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General question on multivariate calculus

  1. Oct 16, 2013 #1
    Just started multivariate course, can't figure out this simple question. If f(u,v,w) is a function of 3 variables. And u, v and w are themselves function of t. Then does f(u,v,w)=0 implies df/dt=0 or df/du=0. or both.
  2. jcsd
  3. Oct 17, 2013 #2


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    df/dt = 0, the partials with respect to u, v, w may not be.
    df/dt = (∂f/∂u)du/dt + (∂f/∂v)dv/dt + (∂f/∂w)dw/dt.
  4. Oct 17, 2013 #3


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    If, as you wrote, F(x, y, z)= 0 for all x, y, and z, it is constant no matter how x, y, and z are changed and all partial derivatives are 0. With x, y, and z functions of t, no matter how t changes, that simply results in x, y, and z changing so F remains constant. dF/dt= 0.

    (This is NOT the question mathman answered. He appears to be thinking you were asking about the derivative being 0, not "f(u,v,w)=0".

    As long as F(x, y, z) has continuous partial derivtives, and x, y, and z are differentiable functions of t,
    [tex]\frac{dF}{dt}= \frac{\partial F}{\partial x}\frac{dx}{dt}+ \frac{\partial F}{\partial y}\frac{dy}{dt}+ \frac{\partial F}{\partial z}\frac{dz}{dt}[/tex]
    so that if all partial derivatives of F, with respect to x, y, and z, the dF/dt= 0 for any parameter, t. But if dF/dt= 0 for some t, it does NOT follow that the partial derivatives are 0.)
  5. Oct 17, 2013 #4


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    There seems to be some ambiguity in the question... is f(u,v,w) = 0 for all (u,v,w) or just for all t?
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