Sand is being poured to form a cone of height \(h\) cm and base radius \(r\) cm. The height remains equal to the base radius at all times. The height of the cone is increasing at a rate of \(0.5{\text{ cm}}\,{\text{mi}}{{\text{n}}^{ - 1}}\).

Find the rate at which sand is being poured, in \({\text{c}}{{\text{m}}^3}\,{\text{mi}}{{\text{n}}^{ - 1}}\), when the height is 4 cm.

METHOD 1

volume of a cone is \(V = \frac{1}{3}\pi {r^2}h\)

given \(h = r,{\text{ }}V = \frac{1}{3}\pi {h^3}\)     M1

\(\frac{{{\text{d}}V}}{{{\text{d}}h}} = \pi {h^2}\)     (A1)

when \(h = 4,{\text{ }}\frac{{{\text{d}}V}}{{{\text{d}}t}} = \pi  \times {4^2} \times 0.5{\text{ (using }}\frac{{{\text{d}}V}}{{{\text{d}}t}} = \frac{{{\text{d}}V}}{{{\text{d}}h}} \times \frac{{{\text{d}}h}}{{{\text{d}}t}})\)     M1A1

\(\frac{{{\text{d}}V}}{{{\text{d}}t}} = 8\pi {\text{ }}( = 25.1){\text{ (c}}{{\text{m}}^3}\,{\text{mi}}{{\text{n}}^{ - 1}})\)     A1

METHOD 2

volume of a cone is \(V = \frac{1}{3}\pi {r^2}h\)

given \(h = r,{\text{ }}V = \frac{1}{3}\pi {h^3}\)     M1

\(\frac{{{\text{d}}V}}{{{\text{d}}t}} = \frac{1}{3}\pi  \times 3{h^2} \times \frac{{{\text{d}}h}}{{{\text{d}}t}}\)     A1

when \(h = 4,{\text{ }}\frac{{{\text{d}}V}}{{{\text{d}}t}} = \pi  \times {4^2} \times 0.5\)     M1A1

\(\frac{{{\text{d}}V}}{{{\text{d}}t}} = 8\pi {\text{ }}( = 25.1){\text{ (c}}{{\text{m}}^3}\,{\text{mi}}{{\text{n}}^{ - 1}})\)     A1

METHOD 3

\(V = \frac{1}{3}\pi {r^2}h\)

\(\frac{{{\text{d}}V}}{{{\text{d}}t}} = \frac{1}{3}\pi \left( {2rh\frac{{{\text{d}}r}}{{{\text{d}}t}} + {r^2}\frac{{{\text{d}}h}}{{{\text{d}}t}}} \right)\)     M1A1

Note:     Award M1 for attempted implicit differentiation and A1 for each correct term on the RHS.

when \(h = 4,{\text{ }}r = 4,{\text{ }}\frac{{{\text{d}}V}}{{{\text{d}}t}} = \frac{1}{3}\pi \left( {2 \times 4 \times 4 \times 0.5 + {4^2} \times 0.5} \right)\)     M1A1

\(\frac{{{\text{d}}V}}{{{\text{d}}t}} = 8\pi {\text{ }}( = 25.1){\text{ (c}}{{\text{m}}^3}\,{\text{mi}}{{\text{n}}^{ - 1}})\)     A1

[5 marks]