An experiment starts and the population of a bacteria culture increases by 10% in the first hour. Assume that the population increases at a rate proportional to the amount of bacteria.
a) What is the doubling time?
b) If the initial population is 2,000, what is the population in 6 hours?

Answer :

Answer:

a) The doubling time is 7.27 hours.

b) The population is 6 hours will be 3,543.

Step-by-step explanation:

The population of the bacteria is modeled by the following equation:

[tex]P(t) = P(0)e^{rt}[/tex]

In which P(0) is the initial population, P(t) is the population after t hours and r is the growth rate.

An experiment starts and the population of a bacteria culture increases by 10% in the first hour.

This means that

[tex]P(1) = 1.1P(0)[/tex]

Which helps us find r.

So

[tex]P(t) = P(0)e^{rt}[/tex]

[tex]1.1P(0) = P(0)e^{r}[/tex]

[tex]e^{r} = 1.1[/tex]

Applying ln to both sides

[tex]\ln{e^{r}} = \ln{1.1}[/tex]

[tex]r = 0.0953[/tex]

So

[tex]P(t) = P(0)e^{0.0953t}[/tex]

a) What is the doubling time?

This is t when [tex]P(t) = 2P(0)[/tex]

So

[tex]P(t) = P(0)e^{0.0953t}[/tex]

[tex]2P(0) = P(0)e^{0.0953t}[/tex]

[tex]e^{0.0953t} = 2[/tex]

Applying ln to both sides

[tex]\ln{e^{0.0953t}} = \ln{2}[/tex]

[tex]0.0953t = \ln{2}[/tex]

[tex]t = \frac{\ln{2}}{0.0953}[/tex]

[tex]t = 7.27[/tex]

The doubling time is 7.27 hours.

b) If the initial population is 2,000, what is the population in 6 hours?

This is P(6) when [tex]P(0) = 2000[/tex]. So

[tex]P(t) = P(0)e^{0.0953t}[/tex]

[tex]P(6) = 2000e^{0.0953*6} = 3543[/tex]

The population is 6 hours will be 3,543.

Other Questions