Question

A bacteria culture starts with 520 bacteria and grows at a rate proportional to its size. After 6 hours there will be 3120 bacteria. ?

1)Express the population after t hours as a function of t.
population:
2)What will be the population after 9 hours?
3)How long will it take for the population to reach 1250 ?

Answer 1

`1) \ \ N(t) = 520e^(1/6ln6 \ t)`

`2) \ \ 7642` bacteria

`3) \ \ 2.93` hours

Explanation:

Let the number of Bacteria at time `t` (hours) be `N`, so that `N=N(t)`.

Knowing that bacteria grow by constant splitting, we first observe that we would expect the growth to be exponential.

We are given that the the rate of growth of the bacteria is proportional to the number of bacteria, thus:

`(dN)/dt prop N => (dN)/dt = KN`

Where `K` is the constant of proportionality. The equation we formed is a First Order, Separable Ordinary Differential Equation, so we can "separate the variables" to get:

`int \ 1/N \ dN = int \ K \ dt`

We can readily integrate (noting that `N` must be positive) to get:

`ln N = Kt + C => N = e^(Kt+C)`

This indicates that the the bacterial growth is exponential, consistent with our assumption.

Note that we have two unknown in our solution, `K` and `C`, but we also have two initial conditions:

`N=520` when `t=0 => 520=e^C => C=ln520`

And:

`N=3120` when `t=6=> 3120=e^(6K+ln520)`
`:. 3120=e^(6K)e^(ln520)`
`:. e^(6K) = 3120/520`
`:. 6K = ln6`
`:. K = 1/6ln6`

So, we can write the solution to the ODE as:

`N(t) = 520e^(1/6ln6 \ t)`

After `9` hours, `t=9` gives us:

`N = 520e^(9/6ln6)`
`\ \ \ = 3120sqrt(6)`
`\ \ \ ~~ 7642` (to nearest whole bacteria)

For the population to reach `1250` we require that:

`1250 = 520e^(1/6ln6 \ t)`
`:. e^(1/6ln6 \ t) = 125/52`
`:. 1/6ln6 \ t = ln(125/52)`

`:. t = (6ln(125/52))/(ln6)`
`\ \ \ \ \ \ \ = 2.937012 ...`
`\ \ \ \ \ \ \ ~~ 2.93` hours