Question

A bacteria culture grows with constant relative growth rate. The bacteria count was 400 after 2 hours and 25,600 after 6 hours.

(a) What is the relative growth rate? Express your answer as a percentage.

(b) What was the initial size of the culture?

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

(d) Find the number of bacteria after 4.5 hours.

(e) Find the rate of growth after 4.5 hours.

(f) When will the population reach 50,000?

Short answer

(a) The relative growth rate is \(104\% \) per year.

(b) The initial size of the culture is 50.

(c) The expression for the number of bacteria is \(P\left( t \right) = 50{\left( 2 \right)^{1.5t}}\).

(d) The population after 4.5 hours is 5382 bacteria.

(e) The rate of growth after 4.5 hours is 5596 per hour.

(f) After 6.64 hours, the population will reach 50,000.

Step-by-step solution

Find the relative growth rate

Using theorem 2 the population function can be expressed as:

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

As the bacteria count after 2 hours is 400, therefore \(P\left( 2 \right) = 400\).

\(\begin{aligned}P\left( 2 \right) & = P\left( 0 \right){e^{k\left( 2 \right)}}\\400 & = P\left( 0 \right){e^{2k}}\end{aligned}\)

As the bacteria count after 6 hours is 25,600, therefore \(P\left( 6 \right) = 25,600\).

\(\begin{aligned}P\left( 2 \right) & = P\left( 0 \right){e^{k\left( 6 \right)}}\\25,600 & = P\left( 0 \right){e^{6k}}\end{aligned}\)

By the above two equations:

\(\begin{aligned}\frac{{{e^{6k}}}}{{{e^{2k}}}} & = \frac{{25600}}{{400}}\\{e^{4k}} & = 64\\4k & = \ln {2^6}\\k & = \frac{3}{2}\ln 2\end{aligned}\)

The relative growth rate can be expressed as:

\(\begin{aligned}\frac{{{\rm{d}}P}}{{{\rm{d}}t}} & = k\\ & = \frac{3}{2}\ln 2\\ & \approx 1.04\end{aligned}\)

So, the relative growth rate is \(104\% \).

Find the initial size of the culture

For \(k = \frac{3}{2}\ln 2\), by using the equation \(400 = P\left( 0 \right){e^{2k}}\).

\(\begin{aligned}400 & = P\left( 0 \right){e^{2\left( {\frac{3}{2}\ln 2} \right)}}\\P\left( 0 \right) & = \frac{{400}}{{{e^{3\ln 2}}}}\\ & = \frac{{400}}{{{2^3}}}\\ & = 50\end{aligned}\)

So, the initial size of the culture is 50.

Find the expression for culture after t hours

Substitute \(\frac{3}{2}\ln 2\) for \(k\) and 50 for \(P\left( 0 \right)\) in the equation \(P\left( t \right) = P\left( 0 \right){e^{kt}}\).

\(\begin{aligned}P\left( t \right) & = 50{e^{\left( {\frac{3}{2}\ln 2} \right)t}}\\ & = 50\left( {{{\left( {{e^{\ln 2}}} \right)}^{\frac{3}{2}t}}} \right)\\ & = 50\left( {{2^{1.5t}}} \right)\end{aligned}\)

So, the expression for a population of culture is \(P\left( t \right) = 50\left( {{2^{1.5t}}} \right)\).

Find the number of bacteria after 4.5 hours

Substitute 4.5 for \(t\) in the equation \(P\left( t \right) = 50\left( {{2^{1.5t}}} \right)\).

\(\begin{aligned}P\left( {4.5} \right) & = 50\left( {{2^{1.5\left( {4.5} \right)}}} \right)\\ & = 50{\left( 2 \right)^{6.75}}\\ & \approx 5382\end{aligned}\)

So, after 4.5 hours, there are 5382 bacteria.

Find the rate of growth after 4.5 hours

The equation gives the rate of growth after 4.5 hours:

\(\frac{{{\rm{d}}P}}{{{\rm{d}}t}} = kP\left( {4.5} \right)\)

Substitute 5382 for \(P\left( {4.5} \right)\) and \(\frac{3}{2}\ln 2\) for \(k\):

\(\begin{aligned}\frac{{{\rm{d}}P}}{{{\rm{d}}t}} & = \left( {\frac{3}{2}\ln 2} \right)\left( {5382} \right)\\ & \approx 5596\end{aligned}\)

So, the rate of growth after 4.5 hours is 5596 per hour.

Find the time when the population will reach 50,000

Substitute 50,000 for \(P\left( t \right)\) in the equation \(P\left( t \right) = 50\left( {{2^{1.5t}}} \right)\).

\(\begin{aligned}50,000 & = 50\left( {{2^{1.5t}}} \right)\\1000 & = {2^{1.5t}}\\1.5t & = \frac{{\ln 1000}}{{\ln 2}}\\t & \approx 6.64\end{aligned}\)

Thus, after 6.64 hours, the population will reach 50,000.