Chapter 9.2, Problem 74STP

a.

To determine

Tofind: the constant $k$ for this type of bacteria under ideal conditions.

Answer to Problem 74STP

The constant $k$ for this type of bacteria under ideal conditions is $0.035$ .

Explanation of Solution

Given information:

In this type of reproduction, one bacteria divides, forming two bacteria. Under ideal conditions, some bacteria reproduce every $20$ minutes.

Calculation: to find the constant under ideal conditions, some bacteria reproduce every $20$ minutes,

So the number of the bacteria represented by the expression $A{e}^{kt}$ .

Where $A$ will be the initial amount, $t$ will be the amount of time that has elapsed in minutes.

After $20$ minutes, one bacteria divides into two bacterias,

  $\begin{array}{l}2=e^{20k}\\ 20k=\ln \left(2\right)\text{ [take ln]}\\ k=\frac{\ln \left(2\right)}{20}\text{ [divide]}\\ k=0.035\text{ [simpliy]}\end{array}$

Thus, the constant is $0.035$ .

b.

To determine

To find: the equation for modelling the exponential growth of the bacterium.

Answer to Problem 74STP

The equation for modelling the exponential growth of the bacterium is $A=e^{\left(\frac{\ln 2}{20}\right)t}$ .

Explanation of Solution

Given information:

In this type of reproduction, one bacteria divides, forming two bacteria. Under ideal conditions, some bacteria reproduce every $20$ minutes.

Calculation: to find the constant under ideal conditions, some bacteria reproduce every $20$ minutes,

So the number of the bacteria represented by the expression $A{e}^{kt}$ .

Where $A$ will be the initial amount, $t$ will be the amount of time that has elapsed in minutes.

So the equation can be represented as,

  $A=e^{\left(\frac{\ln 2}{20}\right)t}$