Answer
$\frac{3}{2\pi} $ ft/min
Work Step by Step
Let volume=V $ft^3/min$
Radius=r
Then
$V=\frac{4}{3} \pi r^3$
Taking derivative with respect to t
$\frac{dV}{dt}=\frac{d( \frac{4}{3} \pi r^3)}{dt}= \frac{4}{3} \pi \frac{d( r^3)}{dt}$
$\frac{dV}{dt}= \frac{4}{3} \pi \frac{d( r^3)}{dr} \frac{dr}{dt}= \frac{4}{3} \pi (3r^2) \frac{dr}{dt} =4\pi r^2 \frac{dr}{dt}$
$ \frac{dV}{dt}= 4\pi r^2 \frac{dr}{dt}$ ......................... eq (1)
Let z denote the diameter.Then
$z=2r$ Or $r=\frac{z}{2}$
Putting $r=\frac{z}{2}$, $\frac{dV}{dt}=3 $ $ft^3/min$, $r=1ft$ in equation (1)
$ 3= 4\pi \times1^2\times \frac{d ( \frac{z}{2}) }{dt}=2\pi \frac{dz}{dt} $
$\frac{dz}{dt}=\frac{3}{2\pi}$ ft/min
