Question

A bacterial population is known to have a logistic growth pattern with initial population 1000 and an equilibrium population of 10,000. A count shows that at the end of 1 hr there are 2000 bacteria present. Determine the population as a function of time. Determine the time at which the population is increasing most rapidly and draw a sketch of the logistic curve.

Answer 1

$\displaystyle N(t) = \frac{L}{1 + \left(\frac{L}{N_0} - 1\right)e^{-kt}}\\ N(1) = 2000 = \frac{10000}{1 + \left(\frac{10000}{1000} - 1\right)e^{-k}} \\ 2000 = \frac{10000}{1 + 9e^{-k}}\\ 1 + 9e^{-k} = 5 \\ 9e^{-k} = 4\\ e^{-k} = \frac{4}{9} = 0.\dot{4} \\ k = \frac{1}{0.\dot{4}} = 0.81093 \\ N(t) = \frac{10000}{1 + 9e^{-0.81093t}}\\ N'(t) = \frac{7298.37}{\left(1 + 9e^{-0.81093t}\right)^2}\\ N''(t) = 10000\left(\frac{106.532e^{-1.62186t}}{\left(1 + 9e^{-0.81093t}\right)^3}- \frac{5.91847e^{-0.81093t}}{\left(1 + 9e^{-0.81093t}\right)^2}\right) \\ \textsf{The time at which the population is increasing}\\ \textsf{most rapidly is the value of}\,\, t \textsf{when}\,\, N''(t) = 0, \\ 10000\left(\frac{106.532e^{-1.62186t}}{\left(1 + 9e^{-0.81093t}\right)^3}- \frac{5.91847e^{-0.81093t}}{\left(1 + 9e^{-0.81093t}\right)^2}\right) = 0\\ 1065320e^{-1.62186t} - 591847e^{-0.81093t}\left(1 + 9e^{-0.81093t}\right) = 0 \\ \textsf{Let}\,\,\, p = e^{-0.81093t} \\ 1065320p^2 - 59184.7p\left(1 + 9p\right) = 0 \\ 1065320p^2 - 59184.7p - 532662.3p^2 = 0 \\ 532657.7p^2 - 59184.7p = 0 \\ p = 0, 0.11111 \\ e^{-0.81093t} = 0.11111,\,\, t = 2.70952 \approx 2.7\,\, \textsf{hours}\\ \therefore \textsf{The time at which the population is increasing}\\ \textsf{most rapidly is}\,\,\, 2.7\,\, \textsf{hours}.$