Question

A culture of the bacterium Salmonella enteritidis initially contains 50 cells. When introduced into a nutrient broth, the culture grows at a rate proportional to its size. After 1.5 hours the population has increased to 975.

(a) Find an expression for the number of bacteria after t hours.

(b) Find the number of bacteria after 3 hours.

(c) Find the rate of growth after 3 hours.

(d) After how many hours will be the population reach 250,000?

Short answer

(a) The expression for a number of bacteria is \(P\left( t \right) = 50{e^{1.9803t}}\).

(b) The number of bacteria after 3 hours is 19,014 cells.

(c) The rate of change of cells after 3 hours is 37,653 cells per hour.

(d) It will take \(4.3\;{\rm{hours}}\) to reach a count of 250,000 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 1.5 hours, the population is 975. Therefore \(t = 1.5\), and \(P = 975\).

\(\begin{aligned}P\left( {1.5} \right) & = 50{e^{k\left( {1.5} \right)}}\\975 & = 50{e^{k\left( {1.5} \right)}}\\{e^{k\left( {1.5} \right)}} & = \frac{{975}}{{50}}\\k\left( {1.5} \right) & = \ln \left( {19.5} \right)\\k & \approx 1.9803\end{aligned}\)

So, the population function can be expressed as:

\(P\left( t \right) = 50{e^{1.9803t}}\)

The expression for the number of bacteria after t hours is \(P\left( t \right) = 50{e^{1.9803t}}\).

Find the number of bacteria after 3 hours

Substitute 3 for t in the equation \(P\left( t \right) = 50{e^{1.9803t}}\).

\(\begin{aligned}P\left( 3 \right) & = 50{e^{1.9803\left( 3 \right)}}\\ & = 50{e^{5.9409}}\\ & \approx 19,014\end{aligned}\)

So, the number of bacteria after 3 hours is 19,014 cells.

Find the rate of growth after 3 hours

As \(\frac{{{\rm{d}}P}}{{{\rm{d}}t}} = \left( {1.9803} \right)P\left( t \right)\), therefore the rate of change of population after 3 hours is:

\(\begin{aligned}\frac{{{\rm{d}}P}}{{{\rm{d}}t}} & = \left( {1.9803} \right)P\left( 3 \right)\\ & = \left( {1.9803} \right)\left( {19,014} \right)\\ & \approx 37,653\end{aligned}\)

So, the rate of change of cells after 3 hours is 37,653 cells per hour.

Find the time required to reach the count of 250,000 cells

Substitute 250,000 for \(P\left( t \right)\) in the function \(P\left( t \right) = 50{e^{1.9803t}}\).

\(\begin{aligned}250,000 & = 50{e^{1.9803t}}\\1.9803t & = \ln 5000\\t & = \frac{{\ln 5000}}{{1.9803}}\\ & \approx 4.30\end{aligned}\)

So, it will take \(4.3\;{\rm{hours}}\) to reach a count of 250,000 cells.