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 775. (a) Find an expression for the number of bacteria after t hours. (Round your numeric values to four decimal places.)

Answer 1

Answer:

The expression for the number of bacteria after t hours is $P(t) = 50e^{0.2922t}$

Step-by-step explanation:

When introduced into a nutrient broth, the culture grows at a rate proportional to its size.

This means that the size of the population, after t hours, is modeled by the following differential equation:

$\frac{dP}{dt} = rP$

In which R is the growth rate.

The solution of this differential equation is:

$P(t) = P(0)e^{rt}$

In which P(0) is the initial population.

A culture of the bacterium Salmonella enteritidis initially contains 50 cells.

This means that $P(0) = 50$, and so:

$P(t) = P(0)e^{rt}$

$P(t) = 50e^{rt}$

After 1.5 hours, the population has increased to 775.

This means that $P(1.5) = 775$. We use this to find r. So

$P(t) = P(0)e^{rt}$

$500e^{1.5r} = 775$

$e^{1.5r} = \frac{775}{500}$

$\ln{e^{1.5r}} = \ln{\frac{775}{500}}$

$1.5r = \ln{\frac{775}{500}}$

$r = \frac{\ln{\frac{775}{500}}}{1.5}$

$r = 0.2922$

The expression is:

$P(t) = 50e^{rt}$

$P(t) = 50e^{0.2922t}$

The expression for the number of bacteria after t hours is $P(t) = 50e^{0.2922t}$