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Smoothness of curve via Implicit Function THeroem

  1. Jul 13, 2012 #1
    So I have to determine whether the set S = {(x,y): F(x,y) = 0} is a smooth curve, draw a sketch, examine the nautrue of ∇F = 0. Near which points of S is S the graph of a function y = f(x) and x = f(y)

    F(x,y) = (x2+y2)(y-x2-1)


    So it is the union of a point and the parabola y = 1+x2. Not an issue.

    S is smooth everywhere except at the point (0,0)

    y=f(x) is smooth all on the parabola

    (here's my issue) x= f(y) near any point except (0,1)

    I know this point is the y-int of my parabola, but why is x = f(y) not solveable here? or does it mean the curve is not smooth there? Because graphically everything looks fine.
  2. jcsd
  3. Jul 13, 2012 #2


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    If [itex]y= x^2+ 1[/itex], then [itex]x= \pm \sqrt{y- 1}[/itex] so f does not have a true "inverse". You can, of course, invert it "near points", that is, in some neigborhood of the point as long as x, for that point, is either positive or negative. For x= 0, The set, S, contains both (0, 1) and (0, 0). In no neighborhood of (0, 1) can we define x as a function of y because, again, there are both positive and negative values of x corresponding to the same value of x. And, of course, about (0, 0) there is no neighborhood in which we can define x as a function of y.
  4. Jul 13, 2012 #3

    So what your saying there is that in that neighborhood of (0,1) any "y" value that I choose is going to give me both a negative and positive answer so it would not be possible for me to construct a function at that point?

    I have another question in terms of parametric equations and the condition f'(t) = 0:

    Same scenario as above but f(t) = (t^3-1, t^3+1)

    So I've done everything, but when I take the derivative I get the point (-1,1) which should imply that the curve is not "smooth" at that point, but it is......why?
    Last edited: Jul 13, 2012
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