C4 January 2010 Q3
3. The curve \(C\) has the equation \[\cos 2x + \cos 3y = 1, \qquad -\frac{\pi}{4} \leqslant x \leqslant \frac{\pi}{4}, \quad 0 \leqslant y \leqslant \frac{\pi}{6}\]
(a) Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(x\) and \(y\). (3)
The point \(P\) lies on \(C\) where \(x = \dfrac{\pi}{6}\).
(b) Find the value of \(y\) at \(P\). (3)
(c) Find the equation of the tangent to \(C\) at \(P\), giving your answer in the form \(ax + by + c\pi = 0\), where \(a\), \(b\) and \(c\) are integers. (3)
| Scheme | Marks |
|---|---|
| \(-2\sin 2x - 3\sin 3y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) | M1 A1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{2\sin 2x}{3\sin 3y}\) Accept \(\dfrac{2\sin 2x}{-3\sin 3y},\ \dfrac{-2\sin 2x}{3\sin 3y}\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| At \(x = \dfrac{\pi}{6}\), \(\cos\left(\dfrac{2\pi}{6}\right) + \cos 3y = 1\) | M1 |
| \(\cos 3y = \dfrac{1}{2}\) | A1 |
| \(3y = \dfrac{\pi}{3} \Rightarrow y = \dfrac{\pi}{9}\) awrt 0.349 | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| At \(\left(\dfrac{\pi}{6}, \dfrac{\pi}{9}\right)\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{2\sin 2\left(\frac{\pi}{6}\right)}{3\sin 3\left(\frac{\pi}{9}\right)} = -\dfrac{2\sin\frac{\pi}{3}}{3\sin\frac{\pi}{3}} = -\dfrac{2}{3}\) | M1 |
| \(y - \dfrac{\pi}{9} = -\dfrac{2}{3}\left(x - \dfrac{\pi}{6}\right)\) | M1 |
| Leading to \(6x + 9y - 2\pi = 0\) | A1 |
| (3) | |
| (9 marks) |