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Using chain rule to derive 2nd derivative

  1. May 1, 2013 #1
    1. The problem statement, all variables and given/known data


    Use [itex]\frac{\partial z}{\partial r}=\cos\theta\frac{\partial z}{\partial x}+\sin\theta\frac{\partial z}{\partial y}[/itex]
    and [itex]\frac{\partial z}{\partial\theta}=-r\sin\theta\frac{\partial z}{\partial x}+r\cos\theta\frac{\partial z}{\partial y}[/itex] to show that

    [itex] \frac{1}{r^{2}}\frac{\partial^{2}z}{\partial\theta^{2}}+\frac{1}{r} \frac{\partial z}{\partial r}=\sin^{2}\theta\frac{\partial^{2}z}{\partial x^{2}}-2\sin\theta\cos\theta\frac{\partial ^ {2}z}{\partial x \partial y}+\cos^{2}\theta\frac{\partial ^ {2}z}{\partial y^{2}} [/itex]
    2. Relevant equations



    3. The attempt at a solution
    [itex]
    \frac{\partial z}{\partial\theta}\frac{\partial z}{\partial\theta}=\frac{\partial z}{\partial\theta}(-r\sin\theta\frac{\partial z}{\partial x}+r\cos\theta\frac{\partial z}{\partial y})
    [/itex]

    [itex]
    \frac{\partial^{2}z}{\partial\theta^{2}}=(-r\sin\theta\frac{\partial z}{\partial x}+r\cos\theta\frac{\partial z}{\partial y})(-r\sin\theta\frac{\partial z}{\partial x}+r\cos\theta\frac{\partial z}{\partial y})[/itex]

    [itex]
    \frac{\partial^{2}z}{\partial\theta^{2}}=r^{2}\sin^{2}\theta\frac{ \partial ^ {2}z}{\partial x^{2}}-2r^{2}\sin\theta\cos\theta\frac{\partial^{2}z}{\partial x\partial y}+r^{2}\cos^{2}\theta\frac{\partial^{2}z}{\partial y^{2}}[/itex]

    [itex]
    \frac{1}{r^{2}}\frac{\partial^{2}z}{\partial\theta^{2}}=\sin^{2}\theta\frac{\partial^{2}z}{\partial x^{2}}-2\sin\theta\cos\theta\frac{\partial^{2}z}{\partial x\partial y}+\cos^{2}\theta\frac{\partial^{2}z}{\partial y^{2}}[/itex]

    but [itex] \frac{1}{r}\frac{\partial z}{\partial r}=\frac{1}{r}\cos\theta\frac{\partial z}{\partial x}+\frac{1}{r}\sin\theta\frac{\partial z}{\partial y}[/itex]

    and if I add that I get:
    [itex] \frac{1}{r^{2}}\frac{\partial^{2}z}{\partial\theta^{2}}+\frac{1}{r} \frac{ \partial z}{\partial r}=\sin^{2}\theta\frac{\partial^{2}z}{\partial x^{2}}-2\sin\theta\cos\theta\frac{\partial^{2}z}{\partial x\partial y}+\cos^{2}\theta\frac{\partial^{2}z}{\partial y^{2}}+\frac{1}{r}\cos\theta\frac{\partial z}{\partial x}+\frac{1}{r}\sin\theta\frac{\partial z}{\partial y}[/itex] so somehow [itex] \frac{1}{r}\frac{\partial z}{\partial r}=\frac{1}{r}\cos\theta\frac{\partial z}{\partial x}+\frac{1}{r}\sin\theta\frac{\partial z}{\partial y}[/itex] is supposed to be 0???
     
    Last edited: May 1, 2013
  2. jcsd
  3. May 1, 2013 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    What? When I saw "[itex]\frac{dz}{d\theta}\frac{dz}{d\theta}[/itex]" above I started to write "No, that's the wrong notation- that means the product" but now it appears that you really are just multiplying the first derivative with itself. You do understand that this is NOT what "[itex]\frac{d^2z}{d\theta^2}" means, don't you?


     
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