Question

Bacterium Population When a bactericide was added to a nutrient broth in which bacteria were growing, the bacterium population continued to grow for a while but then stopped growing and began to decline. The size of the population at time \(t\) (hours) was \(b(t)=10^{6}+10^{4} t-10^{3} t^{2}\) . Find the growth rates at \(t=0, t=5,\) and \(t=10\) hours.

Short answer

The growth rates of the bacteria population are \(10^{4}\) bacteria/hour at \(t = 0\) hours, \(0\) at \(t = 5\) hours and \(-10^{4}\) bacteria/hour at \(t = 10\) hours.

Step-by-step solution

Differentiate the given function

The first step is finding the derivative of the function \(b(t)=10^{6}+10^{4}t-10^{3}t^{2}\). The derivative will give us the instantaneous growth rate of the bacteria population at any given time 't'. Using the power rule of differentiation, the derivative of \(b(t)\), let's call it \(b'(t)\), will be \(b'(t) = 10^{4} - 2\times10^{3}t\).

Find the growth rate at \(t = 0\) hours

Simply substitute \(t = 0\) into \(b'(t)\) equation which is \(b'(0) = 10^{4} - 2\times10^{3}\times 0\). The result is \(b'(0) = 10^{4}\) bacteria/hour. This is the growth rate at the beginning, at time \(t = 0\) hours.

Find the growth rate at \(t = 5\) hours

Substitute \(t = 5\) into \(b'(t)\) equation. So \(b'(5) = 10^{4} - 2\times10^{3}\times 5 = 0\). So, at \(t = 5\) hours, the growth rate is 0. This means the bacteria population is not growing or decreasing at this point.

Find the growth rate at \(t = 10\) hours

Substitute \(t = 10\) into \(b'(t)\) equation. So \(b'(10) = 10^{4} - 2\times10^{3}\times 10 = -10^{4}\) bacteria/hour. A negative growth rate means the population is decreasing at this time.