# Converting partial derivatives between coordinate frames

## 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:
$$\frac{d \phi}{d r} = \frac{d \phi}{d x} \frac{d x}{d r} + \frac{d \phi}{d y} \frac{d y}{d r} = \frac{y}{x^2+y^2} sin{\phi} - \frac{x}{x^2+y^2} cos{\phi}\\ = \frac{y}{x^2+y^2} \frac{x}{\sqrt{x^2+y^2}} - \frac{x}{x^2+y^2} \frac{y}{\sqrt{x^2+y^2}}\\ = 0$$

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

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

Then
$$\phi(r) = atan(x/y)$$

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.

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

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.

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## The Attempt at a Solution

I tried to compute $$\frac{d \phi}{d r}$$ directly and ended up with this:
$$\frac{d \phi}{d r} = \frac{d \phi}{d x} \frac{d x}{d r} + \frac{d \phi}{d y} \frac{d y}{d r} = \frac{y}{x^2+y^2} sin{\phi} - \frac{x}{x^2+y^2} cos{\phi}\\ = \frac{y}{x^2+y^2} \frac{x}{\sqrt{x^2+y^2}} - \frac{x}{x^2+y^2} \frac{y}{\sqrt{x^2+y^2}}\\ = 0$$
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:
$$\begin{array}{l} dx = \sin \phi dr + r\cos \phi d\phi \\ dy = \cos \phi dr - r\sin \phi d\phi \\ \end{array}$$
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

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

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.