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Question: Calculus

  1. Oct 8, 2006 #1

    Given a function

    z = f(x,y), where x = r * cos(\phi) and y = r * sin (\phi)

    First I show that

    [tex]\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x} cos(\phi) + \frac{\partial z}{\partial y} sin (\phi)[/tex]


    [tex]\frac{\partial z}{\partial \phi} = - \frac{\partial z}{\partial x} r \cdot sin(\phi) + \frac{\partial z}{\partial y} r \cdot sin(\phi)[/tex]

    Finally I need to show that

    [tex](\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 = (\frac{\partial z}{\partial r})^2 + \frac{1}{r^2} (\frac{\partial z}{\partial \phi}) ^2[/tex]

    How do I approach this part of the problem?


    Last edited: Oct 8, 2006
  2. jcsd
  3. Oct 8, 2006 #2


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    Hmm..square both previous equations and add them together, mayhap?
  4. Oct 8, 2006 #3
    This is my solution for (b) please look at them at see if I made a mistake:

    solving (b)

    [tex](\frac{\partial z}{\partial x}^2) ^2 + \frac{1}{r^2} (\frac{\partial z}{\partial \phi})^2 = (\frac{\partial z}{\partial x})^2 \cdot cos ^2 (\phi) + (\frac{\partial z}{\partial y})^2 \cdot sin(\phi) + 2 \frac{\partial z}{\partial x} \cdot \frac{\partial z}{\partial y} sin(\phi) \cdot cos(\phi) + (\frac{\partial z}{\partial x}) ^2 \cdot \frac{r^2 \cdot sin^2 (\phi)}{r^2} + (\frac{\partial z}{\partial y})^2 \cdot \frac{r^2 \cdot cos^2 (\phi)}{r^2} - 2 \frac{\partial z}{\partial x} \cdot \frac{\partial z}{\partial y} \cdot \frac{r^2 sin(\phi) \cdot cos(\phi)}{r^2} = [/tex]

    [tex] = (\frac{\partial z}{\partial x})^2 \cdot (cos^2(\phi) + sin^2 (\phi) \cdot (\frac{\partial z}{\partial y})^2 sin^2 (\phi) = (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2[/tex]

    Sincerely Yours
    Last edited: Oct 8, 2006
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