Question

The number of bacteria in a certain culture grows at a rate that is proportional to the number present. If the number increased from 500 to 2000 in 2 hours, determine.

(a) The number present after 12 hours.

(b) The doubling time.

Answer 1

Solution

(a) The number present after 12 hours.

Let t time after hours

      p the original number of bacteria (t=0)

      x the number of bacteria after time.

$\implies \frac{dx}{dt}=kx,\quad k\in \mathbb{R}$

$\implies\frac{1}{x}dx=kdt$

$\implies\int\frac{1}{x}dx=k\int dt$

      $ln|x|+c=kt$

when  $\begin{cases} t=0\implies x=p,p=500 & \quad \\ t=2\implies x=4p & \quad \end{cases}$

$\implies\begin{cases} (1) : ln|p|+c=0\implies c=-ln|p| & \quad \\ (2) : ln|4p|+c=2k & \quad \end{cases}$

(2) : $ln|4p|-ln|p|=2k\iff ln\Big(\frac{4p}{p}\Big)=2k\implies k=\frac{ln4}{2}$

(*) The number present after t=12h

$\implies lnx-lnp=\frac{ln4}{2}\times 12=6ln4$

$\implies lnx=ln\Big(4^6\times p\Big)\implies x=4^6p=4^6\times 500$

(b) The doubling time.

Let T the doubling time , T=2t

$\implies lnx-lnp=kT=2kt$

$\implies lnx=tlnu+lnp$

$\implies lnx=ln\Big(4^tp\Big)\implies x=4^tp$

 

Therefore.

(a). the number of bacteria after time is 500

(b). $x=4^tp$