Question

A bacteria colony increases in size at a rate of 4.0581e1.6t bacteria per hour. If the initial population is 36 bacteria, find the population four hours later. (Round your answer to the nearest whole number.)

Answer 1

Answer:

1560

Step-by-step explanation:

The rate of Increase of the population (P) of the bacteria is given as:

$\frac{dP}{dt} =4.058e^{1.6t}$

$dP =4.058e^{1.6t}}dt\\Taking\: Integrals\\\int dP =4.058 \int e^{1.6t}dt\\P(t)=\frac{4.058}{1.6} (e^{1.6t} +K)\\P(t)=2.53625 (e^{1.6t} +K)$

Where k is a constant of Integration.

At t=0, P(t)=36

$36=2.53625 (e^{1.6*0} +K)\\36=2.53625 (e^{0} +K)\\36=2.53625 (1 +K)\\36=2.53625 +2.53625K\\K=13.19$

Therefore:

$P(t)=2.53625 (e^{1.6t} +13.19)\\At \: t=4\\P(4)=2.53625 (e^{1.6*4} +13.19)\\=1559.88\\P(4)=1560$