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Is a gradient perpendicular to the osculating plane of a regular curve?

  1. Mar 7, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove the following or disprove with a counterexample: Let f be a differentiable function in an open set U in R^3 and (a, b, c) be a point in U where the gradient of the function f isn't zero. If r: I -> U is a regular curve with a regular derivative on an open interval I, which contains the zero point, and satisfies the following conditions:

    r(0) = (a, b, c)

    r(I) is contained in the contour surface of the function f that goes through the point (a, b, c)

    2. Relevant equations
    The osculating plane is given with {r(0) + a*r'(0) + b*r''(0) : a, b are reals}.


    3. The attempt at a solution
    Since r is regular with with a regular derivative then r' dot r'' = 0. Since r is contained in the contour surface then gradient(f) dot r' = 0. If I can show that the gradient of f dot r'' is or isn't 0 then this is done. I can't find any relevant theorems for this problem though.
     
  2. jcsd
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