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Limits and Growth Rate in Logistic Population Model

Created by @nathanedwards
 at November 1st 2023, 5:39:21 pm.
AP_Calculus_AB-Questions
#ap-calculus-ab
#logistic-growth
#population-dynamics

AP Calculus AB Exam Question:

The population of a species is modeled by the logistic growth equation:

P(t)=6001+4e−0.2tP(t) = \frac{600}{1 + 4e^{-0.2t}}

where P(t)P(t) represents the population at time tt in years.

a) Determine the limit of the population as time approaches infinity. Interpret the meaning of this limit in the context of the problem.

b) Find the population when t=5t = 5 years.

c) Determine the rate of population growth when t=5t = 5 years.

Answer:

a) To find the limit of the population as time approaches infinity, we evaluate:

lim⁡t→∞P(t)\lim_{{t \to \infty}} P(t)

Using the given logistic growth equation:

lim⁡t→∞6001+4e−0.2t\lim_{{t \to \infty}} \frac{600}{1 + 4e^{-0.2t}}

As tt approaches infinity, the term e−0.2te^{-0.2t} tends towards zero. Therefore, the denominator of the fraction approaches 1.

Hence, the limit simplifies to:

lim⁡t→∞P(t)=6001+4⋅0=6001=600\lim_{{t \to \infty}} P(t) = \frac{600}{1 + 4 \cdot 0} = \frac{600}{1} = 600

Interpretation: The limit of the population as time approaches infinity is 600. This represents the maximum sustainable population size for this species in the given environment.

b) To find the population when t=5t = 5 years, we substitute t=5t = 5 into the logistic growth equation:

P(5)=6001+4e−0.2(5)P(5) = \frac{600}{1 + 4e^{-0.2(5)}}

Simplifying further:

P(5)=6001+4e−1P(5) = \frac{600}{1 + 4e^{-1}}

Using the value of e≈2.71828e \approx 2.71828:

P(5)=6001+4(2.71828−1)P(5) = \frac{600}{1 + 4(2.71828^{-1})}
P(5)=6001+4(0.36788)P(5) = \frac{600}{1 + 4(0.36788)}
P(5)=6001+1.47152P(5) = \frac{600}{1 + 1.47152}
P(5)=6002.47152P(5) = \frac{600}{2.47152}
P(5)≈242.437P(5) \approx 242.437

Therefore, when t=5t = 5 years, the population is approximately 242.

c) The rate of population growth when t=5t = 5 years can be determined by finding the derivative of the logistic growth function and evaluating it at t=5t = 5.

dPdt=ddt(6001+4e−0.2t)\frac{dP}{dt} = \frac{d}{dt} \left(\frac{600}{1 + 4e^{-0.2t}}\right)

To simplify the process, let's use the quotient rule:

dPdt=(1+4e−0.2t)⋅ddt(600)−600⋅ddt(1+4e−0.2t)(1+4e−0.2t)2\frac{dP}{dt} = \frac{(1 + 4e^{-0.2t}) \cdot \frac{d}{dt}(600) - 600 \cdot \frac{d}{dt}(1 + 4e^{-0.2t})}{(1 + 4e^{-0.2t})^2}

Differentiating further:

dPdt=(1+4e−0.2t)⋅0−600⋅ddt(1+4e−0.2t)(1+4e−0.2t)2\frac{dP}{dt} = \frac{(1 + 4e^{-0.2t}) \cdot 0 - 600 \cdot \frac{d}{dt}(1 + 4e^{-0.2t})}{(1 + 4e^{-0.2t})^2}
dPdt=−600⋅0.2⋅4e−0.2t(1+4e−0.2t)2\frac{dP}{dt} = \frac{-600 \cdot 0.2 \cdot 4e^{-0.2t}}{(1 + 4e^{-0.2t})^2}

Simplifying further:

dPdt=−480e−0.2t(1+4e−0.2t)2\frac{dP}{dt} = \frac{-480e^{-0.2t}}{(1 + 4e^{-0.2t})^2}

Evaluating the rate of population growth when t=5t = 5:

dPdt∣t=5=−480e−0.2(5)(1+4e−0.2(5))2\left.\frac{dP}{dt}\right|_{t=5} = \frac{-480e^{-0.2(5)}}{(1 + 4e^{-0.2(5)})^2}
dPdt∣t=5=−480e−1(1+4e−1)2\left.\frac{dP}{dt}\right|_{t=5} = \frac{-480e^{-1}}{(1 + 4e^{-1})^2}

Using the value of e≈2.71828e \approx 2.71828:

dPdt∣t=5=−480(0.36788)(1+4(0.36788))2\left.\frac{dP}{dt}\right|_{t=5} = \frac{-480(0.36788)}{(1 + 4(0.36788))^2}
dPdt∣t=5≈−33.98\left.\frac{dP}{dt}\right|_{t=5} \approx -33.98

Therefore, when t=5t = 5 years, the rate of population growth is approximately -33.98 individuals per year. The negative sign indicates that the population is decreasing at that point in time.

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