goes from 0 to 1 and back to 0. That is, as $n\theta$ goes form $-\pi+ 2k\pi$ to $\pi/2+ 2k\pi$ for integer k. The reason for the "if n is even, the function traverses when θ goes from 0 to 2pi, and from 0 to pi if n is odd." is that when $cos(n\theta)$ is negative, r is negative and, because r is alwas positive in polar coordinates, is interpreted as positive but with $\pi$ added to $\theta$. When n is odd, that results in one lobe covering another. When n is even, there are 2n lobes, when n is odd there are n lobes.

So you're saying when r goes from 0 to 1 to 0 (and n*θ goes from -pi/2 to pi/2) the function has fully traversed?

Is this a general rule? That when r goes back to where it started, the function has traversed?

 is that when cos(nθ) is negative, r is negative and, because r is alwas positive in polar coordinates, is interpreted as positive but with π added to θ. When n is odd, that results in one lobe covering another. When n is even, there are 2n lobes, when n is odd there are n lobes.
I think I get you, but can you please provide an example?

Recognitions:
Gold Member
The first graph is $r= 5cos(4\theta)$ and the second graph is $r= 5cos(5\theta)$