Results 1 to 4 of 4

Thread: Implicit Function Problem

  1. #1
    Junior Member
    Joined
    Apr 2009
    Posts
    55
    Thanks
    1

    Implicit Function Problem

    If $\displaystyle f(x, y, z) = 0$, then [MATH%
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member abhishekkgp's Avatar
    Joined
    Jan 2011
    From
    India
    Posts
    495
    Thanks
    1
    Quote Originally Posted by armeros View Post
    If $\displaystyle f(x, y, z) = 0$, then $\displaystyle \frac{\partial z}{
    \partial y} \frac{\partial y}{
    \partial x} \frac{\partial x}{
    \partial z} = -1$

    Make sense out of this nonsense and prove.

    I am totally clueless.

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

    $\displaystyle x(z(y)), y(x(z)), z(y(x))$.

    Anyone has a clue ?

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

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

    For the sake of understanding suppose we could write $\displaystyle z=g(x,y)$ we have $\displaystyle dz= \frac{\partial g}{\partial x}dx + \frac{\partial g}{\partial y}dy$.
    if $\displaystyle z$ is a constant then $\displaystyle 0=\frac{\partial g}{\partial x}+\frac{\partial g}{\partial y} \frac{\partial y}{\partial x}$
    which on using $\displaystyle \frac{\partial g}{\partial x}=\frac{1}{\frac{\partial x}{\partial g}}$ we have,
    $\displaystyle \frac{\partial y}{\partial x} \frac{\partial g}{\partial y} \frac{\partial x}{\partial g}=-1$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2009
    Posts
    55
    Thanks
    1
    The thought that $\displaystyle f(x,y,z(x,y)) = 0$ 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 $\displaystyle x$ and get $\displaystyle \frac{\partial f}{\partial x} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial x} = 0$ ...

    I never have thought that $\displaystyle dz = 0$

    I really appreciate your help.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Apr 2009
    Posts
    55
    Thanks
    1
    Sorry double post
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Problem involving implicit function theorem.
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Oct 20th 2010, 12:35 PM
  2. Replies: 0
    Last Post: Oct 19th 2009, 07:58 PM
  3. Replies: 6
    Last Post: Oct 5th 2009, 01:33 AM
  4. Implicit function
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Jan 28th 2009, 08:37 PM
  5. Implicit function
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Aug 13th 2008, 09:30 PM

Search Tags


/mathhelpforum @mathhelpforum