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

  • #1

Homework Statement


Consider z=f(x,y), where x=rcosθ and y=rsinθ

Show that [itex]\frac{\partial z}{\partial x}=\cos\theta\frac{\partial z}{\partial r}-\frac{1}{r}\sin\theta\frac{\partial z}{\partial\theta}[/itex]



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?
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,874
1,655
Evaluate ##\frac{\partial z}{\partial r}##
 
  • #3
[itex]\frac{\partial z}{\partial r}=\cos\theta\frac{\partial z}{\partial x}+\sin\theta\frac{\partial z}{\partial y}
[/itex]

then rearranging this [itex]\frac{\partial z}{\partial x}=\sec\theta\frac{\partial z}{\partial r}-\tan\theta\frac{\partial z}{\partial y}[/itex]

I'm not doing something right or not understanding some underlying concept here. Why was my dz/dx derivation wrong?
 
  • #4
Simon Bridge
Science Advisor
Homework Helper
17,874
1,655
Don't rearrange yet. Do the other partial - the one with theta.
 
  • #5
[itex]\frac{\partial z}{\partial\theta}=-r\sin\theta\frac{\partial z}{\partial x}+\frac{1}{r}\cos\theta\frac{\partial z}{\partial y}[/itex]

I've re-arranged this and get [itex]\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}[/itex]

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

Now I'm more confused...

Isn't it just [itex]r-r\sin\theta[/itex]?
 
  • #6
Simon Bridge
Science Advisor
Homework Helper
17,874
1,655
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.
 
  • #7
Hey Simon, I got it!

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


However, I still don't get why the solution is simply not [itex] \frac{\partial z}{\partial x}= \frac{\partial x}{\partial r}+ \frac{\partial x}{\partial \theta}=\cos\theta - r\sin\theta[/itex]
 
  • #8
Simon Bridge
Science Advisor
Homework Helper
17,874
1,655
Hey Simon, I got it!
Well done.

However, I still don't get why the solution is simply not [tex] \frac{\partial z}{\partial x}= \frac{\partial x}{\partial r}+ \frac{\partial x}{\partial \theta}=\cos\theta - r\sin\theta[/tex]
... but didn't you just demonstrate that it isn't?
Looks like you may need to go back to what the partial means.
 

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

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
20K
Replies
15
Views
1K
Replies
4
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
6K
Replies
4
Views
1K
Replies
2
Views
322
  • Last Post
Replies
3
Views
2K
Top