Change of Variables using Inverse Mapping

Question

If we have a function $w=f(x,y)$ and we change the variables using polar coordinates $x=r\cos\theta$ and $y=r\sin\theta$ we get a new function $f(r,\theta)$.

In order to find the partial derivative

$$\frac{\partial w}{\partial x}=\frac{\partial w}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial w}{\partial \theta}\frac{\partial \theta}{\partial x}$$

we use the inverse mapping

$$r=(x^2+y^2)^{1/2}$$

and

$$\theta=\tan^{-1}\frac{y}{x}$$

and we get

$$\frac{\partial r}{\partial x}=\frac{\partial}{\partial x}(x^2+y^2)^{1/2}=\cos\theta$$

etc...

Why do we have to use the inverse mapping in order to find the derivative? Why can't we simply do

$$x=r\cos\theta\Rightarrow r=\frac{x}{\cos\theta}$$

and then differentiate that? Doing so is a mistake but why?

Answer 1

You could do that, but you have to take into account that $\theta$ also depends on $x$: $${\partial\over\partial x}\left(x \over \cos\theta\right) = {1\over\cos\theta}+{x\sin\theta \over \cos^2\theta}{\partial\theta\over\partial x} \tag 1$$ and you’re still stuck with having to compute $\partial\theta/\partial x$ somehow, so it doesn’t really get you very far.

However, you can bring the Inverse Function theorem to bear: under suitable conditions, if $f:\mathbb R^m\to\mathbb R^n$ is differentiable at a point $\mathbf p\in\mathbb R^m$ and its differential $\mathrm d f_{\mathbf p}$ at that point is nonzero, then $f$ is invertible in a neighborhood of $f(\mathbf p)$ and $f^{-1}$ is differentiable there with $\mathrm d(f^{-1})_{f(\mathbf p)}=(\mathrm df_{\mathbf p})^{-1}$. In other words, the Jacobian of the inverse of a function is the inverse of its Jacobian.

So, from $x=r\cos\theta$, $y=r\sin\theta$, we get $${\partial(x,y)\over\partial(r,\theta)} = \pmatrix{\cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta}.$$ Inverting this matrix produces $${\partial(r,\theta)\over\partial(x,y)} = \pmatrix{\cos\theta & \sin\theta \\ -\frac1r\sin\theta & \frac1r\cos\theta},$$ from which we can read $\partial r/\partial x = \cos\theta$. To check that equation (1) gave the correct derivative, substitute $\partial\theta/\partial x = -1/r\,\sin\theta$ from the Jacobian to get $${1\over\cos\theta}-\frac xr{\sin^2\theta\over\cos^2\theta} = {1-\sin^2\theta \over \cos\theta} = \cos\theta.$$

Answer 2

Because you want to express $r = r(x, y)$, but in your last expression you have $r = r(x,\theta)$ which will break the dependencies for the chain rule, because the old coordinates and new coordinates are all mixed up.

More explicitly you need

$$ \frac{\partial w(x, y)}{\partial x} = \frac{\partial w(r, \theta)}{\partial r}\frac{\partial r(x, y)}{\partial x} + \frac{\partial w(r, \theta)}{\partial \theta}\frac{\partial \theta(x, y)}{\partial x} $$