Modeling Population Growth with a Logistic Function
A small town's population is modeled by the logistic function ( P(t) = \frac{12000}{1 + 5e^{-0.08t}} ), where ( t ) is the number of years since 2000. In 2000, the population was 2,000. By 2010, it had grown to approximately 4,800. A town planner wants to predict when the population will reach 10,000 and what the maximum sustainable population is.
1. Verify that the model matches the given population values at t=0 and t=10. What is the carrying capacity?
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At ( t=0 ): ( P(0) = \frac{12000}{1+5e^{0}} = \frac{12000}{6} = 2000 ), which matches. At ( t=10 ): ( P(10) = \frac{12000}{1+5e^{-0.8}} ). Compute ( e^{-0.8} \approx 0.4493 ), so ( 5e^{-0.8} \approx 2.2465 ), denominator ( \approx 3.2465 ), and ( P(10) \approx \frac{12000}{3.2465} \approx 3696 ). This does not match the given 4,800. So the model does not fit the 2010 data well. The carrying capacity is the limit as ( t \to \infty ), which is ( \frac{12000}{1+0} = 12000 ).
2. Using the model, solve for when the population will reach 10,000. Is this realistic given the mismatch at t=10?
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Set ( P(t) = 10000 ): ( 10000 = \frac{12000}{1+5e^{-0.08t}} ). Then ( 1+5e^{-0.08t} = 1.2 ), so ( 5e^{-0.08t} = 0.2 ), ( e^{-0.08t} = 0.04 ). Take natural log: ( -0.08t = \ln(0.04) \approx -3.2189 ), so ( t \approx \frac{3.2189}{0.08} \approx 40.24 ) years after 2000, i.e., around 2040. Since the model underestimates the 2010 population, the actual time to reach 10,000 may be sooner. The model's prediction is not reliable because it does not fit the observed data at t=10.
3. Suggest a modification to the model to better fit the data, and explain how you would determine the parameters.
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The logistic model has three parameters: carrying capacity ( L ), growth rate ( k ), and the parameter ( t_0 ) (or the constant in the denominator). The given model has ( L=12000 ), ( k=0.08 ), and the constant 5 corresponds to ( e^{k t_0} ) with ( t_0 = \frac{\ln 5}{0.08} \approx 20.12 ). To better fit the data, we could adjust ( L ) and ( k ) to match both the 2000 and 2010 populations. Using the two data points ( (0,2000) ) and ( (10,4800) ), and assuming the same form ( P(t) = \frac{L}{1+A e^{-kt}} ), we have ( 2000 = \frac{L}{1+A} ) and ( 4800 = \frac{L}{1+A e^{-10k}} ). This gives two equations with three unknowns, so we need another data point or assumption. For example, if we assume the carrying capacity is 15000, then solve for ( A ) and ( k ). Alternatively, use regression on more data points. The key is that the model should be calibrated to the available data.