2015 H2 Maths Paper 1 Q11 — Worked Solution

Part (i)

3 marks

Since $x = \sin^{3}\theta$ and $y = 3\sin^{2}\theta\cos\theta$ ,

\frac{dx}{d\theta} = 3\sin^{2}\theta\cos\theta.

\frac{dy}{d\theta} = 3\left( 2\sin\theta\cos\theta \cdot \cos\theta + \sin^{2}\theta \cdot \left( - \sin\theta \right) \right)

= 6\sin\theta\cos^{2}\theta - 3\sin^{3}\theta.

By the chain rule,

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

= \frac{6\sin\theta\cos^{2}\theta - 3\sin^{3}\theta}{3\sin^{2}\theta\cos\theta}

= \frac{2\cos\theta}{\sin\theta} - \frac{\sin\theta}{\cos\theta}

= 2\cot\theta - \tan\theta.\ \ \left( \text{shown} \right)

Part (ii)

6 marks

Approach

A turning point occurs where $\frac{dy}{dx} = 0$ . We solve $2\cot\theta = \tan\theta$ for $\tan\theta$ directly, since the required form $\tan\theta = \sqrt{k}$ asks for it.

2\cot\theta = \tan\theta

\Rightarrow \frac{2\cos\theta}{\sin\theta} = \frac{\sin\theta}{\cos\theta}

\Rightarrow 2\cos^{2}\theta = \sin^{2}\theta.

Dividing both sides by $\cos^{2}\theta$ , which is non-zero since $0 \leq \theta \leq \frac{1}{2}\pi$ excludes $\theta = \frac{1}{2}\pi$ at a turning point (there $\frac{dx}{d\theta} = 0$ too, so it is not a genuine turning point of the curve):

2 = \tan^{2}\theta.

Since $0 \leq \theta \leq \frac{1}{2}\pi$ gives $\tan\theta \geq 0$ , we take the positive root.

$\therefore\tan\theta = \sqrt{2}$ , so $k = 2$ .

Approach

With $\tan\theta = \sqrt{2}$ , we read $\sin\theta$ and $\cos\theta$ off a right triangle with opposite $\sqrt{2}$ , adjacent $1$ , hypotenuse $\sqrt{3}$ , then substitute into $x$ and $y$ .

Since $\tan\theta = \sqrt{2}$ , $\sin\theta = \frac{\sqrt{2}}{\sqrt{3}}$ and $\cos\theta = \frac{1}{\sqrt{3}}$ .

x = \sin^{3}\theta = \left( \frac{\sqrt{2}}{\sqrt{3}} \right)^{3}

= \frac{2\sqrt{2}}{3\sqrt{3}}

= \frac{2\sqrt{6}}{9}.

y = 3\sin^{2}\theta\cos\theta = 3 \cdot \frac{2}{3} \cdot \frac{1}{\sqrt{3}}

= \frac{2}{\sqrt{3}}

= \frac{2\sqrt{3}}{3}.

$\therefore$ the turning point is $\left( \frac{2\sqrt{6}}{9},\ \frac{2\sqrt{3}}{3} \right)$ .

Explanation: $2\cot\theta$ and $- \tan\theta$ are both strictly decreasing on $\left( 0,\ \frac{1}{2}\pi \right)$ , so $\frac{dy}{dx} = 2\cot\theta - \tan\theta$ is strictly decreasing in $\theta$ , changing sign from positive to negative exactly once, at $\tan\theta = \sqrt{2}$ . Since $\frac{dx}{d\theta} = 3\sin^{2}\theta\cos\theta \geq 0$ throughout $0 \leq \theta \leq \frac{1}{2}\pi$ , $x$ increases as $\theta$ increases, so $\frac{dy}{dx}$ also changes from positive to negative as $x$ increases through this point.

$\therefore$ the turning point is a maximum.

Part (iii)

3 marks

Approach

As $\theta$ increases from $0$ to $\frac{1}{2}\pi$ , $x = \sin^{3}\theta$ increases monotonically from $0$ to $1$ (since $\frac{dx}{d\theta} \geq 0$ ), and $y \geq 0$ throughout, so the area under $C$ can be written as $\int y\, dx$ over this range and converted to an integral in $\theta$ .

\text{Area} = \int_{0}^{1}y\, dx = \int_{0}^{\frac{1}{2}\pi}y\,\frac{dx}{d\theta}\, d\theta

= \int_{0}^{\frac{1}{2}\pi}3\sin^{2}\theta\cos\theta \cdot 3\sin^{2}\theta\cos\theta\, d\theta

= \int_{0}^{\frac{1}{2}\pi}9\sin^{4}\theta\cos^{2}\theta\, d\theta.\ \ \left( \text{shown} \right)

By GC, Area $\text{=}0.884$ (3 d.p.).

Note: this integral has the exact closed form $\frac{9\pi}{32}$ , since $0.884 \approx \frac{9\pi}{32} = 0.8836\ldots$ ; the question only asks for the 3 d.p. calculator value.

Part (iv)

3 marks

Approach

Substitute $y = ax$ directly into the parametric equations of $C$ to find where the line meets the curve, then divide out the root $\theta = 0$ (which gives the origin) to isolate $\tan\theta$ .

At a point of intersection, $y = ax$ gives

3\sin^{2}\theta\cos\theta = a\sin^{3}\theta.

For $\theta \neq 0$ (which gives the origin), dividing both sides by $\sin^{2}\theta$ , which is non-zero:

3\cos\theta = a\sin\theta

\Rightarrow \tan\theta = \frac{3}{a}.\ \ \left( \text{shown} \right)

For the line to pass through the maximum point found in (ii), where $\tan\theta = \sqrt{2}$ :

\frac{3}{a} = \sqrt{2}

\Rightarrow a = \frac{3}{\sqrt{2}}

\Rightarrow a = \frac{3\sqrt{2}}{2}.

$\therefore a = \frac{3\sqrt{2}}{2}$ .

2015 H2 Maths Paper 1, Question 11

Part (i)

Part (ii)

Part (iii)

Part (iv)

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