Converting $\left(\frac{\partial f\left(x,y\right)}{\partial x}\right)^2+\left(\frac{\partial f\left(x,y\right)}{\partial y}\right)^2 $ to Polar

Question

I apologize about the title as I would have included the entire identity I am trying to prove analytically but we are limited to 150 characters.

I'm given the following problem in Widder's Advanced Calculus:

11.) $u=f\left(x,y\right),x=r\cos\left(\theta\right),y=r\sin\left(\theta\right).$ show that \begin{align} \left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial u}{\partial y}\right)^2&=\left(\frac{\partial f}{\partial r}\right)^2+\frac{1}{r^2}\left(\frac{\partial f}{\partial\theta}\right)^2. \end{align}

This is what I have done so far: \begin{align} \left(\frac{\partial f\left(x,y\right)}{\partial x}\right)^2+\left(\frac{\partial f\left(x,y\right)}{\partial y}\right)^2&=\left(\frac{\partial f\left(r\cos\left(\theta\right),r\sin\left(\theta\right)\right)}{\partial\left(r\cos\left(\theta\right)\right)}\right)^2+\left(\frac{\partial f\left(r\cos\left(\theta\right),r\sin\left(\theta\right)\right)}{\partial \left(r\sin\left(\theta\right)\right)}\right)^2\\ \end{align} but is there a way to transform the RHS into what he has written? Perhaps via some substitution?

Answer 1

You need to find $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ in terms of $r$ and $\theta$ using chain rule:

$ \begin{align} \frac{\partial u}{\partial x} = \frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}=\cos \theta \left(\frac{\partial u}{\partial r}\right) - \frac{\sin \theta}{r} \left(\frac{\partial u}{\partial \theta}\right) \end{align} $

and

$ \begin{align} \frac{\partial u}{\partial y} = \frac{\partial u}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial y}=\sin \theta \left(\frac{\partial u}{\partial r}\right) + \frac{\cos \theta}{r} \left(\frac{\partial u}{\partial \theta}\right) \end{align} $

Now by taking squares, we have:

$ \begin{align} \left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2 = \left(\frac{\partial u}{\partial r}\right)^2 + \left(\frac{1}{r}\frac{\partial u}{\partial \theta}\right)^2 \end{align} $