| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2012 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find tangent equation at parameter |
| Difficulty | Moderate -0.3 This is a standard C4 parametric differentiation question requiring routine application of dy/dx = (dy/dt)/(dx/dt), followed by straightforward tangent equation and cartesian conversion using double angle identities. All techniques are textbook exercises 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 |
|---|---|
| \(\frac{dx}{dt} = 2\sqrt{3}\cos 2t\) | B1 |
| \(\frac{dy}{dt} = -8\cos t \sin t\) | M1 A1 |
| \(\frac{dy}{dx} = \frac{-8\cos t \sin t}{2\sqrt{3}\cos 2t}\) | M1 |
| \(= -\frac{4\sin 2t}{2\sqrt{3}\cos 2t}\) | |
| \(\frac{dy}{dx} = -\frac{2}{3}\sqrt{3}\tan 2t \quad \left(k = -\frac{2}{3}\right)\) | A1 (5) |
| Answer | Marks | Guidance |
|---|---|---|
| When \(t = \frac{\pi}{3}\): \(x = \frac{3}{2},\, y = 1\) | B1 | can be implied |
| \(m = -\frac{2}{3}\sqrt{3}\tan\left(\frac{2\pi}{3}\right) \; (= 2)\) | M1 | |
| \(y - 1 = 2\left(x - \frac{3}{2}\right)\) | M1 | |
| \(y = 2x - 2\) | A1 (4) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = \sqrt{3}\sin 2t = \sqrt{3} \times 2\sin t\cos t\) | M1 | |
| \(x^2 = 12\sin^2 t\cos^2 t = 12(1 - \cos^2 t)\cos^2 t\) | ||
| \(x^2 = 12\left(1 - \frac{y}{4}\right)\frac{y}{4}\) | M1 A1 (3) | or equivalent |
| *Alternative:* \(y = 2\cos 2t + 2\), using \(\sin^2 2t + \cos^2 2t = 1\): \(\frac{x^2}{3} + \frac{(y-2)^2}{4} = 1\) | M1; M1 A1 (3) |
## Question 6:
### Part (a)
| $\frac{dx}{dt} = 2\sqrt{3}\cos 2t$ | B1 | |
| $\frac{dy}{dt} = -8\cos t \sin t$ | M1 A1 | |
| $\frac{dy}{dx} = \frac{-8\cos t \sin t}{2\sqrt{3}\cos 2t}$ | M1 | |
| $= -\frac{4\sin 2t}{2\sqrt{3}\cos 2t}$ | | |
| $\frac{dy}{dx} = -\frac{2}{3}\sqrt{3}\tan 2t \quad \left(k = -\frac{2}{3}\right)$ | A1 (5) | |
### Part (b)
| When $t = \frac{\pi}{3}$: $x = \frac{3}{2},\, y = 1$ | B1 | can be implied |
| $m = -\frac{2}{3}\sqrt{3}\tan\left(\frac{2\pi}{3}\right) \; (= 2)$ | M1 | |
| $y - 1 = 2\left(x - \frac{3}{2}\right)$ | M1 | |
| $y = 2x - 2$ | A1 (4) | |
### Part (c)
| $x = \sqrt{3}\sin 2t = \sqrt{3} \times 2\sin t\cos t$ | M1 | |
| $x^2 = 12\sin^2 t\cos^2 t = 12(1 - \cos^2 t)\cos^2 t$ | | |
| $x^2 = 12\left(1 - \frac{y}{4}\right)\frac{y}{4}$ | M1 A1 (3) | or equivalent |
| *Alternative:* $y = 2\cos 2t + 2$, using $\sin^2 2t + \cos^2 2t = 1$: $\frac{x^2}{3} + \frac{(y-2)^2}{4} = 1$ | M1; M1 A1 (3) | |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{12fbfe89-60fe-4890-9a22-2b1988d05d33-09_831_784_127_580}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the curve $C$ with parametric equations
$$x = ( \sqrt { } 3 ) \sin 2 t , \quad y = 4 \cos ^ { 2 } t , \quad 0 \leqslant t \leqslant \pi$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = k ( \sqrt { } 3 ) \tan 2 t$, where $k$ is a constant to be determined.
\item Find an equation of the tangent to $C$ at the point where $t = \frac { \pi } { 3 }$.
Give your answer in the form $y = a x + b$, where $a$ and $b$ are constants.
\item Find a cartesian equation of $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2012 Q6 [12]}}