| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Tangent parallel/perpendicular to initial line |
| Difficulty | Challenging +1.8 This is a Further Maths polar curves question requiring implicit differentiation to find dy/dx = 0, solving a transcendental equation (tan 2θ = -2 tan θ), and coordinate conversion. The 9-mark allocation and need for numerical methods indicate substantial technical demand beyond standard A-level, though the concept is well-defined once the approach is identified. |
| Spec | 4.09b Sketch polar curves: r = f(theta) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = r\sin\theta = \sin(2\theta)\sin\theta\) | M1 | |
| \(\frac{dy}{d\theta} = \sin(2\theta)\cos\theta + 2\sin\theta\cos(2\theta)\) | A2 | A1 each term |
| When parallel to initial line: | ||
| \(\frac{dy}{d\theta} = \sin(2\theta)\cos\theta + 2\sin\theta\cos(2\theta) = 0\) | m1 | \(\frac{dy}{d\theta} = 0\) |
| METHOD 1: | ||
| \(\sin(2\theta)\cos\theta = -2\sin\theta\cos(2\theta)\) | m1 | Use of double-angle formulae |
| \(\tan(2\theta) = -2\tan\theta\) | ||
| \(\frac{2\tan\theta}{1-\tan^2\theta} = -2\tan\theta\) | A1 | Solveable form |
| \(2\tan^2\theta - 4\tan\theta = 0\) | A1 | |
| \(2\tan\theta(\tan^2\theta - 2) = 0\) | ||
| \(\tan\theta = \sqrt{2}\) (or \(\tan\theta = 0\) or \(\tan\theta = -\sqrt{2}\)) | A1 | |
| Therefore, \(\theta = 0.955\), \(r = 0.943\) | A2 | A1 each part; FT their \(\tan\theta\) provided \(\theta\) in range and M1m1m1 |
| METHOD 2: | ||
| \(2\sin\theta\cos^2\theta + 2\sin\theta(2\cos^2\theta - 1) = 0\) | (m1) | Use of double-angle formulae; Solveable form |
| \(2\sin\theta(3\cos^2\theta - 1) = 0\) | (A1) | |
| \(\cos\theta = \pm\frac{1}{\sqrt{3}}\) (or \(\sin\theta = 0\) or \(\cos\theta = -\frac{1}{\sqrt{3}}\)) | (A1) | |
| Therefore, \(\theta = 0.955\), \(r = 0.943\) | (A2) | A1 each part; FT their \(\cos\theta\) provided \(\theta\) in range and M1m1m1 |
| Answer | Marks | Guidance |
|---|---|---|
| \((0.544, 0.770)\) | B1 | FT (a) |
## Part a)
$y = r\sin\theta = \sin(2\theta)\sin\theta$ | M1 |
$\frac{dy}{d\theta} = \sin(2\theta)\cos\theta + 2\sin\theta\cos(2\theta)$ | A2 | A1 each term
When parallel to initial line: | |
$\frac{dy}{d\theta} = \sin(2\theta)\cos\theta + 2\sin\theta\cos(2\theta) = 0$ | m1 | $\frac{dy}{d\theta} = 0$
METHOD 1: | |
$\sin(2\theta)\cos\theta = -2\sin\theta\cos(2\theta)$ | m1 | Use of double-angle formulae
$\tan(2\theta) = -2\tan\theta$ | |
$\frac{2\tan\theta}{1-\tan^2\theta} = -2\tan\theta$ | A1 | Solveable form
$2\tan^2\theta - 4\tan\theta = 0$ | A1 |
$2\tan\theta(\tan^2\theta - 2) = 0$ | |
$\tan\theta = \sqrt{2}$ (or $\tan\theta = 0$ or $\tan\theta = -\sqrt{2}$) | A1 |
Therefore, $\theta = 0.955$, $r = 0.943$ | A2 | A1 each part; FT their $\tan\theta$ provided $\theta$ in range and M1m1m1
METHOD 2: | |
$2\sin\theta\cos^2\theta + 2\sin\theta(2\cos^2\theta - 1) = 0$ | (m1) | Use of double-angle formulae; Solveable form
$2\sin\theta(3\cos^2\theta - 1) = 0$ | (A1) |
$\cos\theta = \pm\frac{1}{\sqrt{3}}$ (or $\sin\theta = 0$ or $\cos\theta = -\frac{1}{\sqrt{3}}$) | (A1) |
Therefore, $\theta = 0.955$, $r = 0.943$ | (A2) | A1 each part; FT their $\cos\theta$ provided $\theta$ in range and M1m1m1
## Part b)
$(0.544, 0.770)$ | B1 | FT (a)The curve $C$ has polar equation
$$r = \sin 2\theta, \quad \text{where} \quad 0 < \theta \leqslant \frac{\pi}{2}.$$
\begin{enumerate}[label=(\alph*)]
\item Find the polar coordinates of the point on $C$ at which the tangent is parallel to the initial line. Give your answers correct to three decimal places. [9]
\item Write the coordinates of this point in Cartesian form. [1]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2019 Q8 [10]}}