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Homework Help: Chain rule

  1. Feb 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Find [tex] \frac{\partial z}{\partial u} [/tex] and [tex] \frac{\partial z}{\partial v} [/tex] using the chain rule.

    [tex] z = \arctan(\frac{x}{y}) , x=u^2+v^2 , y=u^2-v^2 [/tex]


    2. Relevant equations



    3. The attempt at a solution

    [tex] \frac{\partial z}{\partial u} = \frac{4uv^2}{v^4 - 2u^2v^2 + u^4} * \frac{1}{1+((u^2+v^2)/(u^2-v^2))^2)} [/tex]

    [tex] \frac{\partial z}{\partial v} = \frac{4u^2v}{v^4 - 2u^2v^2 + u^4} * \frac{1}{(1+((u^2+v^2)/(u^2-v^2))^2)} [/tex]

    why is my answer wrong?
     
    Last edited: Feb 21, 2009
  2. jcsd
  3. Feb 21, 2009 #2

    Tom Mattson

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    Are we supposed to guess at what steps are written on your paper? :confused:
     
  4. Feb 21, 2009 #3

    Dick

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    For one thing d(arctan(x))/dx=1/(1+x^2). I think you missed the square.
     
  5. Feb 21, 2009 #4
    oh you're right, I missed the squares, but it is still wrong
     
  6. Feb 21, 2009 #5

    Dick

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    Tom Mattson is right. It would be a lot easier to say what is wrong if you would show what you've done. Just showing a wrong answer and saying 'what did I do wrong?' is more of a puzzle than a question. I don't think you've done much wrong. I get a different sign for dz/du and you could certainly simplify them more.
     
  7. Feb 21, 2009 #6
    well I think the problem here is just from taking the derivative of

    [tex] \frac{u^2+v^2}{u^2-v^2} [/tex] with respect to u, right. I realize that there should be a - sign in front of it.. am I right?
     
  8. Feb 21, 2009 #7

    Dick

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    Yes. But like I said you can also simplify those expressions a lot. I don't know if the HW checker requires this or not.
     
  9. Feb 21, 2009 #8
    the best simplification I can think of is:

    [tex] \frac{-4uv}{(u^2-v^2)^2+(u^2+v^2)^2} [/tex]
     
    Last edited: Feb 21, 2009
  10. Feb 21, 2009 #9

    Dick

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    Do you mean -4uv/((u^2-v^2)^2+(u^2+v^2)^2)?? I can write that in less space.
     
  11. Feb 21, 2009 #10
    Yes that's what I meant
     
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