Chapter 3.8, Problem 13E

(a)

(i)

To determine

The average rate of change of the area of a circle with respect to its radius $r$ as $r$ change from 2 to 3.

Answer to Problem 13E

The average rate of change of the area of a circle is $\underset{\_}{5\pi }$.

Explanation of Solution

Expression for the area of the circle is given as below.

$A\left(r\right)=\pi r^2$

Calculate the average rate of change of the area of a circle, when the radius changes from 2 to 3.

$\begin{array}{l}A\left(r\right)=\frac{A\left(3\right)-A\left(2\right)}{3-2}\\ A\left(r\right)=\frac{9\pi -4\pi }{1}\\ A\left(r\right)=5\pi \end{array}$

Therefore, the average rate of change of the area of a circle is $\underset{\_}{5\pi }$.

(ii)

To determine

The average rate of change of the area of a circle with respect to its radius $r$ as $r$ change from 2 to 2.5.

Answer to Problem 13E

The average rate of change of the area of a circle is $\underset{\_}{4.5\pi }$.

Explanation of Solution

Calculate the average rate of change of the area of a circle, when the radius changes from 2 to 2.5.

$\begin{array}{l}A\left(r\right)=\frac{A\left(2.5\right)-A\left(2\right)}{2.5-2}\\ A\left(r\right)=\frac{6.25\pi -4\pi }{0.5}\\ A\left(r\right)=4.5\pi \end{array}$

Therefore, the average rate of change of the area of a circle is $\underset{\_}{4.5\pi }$.

(iii)

To determine

The average rate of change of the area of a circle with respect to its radius $r$ as $r$ change from 2 to 2.1.

Answer to Problem 13E

The average rate of change of the area of a circle is $\underset{\_}{4.1\pi }$

Explanation of Solution

Calculate the average rate of change of the area of a circle, when the radius changes from 2 to 2.1.

$\begin{array}{l}A\left(r\right)=\frac{A\left(2.1\right)-A\left(2\right)}{2.1-2}\\ A\left(r\right)=\frac{4.41\pi -4\pi }{0.1}\\ A\left(r\right)=4.1\pi \end{array}$

Therefore, the average rate of change of the area of a circle is $\underset{\_}{4.1\pi }$.

b)

To determine

The instantaneous rate of change when r = 2.

Answer to Problem 13E

When r is 2 the instantaneous rate of change is $\underset{\_}{4\pi }$.

Explanation of Solution

Calculate the instantaneous rate of change when $r$ is $2$ using the below equation.

$A\left(r\right)=\pi r^2$

Differentiate the above area equation,

$A\text{'}\left(r\right)=2\pi r$ (1)

Substitute the value $2$ for r.

$\begin{array}{l}A\text{'}\left(2\right)=2\pi \left(2\right)\\ A\text{'}\left(2\right)=4\pi \end{array}$

Therefore, when the value of $r$ is $2$ the instantaneous rate of change is $\underset{\_}{4\pi }$.

c)

To determine

To show: The rate of change of the area of a circle with respect to its radius is equal to the circumference of the circle and why it is geometrically true.

Answer to Problem 13E

It is geometrically true. Hence, the resulting change in area $\Delta A$ if change in radius $\Delta r$ is small is $\frac{\Delta A}{\Delta r}\simeq 2\pi r$.

Explanation of Solution

Show that the rate of change of the area of a circle with respect to its radius is equal to the circumference of the circle.

Circumference of the circle is given as below.

$c\left(r\right)=2\pi r$ (2)

First derivative of the area of the circle is given as below from the equation (1).

$A\text{'}\left(r\right)=2\pi r$

Compare both the equations (1) and (2).

$c\left(r\right)=A\text{'}\left(r\right)=2\pi r$

Consider a circular ring of radius r and strip of thickness $\Delta r$ along the radial direction representing the change in the radius for the circle as shown in figure (1).

Show that if the value of $\Delta r$ is small, then the change in the circle is approximately equal to its circumference.

Explain why it is geometrically true.

Approximate the resulting change in area $\Delta A$, if the value of $\Delta r$ is very small.

Write the expression for $\Delta A$ in algebraically manner.

$\begin{array}{l}\Delta A=\text{Area}\text{of}\text{ring}\text{with}\text{increased}\text{radius - Area}\text{of}\text{the}\text{ring }\\ \Delta A=A\left(r+\Delta r\right)-A\left(r\right)\\ \Delta A=\pi {\left(r+\Delta r\right)}^2-\pi r^2\\ \Delta A=2\pi r\left(\Delta r\right)+\pi {\left(\Delta r\right)}^2\end{array}$

The value of $\Delta r$ is very small. Therefore, the above equation will become as below.

$\begin{array}{l}\Delta A=2\pi r\left(\Delta r\right)\\ \frac{\Delta A}{\Delta r}\simeq 2\pi r\end{array}$

Hence, it is proved.