Chapter 1.4, Problem 30E

To determine

To find: the rate of change of the volume with respect to radius r at 2 in.

Answer to Problem 30E

The rate of change is $16\pi \text{}{\text{in}}^3\text{/in}$ .

Explanation of Solution

Given information:

Volume of the sphere.

Concept used:

The volume of sphere with radius r is given by: $V=\frac{4}{3}\pi r^3$ .

Instantaneous rate of change of any function f at any point a is given by: $f^{\prime }\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ .

Calculation:

For finding the rate of change of the volume with respect to radius r at 2 in it is enough to find the instantaneous rate of change at $r=2$ .

For the function $\frac{4}{3}\pi r^3$ , the instantaneous rate of change at $r=2$ will be:

  $\begin{array}{c}{V}^{\prime }\left(2\right)=\underset{h\to 0}{\lim }\frac{V\left(2+h\right)-V\left(2\right)}{h}{\text{ in}}^3\text{/in}\\ =\underset{h\to 0}{\lim }\left(\frac{\frac{4}{3}\pi {\left(2+h\right)}^3-\frac{4}{3}\pi {\left(2\right)}^3}{h}\right){\text{ in}}^3\text{/in}\\ =\frac{4}{3}\pi \underset{h\to 0}{\lim }\left(\frac{8+h^3+6h^2+12h-8}{h}\right){\text{ in}}^3\text{/in}\\ =\frac{4}{3}\pi \underset{h\to 0}{\lim }\left(\frac{h^3+6h^2+12h}{h}\right){\text{ in}}^3\text{/in}\\ =\frac{4}{3}\pi \underset{h\to 0}{\lim }\left(h^2+6h+12\right){\text{ in}}^3\text{/in}\\ \text{Apply limit}\\ =\frac{4}{3}\left(12\right)\pi {\text{ in}}^3\text{/in}\\ =16\pi {\text{ in}}^3\text{/in}\end{array}$

Hence the instantaneous rate of change is $16\pi {\text{ in}}^3\text{/in}$ .