June 2018 Paper 1 Q5
5. Given that
\[y = \frac{3\sin\theta}{2\sin\theta + 2\cos\theta} \qquad -\frac{\pi}{4} \lt \theta \lt \frac{3\pi}{4}\]show that
\[\frac{\mathrm{d}y}{\mathrm{d}\theta} = \frac{A}{1 + \sin 2\theta} \qquad -\frac{\pi}{4} \lt \theta \lt \frac{3\pi}{4}\]where \(A\) is a rational constant to be found. (5)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}\theta} = \dfrac{(2\sin\theta + 2\cos\theta)3\cos\theta - 3\sin\theta(2\cos\theta - 2\sin\theta)}{(2\sin\theta + 2\cos\theta)^2}\) | M1 A1 | 1.1b 1.1b |
| Expands and uses \(\sin^2\theta + \cos^2\theta = 1\) at least once in the numerator or the denominator or uses \(2\sin\theta\cos\theta = \sin 2\theta\) in \(\Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}\theta} = \dfrac{\ldots}{\ldots\ldots C\sin\theta\cos\theta}\) | M1 | 3.1a |
| Expands and uses \(\sin^2\theta + \cos^2\theta = 1\) the numerator and the denominator AND uses \(2\sin\theta\cos\theta = \sin 2\theta\) in \(\Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}\theta} = \dfrac{P}{Q + R\sin 2\theta}\) | M1 | 2.1 |
| \(\Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}\theta} = \dfrac{3}{2 + 2\sin 2\theta} = \dfrac{3/2}{1 + \sin 2\theta}\) | A1 | 1.1b |
| (5 marks) |
Notes
M1: For choosing either the quotient, product rule or implicit differentiation and applying it to the given function. Look for the correct form of \(\dfrac{\mathrm{d}y}{\mathrm{d}\theta}\) (condone it being stated as \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)) but tolerate slips on the coefficients and also condone \(\dfrac{\mathrm{d}(\sin\theta)}{\mathrm{d}\theta} = \pm\cos\theta\) and \(\dfrac{\mathrm{d}(\cos\theta)}{\mathrm{d}\theta} = \pm\sin\theta\)
For quotient rule look for \(\dfrac{\mathrm{d}y}{\mathrm{d}\theta} = \dfrac{(2\sin\theta + 2\cos\theta) \times \pm\ldots\cos\theta - 3\sin\theta(\pm\ldots\cos\theta \pm \ldots\sin\theta)}{(2\sin\theta + 2\cos\theta)^2}\)
For product rule look for
\(\dfrac{\mathrm{d}y}{\mathrm{d}\theta} = (2\sin\theta + 2\cos\theta)^{-1} \times \pm\ldots\cos\theta \pm 3\sin\theta \times (2\sin\theta + 2\cos\theta)^{-2} \times (\pm\ldots\cos\theta \pm \ldots\sin\theta)\)
Implicit differentiation look for \((\ldots\cos\theta \pm \ldots\sin\theta)y + (2\sin\theta + 2\cos\theta)\dfrac{\mathrm{d}y}{\mathrm{d}\theta} = \ldots\cos\theta\)
A1: A correct expression involving \(\dfrac{\mathrm{d}y}{\mathrm{d}\theta}\) condoning it appearing as \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
M1: Expands and uses \(\sin^2\theta + \cos^2\theta = 1\) at least once in the numerator or the denominator OR uses \(2\sin\theta\cos\theta = \sin 2\theta\) in \(\Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}\theta} = \dfrac{\ldots}{\ldots\ldots C\sin\theta\cos\theta}\)
M1: Expands and uses \(\sin^2\theta + \cos^2\theta = 1\) in the numerator and the denominator AND uses \(2\sin\theta\cos\theta = \sin 2\theta\) in the denominator to reach an expression of the form \(\dfrac{\mathrm{d}y}{\mathrm{d}\theta} = \dfrac{P}{Q + R\sin 2\theta}\).
A1: Fully correct proof with \(A = \dfrac{3}{2}\) stated but allow for example \(\dfrac{3/2}{1 + \sin 2\theta}\)
Allow recovery from missing brackets. Condone notation slips. This is not a given answer