How Does Exponential Growth Affect a Bacterial Population?

[Tineeyyy]

A bacterial population x is known to have a growth rate proportional to x itself. If between 12 noon and 2:00 pm, the popilation triples.
1. What is the growth rate pf the given problem?
2. Is the number of bacterial population important to the problem?
3. At what time should x become 100 times what it was at 12 noon assuming that no controls are being exerted

 

[HOI]

A bacterial population x is known to have a growth rate proportional to x itself.

So dx/dt= kx for some k.
dx/x= kdt. Integrating, ln(x)= kt+ c
Taking the exponential, $x= e^{kt+ c}= Ce^{kt}$ where $C= e^c$.

If between 12 noon and 2:00 pm, the popilation triples.

Taking t to be the time in hours since noon, $X(0)= C$, $X(2)= Ce^{2k]= 3C$, e^{2k}= 3$.

1. What is the growth rate pf the given problem?

$e^{2k}= 3$ so $2k= ln(3)$ and $k= \frac{1}{2}ln(3)$.

2. Is the number of bacterial population important to the problem?
That's a strange question! Obviously the number is the whole point of the problem! But I suspect they mean the initial number of bacterial population, X(0). No, that cancels so is not at all important.

3. At what time should x become 100 times what it was at 12 noon assuming that no controls are being exerted.

We want to solve X(t)= 100X(0).
$X(t)= X(0)e^{kt}$ and we have determined that $k= \frac{1}{2}ln(3)= ln(\sqrt{3})$ so $X(t)= X(0)e^{ln(\sqrt{3})t}= 100X(0)$.

$e^{ln(\sqrt{3}t}= \sqrt{3}^t= 100$.
$t ln(\sqrt{3})= t\left(\frac{1}{2} ln(3)\right)= 100$.
$t= \frac{200}{ln(3)}$

So about 182 hours or 7 days 14 hours.