Chapter 2.1, Problem 23E

a

To determine

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

Answer to Problem 23E

The average rate of change between times 0 and 1 is

  $=0.5$

Explanation of Solution

Given information: $b\left(t\right)={1.5}^t$

Calculation:

Consider the equation for a population size.

  $b\left(t\right)={1.5}^t$

The population at time $t=0$ is

  $\begin{array}{l}b\left(0\right)={1.5}^0\\ =1\end{array}$

The population at time t = 1 is shown below:

  $\begin{array}{l}b\left(1\right)={1.5}^1\\ =1.5\end{array}$

The population at time t=0.1 is

  $\begin{array}{l}b\left(0.1\right)={1.5}^{0.1}\\ =1.0414\end{array}$

The population at time t=0.01 is as follows:

  $\begin{array}{l}b\left(0.01\right)={1.5}^{0.01}\\ =1.00406\end{array}$

The population at time t=0.001 is

  $\begin{array}{l}b\left(0.001\right)={1.5}^{0.001}\\ =1.000405\end{array}$

For a given formula, the average rate of Change between times $t_1$ and $t_2$ is

average rate of change $=\frac{△b}{△t}$

  $=\frac{b\left(t_1\right)-b\left(t_2\right)}{t_1-t_2}$

The average rate of change between times 0 and 1 is

  $\begin{array}{l}\text{average rate of change =}\frac{b\left(t_1\right)-b\left(t_2\right)}{t_1-t_2}\\ =\frac{b\left(0\right)-b\left(1\right)}{0-1}\\ =\frac{1-1.5}{-1}\\ =0.5\\ \end{array}$

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 23E

The average rate of change between times 0 and 0.1 is $=0.414$

Explanation of Solution

Given information: $b\left(t\right)={1.5}^t$

Calculation:

The average rate of change between times 0 and 0.1 is shown below:

  $\begin{array}{l}\text{average rate of change =}\frac{b\left(t_1\right)-b\left(t_2\right)}{t_1-t_2}\\ =\frac{b\left(0\right)-b\left(0.1\right)}{0-0.1}\\ =\frac{1-1.0414}{-0.1}\\ =0.414\end{array}$

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 23E

The average rate of change between times 0 and 0.01 is $0.406$

Explanation of Solution

Given information: $b\left(t\right)={1.5}^t$

Calculation:

The average rate of change between times 0 and 0.01 is

  $\begin{array}{l}\text{average rate of change =}\frac{b\left(t_1\right)-b\left(t_2\right)}{t_1-t_2}\\ =\frac{b\left(0\right)-b\left(0.01\right)}{0-0.01}\\ =\frac{1-1.00406}{-0.01}\\ =0.406\end{array}$

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 23E

The average rate of change between times 0 and 0.001 is $0.405$

Explanation of Solution

Given information: $b\left(t\right)={1.5}^t$

Calculation:

The average rate of change between times 0 and 0.001 is

  $\begin{array}{l}\text{average rate of change =}\frac{b\left(t_1\right)-b\left(t_2\right)}{t_1-t_2}\\ =\frac{b\left(0\right)-b\left(0.001\right)}{0-0.001}\\ =\frac{1-1.000405}{-0.001}\\ =0.405\end{array}$

e

To determine

To calculate: To find the limit using the obtained values

Answer to Problem 23E

The limit is 0.405.

Explanation of Solution

Given information: $b\left(t\right)={1.5}^t$

Calculation:

(e) The limit is 0.405.

f

To determine

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

Answer to Problem 23E

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

  $\begin{array}{l}\hat{b}\left(t\right)=b\left(0\right)+m\left(t-0\right)\\ 1+0.405t\end{array}$

Explanation of Solution

Given information: $b\left(t\right)={1.5}^t$

Calculation:

Since the limit is 0.405, then the slope of the tangent is 0.405.

The equation of the tangent line is

  $\begin{array}{l}\hat{b}\left(t\right)=b\left(0\right)+m\left(t-0\right)\\ 1+0.405t\end{array}$

The graph of the tangent line is shown below: