Expert: Paul Klarreich - 12/7/2011

Question
My calculus professor is kind of in his old world and just creates problems out of the blue. If you could help me at all with this problem it would be greatly appreciated. I have attempted to start on it, but I got stuck on part a and have no clue how to even go about differentiating this problem.

P(t+1)=(rP(t))/(1+P(t)/M) this equation represents the model.

a) Suppose P(0)=2500,r=2.1,M=3000. Calculate P(1) and P(2)

b) Find the positive equilibrium population size E for the law above(with arbitrary values for r and M).

c) What is the equilibrium population size E for the case r=2.1 and M=3000?

d) Is the equilibrium found in part b stable or not?

e) What is the breeding population for the maximal sustainable harvest?

f) What is the maximal sustainable harvest?

Answer
Questioner: Austin
Country: Iowa, United States
Category: Calculus
Private: No
Subject: Calculus (Difference Equations)
Question: My calculus professor is kind of in his old world

>> you mean he's from Poland?

and just creates problems out of the blue. If you could help me at all with this problem it would be greatly appreciated. I have attempted to start on it, but I got stuck on part a
and have no clue how to even go about differentiating this problem.

P(t+1)=(rP(t))/(1+P(t)/M) this equation represents the model.

a) Suppose P(0)=2500,r=2.1,M=3000. Calculate P(1) and P(2)

b) Find the positive equilibrium population size E for the law above(with arbitrary values for r and M).

c) What is the equilibrium population size E for the case r=2.1 and M=3000?

d) Is the equilibrium found in part b stable or not?

e) What is the breeding population for the maximal sustainable harvest?

f) What is the maximal sustainable harvest?
.................................
If
         r P(t)
P(t+1)=  ---------------
         1 + P(t) / M

and you have r = 2.1, M = 1000, then

         2.1 P(t)
P(t+1)=  ---------------
         1 + P(t) / 3000
..........................................
         2.1 P(0)
P(1)=  ------------------
         1 + P(0) / 3000

         2.1(2500)
P(1)=  ---------------
         1 + 2500 / 3000

         2.1(2500)
P(1)=  ---------------
         1 + 5 / 6

         2.1(2500)
P(1)=  --------------
         11 / 6

Now I think you can use Excel for the calculations.  When you do, you will see these P(t) values:
t        P(t)
-----------------
0   2500
1   2863.636364
2   3076.744186
3   3189.78186
4   3246.580602
5   3274.344653
6   3287.733214
7   3294.147278
8   3297.210398
9   3298.67103
10   3299.367023
11   3299.698552
12   3299.856447
13   3299.93164

Aha! They seem to converge to 3300, don't they?  That suggests making this assumption:

For large enough values of t, P(t+1) = P(t).  And we think it will come out to 3300.

Write:

         2.1 P(t)
P(t)=  -----------------   << P(t) equals P(t+1)  
       1 + P(t) / 3000

Do some algebra:

         2.1 P(t)        << Times 3000
P(t)=  -----------------
       1 + P(t) / 3000    << Times 3000

        6300 P(t)     
P(t)=  -----------------
       3000 + P(t)

Cross-multiply:

P(3000 + P) = 6300 P

P^2 + 3000P - 6300P = 0

P^2  - 3300P = 0

and P = 3300 is an obvious solution to that.

As to your other questions, like 'maximal sustainable harvest', I am not from Iowa, so I have no idea what that is.  That's up to you.