2016 H2 Maths Paper 2 Q3 — Worked Solution

Part (i)

4 marks

Approach

Before sketching we find where the curve crosses the $x$ -axis, its maximum point, and how $x$ behaves as $t$ increases, since $D$ is given parametrically and is not a graph of a simple function.

$D$ meets the $x$ -axis where $y = 0$ :

y = 1 - \cos t = 0

\cos t = 1

t = 0\ \text{or}\ t = 2\pi\ \ (0 \leq t \leq 2\pi)

At $t = 0$ : $x = 0 - \cos 0 = - 1$ . At $t = 2\pi$ : $x = 2\pi - \cos 2\pi = 2\pi - 1$ .

$\therefore$ $D$ meets the $x$ -axis at $( - 1,\ 0)$ and $(2\pi - 1,\ 0)$ .

Since $y = 1 - \cos t$ and $- 1 \leq \cos t \leq 1$ , $y$ is greatest when $\cos t = - 1$ , i.e. at $t = \pi$ , giving $y = 2$ . At $t = \pi$ : $x = \pi - \cos\pi = \pi + 1$ .

$\therefore$ the maximum point is $(\pi + 1,\ 2)$ .

Note: $\frac{dx}{dt} = 1 + \sin t \geq 0$ for all $t$ , with equality only at $t = \frac{3\pi}{2}$ , so $x$ increases throughout $\lbrack 0,2\pi\rbrack$ and $D$ does not double back on itself. At $t = \frac{3\pi}{2}$ the curve has a vertical tangent, at the point $\left( \frac{3\pi}{2},\ 1 \right)$ .

Part (ii)

3 marks

Approach

The area under a parametric curve is $\int y\, dx = \int y\,\frac{dx}{dt}\, dt$ , taken over the corresponding range of $t$ .

\text{Area} = \int_{0}^{a}y\,\frac{dx}{dt}\, dt

= \int_{0}^{a}\left( 1 - \cos t \right)\left( 1 + \sin t \right)\, dt

Expanding the integrand:

\left( 1 - \cos t \right)\left( 1 + \sin t \right) = 1 + \sin t - \cos t - \sin t\cos t

So

\text{Area} = \int_{0}^{a}\left( 1 + \sin t - \cos t - \sin t\cos t \right)dt

= \left\lbrack t - \cos t - \sin t - \frac{1}{2}\sin^{2}t \right\rbrack_{0}^{a}

= \left( a - \cos a - \sin a - \frac{1}{2}\sin^{2}a \right) - (0 - 1 - 0 - 0)

= a + 1 - \sin a - \cos a - \frac{1}{2}\sin^{2}a

$\therefore$ Area $\text{=}a + 1 - \sin a - \cos a - \frac{1}{2}\sin^{2}a$ .

Part (iii)

4 marks

Approach

We find the point on $D$ at $t = \frac{\pi}{2}$ and the gradient of the tangent there using $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ , then take the negative reciprocal for the normal. Since $E$ and $F$ are the $x$ - and $y$ -intercepts of the normal, triangle $\text{OEF}$ is right-angled at $O$ , so its area is simply $\frac{1}{2}\,|OE|\,|OF|$ .

At $t = \frac{\pi}{2}$ : $x = \frac{\pi}{2} - \cos\frac{\pi}{2} = \frac{\pi}{2}$ , and $y = 1 - \cos\frac{\pi}{2} = 1$ . So the point is $\left( \frac{\pi}{2},\ 1 \right)$ .

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\sin t}{1 + \sin t}

At $t = \frac{\pi}{2}$ : $\frac{dy}{dx} = \frac{1}{1 + 1} = \frac{1}{2}$ .

Since the normal is perpendicular to the tangent, its gradient is $- 2$ .

Equation of the normal:

y - 1 = - 2\left( x - \frac{\pi}{2} \right)

At $E$ (on the $x$ -axis), $y = 0$ :

0 - 1 = - 2\left( x - \frac{\pi}{2} \right)

x = \frac{\pi}{2} + \frac{1}{2}

So $E = \left( \frac{\pi + 1}{2},\ 0 \right)$ .

At $F$ (on the $y$ -axis), $x = 0$ :

y - 1 = - 2\left( 0 - \frac{\pi}{2} \right)

y = 1 + \pi

So $F = (0,\ 1 + \pi)$ .

Since $\text{OEF}$ is right-angled at $O$ :

\text{Area of }\text{OEF} = \frac{1}{2}\,|OE|\,|OF|

= \frac{1}{2} \cdot \frac{\pi + 1}{2} \cdot (\pi + 1)

= \frac{(\pi + 1)^{2}}{4}

To check your answer: numerically

\frac{(\pi + 1)^{2}}{4} \approx 4.29

, and directly multiplying the intercepts

\frac{1}{2}(2.071)(4.142) \approx 4.29

. ✓

$\therefore$ the exact area of triangle $\text{OEF}$ is $\frac{(\pi + 1)^{2}}{4}\ \text{units}^{2}$ .

2016 H2 Maths Paper 2, Question 3

Part (i)

Part (ii)

Part (iii)

This is how we mark, and how we teach.