C4 June 2014 (R) Q8

Notes

M1: Sets \(y = 1\) to find \(t\) and uses their \(t\) to find \(x\).

Note: M1 can be implied by either \(x\) or \(k = 4 - \dfrac{\pi}{2}\) or 2.429… or \(\dfrac{\pi}{2} - 4\) or \(-2.429\ldots\)

A1: \(x\) or \(k = 4 - \dfrac{\pi}{2}\) or \(\dfrac{8 - \pi}{2}\)

Note: A decimal answer of 2.429… (without a correct exact answer) is A0.

Note: Allow A1 for a candidate using \(t = \dfrac{\pi}{2}\) to find \(x = \dfrac{\pi}{2} - 4\) and then stating that \(k\) must be \(4 - \dfrac{\pi}{2}\) o.e.

Notes

B1: At least one of \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) correct. Note: that this mark can be implied from their working.

B1: Both \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) are correct. Note: that this mark can be implied from their working.

M1: Applies their \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) divided by their \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) and attempts to substitute their \(t\) into their expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\).

Note: This mark may be implied by their final answer.
i.e. \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2\sin t}{1 - 4\cos t}\) followed by an answer of \(-2\) (from \(t = -\dfrac{\pi}{2}\)) or 2 (from \(t = \dfrac{\pi}{2}\))

Note: Applying \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) divided by their \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) is M0, even if they state \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t} \div \dfrac{\mathrm{d}x}{\mathrm{d}t}\).

A1: Using \(t = -\dfrac{\pi}{2}\) \(\left(\text{and not } t = \dfrac{3\pi}{2}\right)\) to find a correct \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) of \(-2\) by correct solution only.

Notes

VERY IMPORTANT NOTE FOR PART (c)

NOTE: Candidates who state \(t = 0.6077\) with no intermediate working from \(4\sin t - 4\cos t = -1\) will get 2^nd M0, 2^nd A0, 3^rd M0, 3^rd A0.
They will not express \(4\sin t - 4\cos t\) as either \(4\sqrt{2}\sin\left(t - \dfrac{\pi}{4}\right)\) or \(-4\sqrt{2}\cos\left(t + \dfrac{\pi}{4}\right)\).
OR use any acceptable alternative method to achieve \(t = 0.6077\)

NOTE: Alternative methods for part (c) are given below.

NOTE: If a candidate uses an incorrect \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) expression in part (c) then the accuracy marks are not obtainable.

1^st M1: Sets their \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{1}{2}\)

1^st A1: Rearranges to give the correct equation with \(\sin t\) and \(\cos t\) on the same side.
eg. \(4\sin t - 4\cos t = -1\) or \(4\cos t - 4\sin t = 1\) or \(\sin t - \cos t = -\dfrac{1}{4}\) or \(\cos t - \sin t = \dfrac{1}{4}\)
or \(4\sin t - 4\cos t + 1 = 0\) or \(4\cos t - 4\sin t - 1 = 0\) or \(\sin t - \cos t + \dfrac{1}{4} = 0\) etc. are fine for A1.

2^nd M1: Rewrites \(\pm\lambda\sin t \pm \mu\cos t\) in the form of either \(R\cos(t \pm \alpha)\) or \(R\sin(t \pm \alpha)\) where \(R \neq 1\) or 0 and \(\alpha \neq 0\)

2^nd A1: Correct equation. Eg. \(4\sqrt{2}\sin\left(t - \dfrac{\pi}{4}\right) = -1\) or \(-4\sqrt{2}\cos\left(t + \dfrac{\pi}{4}\right) = -1\)
or \(\sqrt{2}\sin\left(t - \dfrac{\pi}{4}\right) = -\dfrac{1}{4}\) or \(\sqrt{2}\cos\left(t + \dfrac{\pi}{4}\right) = \dfrac{1}{4}\), etc.

Note: Unless recovered, give A0 for \(4\sqrt{2}\sin(t - 45^\circ) = -1\) or \(-4\sqrt{2}\cos(t + 45^\circ) = -1\), etc.

3^rd M1: which is dependent on the 2^nd M1 mark. Uses correct algebraic processes to give \(t = \ldots\)

4^th A1: anything that rounds to 0.6077

Note: Do not give the final A1 mark in (c) if there any extra solutions given in the range \(-\dfrac{2\pi}{3} \leqslant t \leqslant \dfrac{2\pi}{3}\).

Note: You can give the final A1 mark in (c) if extra solutions are given outside of \(-\dfrac{2\pi}{3} \leqslant t \leqslant \dfrac{2\pi}{3}\).