Chapter 3.9, Problem 28E

(a)

To determine

To estimate: The maximum error in the calculated area of circular disk.

Answer to Problem 28E

The maximum error in the area of the circular disk is $\Delta S\approx 9.6\pi {\text{cm}}^2$.

Explanation of Solution

Given:

The radius of the circular disk is 24 cm and the maximum error in measurement is 0.2 cm.

Calculation:

Suppose r be a radius of the circle.

Note that, the area of the circular disk is $S=\pi r^2$.

The differential is $dS=f^{\prime }\left(r\right)dr$ (1)

Derivative of the function $f\left(r\right)=\pi r^2$ is $f^{\prime }\left(r\right)=2\pi r$,

Substitute $f^{\prime }\left(r\right)=2\pi r$ in equation (1),

$dS=2\pi rdr$

Substitute $r=24\text{ and }dr=0.2$,

$\begin{array}{c}dS=2\pi \left(24\right)\left(0.2\right)\\ =9.6\pi {\text{cm}}^2\end{array}$

The maximum error in the area of the circular disk is $\Delta S$ which is approximately same as $dS$.

Therefore, the maximum error in the area of the circular disk is $\Delta S\approx 9.6\pi {\text{cm}}^2$.

(b)

To determine

To find: The relative error and percentage error.

Answer to Problem 28E

The relative error and the percentage error are $0.01\overline{6}$ and $1.\overline{6}\%$ respectively.

Explanation of Solution

Calculation:

The relative error is computed by dividing the change of the area $\left(\Delta S\right)$ with the area S.

That is, the relative error is $\frac{\Delta S}{S}$.

Note that, the value $\Delta S$ is approximately equal to $dS$.

$\begin{array}{c}\frac{\Delta S}{S}\approx \frac{dS}{S}\\ =\frac{2\pi rdr}{\pi r^2}\\ =2\frac{dr}{r}\end{array}$

Substitute $r=24\text{ and }dr=0.2$,

$\begin{array}{c}\frac{\Delta S}{S}\simeq 2\frac{0.2}{24}\\ =0.01\overline{6}\end{array}$

Therefore, the relative error is $0.01\overline{6}$.

Note that, the percentage error is the product of the relative error and 100.

The percentage error is computed as follows,

$\begin{array}{c}\text{Percentage error}=0.01\overline{6}\times 100\%\\ =1.\overline{6}\%\end{array}$

Therefore, the percentage error is $1.\overline{6}\%$.