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Directional derivatives

  1. Feb 24, 2014 #1
    1. The problem statement, all variables and given/known data
    Suppose you are standing at the point (-100,-100,360) on a hill that has the shape of the graph of z=500-0.006x2-0.008y2.

    In what direction should you head to maintain a constant altitude?

    2. Relevant equations
    Duf = ∇f[itex]\bullet[/itex]u
    formula for directional derivative


    3. The attempt at a solution
    Here's my idea. If the directional derivative at the point specified is zero, then that's equivalent to maintaining constant altitude. Therefore, I must find the corresponding gradient vector ∇f when the directional derivative is zero right? I have no idea how to carry this out however.
     
  2. jcsd
  3. Feb 24, 2014 #2

    Dick

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    Yes, find the gradient vector. But you can't choose what it is, it's fixed by the problem statement. What you need to find is a vector u such that the directional derivative is zero.
     
  4. Feb 24, 2014 #3
    Ok, I'm confused on how to go about this.

    0 = ∇f[itex]\bullet[/itex]u

    You say ∇f is fixed:

    ∇f = <zx,zy>

    = < -0.012x, -0.016y >

    = < 1.2, 1.6 > Subs. coordinates given

    Then solve for u?
     
  5. Feb 24, 2014 #4

    Dick

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    Yes, write u=<ux, uy>. Since the dot product of u with the gradient is 0 you should be able to find a relation between ux and uy. You won't be able to find a unique u. But the question is only asking for a direction.
     
  6. Feb 24, 2014 #5
    So u could just equal < 0 , 0 > ?
     
  7. Feb 24, 2014 #6

    Dick

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    It could. That means you just stay in the same place and don't go anywhere. But the question is asking for a direction in which you could move and not change altitude. Find another solution.
     
  8. Feb 24, 2014 #7
    Oh I get it.

    I'll use u = < 1 , -0.75 > , that comes out to zero.

    Thanks Dick.
     
  9. Feb 24, 2014 #8

    Dick

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    Right and you're welcome. Actually there are lots of solutions but they all point in either that direction or the opposite direction.
     
  10. Feb 24, 2014 #9
    u is supposed to be a unit vector. So take your u and divide by its magnitude.
     
  11. Feb 24, 2014 #10

    Dick

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    Or they might want an angular bearing. Hard to say. reddawg might know based on previous questions.
     
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