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Homework Help: Directional derivatives

  1. Aug 4, 2010 #1
    1. The problem statement, all variables and given/known data
    See first figure.

    2. Relevant equations

    3. The attempt at a solution

    See second figure. I defined direction of the line by which the two planes intersect as,


    and found that the point they are asking about is when,


    and I'm stuck here. This is my first attempt at a question like this so I applogize in advance if I'm missing something obvious.

    How do I figure out what direction is the direction of increasing x?

    [tex]D_{u^{+}}f(x,y,z) = f_{x}(x,y,z)a + f_{y}(x,y,z)b + f_{z}(x,y,z)c[/tex]


    [tex]D_{u^{+}}f(x,y,z) = ax^{2} + b2y -c2z [/tex]

    At the point (-1,1,3),

    [tex]D_{u^{+}}f(-1,1,3) = a + 2b -6c[/tex]

    This is as far as I got.

    Any suggestions?

    Attached Files:

  2. jcsd
  3. Aug 4, 2010 #2


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    To take a simpler example: If you have the function f(x,y)=x2+y2, and I told you to take the directional derivative along the line y=x....

    The obvious vector to take is the vector (1,1) (scaled to be a unit vector). But that's not the only direction you could pick, (-1,-1) also is a vector pointing along the line y=x. If I addeed: take the direction of increasing x, then you would want to pick the vector that points along the direction in which x increases. This means that the vector pointing in the direction of the line has a positive x coordinate, so the right direction would be (1,1)
  4. Aug 4, 2010 #3

    Okay so the line I'm given is as follows,

    [tex]\vec{l} = <0, -1, 2> + <-1,2,1>t[/tex]

    and for x to increase along the line l,

    [tex]t \leq -1[/tex].

    so if I simply compute t = -1, I'll get a vector with the direction,

    [tex]\vec{u} = <1,-3,1>[/tex]

    Scaling this to be a unit vector,

    [tex]\hat{u} = \frac{1}{\sqrt{11}}<1,-3,1>[/tex]

    Would this be a vector of magnitude 1 pointing in the direction of increasing x?
  5. Aug 5, 2010 #4
    Bump, still looking for some help on this one.
  6. Aug 8, 2010 #5
    Bumpity bump!
  7. Aug 8, 2010 #6
    That's not right. Your [tex]\vec{u} [/tex] is the vector from the origin to the point (1, -3, 1) on line l where t = -1. You want the direction of your line itself for which x is increasing. So you should just have [tex] \vec{u} = <1, -2, -1>. [/tex] Make that into a unit vector and finish up.
  8. Aug 9, 2010 #7

    Okay so scaling it as a unit vector,

    [tex]\hat{u} = \frac{1}{\sqrt{6}} < 1, -2, -1>[/tex]

    Now how can I use this to solve for a, b and c? That's what I want to do next right?
  9. Aug 9, 2010 #8
    No...that's what you just found. [tex] \vec{u} = <a, b, c> [/tex]
  10. Aug 9, 2010 #9
    Oh okay! I think I get it now.

    Thanks again.
  11. Aug 14, 2010 #10
    Sorry for ressurecting this thread, but I've got another quick question.

    When someone asks me for the directional derivative of something at a given point, is my final answer a scalar?

    In this case would it simply be the scalar provided by,

    [tex] \nabla f(-1,1,3)\cdot \hat{u}[/tex]


    Thanks again!
  12. Aug 14, 2010 #11
    Yes, it's the scalar product of the gradient and a unit vector.
    D_{u}f= \nabla f\cdot\frac{\bold{u}}{\left|\bold{u}\right|}
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