Converting partial derivatives between coordinate frames

  • Thread starter blalien
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  • #1
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Homework Statement


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]

Homework Equations


All given in part 1

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.
 

Answers and Replies

  • #2
679
2
phi and r are independent, it's not surprising that you get 0.
 
  • #3
32
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Sorry, I should clarify. We assume a function [tex]\phi (r)[/tex] exists, but we're not given its definition.
 
  • #4
679
2
Then
[tex]
\phi(r) = atan(x/y)
[/tex]
generally won't be true, your calculation about dphi/dx is wrong.
 
  • #5
32
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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.
 
  • #6
679
2
Could you post the whole question, I'm not really sure about what you want.
 
  • #7
32
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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.
 
  • #8
679
2
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.
[

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]
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
 
  • #9
679
2
But if I understand your question correctly, then actually your title would be inappropriate, because you're not converting partial derivatives
 
  • #10
32
0
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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