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Directional derivative of a surface

  1. Sep 4, 2012 #1
    1. The problem statement, all variables and given/known data
    What is the directional derivative of the function z = x3 - y at the point (1, 2, -1) and in the direction of a vector (1,1,1)?


    2. Relevant equations



    3. The attempt at a solution
    If f(x,y) = x3 - y, then ∇f = (3x2, -1) which equals (3, -1) at the given point. Now I understand I have to take the dot product of the gradient with the unit vector (1/√3, 1/√3, 1/√3) but I'm not quite sure how to...

    Can a function like z = f(x,y) have a directional derivative in the direction of a three dimensional vector?

    Thanks for any help!
     
  2. jcsd
  3. Sep 5, 2012 #2
    You are right, you cannot take the directional derivative of f(x,y) in the direction of a vector in three dimensions (unless the z component is zero, in which case you sort of can).
     
  4. Sep 5, 2012 #3

    HallsofIvy

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    The surface [itex]z= x^3- y[/itex] is given by [itex]x\vec{i}+ y\vec{j}+ (x^3- y)\vec{k}[/itex] the gradient of that vector gives the normal vector and you can take the dot product of that with a unit vector in the direction of [itex]\vec{i}+ \vec{j}+ \vec{k}[/itex]- but I would NOT call that a "directional derivative".
     
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