# Z=f(x,y), x=function, y=function. dz/dx=

1. Apr 24, 2013

### mrcleanhands

1. The problem statement, all variables and given/known data
Consider z=f(x,y), where x=rcosθ and y=rsinθ

Show that $\frac{\partial z}{\partial x}=\cos\theta\frac{\partial z}{\partial r}-\frac{1}{r}\sin\theta\frac{\partial z}{\partial\theta}$

2. Relevant equations

3. The attempt at a solution

Z
Connects to X and Y
X Connects to r and θ, Y Connects to r and θ

Is dz/dx not then equal to dx/dr + dx/dθ? because that is equivalent to dz/dx = r -rsinθ.
What am I doing wrong here?

2. Apr 24, 2013

### Simon Bridge

Evaluate $\frac{\partial z}{\partial r}$

3. Apr 24, 2013

### mrcleanhands

$\frac{\partial z}{\partial r}=\cos\theta\frac{\partial z}{\partial x}+\sin\theta\frac{\partial z}{\partial y}$

then rearranging this $\frac{\partial z}{\partial x}=\sec\theta\frac{\partial z}{\partial r}-\tan\theta\frac{\partial z}{\partial y}$

I'm not doing something right or not understanding some underlying concept here. Why was my dz/dx derivation wrong?

4. Apr 24, 2013

### Simon Bridge

Don't rearrange yet. Do the other partial - the one with theta.

5. Apr 25, 2013

### mrcleanhands

$\frac{\partial z}{\partial\theta}=-r\sin\theta\frac{\partial z}{\partial x}+\frac{1}{r}\cos\theta\frac{\partial z}{\partial y}$

I've re-arranged this and get $\frac{\partial z}{\partial x}=-\frac{1}{r}\csc\frac{\partial z}{\partial\theta}+\frac{1}{r^{2}}\cot\theta\frac{\partial z}{\partial y}$

which should be the same as $\frac{\partial z}{\partial x}=\sec\theta\frac{\partial z}{\partial r}-\tan\theta\frac{\partial z}{\partial y}$

Now I'm more confused...

Isn't it just $r-r\sin\theta$?

6. Apr 25, 2013

### Simon Bridge

- you want ∂z/∂x in terms of ∂z/∂r and ∂z/dθ.

You don't need any terms in ∂z/∂y ... but the above relations each have such a term.
This should suggest a course of action to you.

7. Apr 26, 2013

### mrcleanhands

Hey Simon, I got it!

I isolated $\frac{\partial z}{\partial y}$ for both equations of $\frac{\partial z}{\partial r}$ and $\frac{\partial z}{\partial \theta}$ and then equated the two and re-arranged a little to get $\frac{\partial z}{\partial x}=\cos\theta\frac{\partial z}{\partial r}-\frac{1}{r}\sin\theta\frac{\partial z}{\partial\theta}$.

However, I still don't get why the solution is simply not $\frac{\partial z}{\partial x}= \frac{\partial x}{\partial r}+ \frac{\partial x}{\partial \theta}=\cos\theta - r\sin\theta$

8. Apr 27, 2013

### Simon Bridge

Well done.

... but didn't you just demonstrate that it isn't?
Looks like you may need to go back to what the partial means.