It is usually stated in Precalculus textbooks that in polar coordinates when a relation between $r$ and $\theta$ passes a symmetry test then the curve described by that relation has that symmetry. But if it fails a symmetry test it doesn't mean that the curve doesn't have that symmetry - it may still have the symmetry even though it fails the test. The reason for this is a bit subtle and is to do with being able to "start the curve at different points". Or to put it another way, there are an infinite number of ways of specifying any given point in polar coordinates. All well and good, though for this reason I've always felt like the symmetry tests in polar coordinates aren't all that useful.

Anyway, the Archimedes spiral seems to throw an additional wrench into the works, and it's got me very confused. The Archimedes spiral is

$ r = a\theta $

If we do the test for symmetry under reflection through the vertical axis, we replace $(r, \theta)$ with $(-r, -\theta)$. Doing this we get

$-r = a(-\theta) \\ r = a \theta$

So we recover the original relation and it passes the test. This should mean that the Archimedes spiral is symmetric under reflection in the vertical axis. But it most certainly doesn't have this symmetry (or any other reflection or inversion symmetry for that matter). So what's up? How can this relation pass the test and yet not have the symmetry?


Did you draw the whole thing, or just the portion for positive $\theta$?

The usual Archimedean spiral, for $\theta$ from zero to $20$ (radians):


The spiral, for $\theta$ from $-20$ to $20$:

Double Spiral

Do you see the symmetry now?

Most polar graphs you deal with, if you graphed them over a full period from zero to $2\pi$, wouldn't have this problem; the symmetry would show, if it's there. But the Archimedean spiral isn't periodic. In order for that symmetry to show up, we have to actually graph the parts with negative $\theta$.

  • $\begingroup$ Ohhhhhh...Now I feel foolish. The answer was "obvious". Thanks! $\endgroup$ – gleedadswell Feb 7 at 4:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.