Polar form $\frac{dy}{dx}$

Question

Trying to find the derivative $\dfrac{dy}{dx}$ in polar form, where:
$$x=r\cos\theta \,\text{ and } \, y=r\sin\theta$$ Seems like the common approach (on Wikipedia and other sites) is to assume that $r$ is a function of $\theta$. Finding the derivative assuming $r(\theta)$ is no problem for me.
But why is this assumption true? Why can't $r$ change independently from $\theta$?

Answer 1

The expression $\frac{dy}{dx}$ doesn't even make sense unless $x$ and $y$ are related by a (suitable) differentiable function. The question you are asked is presumably implicitly (or explicitly and you just haven't written it) assuming that $x$ and $y$ are so related.

Therefore $r$ and $\theta$ are implicitly related as well, by putting all of the relations together.

Even without the variables being functionally related, we can still say in terms of differential forms:

$$ dx = \cos \theta \, dr - r \sin \theta \, d\theta $$ $$ dy = \sin \theta \, dr + r \cos \theta \, d\theta $$

and we can even solve for $dr$ and $d\theta$ in terms of $dx$ and $dy$.

But as soon as we have a relation between $dx$ and $dy$, we can use that to solve for a relation between $dr$ and $d\theta$. (barring a degeneracy causing problems)