Finding Approximate Location of Particle at Time t=1.05

[Kokordilos]

Homework Statement

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

Homework Equations

n/a

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.

 

[Kokordilos]

anyone?

 

[Mark44]

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 $\Delta t$, F(x, y) $\Delta t$ should give the approximate changes in x and y.

IOW, (dx/dt, dy/dt) $\Delta t \approx (\Delta x, \Delta y)$
or (x, y) + (dx/dt, dy/dt)*dt $\approx (x, y) + (\Delta x, \Delta y)$, 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).