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Question about grad, nabla (vector operator)

  1. Oct 22, 2011 #1
    In my notes it says that grad F will give you a vector normal to the contour. Howver I thought grad F would give you a vector tangent because the path is aligned with the vector field. Is it different when talking about contours and paths?

    If you find grad F of a function F does that give you a vector tangent or normal to the contour/path?
     
  2. jcsd
  3. Oct 22, 2011 #2

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    Hi aero&astro! :smile:

    Yep.
    grad F on a path will give you a vector tangent to the path.
    grad F on an equation will give you a normal to a contour satisfying the equation.

    Actually, the gradient gives you the vector in which F increases.
    A path "increases" in the direction of the path.
    The equation-value of a contour increases perpendicular to the contour.

    Consider for instance y=x2.
    Its path is given by f(x) = x2 and its contour-equation is F(x,y) = x2 - y = 0.

    grad f(x) =2x, which gives you the direction f(x) increases with increasing x.
    grad F(x,y)= (2x, -1), which gives you the direction F(x,y) increases with changing (x,y).

    In the first case you "walk the path".
    In the second case you move to a higher level contour given by F(x,y)=c.
     
    Last edited: Oct 22, 2011
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