| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Show dy/dx equals expression |
| Difficulty | Standard +0.3 This is a standard C4 parametric differentiation question covering routine techniques: finding dy/dx using the chain rule, locating horizontal tangents by setting dy/dx = 0, finding a specific tangent equation, and converting to Cartesian form using trigonometric identities. All parts follow predictable methods with no novel problem-solving required, making it slightly easier than the average A-level question. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dx}{dt} = 6\cos t \times (-\sin t), \frac{dy}{dt} = 2\cos 2t\) | M1 A1 | |
| \(\frac{dy}{dx} = \frac{2\cos 2t}{-6\cos t \sin t} = \frac{2\cos 2t}{-3\sin 2t} = -\frac{2}{3}\cot 2t\) | M1 A1 | |
| \(-\frac{2}{3}\cot 2t = 0 \Rightarrow 2t = \frac{\pi}{2}, \frac{3\pi}{2} \Rightarrow t = \frac{\pi}{4}, \frac{3\pi}{4}\) | M1 A1 | |
| \(\therefore (\frac{3}{2}, 1), (\frac{3}{2}, -1)\) | A1 | |
| \(t = \frac{\pi}{6}, x = \frac{9}{4}, y = \frac{\sqrt{3}}{2}\), grad \(= -\frac{2}{\sqrt{3}}\) | B1 | |
| \(\therefore y - \frac{\sqrt{3}}{2} = -\frac{2}{\sqrt{3}}(x - \frac{9}{4})\) | M1 | |
| \(6\sqrt{3}y - 9 = -4x + 9\) | A1 | |
| \(2x + 3\sqrt{3}y = 9\) | A1 | |
| \(y^2 = \sin^2 2t = 4\sin^2 t\cos^2 t = 4(1 - \cos^2 t)\cos^2 t = 4\cos^2 t - 4\cos^4 t\) | M2 | |
| \(\cos^2 t = \frac{x}{3} \therefore y^2 = 4(1 - \frac{x}{3})\frac{x}{3}, y^2 = \frac{x}{3}(3 - x)\) | M1 A1 | (14 marks) |
$\frac{dx}{dt} = 6\cos t \times (-\sin t), \frac{dy}{dt} = 2\cos 2t$ | M1 A1 |
$\frac{dy}{dx} = \frac{2\cos 2t}{-6\cos t \sin t} = \frac{2\cos 2t}{-3\sin 2t} = -\frac{2}{3}\cot 2t$ | M1 A1 |
$-\frac{2}{3}\cot 2t = 0 \Rightarrow 2t = \frac{\pi}{2}, \frac{3\pi}{2} \Rightarrow t = \frac{\pi}{4}, \frac{3\pi}{4}$ | M1 A1 |
$\therefore (\frac{3}{2}, 1), (\frac{3}{2}, -1)$ | A1 |
$t = \frac{\pi}{6}, x = \frac{9}{4}, y = \frac{\sqrt{3}}{2}$, grad $= -\frac{2}{\sqrt{3}}$ | B1 |
$\therefore y - \frac{\sqrt{3}}{2} = -\frac{2}{\sqrt{3}}(x - \frac{9}{4})$ | M1 |
$6\sqrt{3}y - 9 = -4x + 9$ | A1 |
$2x + 3\sqrt{3}y = 9$ | A1 |
$y^2 = \sin^2 2t = 4\sin^2 t\cos^2 t = 4(1 - \cos^2 t)\cos^2 t = 4\cos^2 t - 4\cos^4 t$ | M2 |
$\cos^2 t = \frac{x}{3} \therefore y^2 = 4(1 - \frac{x}{3})\frac{x}{3}, y^2 = \frac{x}{3}(3 - x)$ | M1 A1 | (14 marks)
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**Total: 75 marks**A curve has parametric equations
$$x = 3 \cos^2 t, \quad y = \sin 2t, \quad 0 \leq t < \pi.$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac{dy}{dx} = -\frac{2}{3} \cot 2t$. [4]
\item Find the coordinates of the points where the tangent to the curve is parallel to the $x$-axis. [3]
\item Show that the tangent to the curve at the point where $t = \frac{\pi}{6}$ has the equation
$$2x + 3\sqrt{3} y = 9.$$ [3]
\item Find a cartesian equation for the curve in the form $y^2 = \text{f}(x)$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q7 [14]}}