2017 NJC PROMO Q12

An architect designs and constructs a hump such that when a square wheel of dimensions 2 cm by 2 cm is rolled over the hump, the height of its centre above the horizontal ground (represented by the \(x\)-axis) is always constant. To test his invention, he places a square wheel against the hump, as shown in the diagram below. The points \(R\), \(S\) and \(Q\) represent the centre of the square, the vertex of the square that is initially on the hump (at the origin \(O\)), and another vertex of the square which has a common side as \(S\), respectively.

The architect begins pushing the square wheel over the hump in the clockwise direction. He discovers that the centre of the square is always vertically above the variable point of contact between the square and the hump, \(P\), as shown in the diagram below. It is further given that the angle \(\theta\) is the angle between the side \(QS\) of the square and the horizontal line passing through \(S\) , measured in the positive \(x\)-direction. (\(\theta\) is taken to be positive if the vertex \(Q\) lies above the horizontal line passing through \(S\), and negative if otherwise.)

1. By considering the length of \(RP\) or otherwise, show that the \(y\)-coordinate of \(P\) can be expressed as \(\sqrt{2}-\sec \theta \), for \(-\frac{\pi }{4}\le \theta \le \frac{\pi }{4}\).[3]
2. State the gradient of \(QS\) in terms of \(\theta\) for \(-\frac{\pi }{4}\le \theta \le \frac{\pi }{4}\). Deduce that \(\frac{\mathrm{d}x}{\mathrm{d}\theta }=-\sec \theta \), where \(x\) is the \(x\)-coordinate of \(P\).[3]
3. Find the area bounded between the hump and the \(x\)-axes, correct to 4 decimal places.[3]