Q 78. - Precalculus Enhanced with Graphing Utilities

Question

E.coli Growth: A strain of E.coli Beu 397-recA441 is placed into a nutrient broth at $30°$Celsius and allowed to grow.

The data shown in the table are collected. The population is measured in grams and the time in hours. Since population P depends on time $t$ and each input corresponds to exactly one output, we can say that population is a function of time; so $P\left(t\right)$ represents the population at time width="6" style="max-width: none; vertical-align: -4px;" $t$.

Time(hours), t	Population(grams), P
$0$	$0.09$
$2.5$	$0.18$
$3.5$	$0.26$
$4.5$	$0.35$
$6$	$0.50$

(a) Plot the points $(0, 0.09), (2.5, 0.18),$ and so on in a Cartesian plane.

(b) Draw a line segment from the point $(0, 0.09)$ to $(2.5, 0.18)$. What does the slope of this line segment represent?

(c) Find the average rate of change of the population from $0$ to $2.5$ hours.

(d) Find the average rate of change of the population from $4.5$ to $6$ hours.

Step-by-Step Solution

Answer

Part (a). The required graph is shown below:

Part (b). The graph with the line segment is shown below:

The slope of the line segment represents the average rate of change of the population from $0$ to $2.5$ hours.

Part (c). The average rate of change from $0$ to $2.5$ is $0.036\text{grams/hr}$.

Part (d). The average rate of change from $4.5$ to $6$ is $0.1\text{grams/hr}$.

Part (e). The average rate of change of the population is increasing as time passes. 

Part (a). Step 1. Given information

The data that represent the time in hours and population in grams is shown in the table below:

 

Time(hours), t	Population(grams), P
$0$	$0.09$
$2.5$	$0.18$
$3.5$	$0.26$
$4.5$	$0.35$
$6$	$0.50$

Part (a). Step 2. Plot the points ( 0 , 0 . 09 ) ,   ( 2 . 5 , 0 . 18 ) , and so on in a Cartesian plane.

The required graph is shown below:

Part (b) Step 1. Draw a line segment from the point ( 0 , 0 . 09 ) to ( 2 . 5 , 0 . 18 ) to determine the representation of the slope of the line segment.

The graph with the line segment from the point $(0,0.09)$ to $(2.5, 0.18)$ is shown below:

 

From the graph, observe that the slope of the line segment represents the average rate of change of the population from $0$ to $2.5$ hours.

Part (c). Step 1. Determine the average rate of change of the population from 0 to 2 . 5 hours.

The average rate of change from $a$ to $b$ is defined as folllows:

$\frac{∆y}{∆x}=\frac{f\left(b\right)-f\left(a\right)}{b-a}\dots \left(1\right)$, where $a\ne b$.

Substitute $a=0$ and $b=2.5$ into $\left(1\right)$.

$\begin{array}{rcl}\frac{∆y}{∆x}& =& \frac{f(2.5)-f\left(0\right)}{2.5-0}\end{array}$

Substitute $\begin{array}{rcl}f\left(0\right)& =& 0.09\end{array}$ and $f(2.5)=0.18$ from the given table.

$\begin{array}{rcl}\frac{∆y}{∆x}& =& \frac{0.18-0.09}{2.5}\\ & =& \frac{0.09}{2.5}\\ & =& 0.036\end{array}$

Thus, the average rate of change of the population is $0.036\text{grams/hr}$.

Part (d). Step 1. Determine the average rate of change of the population from 4 . 5 to 6 hours.

Substitute $a=4.5$ and $b=6$ into $\left(1\right)$.

$\begin{array}{rcl}\frac{∆y}{∆x}& =& \frac{f\left(6\right)-f(4.5)}{6-4.5}\end{array}$

Substitute $f(4.5)=0.35$ and $f\left(6\right)=0.50$ from the given table.

$\begin{array}{rcl}\frac{∆y}{∆x}& =& \frac{0.50-0.35}{1.5}\\ & =& \frac{0.15}{1.5}\\ & =& 0.1\end{array}$

Thus, the average rate of change of the population is $0.1\text{grams/hr}$.

Part (e) Step 1. Determine the state to the average rate of change as time passes.

The average rate of change of the population is increasing as time passes.