Chapter 4.5, Problem 42E

a.

To determine

To find: To find the estimate of the resulting change in area.

Answer to Problem 42E

The change in area is $0.2513{\text{ m}}^2$ .

Explanation of Solution

Given: The radius is changed from $2.00m$ to $2.02\text{ m}$ .

Formula used: $dy=f\text{'}(x)dx$

Calculation:

Let $y=f(x)$ be the differential function. The differential $dx$ is an independent variable. The differential $dy$ is,

  $dy=f\text{'}(x)dx$

Change in area with respect to radius,

  $\begin{array}{l}\frac{dA}{dr}=\frac{d}{dr}(\pi r^2)\\ da=2\pi rdr\end{array}$

Put $r=2\text{ and }dr=0.2$ ,

  $\begin{array}{l}da=2\pi rdr\\ da=2\pi (2)(0.02)\\ da=0.2513{\text{ m}}^2\end{array}$

Conclusion:

The change in area is $0.2513{\text{ m}}^2$ .

b.

To determine

To find: To find the estimate in the percentage of the circle’s original area.

Answer to Problem 42E

The percentage of the circle’s original area is $2\%$ .

Explanation of Solution

Given: The radius is changed from $2.00m$ to $2.02\text{ m}$ .

Formula used: The differential $dx$ is an independent variable. The differential $dy$ is

  $dy=f\text{'}(x)dx$

Calculation:

Original area is given by

  $\begin{array}{l}A=\pi r^2\\ $ A=\pi {(2)}^2$A=4\pi \end{array}$

Let $y=f(x)$ be the differential function. The differential $dx$ is an independent variable. The differential $dy$ is

  $dy=f\text{'}(x)dx$

Change in area with respect to radius

  $\begin{array}{l}\frac{dA}{dr}=\frac{d}{dr}(\pi r^2)\\ da=2\pi rdr\end{array}$

  $r=2\text{ and }dr=0.2$

  $\begin{array}{l}da=2\pi rdr\\ da=2\pi (2)(0.02)\\ da=0.2513{\text{ m}}^2\end{array}$

  $\begin{array}{l}\frac{dA}{A}=\frac{0.08\pi }{4\pi }\times 100\\ \frac{dA}{A}=2\%\end{array}$

Conclusion:

The percentage of the circle’s original area is $2\%$ .