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

[mrcleanhands]

Homework Statement

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}

Homework Equations

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?

 

[Simon Bridge]

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

 

[mrcleanhands]

\frac{\partial z}{\partial r}=\cos\theta\frac{\partial z}{\partial x}+\sin\theta\frac{\partial z}{\partial y}<br />

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?

 

[Simon Bridge]

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

 

[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?

 

[Simon Bridge]

Start with the end in mind or you won't get anywhere:
- 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.

 

[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

 

[Simon Bridge]

Hey Simon, I got it!

Well done.

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

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