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

  1. Apr 24, 2013 #1
    1. The problem statement, all variables and given/known data
    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]



    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. jcsd
  3. Apr 24, 2013 #2

    Simon Bridge

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    Evaluate ##\frac{\partial z}{\partial r}##
     
  4. Apr 24, 2013 #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?
     
  5. Apr 24, 2013 #4

    Simon Bridge

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    Don't rearrange yet. Do the other partial - the one with theta.
     
  6. Apr 25, 2013 #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]?
     
  7. Apr 25, 2013 #6

    Simon Bridge

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    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.
     
  8. Apr 26, 2013 #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]
     
  9. Apr 27, 2013 #8

    Simon Bridge

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    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.
     
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