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I am a math educator preparing a unit on the calculus of polar curves. This is my first time teaching this particular unit, so it was also the first time I noticed that the "periods" of different polar curves were not what I expected. By "period" of a polar curve I mean the minimum angle that must be spanned by $\theta$ in order to trace the polar curve such that it may intersect itself but not overlap itself.

Here are some examples:

The period of $r(\theta)=cos(2\theta)$ is $2\pi$, even though the period of $y=cos(2x)$ is $\pi$.

The period of $r(\theta)=cos(\theta/2)$ is $4\pi$, and so also is the period of $y=cos(x/2)$.

The period of $r(\theta)=tan(\theta/3)$ is $6\pi$, even though the period of $y=tan(x/3)$ is $3\pi$.

Can anyone explain a general method for determining the "period" of a polar curve and perhaps even its relation to the period of the related trigonometric function?

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So as you have probably seen geometrically, the periodicity of the function $r(\theta)$ tells you that you're going to get the same "shape" every so often, but it might be rotated based on what the period actually is; for example, the shape of $r = \cos(2\theta)$ for $\theta\in[\pi,2\pi]$ is the same shape as $r=\cos(2\theta)$ for $\theta\in[0,\pi]$, but rotated to be upside down (i.e. rotated by $\pi$) and thus doesn't overlap.

If you want the points $(x,y)$ to overlap, you should look at the points $(x,y)$, that is $(r\cos\theta,r\sin\theta)$, and see what the period of those are. So in your first example, the $(x,y)$ coordinates are given by $(\cos(2\theta)\cos\theta, \cos(2\theta)\sin\theta)$ which have period $2\pi$. (Of course if your curve isn't given as $r=f(\theta)$ you maybe have a bit more work to do.)

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  • $\begingroup$ Thank you! You've answered my question. The period of a polar function is equal to the period of the corresponding trigonometric function multiplied by either cos 𝜃 or sin 𝜃. $\endgroup$
    – NC1208
    Commented Jul 11 at 11:02

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