Converting partial derivatives between coordinate frames

[blalien]

Homework Statement

Given Cartesian coordinates x, y, and polar coordinates r, phi, such that
$$r=\sqrt{x^2+y^2}, \phi = atan(x/y)$$ or
$$x=r sin(\phi), y=r cos(\phi)$$
(yes, phi is defined differently then you're used to)
I need to find $$\frac{d\phi}{dr}$$ in terms of $$\frac{dy}{dx}$$

Homework Equations

All given in part 1

The Attempt at a Solution

I tried to compute $$\frac{d \phi}{d r}$$ directly and ended up with this:
$$\frac{d \phi}{d r} = \frac{d \phi}{d x} \frac{d x}{d r} + \frac{d \phi}{d y} \frac{d y}{d r}<br /> = \frac{y}{x^2+y^2} sin{\phi} - \frac{x}{x^2+y^2} cos{\phi}\\<br /> = \frac{y}{x^2+y^2} \frac{x}{\sqrt{x^2+y^2}} - \frac{x}{x^2+y^2} \frac{y}{\sqrt{x^2+y^2}}\\<br /> = 0$$
Obviously this isn't correct, so I must be going about this the wrong way.

 

[kof9595995]

phi and r are independent, it's not surprising that you get 0.

 

[blalien]

Sorry, I should clarify. We assume a function $$\phi (r)$$ exists, but we're not given its definition.

 

[kof9595995]

Then
$$\phi(r) = atan(x/y)$$
generally won't be true, your calculation about dphi/dx is wrong.

 

[blalien]

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.

 

[kof9595995]

Could you post the whole question, I'm not really sure about what you want.

 

[blalien]

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.

 

[kof9595995]

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 $$\frac{d \phi}{d r}$$ directly and ended up with this:
$$\frac{d \phi}{d r} = \frac{d \phi}{d x} \frac{d x}{d r} + \frac{d \phi}{d y} \frac{d y}{d r}<br /> = \frac{y}{x^2+y^2} sin{\phi} - \frac{x}{x^2+y^2} cos{\phi}\\<br /> = \frac{y}{x^2+y^2} \frac{x}{\sqrt{x^2+y^2}} - \frac{x}{x^2+y^2} \frac{y}{\sqrt{x^2+y^2}}\\<br /> = 0$$

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:
$$\begin{array}{l}<br /> dx = \sin \phi dr + r\cos \phi d\phi \\ <br /> dy = \cos \phi dr - r\sin \phi d\phi \\ <br /> \end{array}$$
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

 

[kof9595995]

But if I understand your question correctly, then actually your title would be inappropriate, because you're not converting partial derivatives

 

[blalien]

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:
$$\frac{d\phi}{dr}=\frac{y-x y'}{(x+y' y)\sqrt{x^2+y^2}}$$

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