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Max rate of descent , gradient?

  1. Oct 11, 2009 #1
    Can anyone help me with the following question?
    Find the path of the steepest descent along the surface z=x^3 + 3y^2 from the point (1, -2, 13) to (0,0,0)
    Note: the general solution of the differential equation f ' (t)-kf(t) =0 is
    f(t) = ce^kt, where c is an arbitary number
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    I understand that max rate of change occurs in the direction of the gradient, and max descent occurs in the -gradient(f) direction

    But I don't know what to do with the 2 points..and how it has anything to do with the formula given as a hint..

    Can anyone help
    Thanks so much
     
  2. jcsd
  3. Oct 11, 2009 #2
    Research something called directional derivative.
     
  4. Oct 11, 2009 #3
    I do know what that directional derivative is, but can u give a more concrete suggestion?
     
  5. Oct 11, 2009 #4

    zcd

    User Avatar

    [tex]\nabla{f}= \frac{\partial f}{\partial x} \hat{i}+\frac{\partial f}{\partial y} \hat{j}[/tex], and [tex]\hat{u}=-\frac{1}{\sqrt{5}}\hat{i}+\frac{2}{\sqrt{5}}\hat{j}[/tex] for all values along the path.
     
  6. Oct 11, 2009 #5
    thanks , but how did you get the answer??

    I don't undestand it...
     
  7. Oct 11, 2009 #6

    zcd

    User Avatar

    That's not the answer, but you can get direction by subtracting P-P0 and dividing by its magnitude (to make it a unit vector).
     
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