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That given a continuous surface, contour lines exist

  1. Apr 11, 2007 #1
    Can you guys help me prove: Given a continuous and differentiable function (or surface) f: R^2 -> R, such that f(x,y) = z ... contour lines can always be drawn... the function is NOT bijective.

    I've been thinking of choosing any arbitrary point and showing that the curves that intersect to form that point (over y-axis and x-axis) always manage to have certain z-values that are equal.

    How do I point this is in mathematical terms?
    Last edited: Apr 11, 2007
  2. jcsd
  3. Apr 11, 2007 #2
    Here is a start of my solution which I am stuck with.

    Consider this continuous surface [tex]f(x,y) = z[/tex]. Consider arbitrary surface point [tex](i,j,z _i_j)[/tex]. This point must exist (given [tex]f(i,j) = z _i_j[/tex]).

    Proof by contradiction:
    Consider continuous functions [tex]f(x,j)[/tex] and [tex]f(i,y)[/tex] which are curves that are part of the surface. [tex]f(x,j) = f(i,y) = z_i_j[/tex] are equal IFF x = i and y = j.

    If I can disprove the IFF part, then I should be done right?
    Last edited: Apr 11, 2007
  4. Apr 11, 2007 #3


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    Let P = (a, b, f(a, b)) be a point on the surface.

    Consider the curves C = (x, b, f(x, b)) and D = (a, y, f(a, y)) that lie on the surface.

    There exists an s and a t such that (s, b, f(s, b)) and (a, t, f(a, t)) have the same z-coordinate.

    Incidentally, this hypothesis is false. For example, consider the point (0, 0, 0) on the surface:
    z = x^2 - y^2​
    The curve over the x-axis never shares a z-coordinate with a point over the y-axis. (Except at the origin when both z-coordinates are zero)

    Hint: it might help to think, not of a surface in R^3, but of a function R^2-->R. (And I think you have one of the right ideas for proving this)
    Last edited: Apr 11, 2007
  5. Apr 11, 2007 #4
    Hmm.... I can't just use the curves along x and y axes. Because there are bound to be such R^2 -> R functions which elude only those particular axes?
  6. Apr 11, 2007 #5
    The function is everywhere differentiable and continuous/
  7. Apr 12, 2007 #6


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    Therefore at any point you can define a tangent plane. If the tangent plane is not horizontal then you can define the direction of the contour line using it.

    If the tangent plane is horizontal then you need to refine what you mean by "drawing contours". There may be a finite number (including zero) or an infinite number of "contour lines" through such a point.

    Examples: functions f(x,y) = x^2 + y^2. f(x,y) = x^2 - y^2, f(x,y) = 0 at the origin.
  8. Apr 13, 2007 #7


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    Well, you hypothesized something much stronger than what you actually wanted to prove.

    Maybe your idea is salvagable? Maybe if you examine all the ways things might behave, you'll come up with a different statement that you can prve, about those two curves?
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