I was working with a simple transformation of rectangular coordinates - symmetry around the y-axis, i.e. $$f(x,y) = (x, -y)$$ I wanted to express the identical concept in polar coordinates. After some messing around I remembered f(r, $\theta$) = (r, -$\theta$) gives symmetry around the x-axis, and thus translating the angle $\theta$ both before and after the flip, you could create $$f(r, \theta) = (r, -(\theta - \pi/2) + \pi/2) = (r, \pi - \theta)$$ for the desired result.
My question is, is there a mechanical technique for converting the expression of a function on the plane in rectangular coordinates to an expression of that same function in polar coordinates?
Or do I simply have to 'mess around' each time? (I went through the standard $x=r* cos (\theta)$ type formulas, but they aren't really the same, are they? I think I'm talking about operating "on" the plane, instead of operating "in" it? Or am I overthinking and confusing myself)
Thanks.
You know there is a map, or function, $(r, \theta) \rightarrow (x,y)$, given by $x=rcos(\theta), y=rsin(\theta)$. There is also a function $(x,y)\rightarrow (r,\theta)$, although the formulas are hard to write down, and they're made especially complicated by the fact that $\theta$ is not well-defined at the origin.
So one "mechanical" approach to your problem would be to express $f(r,\theta)$ as a composition of three functions:
$(r,\theta)\rightarrow (x,y)$
$f$
$(x,y) \rightarrow (r,\theta)$.
But try to compute this for any reasonable function, and you will probably conclude that a more informal approach is usually faster. :)