Question

The surface area S of a sphere of radius r feet is$S=S\left(r\right)=4{\mathrm{\pi r}}^2$. Find the instantaneous rate of change of the surface area with respect to the radius r at$r=2$.

Short answer

If the Function for the surface area of a sphere is $S=S\left(r\right)=4{\mathrm{\pi r}}^2$then the instantaneous rate of change of the surface area with respect to the radius r at$2$will be$16\mathrm{\pi }\frac{{\mathrm{ft}}^2}{\mathrm{ft}}$

Step-by-step solution

Step 1. Given information 

Function for the surface area of a sphere is

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

Radius$r=2$

Step 2. The surface area of a sphere

Substitute $r=2$in the function for the surface area of a sphere

$$\begin{gathered} S\left(r\right)=4{\mathrm{\pi r}}^2 \\ S\left(2\right)=4\pi {\left(2\right)}^{2}S\left(2\right)=16\pi \end{gathered}$$

So surface area of the sphere at radius$r=2$is$16\mathrm{\pi }{\mathrm{ft}}^2$

Step 3. Instantaneous rate of change of surface area

Use instantaneous rate of change formula with respect r at $2$

localid="1647033413018" $$\begin{gathered} S\text{'}\left(c\right)=\underset{r\to c}{\lim }\frac{S\left(r\right)-S\left(c\right)}{r-c} \\ S\text{'}\left(2\right)=\underset{r\to 2}{\lim }\frac{S\left(r\right)-S\left(2\right)}{r-2} \\ S\text{'}\left(2\right)=\underset{r\to 2}{\lim }\frac{4{\mathrm{\pi r}}^2-16\mathrm{\pi }}{r-2} \\ S\text{'}\left(2\right)=\underset{r\to 2}{\lim }\frac{4\mathrm{\pi }({\mathrm{r}}^2-2^2)}{r-2} \\ S\text{'}\left(2\right)=\underset{r\to 2}{\lim }\frac{4\mathrm{\pi }(\mathrm{r}+2)(\mathrm{r}-2)}{r-2} \\ S\text{'}\left(2\right)=\underset{r\to 2}{\lim }4\mathrm{\pi }(\mathrm{r}+2) \\ S\text{'}\left(2\right)=4\mathrm{\pi }(2+2) \\ S\text{'}\left(2\right)=16\mathrm{\pi } \end{gathered}$$

So Instantaneous rate of change of surface area is$16\mathrm{\pi }\frac{{\mathrm{ft}}^2}{\mathrm{ft}}$