Question

The volumeV of a sphere of radius r feet is$V=V\left(r\right)=\frac{4}{3}{\mathrm{\pi r}}^3$. Find the instantaneous rate of change of the volume with respect to the radius r at$r=2.$

Short answer

If the function for the volume of a sphere is$V=V\left(r\right)=\frac{4}{3}{\mathrm{\pi r}}^3$then the instantaneous rate of change of the volume with respect to the radius r at$r=2$will be$16\mathrm{\pi }\frac{{\mathrm{ft}}^3}{\mathrm{ft}}.$

Step-by-step solution

Step 1. Given information  

Function for the volume of a sphere is

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

Radius$r=2$

Step 2. The Volume of a sphere 

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

$$\begin{gathered} V\left(r\right)=\frac{4}{3}{\mathrm{\pi r}}^3 \\ V\left(2\right)=\frac{4}{3}\pi {\left(3\right)}^{3}V\left(2\right)=\frac{4\times 9}{3}\pi \\ V\left(2\right)=12\mathrm{\pi } \end{gathered}$$

So the volume of the sphere at the radius $r=2$is $12\mathrm{\pi }{\mathrm{ft}}^3$

Step 3. Instantaneous rate of change of volume

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

localid="1647033372487" $$\begin{gathered} V\text{'}\left(c\right)=\underset{r\to c}{\lim }\frac{V\left(r\right)-V\left(c\right)}{r-c} \\ V\text{'}\left(2\right)=\underset{r\to 2}{\lim }\frac{V\left(r\right)-V\left(2\right)}{r-2} \\ V\text{'}\left(2\right)=\underset{r\to 2}{\lim }\frac{\frac{4}{3}{\mathrm{\pi r}}^3-12\mathrm{\pi }}{r-2} \\ V\text{'}\left(2\right)=\underset{r\to 2}{\lim }\frac{\frac{4}{3}\mathrm{\pi }({\mathrm{r}}^3-8)}{r-2} \\ V\text{'}\left(2\right)=\underset{r\to 2}{\lim }\frac{\frac{4}{3}\mathrm{\pi }({\mathrm{r}}^2+2r+4)(\mathrm{r}-2)}{r-2} \\ V\text{'}\left(2\right)=\underset{r\to 2}{\lim }\frac{4}{3}\mathrm{\pi }({\mathrm{r}}^2+2r+4) \\ V\text{'}\left(2\right)=\frac{4}{3}\mathrm{\pi }(2^2+2(2)+4) \\ V\text{'}\left(2\right)=16\mathrm{\pi } \end{gathered}$$

So Instantaneous rate of change of volume islocalid="1647032540665" $16\mathrm{\pi }\frac{{\mathrm{ft}}^3}{\mathrm{ft}}$