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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
    Then
    [tex]
    \phi(r) = atan(x/y)
    [/tex]
    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:
    [tex]\begin{array}{l}
    dx = \sin \phi dr + r\cos \phi d\phi \\
    dy = \cos \phi dr - r\sin \phi d\phi \\
    \end{array}[/tex]
    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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