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Math Help - Differentiation of Multivariable Functions

  1. #1
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    Differentiation of Multivariable Functions

    I need help with finding the derivative of z, with respect to t, in the following questions:

    a)  z\; =\; x^{2}\; -\; y^{2} , with x\; =\; t and  y\; =\; t^{2}

    b)  z\; =\; \frac{x}{y} , with  x\; =\; \sin \left( t \right)\; and  y\; =\; \cos \left( t \right)

    I believe the method for doing both questions is:

    \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}\; +\; \frac{\partial z}{\partial y}\frac{dy}{dt}

    Is this correct? Because no matter how many times I try to do the above questions using that method, I get answers that don't match the ones in the text book.

    My answer for a) is \frac{dz}{dt}=2t-t^{4}-2t^{3} and my answer for b) is \frac{dz}{dt}=1\; +\; \frac{\sin ^{2}t}{\cos ^{2}t}
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  2. #2
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    Quote Originally Posted by DJ Hobo View Post
    I need help with finding the derivative of z, with respect to t, in the following questions:

    a)  z\; =\; x^{2}\; -\; y^{2} , with x\; =\; t and  y\; =\; t^{2}

    b)  z\; =\; \frac{x}{y} , with  x\; =\; \sin \left( t \right)\; and  y\; =\; \cos \left( t \right)

    I believe the method for doing both questions is:

    \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}\; +\; \frac{\partial z}{\partial y}\frac{dy}{dt}

    Is this correct? Because no matter how many times I try to do the above questions using that method, I get answers that don't match the ones in the text book.

    My answer for a) is \frac{dz}{dt}=2t-t^{4}-2t^{3} and my answer for b) is \frac{dz}{dt}=1\; +\; \frac{\sin ^{2}t}{\cos ^{2}t}
    Your formula is correct but I don't see how you could have gotten that answer for (a) from that formula!

    For (a) \frac{\partial z}{\partial x}= 2x, \frac{\partial z}{\partial y}= -2y, \frac{dx}{dt}= 1, and [tex]\frac{dy}{dt}= 2t.

    \frac{dz}{dt}= \frac{\partial z}{\partial x}\frac{dx}{dt}+ \frac{\partial z}{\partial y}\frac{dy}{dt}
    =  (2x)(1)+ (-2y)(2t)= (2t)(1)+ (-2t^2)(2t)

    Your answer for (b) is correct. Is it possible that the answer in the book is 1+ tan^2(t()?
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Your formula is correct but I don't see how you could have gotten that answer for (a) from that formula!

    For (a) \frac{\partial z}{\partial x}= 2x, \frac{\partial z}{\partial y}= -2y, \frac{dx}{dt}= 1, and [tex]\frac{dy}{dt}= 2t.

    \frac{dz}{dt}= \frac{\partial z}{\partial x}\frac{dx}{dt}+ \frac{\partial z}{\partial y}\frac{dy}{dt}
    =  (2x)(1)+ (-2y)(2t)= (2t)(1)+ (-2t^2)(2t)

    Your answer for (b) is correct. Is it possible that the answer in the book is 1+ tan^2(t()?
    Heh I'm gonna have to keep trying with a), then. And no, the answer in the book isn't 1+ tan^{2}t. It's \frac{dz}{dt}=\sec ^{2}t.

    EDIT - I understand why the answer for a) is so, now. Thanks for your help
    Last edited by DJ Hobo; March 6th 2010 at 04:32 AM.
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