Chapter 4.5, Problem 50E

To determine

To find: To find sensitivity of the volume and circumference.

Answer to Problem 50E

Sphere type	True radius	Tape error	Radius error	Volume error
Orange	$2\text{ in}$	$\frac{1}{8}\text{ in}$	$\frac{1}{16\pi }\text{ in}$	$1{\text{ in}}^3$
Melon	$\text{4 in}$	$\frac{1}{8}\text{ in}$	$\frac{1}{16\pi }\text{ in}$	${\text{4 in}}^3$
Beach ball	$\text{7 in}$	$\frac{1}{8}\text{ in}$	$\frac{1}{16\pi }\text{ in}$	$12.25{\text{ in}}^3$

Explanation of Solution

Given: The values are given in the table.

Calculation:

Let V is the volume, r is radius and C is circumference.

Find $\frac{dC}{dr}$ .

  $\begin{array}{c}\frac{dC}{dr}=\frac{d}{dr}(2\pi r)\\ \frac{dC}{dr}=2\pi \\ dC=2\pi dr\\ dC=\frac{1}{8}\\ \frac{1}{8}=2\pi dr\\ dr=\frac{1}{16\pi }\text{ inch }\end{array}$

Change in volume will be,

  $\begin{array}{l}\frac{dS}{dr}=\frac{d}{dr}(4\pi r^2)\\ \frac{dS}{dr}=8\pi r\\ dS=8\pi rdr\\ dr=\frac{1}{16\pi }\\ dS=8\pi r\times \frac{1}{16\pi }\\ dS=\frac{r}{2}{\text{ in}}^2\end{array}$

Conclusion:

Sphere type	True radius	Tape error	Radius error	Surface area error
Orange	$2\text{ in}$	$\frac{1}{8}\text{ in}$	$\frac{1}{16\pi }\text{ in}$	$1{\text{ in}}^2$
Melon	$\text{4 in}$	$\frac{1}{8}\text{ in}$	$\frac{1}{16\pi }\text{ in}$	${\text{2 in}}^2$
Beach ball	$\text{7 in}$	$\frac{1}{8}\text{ in}$	$\frac{1}{16\pi }\text{ in}$	$3.5{\text{ in}}^2$