Question

A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 590. (a) Find an expression for the number of bacteria after t hours. P(t)

Answer 1

Answer:

Step-by-step explanation:

Given that a bacteria culture initially contains 100 cells and grows at a rate proportional to its size.

i.e. P' = kP where k = constant of proportionality

We separate variables and solve this Differential equation.

$\frac{dP}{dt} =kP\\\frac{dP}{P} =kdt\\ln P = kt+C\\P(t) = Ae^{kt}$

When t =0 P(0) = A = 100

So equatin is

$P(t) = 100e^{kt}$

to find k

When t=1, P =590

$P(t) =590= 100e^{k*1}\\e^k = 5.9 \\ k=ln 5.9 = 1.775$

So

$P(t) = 100e^{1.775t}$