(a)     show that the rate of change of ${\rm{H}}\hat {\text{P}}{\text{Q}}$ is $0.16$ radians per second;

(b)     find the rate of change of PH.

(a)     let ${\text{H}}\hat {\text{P}}{\text{Q}} = \theta$

$\tan \theta  = \frac{h}{{40}}$

${\sec ^2}\theta \frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{1}{{40}}\frac{{{\text{d}}h}}{{{\text{d}}t}}$     M1

$\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{1}{{4{{\sec }^2}\theta }}$     (A1)

$= \frac{{16}}{{4 \times 25}}$   $\left( {\sec \theta  = \frac{5}{4}{\text{ or }}\theta  = 0.6435} \right)$     A1

$= 0.16$ radians per second     AG

(b)     ${x^2} = {h^2} + 1600$, where PH $= x$

$2x\frac{{{\text{d}}x}}{{{\text{d}}t}} = 2h\frac{{{\text{d}}h}}{{{\text{d}}t}}$     M1

$\frac{{{\text{d}}x}}{{{\text{d}}t}} = \frac{h}{x} \times 10$     A1

$= \frac{{10h}}{{\sqrt {{h^2} + 1600} }}$     (A1)

$h = 30$ , $\frac{{{\text{d}}x}}{{{\text{d}}t}} = 6$ ms^–1     A1

Note: Accept solutions that begin$x = 40\sec \theta$ or use $h = 10t$ .

[7 marks]

For those candidates who realized this was an applied calculus problem involving related rates of change, the main source of error was in differentiating inverse tan in part (a). Some found part (b) easier than part (a), involving a changing length rather than an angle. A number of alternative approaches were reported by examiners.