Chapter 8.8, Problem 2CYU

(a)

To determine

To Find: The value of $k$ for the bacteria.

Answer to Problem 2CYU

The value of the k for the bacteria is $k=0.0693$ .

Explanation of Solution

Given:

The number of cells is 2 initially and 8 cells after 20 seconds.

The given function is $y=2e^{kt}$ .

Calculation:

Consider the given function is,

  $y=2e^{kt}$

Then,

  $\begin{array}{l}8=2e^{k20}\\ 4=e^{k20}\\ 1.386=20k\\ k=\frac{1.386}{20}\end{array}$

Solve further as,

  $k=0.0693$

Thus, the value of the k for the bacteria is $k=0.0693$ .

(b)

To determine

To Find: The number of bacteria on the food after 5 seconds.

Answer to Problem 2CYU

The number of bacteria is approximately is 3

Explanation of Solution

Given:

The number of cells is 2 initially and 8 cells after 20 seconds.

Calculation:

Consider the given function is,

  $y=2e^{kt}$

Then,

  $\begin{array}{c}y=2e^{\left(0.0693\right)5}\\ =2e^{0.3465}\\ =2\left(1.414\right)\\ =2.828\end{array}$

Thus, the number of bacteria is approximately is 3

(c)

To determine

To Find: Whether the food on the floor for 5 seconds can be eaten and give the reasonable explanation.

Answer to Problem 2CYU

The food must not be good to eaten as it is on the floor for more than 5 seconds.

Explanation of Solution

No the food on the food cannot be eaten after the food is on the floor after 5 seconds as the cells start growing on the food. Consider the result of part (b) and how the number of bacteria increases in 5 second. The food must not be good to eaten as it is on the floor for more than 5 seconds.