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Math Help - Ratio of differentials to partial derivatives of two (implicit?) functions

  1. #1
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    Ratio of differentials to partial derivatives of two (implicit?) functions

    Hi.

    So, given  f(x,y,z) = 0 and  g(x,y,z) = 0

    Show that \frac{dx}{\frac{\delta (f,g)}{\delta (y,z)}} = \frac{dy}{\frac{\delta (f,g)}{\delta (x,z)}} = \frac{dz}{\frac{\delta (f,g)}{\delta (x,y)}}

    I think it makes sense (I can "see" why the equality holds) but can't think of a way to derive it. To begin, I wrote  f(x,y,z) = 0 as  f(x_f(y,z),y_f(x,z),z_f(x,y)) = 0 , assuming each argument was an implicit function of the others. Similarly,  g(x_g(y,z),y_g(x,z),z_g(x,y)) = 0

    I feel like some version of the chain rule would be useful ( does the cross product of the gradients have some relationship to differential??), and that the above equality could be easier to derive if written as \frac{dy}{dx} = \frac{\frac{\delta (f,g)}{\delta (x,z)}}{\frac{\delta (f,g)}{\delta (y,z)}}  = .... . I seem to have trouble taking the gradient,or any partial derivatives of implicit functions when using the chain rule... ie, is \frac{\delta f}{\delta x_f(y,z)}\frac{\delta x_f(y,z)}{\delta (y,z)} the correct derivative of f w.r.t x?

    Any help appreciated, thanks.
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  2. #2
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    Progress? I think that

    \nabla f = \frac{\delta f}{\delta x_f} \frac{\delta x_f}{\delta (y,z)} + \frac{\delta f}{\delta y_f} \frac{\delta y_f}{\delta (x,z)} + \frac{\delta f}{\delta z_f} \frac{\delta z_f}{\delta (x,y)}

    and

    \nabla g = \frac{\delta g}{\delta x_g} \frac{\delta x_g}{\delta (y,z)} + \frac{\delta g}{\delta y_g} \frac{\delta y_g}{\delta (x,z)} + \frac{\delta g}{\delta z_g} \frac{\delta z_g}{\delta (x,y)}

    I think that is how the gradients look, anyways.

    When we take their cross products, are the numerators of their components denoted by \delta (f,g)? if so, then it looks like the solution is that the cross product of the gradients is proportional to the differentials?
    Last edited by matt.qmar; March 15th 2011 at 05:06 PM.
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  3. #3
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    Whoops, I meant to write the gradients as vectors, not as a sum.

    \nabla f = (\frac{\delta f}{\delta x_f} \frac{\delta x_f}{\delta (y,z)} , \frac{\delta f}{\delta y_f} \frac{\delta y_f}{\delta (x,z)} , \frac{\delta f}{\delta z_f} \frac{\delta z_f}{\delta (x,y)})

    and

    \nabla g = (\frac{\delta g}{\delta x_g} \frac{\delta x_g}{\delta (y,z)} , \frac{\delta g}{\delta y_g} \frac{\delta y_g}{\delta (x,z)} , \frac{\delta g}{\delta z_g} \frac{\delta z_g}{\delta (x,y)})
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  4. #4
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    Another update, I just went to double check that I had computed the gradient correctly but got something different, and the Wolfram gradient is messier but I suppose it must be correct? I guess I didn't really know the chain rule properly...
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