The curve shown in Figure 3 has parametric equations $x = 6 \, ext{sin} \, t$ $y = 5 \, ext{sin} \, 2t$ $0 \leq t \leq \frac{\pi}{2}$ The region $R$, shown shaded in Figure 3, is bounded by the curve and the x-axis - Edexcel - A-Level Maths Pure - Question 13 - 2020 - Paper 2

From Figure 4, the vertical height of the dam is given as 4.2 metres. From the parametric equation, we know:

$y = 5 \text{sin} \, 2t$

We find $t$ such that $y = 4.2$:

$4.2 = 5 \text{sin} \, 2t \implies \text{sin} \, 2t = \frac{4.2}{5} = 0.84$

Thus: $2t = \text{arcsin}(0.84)$

Calculating: $2t = 0.973$ So, $t = 0.4865$

Now, we can compute the width of the walkway:

For the horizontal distance at $t = 0.4865$:

$x = 6 \text{sin} \, t \implies x = 6 \text{sin}(0.4865) \approx 4.24\, ext{metres}$

The width of the walkway $MN$ is calculated using:

$\text{width} = x - \text{other side}\approx 4.24 - 0 \approx 4.24 \text{ metres}$