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Change of variables to polar coordinates

  1. Aug 1, 2008 #1
    I thought I grasped coordinate changes well, but now I've run into some problems. Usually I would have some function [itex] f(x,y) [/itex] and transformation equations like [itex] s = a*x+b*y [/itex]. I would apply chain rule and stayed left with new equations in new variables. (old ones get away through differentiation).

    My question is, what if are transformation equations more complex and old variables don't fade out? More specificaly, how can I convert [itex] f_x(x,y) [/itex] to polar coordinates ?

    my attempt:

    [tex]\theta = Arctan(y/x)[/tex]
    [tex]r = x^2+y^2[/tex]
    [tex]\phi(r,theta) = f(x,y)[/tex]
    [tex]\phi_x = \phi_rr_x + \phi_\theta\theta_x[/tex]
    [tex]\phi_x = \phi_rx\sqrt{x^2+y^2} + \phi_\theta(\frac{y}{x^2+y^2})[/tex]
    now I got the idea to solve for x and y in trans. equations and substitute, but I'm not sure
    [tex]\phi_x = rcos(\theta)\phi_r + \frac{sin(\theta)}{r^2}\phi_\theta[/tex]
  2. jcsd
  3. Aug 1, 2008 #2


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    [itex]x= r cos(\theta)[/itex] and [itex]y= r sin(\theta)[/itex] so your terms are missing a factor of r;
    [tex]x\sqrt{x^2+ y^2}= (r cos(\theta))r= r^2 cos(\theta)[/tex]
    [tex]\frac{y}{x^2+ y^2}= \frac{ r sin(\theta)}{r^2}= \frac{sin(\theta)}{r}[/tex]
    but that's the right way to proceed.
    Last edited by a moderator: Aug 1, 2008
  4. Aug 1, 2008 #3
    Thanks, I've been sudying PDE's few days in a row and I'm really getting tired..I should take a break and yet I can't, it's so interesting..
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