| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2018 |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find tangent equation at parameter |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring the chain rule (dy/dx = dy/dt รท dx/dt) and standard trigonometric derivatives. Part (a) involves routine algebraic manipulation with double angle formulas, and part (b) is a standard tangent line calculation. While it requires multiple steps, all techniques are standard P4 material with no novel insight needed, 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{dt} = 2\sqrt{3}\cos 2t\) | B1 | The correct \(\frac{dx}{dt}\) |
| \(\frac{dy}{dt} = -8\cos t \sin t\) | M1 A1 | M1: \(\frac{dy}{dt} = \pm k\cos t\sin t\) or \(\pm k\sin 2t\), where \(k\) is non-zero. A1: \(-8\cos t\sin t\) or \(-4\sin 2t\) or equivalent |
| \(\frac{dy}{dx} = \frac{-8\cos t\sin t}{2\sqrt{3}\cos 2t}\) | M1 | Their \(\frac{dy}{dt}\) divided by their \(\frac{dx}{dt}\); answer must be function of \(t\) only |
| \(= -\frac{4\sin 2t}{2\sqrt{3}\cos 2t}\) | ||
| \(\frac{dy}{dx} = -\frac{2}{3}\sqrt{3}\tan 2t\) \(\left(k = -\frac{2}{3}\right)\) | A1 | Constant must be \(-\frac{2}{3}\); accept equivalent fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(t = \frac{\pi}{3}\): \(x = \frac{3}{2}\), \(y = 1\) | B1 | Exact numerical values required but can be implied by correct final answer |
| \(m = -\frac{2}{3}\sqrt{3}\tan\left(\frac{2\pi}{3}\right)\) \((= 2)\) | M1 | Substituting \(t = \frac{\pi}{3}\) into their \(\frac{dy}{dx}\); \(\tan\frac{2\pi}{3}\) need not be evaluated |
| \(y - 1 = 2\left(x - \frac{3}{2}\right)\) | dM1 | Dependent on previous M. Finding tangent equation with point and numerical gradient. Must evaluate \(\tan\frac{2\pi}{3}\). Equation must be linear |
| \(y = 2x - 2\) | A1 | cao, correct form |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = 2\sqrt{3}\cos 2t$ | B1 | The correct $\frac{dx}{dt}$ |
| $\frac{dy}{dt} = -8\cos t \sin t$ | M1 A1 | M1: $\frac{dy}{dt} = \pm k\cos t\sin t$ or $\pm k\sin 2t$, where $k$ is non-zero. A1: $-8\cos t\sin t$ or $-4\sin 2t$ or equivalent |
| $\frac{dy}{dx} = \frac{-8\cos t\sin t}{2\sqrt{3}\cos 2t}$ | M1 | Their $\frac{dy}{dt}$ divided by their $\frac{dx}{dt}$; answer must be function of $t$ only |
| $= -\frac{4\sin 2t}{2\sqrt{3}\cos 2t}$ | | |
| $\frac{dy}{dx} = -\frac{2}{3}\sqrt{3}\tan 2t$ $\left(k = -\frac{2}{3}\right)$ | A1 | Constant must be $-\frac{2}{3}$; accept equivalent fractions |
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $t = \frac{\pi}{3}$: $x = \frac{3}{2}$, $y = 1$ | B1 | Exact numerical values required but can be implied by correct final answer |
| $m = -\frac{2}{3}\sqrt{3}\tan\left(\frac{2\pi}{3}\right)$ $(= 2)$ | M1 | Substituting $t = \frac{\pi}{3}$ into their $\frac{dy}{dx}$; $\tan\frac{2\pi}{3}$ need not be evaluated |
| $y - 1 = 2\left(x - \frac{3}{2}\right)$ | dM1 | Dependent on previous M. Finding tangent equation with point and numerical gradient. Must evaluate $\tan\frac{2\pi}{3}$. Equation must be linear |
| $y = 2x - 2$ | A1 | cao, correct form |4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4de08317-5fb9-4789-8d57-ccf463224c78-10_899_759_127_621}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 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 found.
\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.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2018 Q4 [9]}}