A2 October 2020 Paper 2 Q6
6 The equation of a curve in polar coordinates is \(r = \ln(1 + \sin\theta)\) for \(\alpha \leqslant \theta \leqslant \beta\) where \(\alpha\) and \(\beta\) are non-negative angles. The curve consists of a single closed loop through the pole.
| Scheme | Marks | AO |
|---|---|---|
| \(\ln(1 + \sin\theta) = 0 \Rightarrow 1 + \sin\theta = 1 \Rightarrow \sin\theta = 0\) | M1 | 1.1a |
| so \(\alpha = 0\) and \(\beta = \pi\) | A1 | 2.2a |
| [2] |
| Scheme | Marks | AO |
|---|---|---|
| \(A = \dfrac{1}{2}\displaystyle\int_0^{\pi} (\ln(1 + \sin\theta))^2\,\mathrm{d}\theta\) | M1 | 1.2 |
| \(= 0.4162\) (4 sf) cao | A1 | 1.1 |
| [2] |
Notes
M1: Correct formula for area with \(r\) correctly substituted and their limits. Must be unambiguous but can be implied by correct answer/later work
Incorrect formula = M0A0
Condone missing \(\mathrm{d}\theta\)
A1: BC
| Scheme | Marks | AO |
|---|---|---|
| \(\theta = \dfrac{\pi}{2} \Rightarrow r = \ln 2 = 0.6931\) (4 sf) which would be the diameter, \(D\), of the circle | M1 | 3.1a |
| But \(A = 0.4162\) (4 sf) \(\Rightarrow D = 0.7280\) (4 sf) or \(R = 0.3640\) (4 sf) so the curve is not circular | A1 | 3.2a |
| [2] |
Notes
M1: or radius \(R = 0.3466\) (4 sf); condone correct \(R\) or \(D\) without reasoning
It must be clear that the \(r\) value would be the diameter of the circle; the calculation alone is insufficient for M1.
M1 can be implied by area given as \(\pi\left(\dfrac{\ln 2}{2}\right)^2\)
A1: or \(R = 0.3466\) (4 sf) (or \(D = 0.6931\)) \(\Rightarrow A = 0.3773\) (4 sf) which is not 0.4162 (4 sf)
Explanation must include comparison of \(R\)’s, \(D\)’s or \(A\)’s and conclusion. Allow correct working to 3 sf.