2018 H2 Maths Paper 1 Q9 — Worked Solution

Part (i)

4 marks

Since $x = 2\theta - \sin 2\theta$ ,

\frac{dx}{d\theta} = 2 - 2\cos 2\theta

= 2\left( 1 - \cos 2\theta \right)

= 2\left( 2\sin^{2}\theta \right)

= 4\sin^{2}\theta,

using $1 - \cos 2\theta = 2\sin^{2}\theta$ . Since $y = 2\sin^{2}\theta$ ,

\frac{dy}{d\theta} = 4\sin\theta\cos\theta

= 2\sin 2\theta.

Then

\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}

= \frac{4\sin\theta\cos\theta}{4\sin^{2}\theta}

= \frac{\cos\theta}{\sin\theta}

= \cot\theta\ \ \left( \text{shown} \right)

Part (ii)

4 marks

Approach

The normal at $\theta = \alpha$ has gradient $- \tan\alpha$ , the negative reciprocal of the gradient $\cot\alpha$ found in (i). Writing its equation and setting $y = 0$ gives the $x$ -coordinate of $A$ directly in terms of $\alpha$ ; the point of the question is that the $\sin 2\alpha$ terms cancel, leaving the clean multiple $k\alpha$ .

At $\theta = \alpha$ , the point on $C$ is $\left( 2\alpha - \sin 2\alpha,\ 2\sin^{2}\alpha \right)$ , and by (i) the gradient of $C$ there is $\cot\alpha$ . Since the normal is perpendicular to the curve, its gradient is $- \tan\alpha$ . The normal is

y - 2\sin^{2}\alpha = - \tan\alpha\left\lbrack x - \left( 2\alpha - \sin 2\alpha \right) \right\rbrack.

Setting $y = 0$ to find where the normal meets the $x$ -axis,

- 2\sin^{2}\alpha = - \tan\alpha\left\lbrack x - \left( 2\alpha - \sin 2\alpha \right) \right\rbrack

2\sin^{2}\alpha = \tan\alpha\left\lbrack x - \left( 2\alpha - \sin 2\alpha \right) \right\rbrack

x - \left( 2\alpha - \sin 2\alpha \right) = 2\sin^{2}\alpha\cot\alpha

x - \left( 2\alpha - \sin 2\alpha \right) = 2\sin\alpha\cos\alpha

x - \left( 2\alpha - \sin 2\alpha \right) = \sin 2\alpha.

Therefore

x = 2\alpha - \sin 2\alpha + \sin 2\alpha

x = 2\alpha.

To check your answer: the

\sin 2\alpha

terms cancel exactly, leaving a linear multiple of

\alpha

alone, consistent with the question stating the answer has the form

k\alpha

. ✓

$\therefore$ the $x$ -coordinate of $A$ is $2\alpha$ , so $k = 2$ .

Part (iii)

5 marks

Approach

The arc-length integrand looks unpleasant until $\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2}$ is expanded: both terms carry a common factor of $16\sin^{2}\theta$ , and $\sin^{2}\theta + \cos^{2}\theta = 1$ collapses the sum under the square root to a single power of $\sin\theta$ , which integrates directly — no calculator needed.

From (i), $\frac{dx}{d\theta} = 4\sin^{2}\theta$ and $\frac{dy}{d\theta} = 4\sin\theta\cos\theta$ . Then

\left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2}

= 16\sin^{4}\theta + 16\sin^{2}\theta\cos^{2}\theta

= 16\sin^{2}\theta\left( \sin^{2}\theta + \cos^{2}\theta \right)

= 16\sin^{2}\theta.

Since $0 \leq \theta \leq \pi$ , $\sin\theta \geq 0$ , so $\sqrt{16\sin^{2}\theta} = 4\sin\theta$ . The total length of $C$ is

\int_{0}^{\pi}4\sin\theta\, d\theta

= \left\lbrack - 4\cos\theta \right\rbrack_{0}^{\pi}

= \left( - 4\cos\pi \right) - \left( - 4\cos 0 \right)

= 4 + 4

= 8.

$\therefore$ the total length of $C$ is $8$ units.

2018 H2 Maths Paper 1, Question 9

Part (i)

Part (ii)

Part (iii)

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