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Vector Analysis Help

  1. Feb 12, 2008 #1
    1. The problem statement, all variables and given/known data

    By vector methods, find the point on the curve x = t, y = t^2, z = 2 at which the temperature p(x,y,z) = x^2 - 6x + y^2 takes its minimum value

    2. Relevant equations

    3. The attempt at a solution
    As far as I got was finding grad(p). From there I'm not sure where to go -- If I take the magnitude of grad(p) i can find the maximum at any point in space, but how to find the minimum I have no clue. I'm also just having trouble visualizing what exactly is being asked and how a curve and the surface relate to a physical quantity like the temperature. If anyone could help me out that would be much appreciated. Thanks a lot
  2. jcsd
  3. Feb 12, 2008 #2
    I believe this calls for the method of Lagrange multipliers?
  4. Feb 12, 2008 #3


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    If r(t) is your curve, the 'vector' way is to find points where r'(t).grad(p)=0. Do you see why? grad(p) points in direction of steepest change for p, the direction normal to that is the derivative 0 direction. The direct way to do it is just to substitute the r(t) coordinate values into p, and then find the minimum as a function of t. Try is both ways. Algebraically they are identical.
  5. Feb 12, 2008 #4
    doing that i get the following:

    grad(p) = <2x - 6, 2y, 0>
    dr/dt = <1, 2t, 0>

    grad(p) . dr/dt = 2x - 6 + 4ty = 0.

    From this i see that the point where this is true is (1,1,2) with t = 1, which is the right answer. However, I am still somewhat confused because I have no clue how I would pick a correct value for the parameter given a more complicated situation since I chose it because it was trivial. I feel like i just got lucky in picking the correct answer. Is there a more "algorithmic" way to do this? Thank you for all the help!
  6. Feb 12, 2008 #5


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    In a more complicated situation, the situation is more complicated. Then you might have to resort to numerical methods instead of just guessing the answer. That's life. It can be complicated. This is an exercise. It's not complicated.
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