I'm having troubles with solving problems with partial differentiations... and this one is double. I don't thing we've even learned this in class...
Question: If $z=f(x,y)$, where $x=r\cos(\theta), y=r\sin(\theta)$, show that
$$ \frac{ \partial ^2 z}{ \partial x^2}+\frac{ \partial ^2 z}{ \partial y^2}= \frac{ \partial ^2 z}{ \partial r^2} + \frac{1}{r^2}\frac{ \partial ^2 z}{ \partial \theta ^2} +\frac{1}{r}\frac{ \partial z}{ \partial \theta}$$
I would be grateful for a little help.
okay, so I've worked it out a bit more. so I'm trying to find out $z_{xx}+z_{yy}=z_{rr}+\frac{1}{r^2}z_{\theta\theta} + \frac{1}{r}z_{\theta}$.
$z_r=z_xx_r + z_yy_r=z_x\cos(\theta) + z_y\sin(\theta)$
$z_{rr}=(z_r)_xx_r+(z_r)_yy_r=z_{xx}\cos^2(\theta)+2z_{xy}\sin(\theta)\cos(\theta)+z_{yy}\sin^2(\theta)$
and i did a similar thing with $z_\theta$ and got $z_{\theta\theta}=z_{xx}r^2\sin^2(\theta)-2z_{xy}r^2\cos(\theta)\sin(\theta)+z_{yy}r^2\cos^2(\theta)$
and the addition of them was
$z_{rr}+z_{\theta\theta}=z_{xx}[\cos^2(\theta)+r^2\sin^2(\theta)]+2z_{xy}[sin(\theta)\cos(\theta)-r^2\sin(\theta)\cos(\theta)]+z_{yy}[sin^2(\theta)+r^2\cos(\theta)].$
Am I on right track? I'm a beginner at learning this and I'm not so sure of myself. How should I continue from here? I don't really see how i can change this to the form required by the question...
If $z=f(x,y)$, where $x=rcos(\theta), y=rsin(\theta)$, then we have $z=f(x(r,\theta), y(r,\theta))$.
We start with $\partial z/ \partial r$:
$\partial z/ \partial r=\partial z(x( r, \theta))/\partial r=(\partial z/ \partial x)(\partial x/ \partial r)+ (\partial z/\partial y)( \partial y/\partial r)$ Now, $\partial^2 = \partial \circ \partial $, so we next do:
$( \partial/\partial r)[(\partial z/ \partial x)(\partial x/ \partial r)]$ , where we use the chain rule: $$( \partial/\partial r)[ \partial z/ \partial x)(\partial x/ \partial r)]= (\partial^2z /\partial x \partial r)(\partial x/\partial r)+(\partial ^2x /\partial^2 r )(\partial z/\partial x)$$
Where, e.g. :
$(\partial^2z /\partial^2 r)$
Means you differentiate twice with respect to r;
While,
$(\partial^2z /\partial x \partial r)$ Means you differentiate twice, but once with respect to $x$, and once with respect to $r$ (in most cases you will find, the order in which you differentiate does not matter)
This is more tedious than difficult ( the theory is actually interesting, but learning the computation techniques can be tedious). I would suggest one of the Schaum's Advance Calculus books, where they solve a lot of these problems step-by-step.