Example 4 is a model of uninhibited growth; it accurately reflects growth in early stages. After a time, growth no longer continues at a rate proportional to the number present. Factors, such as disease, lack of space, and dwindling food supply, begin to affect the rate of growth.

Assume that a colony of bacteria grows at a rate proportional to the number of bacteria present. If the number of bacteria doubles in \(5\) hours (h), how long will it take for the number of bacteria to triple?

Solution Let \(N(t) \) be the number of bacteria present at time \(t\). Then the assumption that this colony of bacteria grows at a rate proportional to the number present can be modeled by \[ \dfrac{dN}{dt}=kN \]

where \(k\) is a positive constant of proportionality. To find \(k,\) we write the differential equation as \(\dfrac{dN}{N}=kdt\) and integrate both sides. This differential equation is of form (4), and its solution is given by (5). So, we have \[ N(t) =N_{0}e^{kt} \]

where \(N_{0}\) is the initial number of bacteria in the colony. Since the number of bacteria doubles to \(2N_{0}\) in 5h, \[ \begin{eqnarray*} N(5) =N_{0}\,e^{5k} &=& 2N_{0} \\ e^{5k} &=&2 \\ k &=&\dfrac{1}{5}\ln 2 \end{eqnarray*} \]

The time \(t\) required for this colony to triple obeys the equation \[ \begin{eqnarray*} N(t) &=&3N_{0} \\ N_{0}e^{kt} &=&3N_{0} \\ e^{kt} &=&3 \\ t &=&\dfrac{1}{k}\ln 3 \underset{\underset{{\color{#0066A7}{\hbox{\( k=\tfrac{1}{5}\ln 2\)}}}}{\color{#0066A7}{\uparrow }}}{=}5\dfrac{\ln 3}{\ln 2}\approx 8 \end{eqnarray*} \]

The number of bacteria will triple in about 8h.