| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2023 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Tangent parallel/perpendicular to initial line |
| Difficulty | Challenging +1.8 This is a challenging Further Maths polar coordinates question requiring multiple sophisticated techniques: deriving the t-formula for tan θ, finding tangent conditions in polar form (involving dr/dθ), solving a transcendental equation using the substitution from part (a), and computing a polar area integral. The multi-step reasoning and integration of several advanced topics places it well above average difficulty, though the individual techniques are standard for Further Maths students. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| a) | METHOD 1: | |
| \(\tan\theta = \tan\left(\frac{\theta}{2} + \frac{\theta}{2}\right)\) | ||
| \(\tan\left(\frac{\theta}{2} + \frac{\theta}{2}\right) = \frac{\tan\left(\frac{\theta}{2}\right) + \tan\left(\frac{\theta}{2}\right)}{1 - \tan\left(\frac{\theta}{2}\right)\tan\left(\frac{\theta}{2}\right)}\) | ||
| \(\tan\left(\frac{\theta}{2} + \frac{\theta}{2}\right) = \frac{t + t}{1 - t \times t}\) | E1 | Convincing |
| \(\tan\theta = \frac{2t}{1 - t^2}\) | ||
| METHOD 2: | ||
| For \(t = \tan\left(\frac{\theta}{2}\right)\): | ||
| \(\sin\theta = \frac{2t}{1+t^2}\) and \(\cos\theta = \frac{1-t^2}{1+t^2}\) | ||
| Therefore: \(\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}} = \frac{2t}{1-t^2}\) | (E1) | Convincing |
| (1) | ||
| b) | \(x = r\cos(\theta) = \cos\left(\frac{\theta}{2}\right)\cos\theta\) | M1 |
| \(\frac{dx}{d\theta} = -\cos\left(\frac{\theta}{2}\right)\sin\theta - \frac{1}{2}\cos\theta\sin\left(\frac{\theta}{2}\right)\) | m1 | Use of product rule |
| When perpendicular to initial line: | ||
| \(\frac{dx}{d\theta} = -\cos\left(\frac{\theta}{2}\right)\sin\theta - \frac{1}{2}\cos\theta\sin\left(\frac{\theta}{2}\right) = 0\) | A1 | \(\frac{dx}{d\theta} = 0\) |
| \(-\cos\left(\frac{\theta}{2}\right)\sin\theta = \frac{1}{2}\cos\theta\sin\left(\frac{\theta}{2}\right)\) | ||
| \(\tan(\theta) = -\frac{1}{2}\tan\left(\frac{\theta}{2}\right)\) | A1 | Convincing |
| (4) | ||
| c) | \(\frac{2t}{1-t^2} = -\frac{1}{2}t\) | M1 |
| \(2t = -\frac{1}{2}t + \frac{1}{2}t^3\) | ||
| \(4t = -t + t^3\) | ||
| \(t^3 - 5t = 0\) | A1 | |
| \(t(t^2 - 5) = 0\) | ||
| \(t = 0, \quad t = \pm\sqrt{5}\) | A1 | For at least 2 solutions |
| \(\tan\left(\frac{\theta}{2}\right) = 0 \quad \text{or} \quad \tan\left(\frac{\theta}{2}\right) = \pm\sqrt{5}\) | ||
| \(\frac{\theta}{2} = 0 \quad \frac{\theta}{2} = 1.15\ldots \quad \frac{\theta}{2} = -1.15\ldots\) | A1 | Ignore angles outside range |
| \(\theta = 0, \quad r = 1\) | B2 | FT their \(\frac{\theta}{2}\) values provided \(r > 0\) |
| \(\theta = 2.30, \quad r = 0.408\) | B2 All correct, no additional sets | |
| \(\theta = -2.30, \quad r = 0.408\) | B1 for 2 correct sets of coordinates | |
| \((1, 0), (0.408, 2.30), (0.408, -2.30)\) | (6) | |
| d) | Area \(= \frac{1}{2}\int_0^{\pi}\cos^2\left(\frac{\theta}{2}\right)d\theta\) | M1 |
| \(= \frac{1}{2}\int_0^{\pi}\frac{\cos\theta + 1}{2}d\theta\) | M1, A1 | Use of double-angle formula; All correct, including limits |
| \(= \frac{1}{4}\int_0^{\pi}(\cos\theta + 1)d\theta\) | ||
| \(= \frac{1}{4}\left[\sin\theta + \theta\right]_0^{\pi}\) | A1 | |
| \(= \frac{1}{\pi}\left[(0 + \pi) - (0 + 0)\right]\) | A1 | |
| \(= \frac{\pi}{4}\) | ||
| (5) | ||
| [16] |
a) | **METHOD 1:** | | |
| $\tan\theta = \tan\left(\frac{\theta}{2} + \frac{\theta}{2}\right)$ | | |
