(1 pt) Biologists stocked a lake with 500 fish and estimated the carrying capaci

Question

(1 pt) Biologists stocked a lake with 500 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 8800. The number of fish doubled in the first year. (a) Assuming that the size of the fish population satisfies the logistic equation dP=kP(L-P), dt determine the constant kL -a, and then solve the equation to find an expression for the size of the population after t years. dP Hint: P(t) = 1+LCe-kr or P)- 1 +He-at for the Int. ( kl=a= P(t) = (b) How long will it take for the population to increase to 4400 (half of the carrying capacity)? It will take years.

Explanation / Answer

part (a) soution-

given:
L = 8800
dP/dt=kP(L-P)

dP/dt=P(8800 - P) k
1/ [ P(8800 - P) ] dP = k dt

convert 1/ [ P(8800 - P) ] to partial fractions
(1/ 8800P) + [ (1/8800) 1/ (8800 - P) ] dP = k dt
1/P+ 1/ (8800 - P) ] dP = k.8800 dt

integrate to get
ln(P) - ln(8800 - P) = kt.8800 + C
ln[ P/ (8800 - P)] = kt.8800 + C

also given:
t = 0, 500
ln[ P/ (8800 - P)] = kt.8800 + C
ln[ 500/ (8800 - 500)] = C
C = -ln(16.6)

ln[ P/ (8800 - P)] = kt.8800 - ln(16.6)
ln[ 16.6P / (8800 - P)] = kt.8800
t = 1, P = 2(500) = 1000
ln[ 16.6(1000) / (8800 - 1000)] = k
k = ln(166/78)
k = ln(2.128)
k 0.33
`````````````
ln[ 16.6P / (8800 - P)] = kt .8800
16.6P / (8800 - P) = e^(kt.8800)
16.6P = 8800e^(kt.8800) - Pe^(kt.8800)
16.6P + Pe^(kt.8800) = 8800e^(kt.8800)
P = 8800e^(kt.8800) / [ 16.6 + e^(kt.8800) ]
P = 8800e^(0.33t.8800) / [ 16.6 + e^(0.0.33t.8800) ]
----------------- ----------------- ---------------

There seems to be a mistake in the differential equation given in question. The answer is going to infinity, the method will be same for the next part.

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