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Proof inverse f'n th'mimplicit f'n th'm?

  1. Nov 24, 2005 #1
    Show that if g: [0,1] x [0,1]-> R^3 is continuously differentiable with
    det [Dg] (s',t') not equal to 0, then there exists S>0 such that
    {g(s,t): (s,t) are elements of the ball, radius S, centred at (s',t')} is the graph of some function.
    (s',t') is just some point.

    I'm just not sure how to start. The prof said that it might be best to use inverse function theorem, but yah...any help would be great.
  2. jcsd
  3. Nov 24, 2005 #2
    what does the inverse function thm say
  4. Nov 24, 2005 #3
    The inverse function theorem says:

    Suppose U, a proper subset of R^n, is open, and x' is an element of U. F: U -> R^n is C1, and DF(x') is invertible. Then there is a neighbourhood V, a proper subset of U, of x' on which F has a C1 inverse function. So there are neighbourhoods V of x' and W of F(x'), and a function g: W-> V so that

    F(g(y))=y for all y in W, and g(F(x))=x for all x in V.

    Moreover, if F(x)=y, then

    Dg(y)= (DF(x))^-1

    [note: we're taking everything as vector-valued, although it should't really matter]
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