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Equation of a line that lies on a z=f(x,y) function

  1. Oct 12, 2015 #1
    • Member warned about posting with no effort shown
    1. The problem statement, all variables and given/known data
    Hello,
    I have this function z(x,y) = sin(2*x)*cos(2*y)*e-(x^2+y^2)/6

    I need to find an equation of any curve line that lies on that z(x,y) function.

    2. Relevant equations


    3. The attempt at a solution
    Sorry, I really have no idea how to proceed on this, been looking on my note book and google and have not find anything.
     
  2. jcsd
  3. Oct 12, 2015 #2

    SteamKing

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    What do you mean a line "that lies on that ... function"?

    Are you talking about a line which is tangent to the function's surface?
     
  4. Oct 12, 2015 #3
    Sorry edited the post, its a curve line.

    To clarify, the curve line has to be part of the z(x,y) function. The curve line has to be completely on the function. If I didn't clarify anything, let me know.
     
  5. Oct 12, 2015 #4

    Ray Vickson

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    A curve ##(x(t),y(t),z(t))## lying in the surface must project down to a curve on the ##x\,y## plane, obtained by just ignoring the ##z##-component ##z(t)##.
     
  6. Oct 12, 2015 #5
    how do I proceed on this? what are the steps I have to take.

    Do I take a random number of x and y and plot it in my equation to find z?
    sorry I have never done this before.
     
  7. Oct 12, 2015 #6

    Mark44

    Staff: Mentor

    It might be that you need to find level curves on this surface. As already mentioned, the graph of ##z = f(x, y) = \sin(2x) \cos(2y) e^{-(x^2+y^2)/6}## represents a surface in three dimensions. If you set z to some specific value, that defines a curve that lies on the surface. On level curve is when z = 0, or ##0 = \sin(2x) \cos(2y) e^{-(x^2+y^2)/6}##. Geometrically, this level curve is the intersection of the surface with the x-y plane.
     
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