Chapter 14, Problem 72RE

To determine

To find:

The rate of change of area with respect to r at $r=2,3,2.1$

Answer to Problem 72RE

  $\frac{\Delta A}{\Delta r}=5\pi$ , $\frac{\Delta A}{\Delta r}=4.5\pi$ , $\frac{\Delta A}{\Delta r}=4.1\pi$

Explanation of Solution

Given:

The area of a circle is

  $A\left(r\right)=\pi r^2$

Concept used:

The area of a circle is a function of the radius

  $A\left(r\right)=\pi r^2$

The average rate of change of the area of a circle with respect to radius

  $\frac{\Delta A}{\Delta r}=\frac{\pi r_1^2-\pi r_2^2}{r_1-r_2}$

Calculation:

The average rate of change of the area of a circle with respect to radius

  $\begin{array}{l}\frac{\Delta A}{\Delta r}=\frac{\pi r_1^2-\pi r_2^2}{r_1-r_2}\\ \frac{\Delta A}{\Delta r}=\frac{\pi {\left(2\right)}^2-\pi {\left(3\right)}^2}{2-3}\\ \frac{\Delta A}{\Delta r}=5\pi \end{array}$

  $\begin{array}{l}\frac{\Delta A}{\Delta r}=\frac{\pi r_1^2-\pi r_2^2}{r_1-r_2}\\ \frac{\Delta A}{\Delta r}=\frac{\pi {\left(2\right)}^2-\pi {\left(2.5\right)}^2}{2-2.5}\\ \frac{\Delta A}{\Delta r}=4.5\pi \end{array}$

  $\begin{array}{l}\frac{\Delta A}{\Delta r}=\frac{\pi r_1^2-\pi r_2^2}{r_1-r_2}\\ \frac{\Delta A}{\Delta r}=\frac{\pi {\left(2\right)}^2-\pi {\left(2.1\right)}^2}{2-2.1}\\ \frac{\Delta A}{\Delta r}=4.1\pi \end{array}$