Question

Bacterial cells grow at a rate proportional to the volume of dividing cells at that moment. If V0



is the volume of dividing cells at any time t . Find the time at which the volume of the cells will

be double its original size.

Answer 1

$\frac{dV(t)}{dt}=V(t)$ then $\ln$ |$V$|$=t+\ln C$ then $V=Ce^t$ . As $V(0)=V_0=C,$ then

$V(t)=V_0e^t$ hence $V(\tau)=2V_0=V_0e^\tau$ hence $e^\tau=2,$ hence $\tau= ln 2\approx0.69 s$