Question

1. (8 points) in a biology experiment, let ( n(t) ) be the number of bacteria in a colony after ( t ) hours, where ( t = 0 ) corresponds to the time the experiment begins. suppose that during the period from ( t = ) to ( t = 8 ) hours the number of bacteria is modeled by the exponential function ( n(t)=e^{t} ). (see below).

graphical perspective

( n(t)=e^{t}, 4 leq t leq 8 )

(a) (4 points) find the average rate of change of the population over the time period ( 5 leq t leq 7 ).

Explanation:

Step1: Recall the formula for average rate of change

The formula for the average rate of change of a function \(y = f(x)\) over the interval \([a,b]\) is \(\frac{f(b)-f(a)}{b - a}\). Here, \(f(t)=N(t)=e^{t}\), \(a = 5\), and \(b=7\).

Step2: Calculate \(N(7)\) and \(N(5)\)

Substitute \(t = 7\) into \(N(t)\): \(N(7)=e^{7}\). Substitute \(t = 5\) into \(N(t)\): \(N(5)=e^{5}\).

Step3: Apply the formula

The average rate of change \(\frac{N(7)-N(5)}{7 - 5}=\frac{e^{7}-e^{5}}{2}\).
Using the property \(e^{m}-e^{n}=e^{n}(e^{m - n}-1)\), we can rewrite it as \(\frac{e^{5}(e^{2}-1)}{2}\).
Since \(e\approx2.718\), \(e^{2}\approx7.389\), \(e^{5}\approx148.413\). Then \(\frac{e^{5}(e^{2}-1)}{2}=\frac{148.413\times(7.389 - 1)}{2}=\frac{148.413\times6.389}{2}\approx\frac{948.3}{2}=474.15\).

Answer:

The average rate of change of the population over the time period \(5\leq t\leq7\) is \(\frac{e^{7}-e^{5}}{2}\approx474.15\)