Biologists stocked a lake with 274 fish and estimated the carrying capacity (the

Question

Biologists stocked a lake with 274 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 8700. The number of fish tripled 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. k = P(t) = How long will it take for the population to increase to 4350 (half of the carrying capacity)? It will take years.

Explanation / Answer

1.

274 fish fish with a carrying capacity of 8,700

our logistic equation is:

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

a) determine the constant k and solve for P(t)

First, we integrate:

dP = (k)(P)(1 - P/K) dt

1/(P)(1 - P/K) dP = k dt

Plug in K = 8700, beause the carrying capacity will never change:

1/(P - P^2/8700) dP = k dt

and integrate:

1/(P - P^2/8700) dP = k dt

ln(P) - ln(8700 - P) = kt

simplifying:

ln(P/(8700 - P)) = kt

take e^ of both sides:

P/(8700 - P) = e^(kt) = (e^k)(e^t) = e^k because t = 0 at the starting population

Plug in 274 = P and solve for k, our constant:

247/(8700 - 274) = e^(k)

(ln(247/8426)) = k [our first answer]

So we have:

ln(P/(8700 - P)) = (ln(247/8426))t

in other words:

P(t) = (247/8426)e^t

b) How long will it take to increase to 4350?

Plug in

4350 = (247/8426)e^t

and solve for t

t = [approx] 6.2319

or simply 6.23 years.

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