# predator prey model

• Apr 13th 2009, 01:47 PM
ardam
predator prey model
How can you find the critical points of the a predator prey model and solve the linear comparisonsystem corresponding to each critical point?

Then sketch thetrajectories in the vicinity of each?
• Apr 13th 2009, 01:50 PM
icemanfan
Presumably, the predator-prey model has an equation that models a relationship. Without that, you can't do anything.
• Apr 13th 2009, 01:52 PM
ardam
yer can use
x' = 5x − xy, y' = −2y + 3xy.
• Apr 13th 2009, 01:58 PM
icemanfan
Given a function $f(x,y)$, the critical points occur at $\frac{df}{dx} = 0, \frac{df}{dy} = 0$. Using your notation, this is where x' = 0 and y' = 0. So you have to solve that system of equations to find the critical points.
• Apr 13th 2009, 02:05 PM
ardam
so you sub these in and them find out x and y to be
x=2/3 and y=5?
• Apr 13th 2009, 02:26 PM
icemanfan
Quote:

Originally Posted by ardam
so you sub these in and them find out x and y to be
x=2/3 and y=5?

That is correct. The one critical point is (2/3, 5).
• Apr 13th 2009, 02:40 PM
ardam
Quote:

Originally Posted by icemanfan
That is correct. The one critical point is (2/3, 5).

so there is only one?

so how can i Solve the linear comparison?
• Apr 13th 2009, 02:49 PM
icemanfan
Quote:

Originally Posted by ardam
so there is only one?

so how can i Solve the linear comparison?

Yes, there is only one critical point, which is the one you found. I'm afraid I don't know what "solving the linear comparison" refers to. The tangent plane to the surface $z(x, y)$ at that point is going to be $z = c$ for some constant c, because the surface is flat at a critical point. I don't know what else to say about it.
• Apr 13th 2009, 02:53 PM
ardam
Quote:

Originally Posted by icemanfan
Yes, there is only one critical point, which is the one you found. I'm afraid I don't know what "solving the linear comparison" refers to. The tangent plane to the surface $z(x, y)$ at that point is going to be $z = c$ for some constant c, because the surface is flat at a critical point. I don't know what else to say about it.

it means to determine a relationship
between x and y (or between u and v for translated critical points).
Determine the type and stability of each critical point, and sketch the
trajectories in the vicinity of each.

Which im not too sure about
Its probably obvious