Chapter 2.4, Problem 30E

To determine

To find: The rate of change of volume of sphere with respect to radius when $r=2\text{in}\text{.}$.

Answer to Problem 30E

The rate of change of volume of sphere with respect to radius when $r=2\text{in}\text{.}$ is equal to $16\text{in}{\text{.}}^{\text{3}}/\text{in}\text{.}$.

Explanation of Solution

Given information:

The radius of the sphere is $r=2\text{in}\text{.}$.

Calculation:

The formula for the volume of the sphere is equal to $\frac{4}{3}\pi r^3$ . The rate of change of volume at $r=2\text{in}\text{.}$ is the instantaneous rate of change of volume function at $r=2\text{in}\text{.}$.

The formula for the instantaneous rate of change of volume with respect to radius at $r\text{in}\text{.}$ is equal to $\underset{h\to 0}{\lim }\frac{f\left(r+h\right)-f\left(r\right)}{h}$.

The formula for the instantaneous rate of change of $y=\frac{4}{3}\pi r^3\text{in}{\text{.}}^3$ at $r=2\text{in}\text{.}$ is equal to $\underset{h\to 0}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}$.

Substitute $2+h$ for $r$ in the function.

  $\begin{array}{c}f\left(r\right)=\frac{4}{3}\pi r^3\\ f\left(2+h\right)=\frac{4}{3}\pi {\left(2+h\right)}^3\\ f\left(2+h\right)=\frac{4}{3}\pi \left(8+h^3+6h^2+12h\right)\\ f\left(2+h\right)=\frac{4}{3}\pi \left(8+h^3+6h^2+12h\right)\end{array}$

Substitute $2$ for $r$ in the function.

  $\begin{array}{c}f\left(r\right)=\frac{4}{3}\pi r^3\\ f\left(2\right)=\frac{4}{3}\pi {\left(2\right)}^3\\ f\left(2\right)=\frac{4}{3}\pi \left(8\right)\\ f\left(2\right)=\frac{32}{3}\pi \end{array}$

Substitute $\frac{4}{3}\pi \left(8+h^3+6h^2+12h\right)$ for $f\left(2+h\right)$ and $\frac{32}{3}\pi$ for $f\left(2\right)$ in the formula.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}=\underset{h\to 0}{\lim }\frac{\frac{4}{3}\pi \left(8+h^3+6h^2+12h\right)-\frac{32}{3}\pi }{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{4}{3}\pi \left(h^3+6h^2+12h\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{4}{3}\pi h\left(h^2+6h+12\right)}{h}\\ =\frac{4}{3}\pi \underset{h\to 0}{\lim }\left(h^2+6h+12\right)\end{array}$

Further simplify.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}=\frac{4}{3}\pi \left(0^2+6\left(0\right)+12\right)\\ =16\pi \end{array}$

Therefore, the rate of change of volume of sphere with respect to radius when $r=2\text{in}\text{.}$ is equal to $16\text{in}{\text{.}}^{\text{3}}/\text{in}\text{.}$.