Second derivative of polar coordinates How do I express $\dfrac{\partial^2z}{\partial\theta^2}$ in terms of Cartesian coordinates given that $(x,y)$ are Cartesian coordinates and $(r,\theta)$ are polar coordinates.
Attempt:
$$
\frac{\partial^2z}{\partial\theta^2} = 
\frac{\partial }{\partial \theta}\left[\frac{\partial z }{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial \theta}\right]
$$
I'm not entirely sure if I am on the right track because z isn't specified so simplifying that expression down isn't possible. 
 A: Let's go ahead and let $z=f(x,y)$. Then since $$x=rcos(\theta)$$ and $$y=rsin(\theta)$$ then we have $$z=f(rcos(\theta), rsin(\theta))$$ We thus apply the chain rule to obtain $$\displaystyle \frac{\partial z}{\partial \theta}= \frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta}$$ which thus gives: $$\displaystyle \frac{\partial z}{\partial \theta}=-\frac{\partial f}{\partial x}rsin(\theta)+\frac{\partial f}{\partial y}rcos(\theta)$$ Now differentiate a second time with respect to $\theta$ to obtain:
$$\displaystyle \frac{\partial^2z}{\partial \theta^2}=-\left(\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial x}\right)rsin(\theta)+\frac{\partial f}{\partial x}rcos(\theta)\right)+\left(\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial y}\right)rcos(\theta)-\frac{\partial f}{\partial x}rsin(\theta)\right),$$ by the product rule.
A: Yes, you are on the right track.
If $z=z(x,y)$, then 
$$
\frac{\partial z}{\partial\theta}=\frac{\partial z }{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial \theta}
=(-y)\frac{\partial z }{\partial x} + x\frac{\partial z}{\partial y}:=w
$$
Now you have $w=w(x,y)=(-y)\frac{\partial z }{\partial x} + x\frac{\partial z}{\partial y}$. And you simply repeat what you did for $z$ to $w$. Note that
$$
\frac{\partial^2 z}{\partial\theta^2}=\frac{\partial w}{\partial\theta}.
$$
