Derivative in polar

1. Feb 4, 2009

mysubs

Hello,

I want to convert from cylindrical (r,a,z) --to--> cartesian (x,y,z). However, I'm a not very confident about my level at this.

Say, I have dm/dr, a derivative in polar, and I want to find the equivalent expression in cartesian. (d: is partial derivative here).

I thought about using chain rule, but [r] depends on both [x] and [y]. Is doing it this way: dr=(xdx+ydy)/(x2+y2)1/2, correct? and If not why?

How should I work with this?

2. Feb 7, 2009

CompuChip

Yes, that's the way.
So

$$\frac{\partial m}{\partial x} = \frac{\partial m}{\partial r} \frac{\partial r}{\partial x}$$
where
$$\frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}}$$