Chapter 14.4, Problem 45AYU

To determine

To find: The instantaneous rate of change of the volume of sphere with respect to the radius $r$ at $r=2$.

Answer to Problem 45AYU

Solution:

$16\pi$

Explanation of Solution

Given:

Volume of sphere: $V=V\left(r\right)=\frac{4}{3}\pi r^3$.

$r=2$

Formula Used:

The instantaneous rate of change of volume $V$ of sphere with respect to radius $r$ is given by $\frac{dV}{dr}$.

Calculation:

Here, $V=V\left(r\right)=\frac{4}{3}\pi r^3$.

$V\text{'}\left(r\right)=\frac{dV}{dr}=\frac{d}{dr}\left(\frac{4}{3}\pi r^3\right)$

$V\text{'}\left(r\right)=\frac{4}{3}\pi \times \frac{d}{dr}\left(r^3\right)$

$V\text{'}\left(r\right)=\frac{4}{3}\pi \times 3r^2$

$V\text{'}\left(r\right)=4\pi r^2$

At $r=2$, we get;

$V^{\prime }\left(2\right)=4\pi .\left(2^2\right)$

$V\text{'}\left(2\right)=16\pi$