26 June, 01:02

# During the exponential phase, E. coli bacteria in a culture increase in number at a rate proportional to the current population. If the growth rate is 1.9% per minute and the current population is 172.0 million, what will the population be 7.2 minutes from now?

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1. 26 June, 02:49
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Step-by-step explanation:

Let the current population be p.

The rate of increase per time is dp/dt.

This rate is proportional to the current population meaning that a non zero number c exists so that:

dp/dt = a*p (t).

If we integrate both parts we get that:

ln|p (t) |=a*t + c. We assume that p (t) is positive since it describes population so it only makes sense to be positive in order to increase.

As a result: p (t) = e^ (at+c).

In the moment t1 where p=172 * 10^6 dp/dt = 0.019/60 (we will compute all rates per minute).

This means that:

0.019=a*172*10^6 - > a = 0,000110465116 * 10^-6. minutes^-1

In order to find the population after 7.2 minutes:

p (t1+7.2mins) / p (t1) = e^ (a*t1+a*7.2mins+c) / e^ (a*t1+c) = e^ (a*7.2mins)

After calculations we get that:

p (t1+7.2mins) = e^ (0,000795348837) / 172 million.

Not that fancy.