Question

In Example 6 we considered a bacteria population that doubles every hour. Suppose that another population of bacteria triples every hour and starts with 400 bacteria. Find an expression for the number \(n\) of bacteria after \(t\) hours and use it to estimate the rate of growth of the bacteria population after 2.5 hours.

Short answer

Rate of growth is \(6850{\rm{ bacteria/hour}}\).

Step-by-step solution

Rate of Growth

TheRate of Growth of population can be determined using method of calculus commonly named asdifferentiation, in which the function is taken as its derivative.

Evaluating the function from example 6 using differentiation: 

Thefunction from example 6 for initial population of 400 is:

\(n\left( t \right) = \left( {400} \right){3^t}\)

Now, differentiating with respect to t:

\(\begin{aligned}\frac{{dn\left( t \right)}}{{dt}} &= \frac{d}{{dt}}\left( {\left( {400} \right){3^t}} \right)\\ &= \left( {400} \right){3^t}.\ln 3\end{aligned}\)

After 2.5 hours, rate will be:

\(\begin{aligned}\frac{{dn\left( t \right)}}{{dt}}\left| {_{t = 2.5}} \right. &= \left( {400} \right){3^{2.5}}.\ln 3\\ \approx 6850{\rm{ bacteria/hour}}\end{aligned}\)

Hence, this is the required rate of growth of population.