Chapter 14.4, Problem 44AYU

To determine

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

Answer to Problem 44AYU

Solution:

$16\pi$

Explanation of Solution

Given:

Surface area of sphere: $S=S\left(r\right)=4\pi r^2$.

$r=2$

Formula Used:

The instantaneous rate of change of surface area $S$ of sphere with respect to radius $r$ is given by $\frac{dS}{dr}$.

Calculation:

Here, $S=S\left(r\right)=4\pi r^2$.

$S^{\prime }\left(r\right)=\frac{dS}{dr}=\frac{d}{dr}\left(4\pi r^2\right)$

$S^{\prime }\left(r\right)=4\pi \times \frac{d}{dr}\left(r^2\right)$

$S^{\prime }\left(r\right)=4\pi \times 2r$

$S^{\prime }\left(r\right)=8\pi r$

At $r=2$, we get;

$S^{\prime }\left(2\right)=8\pi \times 2$

$S^{\prime }\left(2\right)=16\pi$