Chapter 4.5, Problem 49E

To determine

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

Answer to Problem 49E

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 $\frac{dC}{dr}$ ,

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

Change in volume will be,

  $\begin{array}{l}\frac{dV}{dr}=\frac{d}{dr}(\frac{4}{3}\pi r^3)\\ \frac{dV}{dr}=4\pi r^2\\ dV=4\pi r^{2}dr\\ dr=\frac{1}{16\pi }\\ dV=4\pi r^2\times \frac{1}{16\pi }\\ dV=\frac{r^2}{4}{\text{ in}}^3\end{array}$

Conclusion:

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	$\text{7 in}$	$\frac{1}{8}\text{ in}$	$\frac{1}{16\pi }\text{ in}$	$12.25{\text{ in}}^3$