The diagram below shows the boundary of the cross-section of a water channel.

The equation that represents this boundary is $y = 16\sec \left( {\frac{{\pi x}}{{36}}} \right) - 32$ where x and y are both measured in cm.

The top of the channel is level with the ground and has a width of 24 cm. The maximum depth of the channel is 16 cm.

Find the width of the water surface in the channel when the water depth is 10 cm.

Give your answer in the form $a\arccos b$ where $a,{\text{ }}b \in \mathbb{R}$ .

10 cm water depth corresponds to $16\sec \left( {\frac{{\pi x}}{{36}}} \right) - 32 = - 6$     (A1)

Rearranging to obtain an equation of the form $\sec \left( {\frac{{\pi x}}{{36}}} \right) = k$ or equivalent

i.e. making a trignometrical function the subject of the equation.     M1

$\cos \left( {\frac{{\pi x}}{{36}}} \right) = \frac{8}{{13}}$     (A1)

$\frac{{\pi x}}{{36}} = \pm \arccos \frac{8}{{13}}$     M1

$x = \pm \frac{{36}}{\pi }\arccos \frac{8}{{13}}$     A1

Note: Do not penalise the omission of ±.

Width of water surface is $\frac{{72}}{\pi }\arccos \frac{8}{{13}}{\text{ (cm)}}$     R1     N1

Note: Candidate who starts with 10 instead of −6 has the potential to gain the two M1 marks and the R1 mark.

[6 marks]

This was a question in context which proved difficult for many candidates. Many appeared not to have fully comprehended the implications of the details of the diagram. A few candidates attempted integration, for no apparent reason.