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Homework Help: Converting partial derivatives between coordinate frames

  1. Jan 7, 2010 #1
    1. The problem statement, all variables and given/known data
    Given Cartesian coordinates x, y, and polar coordinates r, phi, such that
    [tex]r=\sqrt{x^2+y^2}, \phi = atan(x/y)[/tex] or
    [tex]x=r sin(\phi), y=r cos(\phi)[/tex]
    (yes, phi is defined differently then you're used to)
    I need to find [tex]\frac{d\phi}{dr}[/tex] in terms of [tex]\frac{dy}{dx}[/tex]

    2. Relevant equations
    All given in part 1

    3. The attempt at a solution
    I tried to compute [tex]\frac{d \phi}{d r}[/tex] directly and ended up with this:
    [tex]\frac{d \phi}{d r} = \frac{d \phi}{d x} \frac{d x}{d r} + \frac{d \phi}{d y} \frac{d y}{d r}
    = \frac{y}{x^2+y^2} sin{\phi} - \frac{x}{x^2+y^2} cos{\phi}\\
    = \frac{y}{x^2+y^2} \frac{x}{\sqrt{x^2+y^2}} - \frac{x}{x^2+y^2} \frac{y}{\sqrt{x^2+y^2}}\\
    = 0[/tex]
    Obviously this isn't correct, so I must be going about this the wrong way.
  2. jcsd
  3. Jan 7, 2010 #2
    phi and r are independent, it's not surprising that you get 0.
  4. Jan 7, 2010 #3
    Sorry, I should clarify. We assume a function [tex]\phi (r)[/tex] exists, but we're not given its definition.
  5. Jan 7, 2010 #4
    \phi(r) = atan(x/y)
    generally won't be true, your calculation about dphi/dx is wrong.
  6. Jan 7, 2010 #5
    So what should I do? I don't even have a guarantee that there is a general solution, so this could be a wild goose chase.
  7. Jan 7, 2010 #6
    Could you post the whole question, I'm not really sure about what you want.
  8. Jan 7, 2010 #7
    This isn't a homework question out of a textbook. But here's what I'm looking for:
    Given variables phi, r, x, y such that
    x = r sin(phi), y = r cos(phi)
    and a function phi(r)
    I need to find dphi/dr in terms of x, y, and dy/dx.
  9. Jan 7, 2010 #8
    Now I kinda understand your problem. So in the first place your x and y is not independent, instead, it's y=y(x) instead.
    Here you are using chain rule under partial differentiation, but when x,y are not independent, this won't be valid.

    You can try this approach:
    dx = \sin \phi dr + r\cos \phi d\phi \\
    dy = \cos \phi dr - r\sin \phi d\phi \\
    Now view dr, dphi as unknowns you'll be able so solve dr, dphi in terms dx, dy. Then you'll know what to do next
  10. Jan 7, 2010 #9
    But if I understand your question correctly, then actually your title would be inappropriate, because you're not converting partial derivatives
  11. Jan 7, 2010 #10
    Yeah, it might be. But your method works. Thank you so much for your help. You always overlook the simple solutions, right?

    It turns out the answer is:
    [tex]\frac{d\phi}{dr}=\frac{y-x y'}{(x+y' y)\sqrt{x^2+y^2}}[/tex]

    I also need to find [tex]\frac{d^2\phi}{d r^2}[/tex], but I think you just need to apply the chain rule to the result above.
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