| $\tan\left(\frac{\theta}{2} + \frac{\theta}{2}\right) = \frac{\tan\left(\frac{\theta}{2}\right) + \tan\left(\frac{\theta}{2}\right)}{1 - \tan\left(\frac{\theta}{2}\right)\tan\left(\frac{\theta}{2}\right)}$ | | |
| $\tan\left(\frac{\theta}{2} + \frac{\theta}{2}\right) = \frac{t + t}{1 - t \times t}$ | E1 | Convincing |
| $\tan\theta = \frac{2t}{1 - t^2}$ | | |
| | | |
| **METHOD 2:** | | |
| For $t = \tan\left(\frac{\theta}{2}\right)$: | | |
| $\sin\theta = \frac{2t}{1+t^2}$ and $\cos\theta = \frac{1-t^2}{1+t^2}$ | | |
| Therefore: $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}} = \frac{2t}{1-t^2}$ | (E1) | Convincing |
| | (1) | |
b) | $x = r\cos(\theta) = \cos\left(\frac{\theta}{2}\right)\cos\theta$ | M1 | |
| $\frac{dx}{d\theta} = -\cos\left(\frac{\theta}{2}\right)\sin\theta - \frac{1}{2}\cos\theta\sin\left(\frac{\theta}{2}\right)$ | m1 | Use of product rule |
| | | |
| When perpendicular to initial line: | | |
| $\frac{dx}{d\theta} = -\cos\left(\frac{\theta}{2}\right)\sin\theta - \frac{1}{2}\cos\theta\sin\left(\frac{\theta}{2}\right) = 0$ | A1 | $\frac{dx}{d\theta} = 0$ |
| | | |
| $-\cos\left(\frac{\theta}{2}\right)\sin\theta = \frac{1}{2}\cos\theta\sin\left(\frac{\theta}{2}\right)$ | | |
| $\tan(\theta) = -\frac{1}{2}\tan\left(\frac{\theta}{2}\right)$ | A1 | Convincing |
| | (4) | |
c) | $\frac{2t}{1-t^2} = -\frac{1}{2}t$ | M1 | Use of compound-angle; Accept use of $\tan\left(\frac{\theta}{2}\right)$ in place of $t$ throughout |
| | | |
| $2t = -\frac{1}{2}t + \frac{1}{2}t^3$ | | |
| $4t = -t + t^3$ | | |
| $t^3 - 5t = 0$ | A1 | |
| $t(t^2 - 5) = 0$ | | |
| $t = 0, \quad t = \pm\sqrt{5}$ | A1 | For at least 2 solutions |
| | | |
| $\tan\left(\frac{\theta}{2}\right) = 0 \quad \text{or} \quad \tan\left(\frac{\theta}{2}\right) = \pm\sqrt{5}$ | | |
| | | |
| $\frac{\theta}{2} = 0 \quad \frac{\theta}{2} = 1.15\ldots \quad \frac{\theta}{2} = -1.15\ldots$ | A1 | Ignore angles outside range |
| | | |
| $\theta = 0, \quad r = 1$ | B2 | FT their $\frac{\theta}{2}$ values provided $r > 0$ |
| $\theta = 2.30, \quad r = 0.408$ | | B2 All correct, no additional sets |
| $\theta = -2.30, \quad r = 0.408$ | | B1 for 2 correct sets of coordinates |
| | | |
| $(1, 0), (0.408, 2.30), (0.408, -2.30)$ | (6) | |
d) | Area $= \frac{1}{2}\int_0^{\pi}\cos^2\left(\frac{\theta}{2}\right)d\theta$ | M1 | |
| $= \frac{1}{2}\int_0^{\pi}\frac{\cos\theta + 1}{2}d\theta$ | M1, A1 | Use of double-angle formula; All correct, including limits |
| $= \frac{1}{4}\int_0^{\pi}(\cos\theta + 1)d\theta$ | | |
| $= \frac{1}{4}\left[\sin\theta + \theta\right]_0^{\pi}$ | A1 | |
| $= \frac{1}{\pi}\left[(0 + \pi) - (0 + 0)\right]$ | A1 | |
| $= \frac{\pi}{4}$ | | |
| | (5) | |
| | [16] | |\begin{enumerate}[label=(\alph*)]
\item Show that $\tan\theta$ may be expressed as $\frac{2t}{1-t^2}$, where $t = \tan\left(\frac{\theta}{2}\right)$. [1]
\end{enumerate}
The diagram below shows a sketch of the curve $C$ with polar equation
$$r = \cos\left(\frac{\theta}{2}\right), \quad \text{where } -\pi < \theta \leqslant \pi.$$
\includegraphics{figure_6}
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that the $\theta$-coordinate of the points at which the tangent to $C$ is perpendicular to the initial line satisfies the equation
$$\tan\theta = -\frac{1}{2}\tan\left(\frac{\theta}{2}\right).$$ [4]
\item Hence, find the polar coordinates of the points on $C$ where the tangent is perpendicular to the initial line. [6]
\item Calculate the area of the region enclosed by the curve $C$ and the initial line for $0 \leqslant \theta \leqslant \pi$. [5]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2023 Q6 [16]}}