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quackensack

Member

Registered: 2007-02-27

Posts: 47

Calc help

I'm given:

F' = 0.01F(10-F-3R)
R' = 0.0125R(8-R-2F)
F' = 0.1F(1 - F/10) - 0.03FR
R' = .1R (1 - R/8) - 0.025FR

This represents a mathematical model of competition, in which two species, wolves and tigers, compete for the same resources.  F represents the wolf population, in hundreds of wolves, and R represents the tiger population in hundreds of tigers.

I have to find the four equilibrium points for the system. I'm a little confused about how to do this so if anyone could explain, I would REALLY appreciate it.

Also, I was wondering, what is the significance of the term -0.03FR in the F' equatiton and of the term -0.025FR in the R' equation?

Thanks in advance

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Calc help

The equilibrium points will be values of R and F such that R' and F' are both 0.

First let's see what needs to happen for F' to equal 0.
F' = 0.01F(10-F-3R), and if F' is 0, that means that either 0.01F or (10-F-3R) will be 0.
0.01F = 0 if F = 0, and 10-F-3R = 0 if F+3R = 10. Satisfying either of those two conditions will mean that F' = 0.

We can do the same thing for R'.
R' = 0.0125R(8-R-2F), and again, if R' is 0, that means that either 0.0125R or (8-R-2F) will be 0.
0.0125R = 0 if R = 0, and 8-R-2F = 0 if R+2F = 8. So again, satisfying either of these two conditions will mean that F' = 0.

Because F' and R' both have two ways of being equal to 0, that means that there are 4 ways of making them both 0. Using each of these 4 ways will find the 4 equilibrium points.

(i) F=0, R=0. This is in equilibrium because everything is dead. Nothing more can die, but there are no living animals to reproduce either.

(ii) F=0, R+2F = 8. Solving the equation shows that R=8. Here there are no wolves, only tigers.

(iii) F+3R = 10, R=0. Solving this equation shows that F=10. This time there are only wolves.

(iv) F+3R = 10, R+2F = 8. This is a set of simultaneous equations, which is a bit trickier to solve but not too much.
R = 2.4, F = 2.8. This is the only equilibrium system in which both animals exist.

In real life, it's likely that none of these would actually happen. Usually the amounts of each animal oscillate in a 'dynamic equilibrium'.

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Why did the vector cross the road?
It wanted to be normal.

quackensack

Member

Registered: 2007-02-27

Posts: 47

Re: Calc help

Thank you so much.  I was wondering, what is the significance of the term -0.03FR in the F' equatiton and of the term -0.025FR in the R' equation?  Meaning, what do they both represent?

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Calc help

I think it's better to think about it in terms of the top two equations.

So, F' = 0.01F(10-F-3R).

The 10-F-3R represents how much spare resources there are. The wolves use a bit, the tigers use a lot, and that much is taken away from 10 to give the amount that's left.
The 0.01F is just a proportion of how many wolves there are currently, which makes sense because the more wolves there are, the more they can reproduce.
Combining the amount of wolves with the amount of rescources gives you the change in their population.

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Why did the vector cross the road?
It wanted to be normal.