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 3400. The number of fish doubled in the first year. (a) Assuming that the size of the fish population satisfies the logistic equation dPdt=kP(1PK), 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)? It will take __________ years.

Explanation / Answer

given:
Carrying capacity ,K = 3400
dP/dt=kP(1-P/3400)

dP/dt=P(3400 - P) k/3400
1/ [ P(3400 - P) ] dP = (k/3400) dt

convert 1/ [ P(3400 - P) ] to partial fractions
(1/ 3400P) + [ (1/3400) 1/ (3400 - P) ] dP = k/3400 dt
(1/P+ 1/ (3400 - P) ) dP = k dt

integrate to get
ln(P) - ln(3400 - P) = kt + C
ln[ P/ (3400 - P)] = kt + C

also given:
t = 0

no. of fish = 300
ln[ P/ (3400 - P)] = kt + C
ln[ 300/ (3400 - 300)] = C
C = -ln(10.3)

ln[ P/ (7500 - P)] = kt
ln[ 10.3P / (3400 - P)] = kt

a)at t = 1 years, P(the size of fish doubled) = 2(300) = 600
ln[ 10.3(600) / (3400 - 600)] = k
k = ln(2.2)
k 0.7916

Now,

ln[ 10.3P / (3400 - P)] = kt
10.3P / (3400 - P) = e^(kt)
10.3P = 3400e^(kt) - Pe^(kt)
10.3P + Pe^(kt) = 3400e^(kt)
P = 3400e^(kt) / [ 10.3 + e^(kt) ]
P = 3400e^(0.7916t) / [ 10.3 + e^(0.791t) ] (ANSWER)

B)

P = 1700

P = 3400e^(0.7916t) / [ 10.3 + e^(0.791t) ]

1700 = 3400e^(0.7916t) / [ 10.3 + e^(0.791t) ]

1/2 =e^(0.7916t) / [ 10.3 + e^(0.791t) ]

[ 10.3 + e^(0.791t) ]/2 = e^(0.7916t)

10.3 = e^(0.791t) /2

20.6 = e^(0.791t)

Take natural log to both side

T = 3.82 years (ANSWER)

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