Question

The population of a culture of bacteria is modeled by the logistic equation P (t) = frac{14,250}{1 + 29e – 0.62t} To the nearest tenth, how many days will it take the culture to reach 75% of it’s carrying capacity? What is the carrying capacity? What are the virtues of logistic model?

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asked 2021-01-19
The population of a culture of bacteria is modeled by the logistic equation \(P (t) = \frac{14,250}{1 + 29e – 0.62t}\) To the nearest tenth, how many days will it take the culture to reach 75% of it’s carrying capacity? What is the carrying capacity? What are the virtues of logistic model?

Answers (1)

2021-01-20

Step1 \(P(t) = \frac{14,250}{1+29e^(-0*62t)}\)
\(As t \Rightarrow \infty, e^{-0*62t} \Rightarrow 0\)
\(\therefore P(t) \Rightarrow 14,250\).

This is the carrying capacity of the bacten`al culture.

To determine the time it take to reach 75% of thr carlying capacity, we write \(P(t) = \frac{3}{4}(14,250).\)

Step 2

\( \Rightarrow \not{14,250}{1+29e^{(-0*62t)}} = \frac{3}{4}(\not 14,250)\)
\(\Rightarrow 1+29e^{-0*62t} = \frac{3}{4}\)
\(\Rightarrow e^{(-0*62t)} = \frac{1}{87}\)
\(\Rightarrow t=\frac{1}{0*62} In(87) = 7\cdot2\) Hense it takes the bacterial Popylation \(7\cdot2\) days to reach 75% of the calling capacity.

Step 3 The growth of any population is curtailed by the amount of resources available. This is better modeled by a logistic S curve rather than an exponential growth model which only works in ideal situations and not for modeling real life situations.

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