Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 6

At what rate is the volume increasing when the diameter is 80min?

Given: $\displaystyle \frac{dr}{dt} = 4mm/s$

Required: $\displaystyle \frac{dv}{dt}$ when $d = 80mm$

Solution: Let $V =\frac{4}{3} \pi r^3$ be the volume of the sphere

where $r$ = radius, diameter = 2 (radius); $\displaystyle r = \frac{d}{2} = \frac{80}{2} = 40$

It is also stated in the problem that $r$ is a constant so,


$
\begin{equation}
\begin{aligned}

\frac{dv}{dt} =& \frac{dv}{dr} \left( \frac{dr}{dt} \right) = \frac{4}{3} (3) \pi r^2 \left( \frac{dr}{dt} \right)
\\
\\
\frac{dv}{dt} =& 4 \pi r^2 \left( \frac{dr}{dt} \right)
\\
\\
\frac{dv}{dt} =& 4 \pi (40)^2(4)

\end{aligned}
\end{equation}
$


$\fbox{$\large \frac{dv}{dt} = 25600 \pi mm^3/s $}$