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Maximum value of directional derivative (Duf)?

  1. May 25, 2014 #1
    Hi guys,

    I am confused from what I know the max. value of directional derivative at a point is the length of the gradient vector ∇f or grad. f?

    Why does the answer in my book of a question say that
    Max. val of Duf = (√3145)/5
    when ∇f = (56/5) i- (3/5) j
    ???

    Thanks
     
  2. jcsd
  3. May 26, 2014 #2

    pasmith

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    Homework Helper

    [tex]\|\nabla f\| = \sqrt{\left(\frac{56}{5}\right)^2 + \left(-\frac{3}{5}\right)^2} = \dots[/tex]
     
  4. May 27, 2014 #3

    HallsofIvy

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    Science Advisor

    It is the difference between "a vector" and "length of a vector".
     
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