I'm asked to show the following four derivatives from the polar coordinate conversions. ($x = r\cos \theta$, $y=r\sin \theta$, and $r^2=x^2+y^2$)
$$ \frac{\partial r}{\partial x} = \cos \theta \ , \ \frac{\partial r}{\partial y} = \sin\theta \ , \ \frac{\partial \theta}{\partial x} = \frac{-\sin \theta}{r} \ , \frac{\partial \theta}{\partial y} = \frac{\cos \theta}{r}$$
I've shown the fist two relatively easy, but I'm not sure how to show $$\frac{\partial \theta}{\partial x} = \frac{-\sin \theta}{r} \ , \frac{\partial \theta}{\partial y} = \frac{\cos \theta}{r}$$
I know how to do this for $$\frac{\partial x}{\partial \theta} \ \ \text{and} \ \ \frac{\partial y}{\partial \theta}$$
Any help would be appreciated, thank you.
Best Answer
The Jacobian matrix for the map $\phi:(x,y)\mapsto(r\cos\theta,r\sin\theta)$ is $$J_\phi=\pmatrix{{\partial x\over\partial r}&{\partial x\over\partial\theta}\\{\partial y\over\partial r}&{\partial y\over\partial\theta}}=\pmatrix{\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta}.$$ Inverting this matrix yields the Jacobian of the inverse map:$$J_\phi^{-1}=\frac1r\pmatrix{r\cos\theta&r\sin\theta\\-\sin\theta&\cos\theta}=\pmatrix{\cos\theta&\sin\theta\\-{\sin\theta\over r}&{\cos\theta\over r}}=J_{\phi^{-1}}=\pmatrix{{\partial r\over\partial x}&{\partial r\over\partial y}\\{\partial\theta\over\partial x}&{\partial\theta\over\partial y}}.$$