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Partial differentiation

  1. Feb 13, 2005 #1
    If z = f(x,y), where x = rcos([itex]\theta[/itex]) and y = rsin([itex]\theta[/itex]), find [itex]\frac{\partial z}{\partial r}[/itex], [itex]\frac{\partial z}{\partial\theta}[/itex], and [itex]\frac{\partial^2 z}{\partial r\partial\theta}[/itex]

    Here's what I've done:
    (a)
    [tex]\frac{\partial z}{\partial r} = \frac{dz}{dx} \frac{\partial x}{\partial r} + \frac{dz}{dy} \frac{\partial y}{\partial r} = \frac{dz}{dx} \cos{\theta} + \frac{dz}{dy} \sin{\theta}[/tex]
    (b)
    [tex]\frac{\partial z}{\partial\theta} = \frac{dz}{dx} \frac{\partial x}{\partial\theta} + \frac{dz}{dy} \frac{\partial y}{\partial\theta} = -\frac{dz}{dx} r\sin{\theta} + \frac{dz}{dy} r\cos{\theta}[/tex]

    My question is, for parts a and b, is this correct or must something also be done with the dz/dx and dz/dy, and for part c, I don't know how to do it. Can someone help please?
     
  2. jcsd
  3. Feb 13, 2005 #2
    you dont have to put the dz/dx and dz/dy. its just cos + sin.
    for c, you just take the derivative with respect to r and theta. i forget which you are supposed to do first. but you just do one, than take the derivative of the new form with respect to the other variable.
     
  4. Feb 13, 2005 #3

    dextercioby

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    Points a) and b) are solved wonderfully.
    Point c) is a bit tricky,meaning that u'll have to differentiate one of the 2 expressions found at a) & b) wrt the other variable.
    [tex] \frac{\partial^{2}z}{\partial r \partial \theta}=\frac{\partial}{\partial r}(\frac{\partial z}{\partial \theta}) [/tex]

    Try to do it this way and tell where you get stuck.

    Daniel.
     
  5. Feb 13, 2005 #4

    dextercioby

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    One more thing.It's still a partial derivative for "z" (or "f") too,becasue it depends explicitely on 2 variables,namely "x" and "y"...

    Daniel.
     
  6. Feb 13, 2005 #5
    Thank you dexter. I checked both
    [tex]\frac{\partial}{\partial r}(\frac{\partial z}{\partial \theta}) [/tex]
    and
    [tex]\frac{\partial}{\partial \theta}(\frac{\partial z}{\partial r}) [/tex]

    and they both come to the same answer. So it must be right. :biggrin:
     
  7. Feb 13, 2005 #6

    dextercioby

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    It meant that the function "z" is "well behaved".There are functions for which the mixed partial derivatives are different one from another.In a more advanced way,the 2-nd rank hessian is not symmetric...

    Daniel.
     
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