C4 January 2006 Q8
8.
The curve shown in Figure 2 has parametric equations
\[x = t - 2\sin t, \qquad y = 1 - 2\cos t, \qquad 0 \leqslant t \leqslant 2\pi.\]
(a) Show that the curve crosses the \(x\)-axis where \(t = \dfrac{\pi}{3}\) and \(t = \dfrac{5\pi}{3}\). (2)
The finite region \(R\) is enclosed by the curve and the \(x\)-axis, as shown shaded in Figure 2.
(b) Show that the area of \(R\) is given by the integral
\[\int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} (1 - 2\cos t)^2\,\mathrm{d}t.\]
(3)(c) Use this integral to find the exact value of the shaded area. (7)
| Scheme | Marks |
|---|---|
| Solves \(y = 0 \Rightarrow \cos t = \frac{1}{2}\) to obtain \(t = \dfrac{\pi}{3}\) or \(\dfrac{5\pi}{3}\) (need both for A1) Or substitutes both values of \(t\) and shows that \(y = 0\) | M1 A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 1 - 2\cos t\) | M1 A1 |
| Area \(= \displaystyle\int y\,\mathrm{d}x = \int_{\pi/3}^{5\pi/3} (1 - 2\cos t)(1 - 2\cos t)\,\mathrm{d}t = \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} (1 - 2\cos t)^2\,\mathrm{d}t\) \(\ast\) AG | B1 |
| (3) |
| Scheme | Marks |
|---|---|
| Area \(= \displaystyle\int 1 - 4\cos t + 4\cos^2 t\,\mathrm{d}t\) 3 terms | M1 |
| \(= \displaystyle\int 1 - 4\cos t + 2(\cos 2t + 1)\,\mathrm{d}t\) (use of correct double angle formula) | M1 |
| \(= \displaystyle\int 3 - 4\cos t + 2\cos 2t\,\mathrm{d}t\) | |
| \(= \left[3t - 4\sin t + \sin 2t\right]\) | M1 A1 |
| Substitutes the two correct limits \(t = \dfrac{5\pi}{3}\) and \(\dfrac{\pi}{3}\) and subtracts. | M1 |
| \(= 4\pi + 3\sqrt{3}\) | A1A1 |
| (7) | |
| (12 marks) |