Question

How many hours after the start will the instantaneous growth rate of the population be 200 cells per hour?

A population of bacteria is initially 1100 cells. After 2 hours, the population has increased to 1200 cells. Assume that the population is an exponential function of time.
How many hours after the start will the instantaneous growth rate of the population be 200 cells per hour?

Answer 1

`3.84` hours.

Explanation:

We need to find a function of the form: `y(t)=ae^(kt)`. Where `a` is the initial amount, `t` is the time, in this case hours and `k` is the growth/decay factor.

First find `k`. We can do this from the given information as follows.

Initial amount is `1100`, and we know after 2 hours that this increases to `1200`. So:

`1200=1100e^(2k)`

solving for `k`:

`1200/1100=e^(2k)`

`ln(12/11)=2klne=>k=ln(12/11)/2`

So we now have:

`y(t)=1100e^(t/2ln(12/11))`

For population to be 200 cells per hour:

`1300=1100e^(t/2ln(12/11))`

Solve for `t`:

`13/11=e^(t/2ln(12/11)`

`ln(13/11)=t/2ln(12/11)=>t=2*ln(13/11)/ln(12/11)=3.84` ( 2 .d.p)

Check:

`y= 1100e^((3.84)/2ln(12/11))=1300`