Chapter 2.1, Problem 26E

a

To determine

To calculate: To find the average rate of change between times 0 and 1.0 from the population

Answer to Problem 26E

The average rate of change between times 0 and 1 is as follows:

  $\text{average rate of change =}26$

Explanation of Solution

Given information: $b\left(t\right)={\left(1.0+2.0t\right)}^3$

Calculation:

Consider the equation for population size.

  $b\left(t\right)={\left(1.0+2.0t\right)}^3$

The population at time t = 0 is

  $\begin{array}{l}b\left(0\right)={\left(1+2\times 0\right)}^3\\ =1\\ \text{The population at time t = 1 is shown below:}\\ \text{b}\left(1\right)={\left(1+2\times 1\right)}^3\\ =3^3\\ =27\end{array}$

The population at time t = 0.1 is

  $\begin{array}{l}b\left(0.1\right)={\left(1+2\times 0.1\right)}^3\\ ={1.2}^3\\ =1.728\end{array}$

The population at time t = 0.01 is as follows:

  $\begin{array}{l}b\left(0.01\right)={\left(1+2\times 0.01\right)}^3\\ ={1.02}^3\\ =1.061208\end{array}$

The population at time t = 0.001 is

  $\begin{array}{l}b\left(0.001\right)={\left(1+2\times 0.001\right)}^3\\ ={1.002}^3\\ =1.006012\end{array}$

  ${\text{For a given formula, the average rate of change between times t}}_1{\text{ and t}}_2\text{ is}\text{shown below:}$

  $\begin{array}{l}\text{average rate of change =}\frac{𑨀b}{𑨀t}\\ =\frac{b\left(t_1\right)-b\left(t_2\right)}{t_1-t_2}\end{array}$

The average rate of change between times 0 and 1 is as follows:

  $\begin{array}{l}\text{average rate of change =}\frac{b\left(t_1\right)-b\left(t_2\right)}{t_1-t_2}\\ =\frac{b\left(0\right)-b\left(1\right)}{0-1}\\ =\frac{1-27}{-1}\\ =26\end{array}$

b

To determine

To calculate: To find the average rate of change between times 0 and 0.1 from the population size.

Answer to Problem 26E

The average rate of change between times 0 and 0.1 is

  $\text{average rate of change = }7.28$

Explanation of Solution

Given information: $b\left(t\right)={\left(1.0+2.0t\right)}^3$

Calculation:

The average rate of change between times 0 and 0.1 is

  $\begin{array}{l}\text{average rate of change =}\frac{b\left(t_1\right)-b\left(t_2\right)}{t_1-t_2}\\ =\frac{b\left(0\right)-b\left(0.1\right)}{0-0.1}\\ =\frac{1-1.728}{-0.1}\\ =7.28\end{array}$

c

To determine

To calculate: To find the average rate of change between times 0 and 0.01 from the population size.

Answer to Problem 26E

The average rate of change between times 0 and 0.01 is shown below:

  $\text{average rate of change = }6.1208$

Explanation of Solution

Given information: $b\left(t\right)={\left(1.0+2.0t\right)}^3$

Calculation:

The average rate of change between times 0 and 0.01 is shown below:

  $\begin{array}{l}\text{average rate of change =}\frac{b\left(t_1\right)-b\left(t_2\right)}{t_1-t_2}\\ =\frac{b\left(0\right)-b\left(0.01\right)}{0-0.01}\\ =\frac{1-1.061208}{-0.01}\\ =6.1208\end{array}$

d

To determine

To calculate: To find the average rate of change between times 0 and 0.001 from the population size.

Answer to Problem 26E

The average rate of change between times 0 and 0.001 is

  $\text{average rate of change = }6.012$

Explanation of Solution

Given information: $b\left(t\right)={\left(1.0+2.0t\right)}^3$

Calculation:

The average rate of change between times 0 and 0.001 is

  $\begin{array}{l}\text{average rate of change =}\frac{b\left(t_1\right)-b\left(t_2\right)}{t_1-t_2}\\ =\frac{b\left(0\right)-b\left(0.001\right)}{0-0.001}\\ =\frac{1-1.006012}{-0.001}\\ =6.012\end{array}$

e

To determine

To calculate: To find the limit using the obtained values

Answer to Problem 26E

The limit is 6

Explanation of Solution

Given information: $b\left(t\right)={\left(1.0+2.0t\right)}^3$

Calculation:

The limit is 6

f

To determine

To calculate: To graph the tangent line using the obtained values

Answer to Problem 26E

Hence the graph is drawn and The equation of the tangent line is

  $\begin{array}{l}\hat{b}\left(t\right)=b\left(0\right)+m\left(t-0\right)\\ =1+6t\end{array}$

Explanation of Solution

Given information: $b\left(t\right)={\left(1.0+2.0t\right)}^3$

Calculation:

Since the limit is 6, then the slope of the tangent is 6.

The equation of the tangent line is

  $\begin{array}{l}\hat{b}\left(t\right)=b\left(0\right)+m\left(t-0\right)\\ =1+6t\end{array}$

The graph of the tangent line is shown below: