| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find normal equation at parameter |
| Difficulty | Moderate -0.3 This is a standard C4 parametric differentiation question requiring the chain rule dy/dx = (dy/dt)/(dx/dt), followed by routine normal line equations. The derivatives are straightforward (sec²t and 2cos2t), and substituting specific t-values is mechanical. Slightly easier than average due to being a textbook application with no problem-solving required. |
| 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 |
|---|---|---|
| Working | Marks | Notes |
| \(\frac{dx}{dt} = \sec^2 t\), \(\frac{dy}{dt} = 2\cos 2t\) | M1 A1 | Correct derivatives |
| \(\Rightarrow \frac{dy}{dx} = \frac{2\cos 2t}{\sec^2 t}\) | \(\Rightarrow\) M1 | Chain rule applied |
| When \(t = \frac{\pi}{3}\), gradient is \(-\frac{1}{4}\) | B1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(P\) has coordinates \((\sqrt{3}, \frac{\sqrt{3}}{2})\); \(y - \frac{\sqrt{3}}{2} = -\frac{1}{m}(x - \sqrt{3})\) | B1 | |
| \(y - \frac{\sqrt{3}}{2} = 4(x - \sqrt{3})\) | M1 | |
| \(y = 4x - \frac{7}{2}\sqrt{3}\) | A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\frac{dy}{dx} = 0 \Rightarrow\) gradient of tangent \(= 0\), gradient of normal undefined | M1 | |
| \(\therefore x = \tan\frac{\pi}{4}\), i.e. \(x = 1\) | A1 | (2 marks) |
# Question 4:
## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $\frac{dx}{dt} = \sec^2 t$, $\frac{dy}{dt} = 2\cos 2t$ | M1 A1 | Correct derivatives |
| $\Rightarrow \frac{dy}{dx} = \frac{2\cos 2t}{\sec^2 t}$ | $\Rightarrow$ M1 | Chain rule applied |
| When $t = \frac{\pi}{3}$, gradient is $-\frac{1}{4}$ | B1 | (4 marks) |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $P$ has coordinates $(\sqrt{3}, \frac{\sqrt{3}}{2})$; $y - \frac{\sqrt{3}}{2} = -\frac{1}{m}(x - \sqrt{3})$ | B1 | |
| $y - \frac{\sqrt{3}}{2} = 4(x - \sqrt{3})$ | M1 | |
| $y = 4x - \frac{7}{2}\sqrt{3}$ | A1 | (3 marks) |
## Part (c):
| Working | Marks | Notes |
|---------|-------|-------|
| $\frac{dy}{dx} = 0 \Rightarrow$ gradient of tangent $= 0$, gradient of normal undefined | M1 | |
| $\therefore x = \tan\frac{\pi}{4}$, i.e. $x = 1$ | A1 | (2 marks) |
---4.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{0191bf56-a59e-44fe-af8c-bad796156f63-3_458_1552_415_223}
\end{center}
\end{figure}
Figure 1 shows part of the curve with parametric equations
$$x = \tan t , \quad y = \sin 2 t , \quad - \frac { \pi } { 2 } < t < \frac { \pi } { 2 } .$$
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of the curve at the point $P$ where $t = \frac { \pi } { 3 }$.
\item Find an equation of the normal to the curve at $P$.
\item Find an equation of the normal to the curve at the point $Q$ where $t = \frac { \pi } { 4 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q4 [9]}}