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Vector Fields

  1. Feb 11, 2009 #1
    1. The problem statement, all variables and given/known data

    This isn't really even a homework question..i've just been reviewing some general math concepts and this one has been driving me nuts..i haven't done math in a while.

    At time t = 1, a particle is located at position (1,3). If it moves in a velocity field F(x,y) = <xy-2,y^2 - 10>, find its approximate location at time t = 1.05

    2. Relevant equations


    3. The attempt at a solution

    I thought, maybe:
    dx/dt = xy - 2 and dy/dt = y^2 - 10
    So I can integrate, so x(t) = (1/2)x^2 * y - 2x + C and y(t) = (1/3)y^3 - 10Y + C and solving for the constants is trivial based on the initial conditions, but then I get totally confused. How do I put this in terms of T? And did I even do this right? It doesn't make sense to me that the position should be the function of a position. I must have done my integral wrong.

    Can you guys help me out...really stuck.
  2. jcsd
  3. Feb 12, 2009 #2
  4. Feb 12, 2009 #3


    Staff: Mentor

    Here's how I would approach it, and if anyone disagrees, I hope they will jump in and correct my reasoning.

    You have the velocity field F(x, y) = (xy - 2, y^2 - 10). I agree that the components are dx/dt and dy/dt.

    For small values of [itex]\Delta t[/itex], F(x, y) [itex]\Delta t[/itex] should give the approximate changes in x and y.

    IOW, (dx/dt, dy/dt) [itex]\Delta t \approx (\Delta x, \Delta y)[/itex]
    or (x, y) + (dx/dt, dy/dt)*dt [itex] \approx (x, y) + (\Delta x, \Delta y)[/itex], where x, y, dx/dt, and dy/dt are evaluated at t = 1.

    Because the time increment is relatively small, the changes in x and y probably will be relatively small as well, so you should end up at a point not far from (1, 3).
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