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 8000 . The number of fish doubled in the first year. (a) Assuming that the size of the fish population satisfies the logistic equation, determine the constant k , and then solve the equation to find an expression for the size of the population after t years. (b) How long will it take for the population to increase to 4000 (half of the carrying capacity)? Thanks!

Explanation / Answer

The form of the equation is F(t) = Ne^k(t-t0)/(1+e^k(t-t0)) Since the carrying capacity is 8000, N = 8000 You solve for the difference in log-odds at the 2 time-points/time elapsed, and this is the rate k. This is an important property of the logistic equation. There is doubling in the first year. Thus, time elapsed = 1. As N = 8000, the log-odds is ln(300/(8000-300)) at t = 0 and ln(600/(8000-600)) at t = 1 Solving the fractions, we have ln(3/77) at t=0 and ln(3/37) at t = 1 k=(ln(3/37)-ln(3/77))/1 = ln(77/37) is approximately 0.732887509209459 Then, as F(0) = 300, we have 8000e^k(0-t0)/(1+e^k(0-t0))= 300, or e^-kt0/(1+e^-kt0) = 3/80 Letting e^-kt0 = y, we have y/(1+y) = 3/80 Cross-multiplying, 80y = 3 + 3y, or 77y = 3, or y = 3/77 Thus, e^- ln(77/37)t0 = 3/77 Taking the logs of both sides, -ln(77/37)t0 = ln(3/77) Multiplying by-1, we have ln(77/37)t0 = ln(77/3), or t0 = ln(77/3)/ln(77/37) This equals approximately 4.427955303 Thus, our equation is F(t) = 8000 e^ 0.732887509209459(t-4.427955303)/(1+e^ 0.732887509209459(t-4.427955303)) As a check, F(0) = 8000 e^ 0.732887509209459(-4.427955303)/(1+e^ 0.732887509209459(-4.427955303)) = 300, and F(1) = 8000 e^ 0.732887509209459(-3.427955303)/(1+e^ 0.732887509209459(-3.427955303)) = 600 b) We reach half of the carrying capacity when t = t0. This is at 4.427955303 years.

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