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Directional derivative

  1. Dec 10, 2008 #1
    I have a function of 2 variables. I know it increase most rapidly in the direction of the gradient, but how about in wht direction is it not increasing?

    I am thinking that the gradient (dot product)(direction in which it is not increasing) = 0

    Any hints?
     
  2. jcsd
  3. Dec 10, 2008 #2

    CompuChip

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    Well, since you mentioned the word "directional derivative" anyway: you could check for which [itex]\vec v[/itex]
    [tex](\vec\nabla f(x, y)) \cdot \vec v < 0[/tex]
    ?
     
  4. Dec 10, 2008 #3

    HallsofIvy

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    Yes, it is true that [itex]\vector{\nabla f}\cdot \vector v[/itex] is the directional derivative in the directional derivative in the direction of [itex]\vec{v}[/itex] (for [itex]\vec{v}[/itex] of length 1). And that tells you the derivative is 0 perpendicular to the gradient.

    (CompuChip, surely you didn't mean "<"?)
     
  5. Dec 11, 2008 #4

    CompuChip

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    Err, no comment? :)
     
  6. Dec 11, 2008 #5
    That is what is was thinking, since of course cos(pi/2) = 0. So the vector that is at a right angle to the gradient is in the direction of zero increase. But how do I go about finding this vector?
     
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