| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (sin/cos identities) |
| Difficulty | Moderate -0.3 This is a straightforward parametric equations question requiring standard techniques: basic differentiation of trig functions, chain rule for dy/dx, and using the double angle formula (sin 2θ = 2sin θ cos θ) with the Pythagorean identity to eliminate the parameter. All steps are routine for C4 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dx}{d\theta} = -\sin\theta\) | B1 | |
| \(\frac{d\theta}{dy} = 2\cos 2\theta\) | B1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{2\cos 2\theta}{\sin\theta}\) and \(\frac{dy}{dx} = \frac{2\cos\frac{\pi}{6}}{\sin\frac{\pi}{6}} = -2\) | M1 | |
| A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 2\sin\theta\cos\theta = 2\sqrt{1-\cos^2\theta}\cos\theta\) | B1 | |
| \(y = 2\sqrt{1-x^2}x\) | M1 | |
| \(y^2 = 4x^2(1-x^2)\) | A1 | 4 marks |
| Alt | ||
| \(y^2 = \sin^2 2\theta = (2\sin\theta\cos\theta)^2 = (4)(1-\cos^2\theta)\cos^2\theta\) | B1 | |
| \(= (4)(1-x^2)x^2\) | B1 | |
| \(= 4(1-x^2)x^2\) | M1 | |
| \(= 4(1-x^2)x^2\) | A1 | 4 marks |
**6(a)(i)**
| $\frac{dx}{d\theta} = -\sin\theta$ | B1 | | |
| $\frac{d\theta}{dy} = 2\cos 2\theta$ | B1 | 2 marks | |
**6(a)(ii)**
| $\frac{dy}{dx} = \frac{2\cos 2\theta}{\sin\theta}$ and $\frac{dy}{dx} = \frac{2\cos\frac{\pi}{6}}{\sin\frac{\pi}{6}} = -2$ | M1 | | use chain rule; their $\frac{dy}{d\theta}$ / their $\frac{dx}{d\theta}$ and substitute $\theta = \frac{\pi}{6}$ |
| | A1 | 2 marks | |
**6(b)**
| $y = 2\sin\theta\cos\theta = 2\sqrt{1-\cos^2\theta}\cos\theta$ | B1 | | use $\sin 2\theta = 2\sin\theta\cos\theta$; use $\sin^2\theta = 1 - \cos^2\theta$ |
| $y = 2\sqrt{1-x^2}x$ | M1 | | $\sin\theta, \cos\theta$ in terms of $x$ |
| $y^2 = 4x^2(1-x^2)$ | A1 | 4 marks | all correct CSO |
| **Alt** | | | |
| $y^2 = \sin^2 2\theta = (2\sin\theta\cos\theta)^2 = (4)(1-\cos^2\theta)\cos^2\theta$ | B1 | | use of double angle formula |
| $= (4)(1-x^2)x^2$ | B1 | | use of $s^2 + c^2 = 1$ to eliminate $\sin\theta$ |
| $= 4(1-x^2)x^2$ | M1 | | Substitute $\cos\theta$ for $x$ |
| $= 4(1-x^2)x^2$ | A1 | 4 marks | CSO |6 A curve is given by the parametric equations
$$x = \cos \theta \quad y = \sin 2 \theta$$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find $\frac { \mathrm { d } x } { \mathrm {~d} \theta }$ and $\frac { \mathrm { d } y } { \mathrm {~d} \theta }$.\\
(2 marks)
\item Find the gradient of the curve at the point where $\theta = \frac { \pi } { 6 }$.
\end{enumerate}\item Show that the cartesian equation of the curve can be written as
$$y ^ { 2 } = k x ^ { 2 } \left( 1 - x ^ { 2 } \right)$$
where $k$ is an integer.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2007 Q6 [8]}}