Understanding Partial Differentiation with Polar Coordinates

[tandoorichicken]

If z = f(x,y), where x = rcos(\theta) and y = rsin(\theta), find \frac{\partial z}{\partial r}, \frac{\partial z}{\partial\theta}, and \frac{\partial^2 z}{\partial r\partial\theta}

Here's what I've done:
(a)
\frac{\partial z}{\partial r} = \frac{dz}{dx} \frac{\partial x}{\partial r} + \frac{dz}{dy} \frac{\partial y}{\partial r} = \frac{dz}{dx} \cos{\theta} + \frac{dz}{dy} \sin{\theta}
(b)
\frac{\partial z}{\partial\theta} = \frac{dz}{dx} \frac{\partial x}{\partial\theta} + \frac{dz}{dy} \frac{\partial y}{\partial\theta} = -\frac{dz}{dx} r\sin{\theta} + \frac{dz}{dy} r\cos{\theta}

My question is, for parts a and b, is this correct or must something also be done with the dz/dx and dz/dy, and for part c, I don't know how to do it. Can someone help please?

 

[formulajoe]

you don't have to put the dz/dx and dz/dy. its just cos + sin.
for c, you just take the derivative with respect to r and theta. i forget which you are supposed to do first. but you just do one, than take the derivative of the new form with respect to the other variable.

 

[dextercioby]

Points a) and b) are solved wonderfully.
Point c) is a bit tricky,meaning that u'll have to differentiate one of the 2 expressions found at a) & b) wrt the other variable.
\frac{\partial^{2}z}{\partial r \partial \theta}=\frac{\partial}{\partial r}(\frac{\partial z}{\partial \theta})

Try to do it this way and tell where you get stuck.

Daniel.

 

[dextercioby]

One more thing.It's still a partial derivative for "z" (or "f") too,becasue it depends explicitely on 2 variables,namely "x" and "y"...

Daniel.

 

[tandoorichicken]

Thank you dexter. I checked both
\frac{\partial}{\partial r}(\frac{\partial z}{\partial \theta})
and
\frac{\partial}{\partial \theta}(\frac{\partial z}{\partial r})

and they both come to the same answer. So it must be right.

 

[dextercioby]

It meant that the function "z" is "well behaved".There are functions for which the mixed partial derivatives are different one from another.In a more advanced way,the 2-nd rank hessian is not symmetric...

Daniel.