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Perpendicular vector Q

  1. Feb 20, 2006 #1
    The question reads:

    "Consider the surface given by the equation [tex] x^3 + xy^3 - z^2 = -4 [/tex] The point p = (1,2,3) lies on this surface. Give a vector that is perpendicular to the surface at p?"

    I'm not too confident about this question although there is a theorem in my notes saying:

    if p exists within the level surface equal to c and the gradient of the function f is not equal to zero, then the gradient at point p is perpendicular to every path in the level surface equal to c which passes through p.

    Does this apply here?
  2. jcsd
  3. Feb 20, 2006 #2


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    Well, yes, of course, it does!! Let f(x,y,z)= x2+ xy3- z2. Then this plane is a level curve of f: f(x,y,z)= -4. The gradient of f is perpendicular to that plane.
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