A2 June 2023 Paper 2 Q4

EdexcelCurrent spec7 marksPolar Coordinates

(a) Sketch the polar curve \(C\), with equation\[r = 3 + \sqrt{5}\cos\theta \qquad 0 \leqslant \theta \leqslant 2\pi\]On your sketch clearly label the pole, the initial line and the value of \(r\) at the point where the curve intersects the initial line. (2)

The tangent to \(C\) at the point \(A\), where \(0 \lt \theta \lt \dfrac{\pi}{2}\), is parallel to the initial line.

(b) Use calculus to show that at \(A\)\[\cos\theta = \frac{1}{\sqrt{5}}\] (4)

Sketch: closed curve around the pole O with a slight dimple on the left, crossing the initial line at r = 3 + root 5 — Mark scheme (a)

Scheme	Marks	AO
Recalls correct shape for the type of curve, including ‘dimple’	B1	1.2
Correct position with labelling of pole, initial line and point.	B1	1.1b
	(2)

Mark scheme (b)

Scheme	Marks	AO
\(\dfrac{\mathrm{d}}{\mathrm{d}\theta}(r\sin\theta) = \dfrac{\mathrm{d}}{\mathrm{d}\theta}\left(3\sin\theta + \sqrt{5}\sin\theta\cos\theta\right) = A\cos\theta + B\cos 2\theta\) \(\dfrac{\mathrm{d}}{\mathrm{d}\theta}(r\sin\theta) = \dfrac{\mathrm{d}}{\mathrm{d}\theta}\left(\left(3 + \sqrt{5}\cos\theta\right)\sin\theta\right) = A\sin^2\theta + B\cos\theta + C\cos^2\theta\)	M1	1.1b
\(\dfrac{\mathrm{d}}{\mathrm{d}\theta}(r\sin\theta) = \dfrac{\mathrm{d}}{\mathrm{d}\theta}\left(3\sin\theta + \sqrt{5}\sin\theta\cos\theta\right) = 3\cos\theta + \sqrt{5}\cos 2\theta\) \(\dfrac{\mathrm{d}}{\mathrm{d}\theta}(r\sin\theta) = \dfrac{\mathrm{d}}{\mathrm{d}\theta}\left(\left(3 + \sqrt{5}\cos\theta\right)\sin\theta\right) = -\sqrt{5}\sin^2\theta + 3\cos\theta + \sqrt{5}\cos^2\theta\)	A1	1.1b
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow 3\cos\theta + \sqrt{5}\left(2\cos^2\theta - 1\right) = 0\) or \(-\sqrt{5}\left(1 - \cos^2\theta\right) + 3\cos\theta + \sqrt{5}\cos^2\theta = 0\) Leading to a quadratic in \(\cos\theta\) \(\left\{2\sqrt{5}\cos^2\theta + 3\cos\theta - \sqrt{5} = 0\right\}\)	M1	3.1a
\(\cos\theta = \dfrac{1}{\sqrt{5}}\) following a correct quadratic, any extra solutions are rejected \(\left\{\cos\theta = \dfrac{-3 \pm 7}{4\sqrt{5}}\right.\), quadrant 1 needs \(\left.\cos\theta \gt 0\right\}\)	A1	2.3
	(4)

Notes

M1: Uses \(y = r\sin\theta\) with the curve and attempts to differentiate. Accept any correct form but may have slips in coefficients, so e.g. as shown or \(A\cos\theta + B\cos^2\theta + C\sin^2\theta\) can score M1.

A1: Correct differentiation. Accept equivalents, e.g. \(3\cos\theta + \sqrt{5}\cos^2\theta - \sqrt{5}\sin^2\theta\)

M1: Sets their derivative equal to zero (may be implied), using trig identities to form a quadratics for \(\cos\theta\)
Allow this mark form the use of \(r\cos\theta\) leads to \(-3\sin\theta - 2\sqrt{5}\sin\theta\cos\theta = 0\), score M1 for factorising out \(\sin\theta\) and finds a value for \(\cos\theta\)

A1: Solves their quadratic and selects the correct value for \(\cos\theta\). If the other value is given it is A0 unless clearly rejected.

Scheme	Marks	AO
\(r = 4\)	B1	1.1b
	(1)
	(7 marks)