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Directional Derivative at an Angle with a 3d Gradient

  1. Sep 27, 2016 #1

    Drakkith

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    1. The problem statement, all variables and given/known data
    Find the directional derivative using ##f\left(x,y,z\right)=xy+z^2## at the point (4, 2, 1) in the direction of a vector making an angle of ##\frac{3π}{4}## with ##\nabla f(4, 2, 1)##.

    2. Relevant equations
    ##f\left(x,y,z\right)=xy+z^2##


    3. The attempt at a solution
    I found the gradient of ##f(4, 2, 1)## as ##\langle 2,4,2 \rangle##.
    Now I'm not sure what to do next. I have a 3d vector, but they only give me one angle, so I don't know which orientation the new unit vector is at.

    It's taken me about 30 min to type all this up (I keep going back and adding and deleting stuff), and I've got to go to an exam for another class, but I just now thought of this: I think that if I knew which direction I had to rotate I could solve this by finding a vector perpendicular to the gradient vector and a vector opposite of the gradient vector and then adding them together to get a vector at 3π/4 from the gradient vector. Knowing that would allow me to get the unit vector I need to solve this.
     
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  3. Sep 27, 2016 #2

    LCKurtz

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    There is a shorter way than we discussed in that recent similar 2D problem. Use$$
    D_{\hat u}f(P) = \nabla f(P)\cdot \hat u = |\nabla f(P)||\hat u|\cos\theta$$
     
  4. Sep 27, 2016 #3

    Drakkith

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    Wait... does that mean the directional derivative is the same at all orientations of 3π/4?

    Also, I got an answer of ##-2\sqrt{3}##, which is correct. Thanks Kurtz.
     
  5. Sep 27, 2016 #4

    LCKurtz

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    Assuming you mean all orientations from the gradient direction, then yes it does. And you could have done the earlier problem this way too. For some reason in that other problem I got the impression you were to find a vector in the new direction first.
     
  6. Sep 27, 2016 #5

    Drakkith

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    That was my impression too. I thought I was supposed to find a vector first. :frown:
     
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