Chapter 7.8, Problem 2GP

a.

To determine

To Find:

The value k for the bacteria.

Answer to Problem 2GP

The value k is $k=0.06931$ .

Explanation of Solution

Given information:

The given exponential model $y=2e^{kt}$ . Initially there are two cells and $8$ cells after $20\text{ sec}$ .

Consider the exponential model $y=2e^{kt}$ .

  $\begin{array}{l}8=2e^{k\left(20\right)}\\ e^{k\left(20\right)}=4\\ \ln \left(e^{k\cdot 20}\right)=\ln \left(4\right)\\ k\cdot 20\ln \left(e\right)=\ln \left(4\right)\\ k\cdot 20=\ln \left(4\right)\\ k=\frac{\ln \left(2\right)}{10}\\ k\approx 0.06931\end{array}$

Therefore, the value is $k=0.06931$ .

b.

To determine

To Find:

The number of bacteria after 5 seconds.

Answer to Problem 2GP

The approximated value is $y\approx 2.8282$ .

Explanation of Solution

Given information:

The given exponential model $y=2e^{kt}$ . Initially there are two cells and $8$ cells after $20\text{ sec}$ .

From part (a) the value is $k=0.06931$

  $\begin{array}{l}y=2e^{kt}\\ y=2e^{\left(0.0693\right)\left(5\right)}\\ y=2\left(1.414109\right)\\ y=2.8282\end{array}$

Therefore, there will be $y=2.8282$ bacteria cells on the food after 5 seconds.

c.

To determine

To explain:

The reason for having or not having the food that had been on the floor for 5 seconds.

Answer to Problem 2GP

The food is not edible as the bacteria reproduces at a much faster rate than expected.

Explanation of Solution

Given information:

The given exponential model $y=2e^{kt}$ . Initially there are two cells and $8$ cells after $20\text{ sec}$ .

Every food is a good food. But in this case the food dropped for 5 seconds is not editable as the bacteria reproduces at a much faster rate than expected.