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Differentiation, change of variable

  1. Oct 5, 2006 #1
    The question:

    if [tex]x=\rho cos \phi[/tex] and [tex]y=\rho sin\phi[/tex]

    prove that if U is a twice differentiable function of x and y that

    [tex] \frac{d^2U}{dx^2} + \frac{d^2U}{dy^2} = \frac{d^2U}{d\rho^2} + \frac{1}{\rho}\frac{dU}{d\rho} + \frac{1}{\rho^2}\frac{d^2U}{d\phi^2}[/tex]

    i am not sure how to approach this since the function is not given. Is there some assumption i am supposed to make about the function? like f(x,y) = xy or something?
    any help on the first couple steps would be appreciated
     
    Last edited: Oct 5, 2006
  2. jcsd
  3. Oct 5, 2006 #2

    arildno

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    With a rather more transparent notation, let u be a function of the polar coordinates, U the function in Cartesian coordinates, related by:
    [tex]u(\rho(x,y),\theta(x,y))=U(x,y)[/tex]

    With this notation, you would have lower case u's on your right-hand side of the identity.

    You need to compute the partial derivatives of the radial&angular variables with respect to x and y.
     
  4. Oct 5, 2006 #3

    NateTG

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    Have you tried the chain rule?
     
  5. Oct 5, 2006 #4
    so [tex] \rho (x,y)= ((\frac{x}{cos \phi})^2 + (\frac{y}{cos \phi})^2)^\frac{1}{2}[/tex]?

    and [tex] \phi(x,y)=arctan(y/x)[/tex] is this correct?
     
  6. Oct 5, 2006 #5
    to what would i apply the chain rule?
     
  7. Oct 5, 2006 #6

    arildno

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    No, [tex]\rho(x,y)=\sqrt{x^{2}+y^{2}}[/tex]

    That ought to be obvious for anyone claiming to have understood Pythagoras' second most important theorem.
     
  8. Oct 5, 2006 #7
    so after i compute these derivatives of the radial and angular components with respect to x and y i will have everything and just have to put it all together so it looks like the identity i am trying to show?
     
  9. Oct 5, 2006 #8

    arildno

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    No, you must also use the chain rule on u.
    To give you a start on that:
    [tex]\frac{\partial{U}}{\partial{x}}=\frac{\partial{u}}{\partial{r}}\frac{\partial{r}}{\partial{x}}+\frac{\partial{u}}{\partial{\theta}}\frac{\partial{\theta}}{\partial{x}}[/tex]
    and so on..
     
  10. Oct 5, 2006 #9
    where your r is actually a rho i assume then? if so thanks for the help i should be able to solve this one
     
  11. Oct 5, 2006 #10

    arildno

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    Yes, r is rho.
     
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