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 14 - 2020 - Paper 2

To find the area, we need to evaluate the integral:

$\int 60 \text{sin} \; t \text{cos}^2 \; t \, dt$

Using the identity for $\text{cos}^2 \; t$, we can write: $\text{cos}^2 \; t = \frac{1 + \text{cos} \; 2t}{2}$

Thus, the integral becomes: $\int 60 \text{sin} \; t \left( \frac{1 + \text{cos} \; 2t}{2} \right) \, dt = 30 \int \text{sin} \; t \, dt + 30 \int \text{sin} \; t \text{cos} \; 2t \, dt$

Evaluating these integrals gives:

• For the first integral: $\int \text{sin} \; t \, dt = -\text{cos} \; t$
• For the second integral, use integration by parts or substitution methods to solve.

After evaluating the definite integral from $0$ to $\frac{\pi}{2}$, combine the results to find that the area $A = 20$.

In Figure 4, we are given that the vertical wall of the dam is 4.2 meters high. Taking into account the parametric equations, we equate the height at point $M$ to:

$y = 5 \text{sin} \; 2t = 4.2$

Solving for $t$, we first isolate $\text{sin} \; 2t$: $\text{sin} \; 2t = \frac{4.2}{5} = 0.84$

Thus, $2t = \text{sin}^{-1}(0.84)$

Consequently, the angle can be calculated: $t = \frac{1}{2} \text{sin}^{-1}(0.84)$

Now, substituting back into the parametric equations to determine $x$ and ultimately the width $MN$: $x = 6 \text{sin} \; t$

Within the parameters gives: $MN = \text{width} = 6 \text{sin} \; \left( \frac{1}{2} \text{sin}^{-1}(0.84) \right)$

Calculating this will yield the desired width of the walkway.