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Math Help - Implicit Function Problem

  1. #1
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    Implicit Function Problem

    If f(x, y, z) = 0, then [MATH%
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  2. #2
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by armeros View Post
    If f(x, y, z) = 0, then \frac{\partial z}{<br />
\partial y} \frac{\partial y}{<br />
\partial x} \frac{\partial x}{<br />
\partial z} = -1

    Make sense out of this nonsense and prove.

    I am totally clueless.

    I think x, y, z can be written as functions of others

    x(z(y)), y(x(z)), z(y(x)).

    Anyone has a clue ?

    Thanks.
    first think in two dimensions. f(x,y)=0 means y= \phi(x) although it may not always be possible to express y as an explicit function of x.
    similarly f(x,y,z)=0 means z=\phi(x,y)  (\text{ or } y= g(z,x)). Again it may not always be possible to express z explicitly as a function of x and y.

    Geometrically f(x,y,x) is a function of 3 independant variable. there is no constraint on the values of x,y,z(except that these must lie in the domain of the definition of the function). if you set f(x,y,z) equal to zero you have effectively put 1 constraint, that is, if you choose arbitrary values of  x \text{ and } y then the value of z should be so that the function f must become zero. now since z becomes fixed when (x,y) is a point chosen(anywhere in the domain) on x- y plane, the equation f(x,y,z)=0 is that of a surface.

    For the sake of understanding suppose we could write z=g(x,y) we have  dz= \frac{\partial g}{\partial x}dx + \frac{\partial g}{\partial y}dy.
    if  z is a constant then 0=\frac{\partial g}{\partial x}+\frac{\partial g}{\partial y} \frac{\partial y}{\partial x}
    which on using \frac{\partial g}{\partial x}=\frac{1}{\frac{\partial x}{\partial g}} we have,
    \frac{\partial y}{\partial x} \frac{\partial g}{\partial y} \frac{\partial x}{\partial g}=-1
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  3. #3
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    The thought that f(x,y,z(x,y)) = 0 came into my mind as well, but I couldn't go this far. What I was doing was that I try to differentiate with respect to x and get \frac{\partial f}{\partial x} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial x} = 0 ...

    I never have thought that dz = 0

    I really appreciate your help.
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  4. #4
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    Sorry double post
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