Oil is dripping into an empty inverted right circular cone at the rate of 16 cm3 per second. The height, h cm, of the cone is three-quarters its radius, r cm. [Volume of a right circular cone = $\tfrac{1}{3} \times \pi \times r^2 \times h$]

1. Express the height in terms of the radius

Given that the height is three-quarters of the radius, we have:
$h = \frac{3}{4}r$

2. Substitute the expression for height into the volume formula and simplify

$V = \frac{1}{3}\pi r^2 h$
$V = \frac{1}{3}\pi r^2 \left(\frac{3}{4}r\right)$
$V = \frac{1}{4}\pi r^3$ (shown)

1. Identify the given rate of change and the required rate of change.

Given: $\frac{dV}{dt} = 16$ cm$^3$/s.
Required: $\frac{dr}{dt}$ when $r = 8$ cm.

2. Differentiate the volume formula with respect to $r$.

From part (a), $V = \frac{1}{4}\pi r^3$.
$\frac{dV}{dr} = \frac{3}{4}\pi r^2$

3. Set up the chain rule equation connecting the rates.

By the chain rule:
$\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}$

4. Substitute the known values to find the rate of change of the radius.

Substitute $\frac{dV}{dt} = 16$ and $r = 8$:
$16 = \left( \frac{3}{4}\pi (8)^2 \right) \times \frac{dr}{dt}$
$16 = \left( \frac{3}{4}\pi (64) \right) \times \frac{dr}{dt}$
$16 = 48\pi \times \frac{dr}{dt}$
$\frac{dr}{dt} = \frac{16}{48\pi} = \frac{1}{3\pi}$ cm/s