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Use the chain rule to show that... Help please

  1. Sep 16, 2015 #1
    1. The problem statement, all variables and given/known data
    z = ƒ(x,y), x = rcos(θ), y = rsin(θ)
    Use the chain rule to show that:
    [tex] \frac{1}{r^{2}}\frac{\partial ^{2} z}{\partial \theta ^{2}} = sin^{2}(\theta)\frac{\partial ^{2} z}{\partial x^{2}}-2sin(\theta)cos(\theta)\frac{\partial ^{2} z}{\partial x \partial y}+cos^{2}(\theta)\frac{\partial ^{2} z}{\partial y ^{2}} - \frac{1}{r}\frac{\partial z}{\partial r} [/tex]
    2. Relevant equations
    ∂z/∂r = (∂z/∂x)(∂x/∂r) + (∂z/∂y)(∂y/∂r)

    3. The attempt at a solution
    I have worked it through and I have found that:
    [tex] \frac{\partial z}{\partial \theta } = -rsin(\theta)\frac{\partial z}{\partial x}+rcos(\theta)\frac{\partial z}{\partial y}[/tex]
    [tex] \frac{\partial z}{\partial r } = cos(\theta)\frac{\partial z}{\partial x}+sin(\theta)\frac{\partial z}{\partial y}[/tex]
    [tex] \frac{\partial ^{2} z}{\partial \theta ^{2}} = r^{2}(sin^{2}(\theta)\frac{\partial ^{2} z}{\partial x^{2}}-2sin(\theta)cos(\theta)\frac{\partial ^{2} z}{\partial x \partial y}+cos^{2}(\theta)\frac{\partial ^{2} z}{\partial y ^{2}}) [/tex]
    I have checked these a number of times and they seem right to me. The last one listed is the second partial derivative of z in terms of θ, but it doesn't match that of the problem statement. I'm not really sure what I'm doing wrong. Thanks :)
     
  2. jcsd
  3. Sep 16, 2015 #2

    RUber

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    Your second derivative looks like you just squared your first derivative.
    You need to actually take one more derivative to get ## \frac{\partial^2 z}{\partial \theta^2}##.
     
  4. Sep 16, 2015 #3
    I didn't square it, I took the second derivative.
     
  5. Sep 16, 2015 #4

    RUber

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    You have the first part right. But your second derivative is wrong.
    ##\frac{\partial}{\partial \theta } f(x,y) = \frac{\partial f}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial \theta}##
    ##\frac{\partial}{\partial \theta }\left( \frac{\partial}{\partial \theta } f(x,y)\right) = \frac{\partial}{\partial \theta }\left( \frac{\partial f}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial \theta}\right) ##
    Which is
    ## = \left( \frac{\partial}{\partial \theta }\frac{\partial f}{\partial x}\right) \frac{\partial x}{\partial \theta}+\underline{ \frac{\partial f}{\partial x} \left( \frac{\partial}{\partial \theta }\frac{\partial x}{\partial \theta} \right) } + \left( \frac{\partial}{\partial \theta } \frac{\partial f}{\partial y}\right) \frac{\partial y}{\partial \theta}+ \underline{ \frac{\partial f}{\partial y} \left( \frac{\partial}{\partial \theta }\frac{\partial y}{\partial \theta}\right)} ##
    In your expansion, you missed the second derivatives with respect to theta on x and y. Which will give you the equivalent parts you need for the partial with respect to r.
     
  6. Sep 16, 2015 #5
    Ahh. That makes sense. Thanks a lot.
     
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