Biologists stocked a lake with [400] fish and estimated the carrying capacity (t
ID: 2835759 • Letter: B
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 [4400] . The number of fish doubled in the first year.
(a) Assuming that the size of the fish population satisfies the logistic equation
[ rac{dP}{dt} = kP left( 1 - rac{P}{K} ight) ,]
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 [2200] (half of the carrying capacity)?
It will take years.
Explanation / Answer
similar question--
Differential Equations Question: 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 7500. 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 3750 (half of the carrying capacity)?
answer--
dP/dt=kP(1-P/K)
given:
K = 7500
dP/dt=kP(1-P/7500)
dP/dt=P(7500 - P) k/7500
1/ [ P(7500 - P) ] dP = k/7500 dt
convert 1/ [ P(7500 - P) ] to partial fractions
(1/ 7500P) + [ (1/7500) 1/ (7500 - P) ] dP = k/7500 dt
1/P+ 1/ (7500 - P) ] dP = k dt
integrate to get
ln(P) - ln(7500 - P) = kt + C
ln[ P/ (7500 - P)] = kt + C
also given:
t = 0, 500
ln[ P/ (7500 - P)] = kt + C
ln[ 500/ (7500 - 500)] = C
C = -ln(14)
ln[ P/ (7500 - P)] = kt - ln(14)
ln[ 14P / (7500 - P)] = kt
t = 1, P = 2(500) = 1000
ln[ 14(1000) / (7500 - 1000)] = k
k = ln(140/65)
k = ln(28/13)
k ? 0.767
`````````````
ln[ 14P / (7500 - P)] = kt
14P / (7500 - P) = e^(kt)
14P = 7500e^(kt) - Pe^(kt)
14P + Pe^(kt) = 7500e^(kt)
P = 7500e^(kt) / [ 14 + e^(kt) ]
P = 7500e^(0.767t) / [ 14 + e^(0.767t) ]
----------------- ----------------- ---------------
OR
P = 7500(28/13)^t / [ 14 + (28/13)^t ]
----------------- ----------------- ---------------
b)
P = 7500(28/13)^t / [ 14 + (28/13)^t ]
7500(28/13)^t / [ 14 + (28/13)^t ] = 3750
7500(28/13)^t = 52500 + 3750(28/13)^t
3750(28/13)^t = 52500
(28/13)^t = 14
t ln(28/13) = ln(14)
t = 3.44 yrs
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