Question

A common inhabitant of human intestines is the bacterium Escherichia coli, named after the German pediatrician Theodor Escherich, who identified it in 1885. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 50 cells.

(a) Find the relative growth rate.

(b) Find an expression for the number of cells after t hours.

(c) Find the number of cells after 6 hours.

(d) Find the rate of growth after 6 hours.

(e) When will the population reach a million cells?

Short answer

(a) The relative growth of population is \(\ln 8\).

(b) The expression for cells after t hours is given as \(P\left( t \right) = 50\left( {{8^t}} \right)\).

(c) There are \(13,207,200\) cells after 6 hours.

(d) The rate of increase of cells at 6 hours is 27,255,656 cells per hour.

(e) It will take 4.76 hours to reach a million cells.

Step-by-step solution

Write the population function using Theorem 2

Using theorem 2 the population function can be expressed as:

\(P\left( t \right) = P\left( 0 \right){e^{kt}}\)

As the initial population is 50, therefore \(P\left( 0 \right) = 50\):

\(P\left( t \right) = 50{e^{kt}}\)

It is given that after 20 minutes, the population is 100, therefore \(t = \frac{1}{3}({\rm{hours}})\) \(P = 100\).

\(\begin{aligned}P\left( {\frac{1}{3}} \right) & = 50{e^{k\left( {\frac{1}{3}} \right)}}\\100 & = 50{e^{k\left( {\frac{1}{3}} \right)}}\\{e^{\frac{k}{3}}} & = 2\\k & = 3\ln 2\\ & = \ln 8\end{aligned}\)

So, the population function can be expressed as:

\(\begin{aligned}P\left( t \right) & = 50{e^{\ln 8t}}\\ & = 50{\left( 8 \right)^t}\end{aligned}\)

The expression for a number of cells after t hours is \(P\left( t \right) = 50\left( {{8^t}} \right)\).

Find the relative growth rate of the population

By theorem 2, the rate of change of population is:

\(\begin{aligned}\frac{{{\rm{d}}P}}{{{\rm{d}}t}} & = kP\\ & = \left( {\ln 8} \right)P\end{aligned}\)

The expression \(\frac{1}{P}\frac{{{\rm{d}}P}}{{{\rm{d}}t}}\) represents relative growth. It can be calculated as follows:

\(\begin{aligned}\frac{1}{P}\frac{{{\rm{d}}P}}{{{\rm{d}}t}} & = \frac{1}{P}\left( {\left( {\ln 8} \right)P} \right)\\ & = \ln 8\end{aligned}\)

So, the relative growth rate of the population is \(\ln 8\).

Find the number of cells after 6 hours

Substitute 6 for \(t\) in the function \(P\left( t \right) = 50\left( {{8^t}} \right)\), to find the number of cells after 6 hours.

\(\begin{aligned}P\left( 6 \right) & = 50\left( {{8^6}} \right)\\ & = 50\left( {{2^{18}}} \right)\\ & = 13,207,200\end{aligned}\)

So, after 6 hours, the number of cells is \(13,207,200\).

Find the rate of growth after 6 hours

As \(\frac{{{\rm{d}}P}}{{{\rm{d}}t}} = \left( {\ln 8} \right)P\left( t \right)\), therefore the rate of change of population after 6 hours is:

\(\begin{aligned}\frac{{{\rm{d}}P}}{{{\rm{d}}t}} & = \left( {\ln 8} \right)P\left( 6 \right)\\ & = \left( {\ln 8} \right)\left( {13,107,200} \right)\\ & \approx 27,255,656\end{aligned}\)

So, the rate of change of population after 6 hours is 27,255,656 cells per hour.

Find the time required by the cell to reach a count of 1 million

By the equation \(P\left( t \right) = 50\left( {{8^t}} \right)\), the time required to reach one million cells can be calculated as follows:

\(\begin{aligned}{10^6} & = 50\left( {{8^t}} \right)\\{8^t} & = 20,000\\t & = \frac{{\ln 20,000}}{{\ln 8}}\\ & \approx 4.76\;{\rm{h}}\end{aligned}\)

So, it will take 4.76 hours to reach the count of one million.