Biologists stocked a lake with 240 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 6,000. The number of fish tripled in the first year.(a) Assuming that the size of the fish population satisfies the logistic equation, find an expression for the size of the population after t years.
Question
Biologists stocked a lake with 240 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 6,000. The number of fish tripled in the first year.(a) Assuming that the size of the fish population satisfies the logistic equation, find an expression for the size of the population after t years.
Solution
The logistic growth model is given by the differential equation:
dP/dt = rP(1 - P/K)
where:
- P is the size of the population at time t,
- r is the relative growth rate,
- K is the carrying capacity of the environment.
Given that the initial population P(0) = 240 and the carrying capacity K = 6000, we can substitute these values into the equation. However, we don't have a value for r, the relative growth rate.
We know that the population tripled in the first year, so P(1) = 3 * 240 = 720. We can use this information to solve for r.
First, we separate the variables and integrate:
∫dP/P(1 - P/K) = ∫rdt
This gives us:
ln|P/(K - P)| = rt + C
where C is the constant of integration.
We can solve for C using the initial condition P(0) = 240:
ln|240/(6000 - 240)| = C
Now we have a value for C, we can write the general solution:
ln|P/(K - P)| = rt + ln|240/(6000 - 240)|
Finally, we can solve for P(t) to get the size of the population after t years:
P(t) = K / (1 + (K - P(0)) / P(0) * e^(-rt))
This is the expression for the size of the population after t years, assuming that the size of the fish population satisfies the logistic equation.
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