2019 H2 Maths Paper 1 Q10 — Worked Solution

Part (i)

2 marks

We first locate the cusp and the extreme points of $C$ , so the sketch is accurate.

At $\theta = 0$ : $\frac{dx}{d\theta} = 2a\left( \sin 2\theta - \sin\theta \right) = 0$ and $\frac{dy}{d\theta} = 2a\left( \cos\theta - \cos 2\theta \right) = 0$ , so $C$ has a cusp there, at $(a,0)$ .

At $\theta = \pi$ : $x = a\left( 2\cos\pi - \cos 2\pi \right) = a( - 2 - 1) = - 3a$ , $y = 0$ , giving the point $( - 3a,0)$ .

Setting $\frac{dy}{d\theta} = 0$ for the turning points off the axis,

\cos\theta = \cos 2\theta

= 2\cos^{2}\theta - 1,

2\cos^{2}\theta - \cos\theta - 1 = 0

\left( 2\cos\theta + 1 \right)\left( \cos\theta - 1 \right) = 0.

So $\cos\theta = 1$ (the cusp, already found) or $\cos\theta = - \frac{1}{2}$ , giving $\theta = \frac{2\pi}{3}$ or $\theta = \frac{4\pi}{3}$ . At $\theta = \frac{2\pi}{3}$ ,

x = a\left( 2\left( - \frac{1}{2} \right) - \left( - \frac{1}{2} \right) \right) = - \frac{1}{2}a,\quad\quad y = a\left( 2 \cdot \frac{\sqrt{3}}{2} - \left( - \frac{\sqrt{3}}{2} \right) \right) = \frac{3\sqrt{3}}{2}a,

giving the topmost point $\left( - \frac{a}{2},\ \frac{3\sqrt{3}}{2}a \right)$ , with the mirror point $\left( - \frac{a}{2},\ - \frac{3\sqrt{3}}{2}a \right)$ at $\theta = \frac{4\pi}{3}$ below the $x$ -axis.

$C$ is a cardioid, symmetric about the $x$ -axis (shown algebraically in (ii)). Its line of symmetry has Cartesian equation

y = 0.

Part (ii)

2 marks

Approach

$C$ meets the $x$ -axis where $y = 0$ ; factorising the resulting equation in $\theta$ gives all the parameter values at once.

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

a\left( 2\sin\theta - \sin 2\theta \right) = 0

2\sin\theta - 2\sin\theta\cos\theta = 0

2\sin\theta\left( 1 - \cos\theta \right) = 0.

So $\sin\theta = 0$ or $\cos\theta = 1$ . For $0 \leq \theta \leq 2\pi$ , $\sin\theta = 0$ gives $\theta = 0,\ \pi,\ 2\pi$ , and $\cos\theta = 1$ gives $\theta = 0,\ 2\pi$ (already included).

$\therefore$ $C$ meets the $x$ -axis at $\theta = 0,\ \pi,\ 2\pi$ .

Note: $\theta = 0$ and $\theta = 2\pi$ both give the same point $(a,0)$ , the cusp where the curve closes up; $\theta = \pi$ gives the distinct point $( - 3a,0)$ .

Part (iii)

3 marks

Approach

For $0 \leq \theta \leq \pi$ , $y \geq 0$ traces the upper half of $C$ from the cusp $(a,0)$ back to $( - 3a,0)$ , with $x$ decreasing throughout (since the curve is traversed leftwards along the top). The area under this arc is $- \int y\,\frac{dx}{d\theta}\, d\theta$ , the minus sign compensating for $x$ decreasing as $\theta$ increases.

By symmetry about the $x$ -axis, the upper half of $C$ (where $y \geq 0$ ) is traced for $0 \leq \theta \leq \pi$ , from $\theta_{1} = 0$ to $\theta_{2} = \pi$ . Differentiating,

\frac{dx}{d\theta} = a\left( - 2\sin\theta + 2\sin 2\theta \right) = 2a\left( \sin 2\theta - \sin\theta \right).

Since $x$ decreases as $\theta$ runs from $0$ to $\pi$ , the area enclosed by the $x$ -axis and this arc is

\text{Area} = - \int_{0}^{\pi}y\,\frac{dx}{d\theta}\, d\theta.

Now

y\,\frac{dx}{d\theta} = a\left( 2\sin\theta - \sin 2\theta \right) \cdot 2a\left( \sin 2\theta - \sin\theta \right)

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

= 2a^{2}\left( 3\sin\theta\sin 2\theta - 2\sin^{2}\theta - \sin^{2}2\theta \right)

= a^{2}\left( 6\sin\theta\sin 2\theta - 4\sin^{2}\theta - 2\sin^{2}2\theta \right).

So

\text{Area} = - \int_{0}^{\pi}a^{2}\left( 6\sin\theta\sin 2\theta - 4\sin^{2}\theta - 2\sin^{2}2\theta \right)\, d\theta

= \int_{\theta_{1}}^{\theta_{2}}a^{2}\left( 4\sin^{2}\theta - 6\sin\theta\sin 2\theta + 2\sin^{2}2\theta \right)\, d\theta,

with $\theta_{1} = 0$ and $\theta_{2} = \pi$ . $\ \$ (shown)

Part (iv)

5 marks

Approach

Each term of the integrand reduces with a standard identity: $\sin^{2}\theta$ and $\sin^{2}2\theta$ via the double-angle cosine formula, and $\sin\theta\sin 2\theta$ via the product-to-sum formula. The total area is then twice the area of the upper half, by the symmetry found in (i).

Using $\sin^{2}\theta = \frac{1}{2}\left( 1 - \cos 2\theta \right)$ and $\sin^{2}2\theta = \frac{1}{2}\left( 1 - \cos 4\theta \right)$ ,

\int_{0}^{\pi}4\sin^{2}\theta\, d\theta = \int_{0}^{\pi}2\left( 1 - \cos 2\theta \right)\, d\theta = 2\pi,

\int_{0}^{\pi}2\sin^{2}2\theta\, d\theta = \int_{0}^{\pi}\left( 1 - \cos 4\theta \right)\, d\theta = \pi.

Using $\sin\theta\sin 2\theta = \frac{1}{2}\left( \cos\theta - \cos 3\theta \right)$ ,

\int_{0}^{\pi}6\sin\theta\sin 2\theta\, d\theta = \int_{0}^{\pi}3\left( \cos\theta - \cos 3\theta \right)\, d\theta

= 3\left\lbrack \sin\theta - \frac{1}{3}\sin 3\theta \right\rbrack_{0}^{\pi}

= 0.

So the area of the upper half is

\int_{0}^{\pi}a^{2}\left( 4\sin^{2}\theta - 6\sin\theta\sin 2\theta + 2\sin^{2}2\theta \right)\, d\theta = a^{2}(2\pi - 0 + \pi) = 3\pi a^{2}.

By the symmetry of $C$ about the $x$ -axis, the lower half encloses the same area, so the total area enclosed by $C$ is

2 \times 3\pi a^{2} = 6\pi a^{2}.

To check your answer: a numerical shoelace computation on

1000\text{+}

plotted points of

C

with

a = 1

gives area

\text{≈}18.8496

, matching

6\pi \approx 18.8496

. ✓

$\therefore$ the total area enclosed by $C$ is $6\pi a^{2}$ .

2019 H2 Maths Paper 1, Question 10

Part (i)

Part (ii)

Part (iii)

Part (iv)

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