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.