- #1
Defconist
- 7
- 0
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]
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]