曲线 $C$ 由参数方程定义：
$x = 1 - \cos t, \quad y = \sin 2t, \quad \text{where } 0 \le t \le \frac{\pi}{2}.$

1. Find the values of the parameter $t$ where the curve intersects the $x$-axis.

At the $x$-axis, the $y$-coordinate is $0$:
$y = 0$
$\sin 2t = 0$
Given the domain $0 \le t \le \frac{\pi}{2}$, the possible range for $2t$ is $0 \le 2t \le \pi$.
$2t = 0 \text{ or } \pi$
$t = 0 \text{ or } \frac{\pi}{2}$

2. Calculate the exact $x$-coordinates for the corresponding values of $t$.

Substitute the values of $t$ into the $x$-equation:
When $t = 0$:
$x = 1 - \cos(0) = 1 - 1 = 0$
When $t = \frac{\pi}{2}$:
$x = 1 - \cos\left(\frac{\pi}{2}\right) = 1 - 0 = 1$
The exact coordinates of the points where $C$ meets the $x$-axis are $(0, 0)$ and $(1, 0)$.

3. Describe the sketch of the curve $C$ within the given domain.

Using a graphing calculator in parametric mode for $0 \le t \le \frac{\pi}{2}$:

• The curve starts at the origin $(0,0)$ with a vertical tangent (since $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} = 2$ at $t=0$).
• It arcs smoothly in the first quadrant, reaching a maximum $y$-value of $1$ (at $t = \frac{\pi}{4}$, $x = 1 - \frac{\sqrt{2}}{2}$).
• It terminates at $(1,0)$ with a non-zero gradient.

(The student should draw this resulting arch shape, clearly labelling the exact coordinate points $(0,0)$ and $(1,0)$ as found above).

Common mistakes typical students make: Sketching the curve beyond the restricted domain $t = \frac{\pi}{2}$, or drawing a purely symmetrical parabola-like shape (the true curve is skewed to the right).

1. Find the derivatives of $x$ and $y$ with respect to $t$.

$\frac{dx}{dt} = \sin t$
$\frac{dy}{dt} = 2\cos 2t$

2. Evaluate the gradient of the normal at $t = \frac{\pi}{2}$.

At $t = \frac{\pi}{2}$:
$\frac{dx}{dt} = \sin\left(\frac{\pi}{2}\right) = 1$
$\frac{dy}{dt} = 2\cos(\pi) = -2$
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2}{1} = -2$
Gradient of normal $m_n = -\frac{1}{dy/dx} = \frac{1}{2}$

3. Find the coordinates of the point on the curve when $t = \frac{\pi}{2}$.

At $t = \frac{\pi}{2}$:
$x = 1 - \cos\left(\frac{\pi}{2}\right) = 1 - 0 = 1$
$y = \sin\left(2 \cdot \frac{\pi}{2}\right) = \sin(\pi) = 0$
The coordinates of the point are $(1, 0)$.

4. Form the equation of the normal and find its $y$-intercept.

Equation of the normal:
$y - 0 = \frac{1}{2}(x - 1)$
$y = \frac{1}{2}x - \frac{1}{2}$
When the normal cuts the $y$-axis, $x = 0$:
$y = \frac{1}{2}(0) - \frac{1}{2} = -\frac{1}{2}$ (shown)

1. Set up the integral for the total area bounded by $C$, the normal, and the $y$-axis.

From part (b), the normal intersects the $y$-axis at $D(0, -\frac{1}{2})$ and the $x$-axis at the point where $t = \frac{\pi}{2}$, which is $(1, 0)$.
The bounded region is bounded above by $C$ (from $x=0$ to $x=1$), below by the normal, and on the left by the $y$-axis.
We can split the total area into the area under $C$ (above the $x$-axis) and the area of the triangle formed by the normal and the axes (below the $x$-axis).
$\text{Total Area} = \int_{0}^{1} y_C \, dx + \text{Area of triangle}$

2. Formulate and evaluate the parametric integral for the area under the curve $C$.

For the curve $C$, $x = 1 - \cos t \implies dx = \sin t \, dt$.
When $x = 0$, $1 - \cos t = 0 \implies \cos t = 1 \implies t = 0$.
When $x = 1$, $1 - \cos t = 1 \implies \cos t = 0 \implies t = \frac{\pi}{2}$.
Substituting these into the area integral:
$\int_{0}^{1} y_C \, dx = \int_{0}^{\frac{\pi}{2}} \sin 2t (\sin t) \, dt$
$= \int_{0}^{\frac{\pi}{2}} (2 \sin t \cos t) \sin t \, dt$
$= \int_{0}^{\frac{\pi}{2}} 2 \sin^2 t \cos t \, dt$
Integrating with respect to $\sin t$:
$= \left[ \frac{2}{3} \sin^3 t \right]_{0}^{\frac{\pi}{2}}$
$= \frac{2}{3}(1)^3 - \frac{2}{3}(0)^3 = \frac{2}{3}$

3. Calculate the area of the triangle and find the total exact area.

The region bounded by the normal, the $x$-axis, and the $y$-axis in the fourth quadrant forms a triangle with vertices $(0,0)$, $(1,0)$, and $D(0, -\frac{1}{2})$.
$\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (1) \left(\frac{1}{2}\right) = \frac{1}{4}$
(Note: This can also be found via integration: $-\int_{0}^{1} \left(\frac{1}{2}x - \frac{1}{2}\right) dx = \frac{1}{4}$)

Adding both areas together gives the exact total area:
$\text{Total Area} = \frac{2}{3} + \frac{1}{4} = \frac{8}{12} + \frac{3}{12} = \frac{11}{12}$