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Problem 81 Hard Difficulty
Some populations initially grow exponentially but eventually level off. Equations of the form
$$ P(t) = \frac{M}{1 + Ae^{-kt}} $$
where $ M $, $ A $, and $ k $ are positive constants, are called logistic equations and are often used to model such populations. (We will investigate these in detail in Chapter 9.) Here $ M $ is called the carrying capacity and represents the maximum population size that can be supported, and $ A = \frac{M - P_0}{P_0} $, where $ P_0 $ is the initial population.
(a) Compute $ lim_{t\to \infty} P(t) $. Explain why your answer is to be expected.
(b) Compute $ lim_{M\to \infty} P(t) $. (Note that $ A $ is defined in terms of $ M $.) What kind of function is your result?
Answer
(a) $\lim _{t \rightarrow \infty} P(t)=\lim _{t \rightarrow \infty} \frac{M}{1+A e^{-k t}}=\frac{M}{1+A \cdot 0}=M$
It is to be expected that a population that is growing will eventually reach the maximum population size that can be
supported.
(b) $\lim _{M \rightarrow \infty} P(t)=\lim _{M \rightarrow \infty} \frac{M}{1+\frac{M-P_{0}}{P_{0}} e^{-k t}}=\lim _{M \rightarrow \infty} \frac{M}{1+\left(\frac{M}{P_{0}}-1\right) e^{-k t}} \stackrel{\underline{H}}{=} \lim _{M \rightarrow \infty} \frac{1}{\frac{1}{P_{0}} e^{-k t}}=P_{0} e^{k t}$
$P_{0} e^{k t}$ is an exponential function.
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Top Calculus 2 / BC Educators
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Missouri State University
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Watch More Solved Questions in Chapter 4
Video Transcript
Yeah. For this problem here, uh we have key approaching infinity in the logistic equation. So we have em over one plus E to the negative Katie. When t goes to infinity, this thing goes to zero, meaning the numerator goes to one. And then we're just left with Emma's results. And then for part B we have um to calculate it differently. M is going to infinity now. So we're going to treat the variable. We're gonna treat amazon a variable. However, when we send it to infinity, we get the indeterminant form. So we have to use local cows role, but we get it into this form right here. Uh, peanut class mm am minus P not I M E to the negative Katie. So when we evaluate this and we take the derivative, what we're left with is P not times E to the K T. As the final answer.
California Baptist University
Top Calculus 2 / BC Educators
Catherine R.
Missouri State University
Anna Marie V.
Campbell University
Samuel H.
University of Nottingham
Michael J.
Idaho State University
