[Cambridge/1998]
A spherical balloon is being inflated and at the instant when its radius is 3 m, its surface area is increasing at the rate of 2 m$^2$s$^{-1}$. Find the rate of increase, at the same instant of, [The formulae for the surface area and the volume of a sphere are $A = 4\pi r^2$ and $V = \frac{4}{3}\pi r^3$.]

1. Differentiate the given surface area formula with respect to the radius $r$.

$A = 4\pi r^2$
$\frac{dA}{dr} = 8\pi r$

2. Set up the connected rates of change equation using the chain rule.

$\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}$

3. Substitute the given values to calculate the rate of increase of the radius.

Substitute $r = 3$ and $\frac{dA}{dt} = 2$:
$2 = 8\pi (3) \times \frac{dr}{dt}$
$2 = 24\pi \frac{dr}{dt}$
$\frac{dr}{dt} = \frac{2}{24\pi} = \frac{1}{12\pi}$

1. Establish the rate of change of the radius at $r = 3$ (as would be found in part i).

Given $A = 4\pi r^2$, we have $\frac{dA}{dr} = 8\pi r$.
By the chain rule, $\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}$.
$2 = 8\pi(3) \times \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{12\pi}$

2. Differentiate the volume formula with respect to the radius.

$V = \frac{4}{3}\pi r^3 \implies \frac{dV}{dr} = 4\pi r^2$

3. Apply the chain rule to find the rate of change of the volume when $r = 3$.

$\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}$
$\frac{dV}{dt} = 4\pi(3^2) \times \frac{1}{12\pi}$
$\frac{dV}{dt} = 36\pi \times \frac{1}{12\pi} = 3$