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Please help! A question about partial differentiation

  1. May 12, 2005 #1
    hi all!
    I know how to solve the part (i) but not part (ii).
    Could anyone help?
     

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  3. May 12, 2005 #2

    arildno

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    Note that in vector notation, we have, from (i):
    [tex]\nabla{u}=\frac{\partial{u}}{\partial{x}}\vec{i}+\frac{\partial{u}}{\partial{y}}\vec{j}=\frac{\partial{u}}{\partial{r}}\vec{i}_{r}+\frac{\partial{u}}{r\partial\theta}\vec{i}_{\theta}[/tex]
    when the polar unit vectors are:
    [tex]\vec{i}_{r}=\cos\theta\vec{i}+\sin\theta\vec{j},\vec{i}_{\theta}=\frac{\partial}{\partial\theta}\vec{i}_{r}=-\sin\theta\vec{i}+\cos\theta\vec{j}[/tex]
    Thus, the gradient operator has the form, in polar coordinates:
    [tex]\nabla=\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta}\frac{\partial}{r\partial\theta}[/tex]
    In agreement thus far?
     
  4. May 12, 2005 #3

    dextercioby

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    Leave vectors alone and simply differentiate

    [tex]\frac{\partial u}{\partial x}=\cos\theta\frac{\partial u}{\partial r}-\frac{\sin\theta}{r}\frac{\partial u}{\partial \theta} [/tex]

    [tex] \frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial x}\right)[/tex]

    [tex]=\cos\theta\frac{\partial}{\partial r}\left(\cos\theta\frac{\partial u}{\partial r}-\frac{\sin\theta}{r}\frac{\partial u}{\partial\theta}\right)-\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\left(\cos\theta\frac{\partial u}{\partial r}-\frac{\sin\theta}{r}\frac{\partial u}{\partial\theta}\right) [/tex]

    and the same way for "y"...

    Daniel.
     
    Last edited: May 12, 2005
  5. May 12, 2005 #4
    Oh! Thank you very much for your reply!
    But I just manage to get the correct answer!
    Thank you anayway! ^^
     
  6. May 12, 2005 #5

    dextercioby

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    So that's the laplacian in polar plane coordinates.BTW,it's customarily to denote these coordinates by [itex] \left(\rho,\theta\right) [/itex] and physicists prefer [itex] \left(\rho,\phi\right) [/itex].

    Daniel.
     
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