A bicycle cylindrical inner tube has a fixed length of $200\text{ cm}$ and radius $r\text{ cm}$. The radius $r$ increases as the inner tube is pumped up. Air is being pumped into the inner tube so that the volume of air in the tube increases at a constant rate of $30\text{ cm}^3\text{s}^{-1}$.

1. Calculate the total volume of air pumped into the tube after one minute.

Time $t = 1\text{ minute} = 60\text{ s}$.
Volume $V = 30 \times 60 = 1800\text{ cm}^3$.

2. Use the volume formula for a cylinder to find the radius.

The volume of the cylindrical tube is $V = \pi r^2 h$.
Given $h = 200\text{ cm}$, we have:
$200\pi r^2 = 1800$
$\pi r^2 = 9$
$r^2 = \frac{9}{\pi}$
$r = \sqrt{\frac{9}{\pi}} \approx 1.6925 \approx 1.69\text{ cm}$ (to 3 s.f.)

1. Express the volume of the inner tube in terms of $r$

The inner tube is a cylinder with fixed length $h = 200\text{ cm}$.
$V = \pi r^2 h = 200\pi r^2$

2. Differentiate $V$ with respect to $r$

$\frac{\text{d}V}{\text{d}r} = 400\pi r$

3. Use the chain rule to find $\frac{\text{d}r}{\text{d}t}$ when $r = 2$

We are given $\frac{\text{d}V}{\text{d}t} = 30$.
By the chain rule, $\frac{\text{d}V}{\text{d}t} = \frac{\text{d}V}{\text{d}r} \times \frac{\text{d}r}{\text{d}t}$
$30 = (400\pi r) \times \frac{\text{d}r}{\text{d}t}$
When $r = 2$:
$30 = 400\pi(2) \times \frac{\text{d}r}{\text{d}t}$
$30 = 800\pi \times \frac{\text{d}r}{\text{d}t}$
$\frac{\text{d}r}{\text{d}t} = \frac{30}{800\pi} = \frac{3}{80\pi} \approx 0.0119\text{ cm s}^{-1}$