Chapter 4, Problem 66RE

$\text{(a)}$

To determine

To find: The radius.

Answer to Problem 66RE

The answer: $\text{4 \%}$

Explanation of Solution

Given information:

The circumference of the equator of a sphere is measured as $10\text{cm}$ with a possible error of $0.4\text{cm}$ .

Calculation:

  $\begin{array}{c}C=2\pi r\\ \frac{dC}{dr}=2\pi \\ dC=2\pi dr\\ dr=\frac{dC}{2\pi }\\ =\frac{0.4}{2\pi }\\ =\frac{1}{5\pi }\end{array}$

Find the value of $r$ use the circumference formula and the given value of $C=10$ .

  $\begin{array}{c}C=2\pi r\\ 10=2\pi r\\ r=\frac{10}{2\pi }\\ =\frac{5}{\pi }\end{array}$

The percentage error in the radius is $\frac{dr}{r}$ , put in the values $dr$ and $r$ .

  $\begin{array}{c}\frac{dr}{r}=\frac{\frac{1}{5\pi }}{\frac{5}{\pi }}\\ =\frac{1}{5\pi }\cdot \frac{\pi }{5}\\ =0.04\\ =4\text{\%}\end{array}$

Therefore, the required radius is $\text{4 \%}$ .

$\text{(b)}$

To determine

To find: The surface area.

Answer to Problem 66RE

The answer: $8\text{\%}$

Explanation of Solution

Given information:

The circumference of the equator of a sphere is measured as $10\text{cm}$ with a possible error of $0.4\text{cm}$ .

Calculation:

  $\begin{array}{c}S=4\pi r^2\\ \frac{dS}{dr}=8\pi r\\ dS=8\pi rdr\\ =8\pi \cdot \frac{5}{\pi }\cdot \frac{1}{5\pi }\\ =\frac{8}{\pi }\end{array}$

The surface area for a radius of $r=\frac{5}{\pi }$ .

Then,

  $\begin{array}{c}S=4\pi r^2\\ =4\pi {\left(\frac{5}{\pi }\right)}^2\\ =4\pi \cdot \frac{25}{{\pi }^2}\\ =\frac{100}{\pi }\end{array}$

The percent error of the surface area is $\frac{dS}{S}$ .

Put in the values.

  $\begin{array}{c}\frac{dS}{S}=\frac{\frac{8}{\pi }}{\frac{100}{\pi }}\\ =\frac{8}{\pi }\cdot \frac{\pi }{100}\\ =\frac{8}{100}\\ =8\text{\%}\end{array}$

Therefore, the required surface area is $8\text{\%}$ .

$(c)$

To determine

To find: The volume.

Answer to Problem 66RE

The answer: $12\text{\%}$

Explanation of Solution

Given information:

The circumference of the equator of a sphere is measured as $10\text{cm}$ with a possible error of $0.4\text{cm}$ .

Calculation:

  $\begin{array}{c}V=\frac{4}{3}\pi r^3\\ \frac{dV}{dr}=4\pi r^2\\ dV=4\pi r^{2}dr\\ =4\pi {\left(\frac{5}{\pi }\right)}^2\left(\frac{1}{5\pi }\right)\\ =4\pi \cdot \frac{25}{{\pi }^2}\cdot \frac{1}{5\pi }\\ =\frac{20}{{\pi }^2}\end{array}$

The volume for radius for $r=\frac{5}{\pi }$ .

Then,

  $\begin{array}{c}V=\frac{4}{3}\pi r^3\\ =\frac{4}{3}\pi {\left(\frac{5}{\pi }\right)}^3\\ =\frac{4}{3}\pi \cdot \frac{125}{{\pi }^3}\\ =\frac{500}{3{\pi }^2}\end{array}$

The percent error of the surface area is $\frac{dV}{V}$

Put in the values.

  $\begin{array}{c}\frac{dV}{V}=\frac{\frac{20}{{\pi }^2}}{500}\\ =\frac{20}{{\pi }^2}\cdot \frac{3{\pi }^2}{500}\\ =\frac{3}{25}\\ =0.12\\ =12\text{\%}\end{array}$

Therefore, the required volume is $12\text{\%}$ .