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

To integrate the area expression:

$A = \int_{0}^{\frac{\pi}{2}} 60 \text{sin} \, t \cos^{2} \, t \, dt$

We can use the substitution method. Let:

$u = \text{cos} \, t \, \Rightarrow \, du = -\text{sin} \, t \, dt$

Changing the limits:

• When $t = 0$, $u = 1$
• When $t = \frac{\pi}{2}$, $u = 0$

Then the integral becomes:

$A = 60 \int_{1}^{0} u^{2} (-du) = 60 \int_{0}^{1} u^{2} \, du$

Calculating this gives:

$A = 60 \cdot \left[ \frac{u^3}{3} \right]_{0}^{1} = 60 \cdot \frac{1}{3} = 20.$

Using the given parameters for the dam profile, we need to find values at specific points. The vertical wall of the dam is fixed at 4.2 metres. The essential aspect is the geometry of the curve defined in the problem.

At point $M$, we have:

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

Solving for $t$ gives:

$2t = \text{sin}^{-1} \left(\frac{4.2}{5}\right) \Rightarrow t = \frac{1}{2} \text{sin}^{-1} \left(\frac{4.2}{5}\right).$

Substituting back, you can find the corresponding $x$ value, which allows us to determine the width of walkway $MN$ by evaluating the curve from $O$ to $M$ (and also from $O$ to $N$ for total width).

Using geometry or trigonometry, compute:

• $MN = x_N - x_M$ where $x = 6 \text{sin} \, t$ gives the distance of the walkway.