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Math Help - Chain Rule

  1. #1
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    Chain Rule

    Hi I am stuck on this problem. Can someone please explain. Thanks!


    Let
    Compute
    _________
    _________
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    Quote Originally Posted by jffyx View Post
    Hi I am stuck on this problem. Can someone please explain. Thanks!


    Let
    Compute
    _________
    _________
    Here's a hint: by the chain rule

    \frac {\partial w}{\partial s} = \frac {\partial w}{\partial x} \cdot \frac {\partial x}{\partial s} + \frac {\partial w}{\partial y} \cdot \frac {\partial y}{\partial s} + \frac {\partial w}{\partial z} \cdot \frac {\partial z}{\partial s}

    and, similarly,

    \frac {\partial w}{\partial t} = \frac {\partial w}{\partial x} \cdot \frac {\partial x}{\partial t} + \frac {\partial w}{\partial y} \cdot \frac {\partial y}{\partial t} + \frac {\partial w}{\partial z} \cdot \frac {\partial z}{\partial t}

    (2,-1) refers to (s,t) it seems (that is, if the problem described the functions as x(s,t) for instance). you can find the corresponding x, y and z values for this point
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    I still do not understand clearly on how to do the problem. Can someone please explain. Thanks a lot.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jffyx View Post
    I still do not understand clearly on how to do the problem. Can someone please explain. Thanks a lot.
    can you find each of the partial derivatives that you see on the right side of the equations i gave you?
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  5. #5
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    yea i can find the partials, but when i plug into the chain rule formula i get 3 variables... im stuck on that step
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    Quote Originally Posted by jffyx View Post
    yea i can find the partials, but when i plug into the chain rule formula i get 3 variables... im stuck on that step
    ok, so you found all the partials, and you plugged them into the formula. now all that's left is to plug in values for the variables. remember, (s,t) = (-2,-1)

    hence we have s = -2,~t = -1,~x = 2,~y = e^2, \text{ and } z = 1
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  7. #7
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    yay i got thanks for all the help
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