If \(z=f(x,y),\ x=r\cos\theta,\ y=r\sin\theta\) then prove that \(\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2=\left(\frac{\partial z}{\partial r}\right)^2+\frac1{r^2}\left(\frac{\partial z}{\partial\theta…

\(\frac{\partial x}{\partial r}=\cos\theta,\ \frac{\partial y}{\partial r}=\sin\theta,\ \frac{\partial x}{\partial\theta}=-r\sin\theta,\ \frac{\partial y}{\partial\theta}=r\cos\theta\).

Chain rule: \(\frac{\partial z}{\partial r}=\frac{\partial z}{\partial x}\cos\theta+\frac{\partial z}{\partial y}\sin\theta\) ...(1)

\(\frac{\partial z}{\partial\theta}=-r\sin\theta\frac{\partial z}{\partial x}+r\cos\theta\frac{\partial z}{\partial y}\Rightarrow\frac1r\frac{\partial z}{\partial\theta}=-\sin\theta\frac{\partial z}{\partial x}+\cos\theta\frac{\partial z}{\partial y}\) ...(2)

Squaring and adding (1) and (2) (the cross terms \(\pm2\sin\theta\cos\theta\,z_xz_y\) cancel):

\(\left(\frac{\partial z}{\partial r}\right)^2+\frac1{r^2}\left(\frac{\partial z}{\partial\theta}\right)^2=(\cos^2\theta+\sin^2\theta)\left(\frac{\partial z}{\partial x}\right)^2+(\sin^2\theta+\cos^2\theta)\left(\frac{\partial z}{\partial y}\right)^2\)

\(=\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2\) (proved).