Question

bacteria in certain culture increase at a rate proportional to the number present. if the number of bacteria doubles in three hours, in how many hours will it triple

Answer 1

Using a differential equation, it is found that the bacteria's population will triple in 4.75 hours.

What is the differential equation used in this problem?

• The bacteria in certain culture increase at a rate proportional to the number present, which means that the population is described by the following differential equation:

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

• In which k is the growth rate.

Applying separation of variables, the solution is:

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

The number of bacteria doubles in three hours, hence P(3) = 2P(0), which is used to find k.

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

$2P(0) = P(0)e^{3k}$

$e^{3k} = 2$

$\ln{e^{3k}} = \ln{2}$

$3k = \ln{2}$

$k = \frac{\ln{2}}{3}$

$k = 0.23105$

Hence:

$P(t) = P(0)e^{0.23105t}$

It triples when P(t) = 3P(0), hence:

$P(t) = P(0)e^{0.23105t}$

$3P(0) = P(0)e^{0.23105t}$

$e^{0.23105t} = 3$

$\ln{e^{0.23105t}} = \ln{3}$

$0.23105t = \ln{3}$

$t = \frac{\ln{3}}{0.23105}$

$t = 4.75$

The bacteria's population will triple in 4.75 hours.

A similar problem, also involving a differential equation, is given at https://brainly.com/question/14423176