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From cartesian to cylindrical

  1. Mar 1, 2005 #1
    I find this passage [tex] \frac{\partial}{\partial x} = \cos(\phi)\frac{\partial}{\partial \rho } - \frac{\sin(\phi)}{\rho}\frac{\partial}{\partial \phi} [/tex] difficult to understand.

    My teacher wrote this as an explanation:

    [tex] \frac{\partial V}{\partial x} = \frac{\partial\rho}{\partial x}\frac{\partial V}{\partial \rho} + \frac{\partial\phi}{\partial x}\frac{\partial V}{\partial \phi} + \frac{\partial z}{\partial x}\frac{\partial V}{\partial z} [/tex] *

    And then inserting for [tex] \rho [/tex] and [tex] \phi [/tex], which will yield a correct result.

    What I don't understand is how * can be correct? To me it seems that the right side of the equation is equal to [tex] 3\frac{\partial V}{\partial x} [/tex]

    Please enlighten me.
     
    Last edited: Mar 1, 2005
  2. jcsd
  3. Mar 1, 2005 #2
    There, I fixed it :)
     
  4. Mar 1, 2005 #3

    HallsofIvy

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    Could you tell us WHY you think there should be a "3" in that?

    [tex] \frac{\partial V}{\partial x} = \frac{\partial\rho}{\partial x}\frac{\partial V}{\partial \rho} + \frac{\partial\phi}{\partial x}\frac{\partial V}{\partial \phi} + \frac{\partial z}{\partial x}\frac{\partial V}{\partial z} [/tex]
    is simply the chain rule!
    That would be 3 times dV/dx only if all those were equal to dV/dx separately and they are not!

    Take a simple example. Suppose V(u,w)= u2+ 2w2 and
    that u= 3x-1, v= 2x+3.
    One way of finding dV/dx would be to substitute. V(x)= (3x-1)2+ 2(2x+3)2= 9x2- 6x+ 1+ 2(4x2+ 12x+ 9)= 9x2- 6x+ 1+ 8x2+ 24x+ 18= 17x2+ 18x+ 19.

    dV/dx= 34x- 30.

    But a simpler way is to use the chain rule: [itex]\frac{dV}{dx}= \frac{dV}{du}\frac{du}{dx}+ \frac{dV}{dw}\frac{dw}{dx}[/itex]= 2u(3)+ 4w(2)= 6(3x-1)+ 8(2x-3)= 18x- 6+ 16x- 24= 34x- 30 just as before. dV/dx is that sum, not the individual parts.
     
  5. Mar 1, 2005 #4
    [tex] \frac{\partial}{\partial x} = \cos(\phi)\frac{\partial}{\partial \rho } - \frac{\sin(\phi)}{\rho}\frac{\partial}{\partial \phi} [/tex]


    If you understand how chain rule works just as explained by HallsofIvy,

    this is not so difficult.
    Lets say [tex] V=V(\rho, \phi, z)[/tex]

    [tex] \frac{\partial V}{\partial x} = \frac{\partial V}{\partial \rho}\frac{\partial\rho}{\partial x} +\frac{\partial V}{\partial \phi} \frac{\partial\phi}{\partial x}+ \frac{\partial V}{\partial z}\frac{\partial z}{\partial x} [/tex]



    Transformation from cylindrical to cartisian coordinates,
    [tex] x= \rho cos \phi ,\\ y= \rho sin \phi ,\\ z= z [/tex]

    [tex]\rho = \sqrt(x^2 +y^2)[/tex]

    [tex]tan \phi = \frac{y}{x}[/tex]

    From here,

    [tex]\frac{\partial\rho}{\partial x}= \frac{x}{\rho}[/tex]

    [tex]\frac{\partial\phi}{\partial x}= \frac{-sin\phi}{\rho}[/tex]

    [tex]\frac{\partial z}{\partial x}= 0[/tex]


    Substituting you get your result.
     
  6. Mar 1, 2005 #5
    Thank you both very much! I was thinking in the wrong paths entirely. To explain:

    What I figured, not being well versed in this kind of manipulation, was this:


    [tex] \frac{\partial V}{\partial \phi}\frac{\partial \phi}{\partial x} = \frac{\partial V}{\partial x}\frac{\partial \phi}{\partial \phi} = \frac{\partial V}{\partial x} [/tex]

    Because I remember doing something like that once.


    Thanks again for your answers :)
     
  7. Mar 1, 2005 #6
    this is not valid in general, you can only do this in a very special case, i.e.

    [tex] V = V(\phi(x)) [/tex]
     
  8. Mar 1, 2005 #7
    Yes that's the chain rule doing its magic I guess :D
     
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