Biologists stocked a lake with 300 fish and estimated the carrying capacity (the

Question

Biologists stocked a lake with 300 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 5000. The number of fish tripled in the first year.

(a) Assuming that the size of the fish population satisfies the logistic equation

dP/dt=kP(1P/K),

determine the constant k, and then solve the equation to find an expression for the size of the population after t years.

k=

p(t)=

(b) How long will it take for the population to increase to 2500 (half of the carrying capacity)?

It will take   years

Another model for a growth function for a limited population is given by the Gompertz function, which is a solution of the differential equation

dP/dt=cln(K/P)P

where c is a constant and K is the carrying capacity.

(a) Solve this differential equation for c=0.25, K=4000, and initial population P0=500.
P(t)=

b) Compute the limiting value of the size of the population.
limtP(t)=  .

(c) At what value of P does P grow fastest?
P=

Explanation / Answer

.By using separation variables

dP/(kP(1-P/k)) = dt

1
-------------- dP
kP(1-P/k)

= (1 - P/k) + P/k
= --------------------- dP =dt
= kP(1-P/k)

= 1/(kP) dP + 1/(k*(k-P)) dP =dt

integrarte both sides we get

ln(P)/k - ln(k-P)/k = t+c

Using the limit values P(0)=300 and P(t) you obtain the solution
ln(P(t)/(k-P(t))) - ln(300/(k-300)) = kt

ln(P(t)*(k-300)/((k-P(t))*300)) = kt

((k-P(t))*300) = e^(kt

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