| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find tangent equation at point |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question with standard trigonometric functions. Part (a) requires routine application of implicit differentiation rules, part (b) involves simple substitution and solving a basic trigonometric equation, and part (c) is standard tangent line calculation. While it involves multiple steps and trigonometric manipulation, all techniques are standard C4 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(-2\sin 2x - 3\sin 3y \frac{dy}{dx} = 0\) then \(\frac{dy}{dx} = \frac{2\sin 2x}{3\sin 3y}\) | M1 A1; Accept \(\frac{-2\sin 2x}{-3\sin 3y}\) or \(\frac{2\sin 2x}{3\sin 3y}\) A1 |
| (b) | At \(x = \frac{\pi}{6}\): \(\cos\left(\frac{2\pi}{6}\right) + \cos 3y = 1\) then \(\cos 3y = \frac{1}{2}\) then \(3y = \frac{\pi}{3} \Rightarrow y = \frac{\pi}{9}\) | M1; A1; awrt 0.349 A1 |
| (c) | At \(\left(\frac{\pi}{6}, \frac{\pi}{9}\right)\): \(\frac{dy}{dx} = \frac{2\sin 2(\frac{\pi}{6})}{3\sin 3(\frac{\pi}{9})} = \frac{2\sin \frac{\pi}{3}}{3\sin \frac{\pi}{3}} = \frac{2}{3}\) then \(y - \frac{\pi}{9} = \frac{2}{3}\left(x - \frac{\pi}{6}\right)\) leading to \(6x + 9y - 2\pi = 0\) | M1; M1; A1 |
(a) | $-2\sin 2x - 3\sin 3y \frac{dy}{dx} = 0$ then $\frac{dy}{dx} = \frac{2\sin 2x}{3\sin 3y}$ | M1 A1; Accept $\frac{-2\sin 2x}{-3\sin 3y}$ or $\frac{2\sin 2x}{3\sin 3y}$ A1 | (3 marks)
(b) | At $x = \frac{\pi}{6}$: $\cos\left(\frac{2\pi}{6}\right) + \cos 3y = 1$ then $\cos 3y = \frac{1}{2}$ then $3y = \frac{\pi}{3} \Rightarrow y = \frac{\pi}{9}$ | M1; A1; awrt 0.349 A1 | (3 marks)
(c) | At $\left(\frac{\pi}{6}, \frac{\pi}{9}\right)$: $\frac{dy}{dx} = \frac{2\sin 2(\frac{\pi}{6})}{3\sin 3(\frac{\pi}{9})} = \frac{2\sin \frac{\pi}{3}}{3\sin \frac{\pi}{3}} = \frac{2}{3}$ then $y - \frac{\pi}{9} = \frac{2}{3}\left(x - \frac{\pi}{6}\right)$ leading to $6x + 9y - 2\pi = 0$ | M1; M1; A1 | (3 marks)
**Total: [9]**
---\begin{enumerate}
\item The curve $C$ has the equation
\end{enumerate}
$$\cos 2 x + \cos 3 y = 1 , \quad - \frac { \pi } { 4 } \leqslant x \leqslant \frac { \pi } { 4 } , \quad 0 \leqslant y \leqslant \frac { \pi } { 6 }$$
(a) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.
The point $P$ lies on $C$ where $x = \frac { \pi } { 6 }$.\\
(b) Find the value of $y$ at $P$.\\
(c) Find the equation of the tangent to $C$ at $P$, giving your answer in the form $a x + b y + c \pi = 0$, where $a , b$ and $c$ are integers.
\section*{LU}
\hfill \mbox{\textit{Edexcel C4 2010 Q3 [9]}}