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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
    N/A


    3. The attempt at a solution

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

    [tex]\vec{d}[/tex]

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

    [tex]t=1[/tex]

    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]

    So,

    [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

    Office_Shredder

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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.
    [tex]
    D_{u}f= \nabla f\cdot\frac{\bold{u}}{\left|\bold{u}\right|}
    [/tex]
     
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