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Math Help - Multivariable Calc Chain Rule proof problem

  1. #1
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    Multivariable Calc Chain Rule proof problem

    If u=f(x,y), where x = e^s cos(t) and y = e^s sin(t), show that:

    (d^2 u)/(dx^2) + (d^2 u)/(dy^2) = e^(-2s) ((d^2 u)/(ds^2)+(d^2 u)/(dt^2))

    Sorry that's really hard to read...don't know how to use the math tags yet. Those are second partial derivatives. How do I do this problem? It has something to do with the chain rule since it's in that section of the book.
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    Senior Member Sampras's Avatar
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    Quote Originally Posted by Susaluda View Post
    If u=f(x,y), where x = e^s cos(t) and y = e^s sin(t), show that:

    (d^2 u)/(dx^2) + (d^2 u)/(dy^2) = e^(-2s) ((d^2 u)/(ds^2)+(d^2 u)/(dt^2))

    Sorry that's really hard to read...don't know how to use the math tags yet. Those are second partial derivatives. How do I do this problem? It has something to do with the chain rule since it's in that section of the book.
    So  u = f(x,y) where  x = e^{s} \cos t and  y = e^{s} \sin t and you want to show that  \frac{\partial^{2}u}{\partial x^{2}} + \frac{\partial^{2}u}{\partial y^2} = e^{-2s} \left(\frac{\partial^{2}u}{\partial s^2} + \frac{\partial^{2}u}{\partial t^2} \right) ? Compute the LHS, and then the RHS to show that they are equal.

    So  \frac{\partial u}{\partial s} = e^{s} \cos t \frac{\partial u}{\partial x}+ e^{s} \sin t \frac{\partial u}{\partial y} and  \frac{\partial u}{\partial t} = e^{s} \sin t \frac{\partial u}{\partial x}+ e^{s} \cos t \frac{\partial u}{\partial y} .
    Last edited by Sampras; May 27th 2009 at 08:30 PM.
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    Yes, that's right.
    Last edited by Susaluda; May 27th 2009 at 08:28 PM.
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    Ok so then do I take the derivative of those again? After that what do I do?

    Sorry, what did you get when you take the derivative again? For some reason it's not working right...
    Last edited by Susaluda; May 27th 2009 at 08:53 PM.
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    Senior Member Sampras's Avatar
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    Quote Originally Posted by Susaluda View Post
    Ok so then do I take the derivative of those again? After that what do I do?
    Then you add them together and multiply by  e^{-2s} . You should get  (\cos^{2}t + \sin^{2}t) \left( \frac{\partial^{2}u}{\partial x^2}+ \frac{\partial^{2}u}{\partial y^2} \right) = \frac{\partial^{2}u}{\partial x^2}+ \frac{\partial^{2}u}{\partial y^2} .
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    Senior Member Sampras's Avatar
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    Quote Originally Posted by Susaluda View Post
    Ok so then do I take the derivative of those again? After that what do I do?

    Sorry, what did you get when you take the derivative again? For some reason it's not working right...
    So  \frac{\partial^{2}u}{\partial s^2} = e^{s} \cos t \frac{\partial u}{\partial x} +e^{s} \cos t  \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial x} \right)+e^{s} \sin t \frac{\partial u}{\partial y} + e^{s} \sin t \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial y} \right) . We know that  \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial x} \right) = \frac{\partial^{2}u}{\partial x^{2}} \frac{\partial x}{\partial s} + \frac{\partial^{2} u}{\partial y \partial x} \frac{\partial y}{\partial s} = e^{s} \cos t \frac{\partial^{2}u}{\partial x^{2}}+ e^{s} \sin t \frac{\partial^{2}u}{\partial y \partial x} . Also  \frac{\partial}{\partial s} \left(\frac{\partial u}{\partial y} \right) = \frac{\partial^{2}u}{\partial y^{2}} \frac{\partial y}{\partial s} + \frac{\partial^{2} u}{\partial x \partial y} \frac{\partial x}{\partial s} = e^{s} \sin t \frac{\partial^{2}u}{\partial y^{2}}+ e^{s} \cos t \frac{\partial^{2}u}{\partial x \partial y} . By continuity of partials,   \frac{\partial^{2} u}{\partial x \partial y} = \frac{\partial^{2} u}{\partial y \partial x} .

    So the same thing for  \frac{\partial^{2} u}{\partial t^2} .
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    Ok so then I just add them together after that?
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  8. #8
    Senior Member Sampras's Avatar
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    yes and multiply the quantity by  e^{-2s} .
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