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I Question about gradient, tangent plane and normal line

  1. Jan 22, 2017 #1
    Hi All,

    This question is about vector calculus, gradient, directional derivative and normal line.

    If the gradient is the direction of the steepest ascent:

    >> gradient(x, y) = [ derivative_f_x(x, y), derivative_f_y(x, y) ]

    Then it really confuse me as when calculating the normal line perpendicular to the tangent plane, the formula would be:

    >> normal line = (derivative_f_x(x, y), derivative_f_y(x, y), z),

    But both derivative_f_x(x,y) & derivative_f_y(x,y) are gradient (the slope of the tangent plane). I don't think the steepest ascent/descent is the slope of the normal line perpendicular to the tangent plane!

    For example
    Find a vector function for the line normal to x^2 + 2y^2 + 4z^2 = 26 at (2, -3, -1).
    Answer: (2 + 4t, -3 -12t, -1 - 8t).

    Anyone care to give it a shot and show me the step??


    Any information would be much appreciated.

    Thanks.
     
  2. jcsd
  3. Jan 22, 2017 #2

    FactChecker

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    Gold Member

    Draw an example of a function of one variable at points with negative, zero, and positive slopes. That should convince you. Adding more independent variables just combines the logic for more dimensions.
    The perpendicular projection of the normal to the subspace of the independent variables should give you the direction of steepest ascent.
     
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