Biologists stocked a lake with 400 fish and estimated the carrying capacity (the

Question

Biologists stocked a lake with 400 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 3400. The number of fish doubled in the first year

a  Assuming that the size of the fish population satisfies the logistic equation dP/dt=kP(1-P/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 1700 (half of the carrying capacity)?

Explanation / Answer

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

The carrying capacity of the lake is given to be K = 3400

Hence, dP/dt = kP(1-P/3400)

Rearranging this we get,

dP/dt = (kP/3400)(3400-P)

Seperating with respect to P:

3400dP/P(3400-P) = kdt

Using Partial fractions to Left hand side:

dP/P + dP/(3400-P) = kdt

Integrating both sides:

ln|P| - ln|3400-P| = kt + C

or ln|P|/|3400-P| = kt + C

=> |P|/|3400-P| = C1ekt....Using another constant C1 = eC

Now 400 fish in the beginning => P(0) = 400

Hence 400/(3400-400) = C1e0 = C1

=> 400/3000 = C1

=> 4/30 = C1

=> 2/15 = C1

Now fish doubled in first year => P(1) = 800

Hence 800/(3400-800) = (2/15)ek

=> 800/2600 = (2/15) ek

=> (4/13)(15/2) = ek

=> 30/13 = ek

=> ln(30/13) = k => k = 0.8362

P(t) = (3400-P(t))(2/15)e0.8362t

=> P(t) = (3400-P(t))(0.1333)e0.8362t

=> P(t) = 453.333e0.8362t - 0.1333e0.8362tP(t)

=> P(t) (1+0.1333e0.8362t) = 453.333e0.8362t

=> P(t) =  (453.333e0.8362t)/(1+0.1333e0.8362t)

b) Set P=1700;

1/2= C1ekt

=> 1/2 = (2/15)e0.8362t

=> 15/4 = e0.8362t

=> 3.75 = e0.8362t

=> ln(3.75) = 0.8362t

=> (1/0.8362)(1.32176) = t

=> 1.6 = t

=> t = 1.6 years

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