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Derivative in polar

  1. Feb 4, 2009 #1

    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. jcsd
  3. Feb 7, 2009 #2


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    Yes, that's the way.

    [tex]\frac{\partial m}{\partial x} = \frac{\partial m}{\partial r} \frac{\partial r}{\partial x}[/tex]
    [tex]\frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}}[/tex]
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