A bicycle inner tube can be considered as a joined up cylinder of fixed length $200$ cm and radius $r$ 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 $30{\text{ c}}{{\text{m}}^3}{{\text{s}}^{ - 1}}$. Find the rate at which the radius of the inner tube is increasing when $r = 2{\text{ cm}}$.

$V = 200\pi {r^2}$     (A1)

Note:     Allow $V = \pi h{r^2}$ if value of $h$ is substituted later in the question.

EITHER

$\frac{{{\text{d}}V}}{{{\text{d}}t}} = 200\pi 2r\frac{{{\text{d}}r}}{{{\text{d}}t}}$     M1A1

Note:     Award M1 for an attempt at implicit differentiation.

at $r = 2$ we have $30 = 200\pi 4\frac{{{\text{d}}r}}{{{\text{d}}t}}$     M1

OR

$\frac{{{\text{d}}r}}{{{\text{d}}t}} = \frac{{\frac{{{\text{d}}V}}{{{\text{d}}t}}}}{{\frac{{{\text{d}}V}}{{{\text{d}}r}}}}$     M1

$\frac{{{\text{d}}V}}{{{\text{d}}r}} = 400\pi r$     M1

$r = 2$ we have $\frac{{{\text{d}}V}}{{{\text{d}}r}} = 800\pi$     A1

THEN

$\frac{{{\text{d}}r}}{{{\text{d}}t}} = \frac{{30}}{{800\pi }}\;\;\;\left( { = \frac{3}{{80\pi }} = 0.0119} \right){\text{ }}({\text{cm}}\,{{\text{s}}^{ - 1}}{\text{)}}$     A1

[5 marks]

This question was well understood and a large percentage appreciated the need for implicit differentiation although some candidates did not recognise the need to treat h as a constant till late in the question. A number of candidates found the answer $\frac{{3\pi }}{{80}}$ instead of $\frac{3}{{80\pi }}$ due to a basic incorrect use of the GDC.