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 6000. The number of fish doubled in the first year. 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. How long will it take for the population to increase to 3000 (half of the carrying capacity)? It will take s:s years.

Explanation / Answer

Solution:

  dP/dt=kP(1-P/K)

Given

K = 6000

dP/dt=kP(1-P/6000)

dP/dt=P(6000 - P) k/6000

1/ [ P(6000 - P) ] dP

= k/6000 dt

convert 1/ [ P(6000 - P) ] into partial fractions

(1/ 6000P) + [ (1/6000) 1/ (6000 - P) ] dP = k/6000 dt

1/P+ 1/ (6000 - P) ] dP = k dt

integrate to get

ln(P) - ln(6000 - P) = kt + C

ln[ P/ (6000 - P)] = kt + C

also given

t = 0, 400

ln[ P/ (6000 - P)] = kt + C

ln[ 400/ (6000 - 400)] = C

C = -ln(14)

ln[ P/ (6000 - P)] = kt - ln(14)

ln[ 14P / (6000 - P)] = kt

t = 1, P = 2(400) = 800

ln[ 14(800) / (6000 - 800)] = k

k = ln(112/52)

k = ln(28/13)

k 0.767

ln[ 14P / (6000 - P)] = kt

14P / (6000 - P) = e(kt)

14P = 6000e(kt) - Pe(kt)

14P + Pe(kt) = 6000e(kt)

P = 6000e(kt) / [ 14 + e(kt) ]

P = 6000e(0.767t) / [ 14 + e(0.767t) ]

OR
P = 6000(28/13)t / [ 14 + (28/13)t ]

b)
P = 6000(28/13)t / [ 14 + (28/13)t ]

6000(28/13)t / [ 14 + (28/13)t ] = 3000

6000(28/13)t = 42000 + 3000(28/13)t

3000(28/13)t = 42000

(28/13)t = 14

t ln(28/13) = ln(14)

t = 3.44 yrs

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