A2 June 2024 Paper 1 Q9

OCR MEICurrent spec8 marksPolar Coordinates

9 A curve has polar equation \(r = a\sin 3\theta\), for \(0 \leqslant \theta \leqslant \pi\), where \(a\) is a positive constant.

(a) Sketch the curve. Indicate the parts of the curve where \(r\) is negative by using a broken line. [3]

(b) In this question you must show detailed reasoning.
Determine the area of one of the loops of the curve. [5]

Sketch of r = a sin 3 theta: two solid loops above the initial line, one in the first quadrant and one in the second, and one broken-line loop below the pole pointing straight down; the initial line is drawn from the pole to the right — Mark scheme (a)

Scheme	Marks	AO
	B1* B1* B1dep	1.1 1.1 1.1
	[3]

Mark scheme (b)

Scheme	Marks	AO
DR \(\displaystyle A = \int_0^{\frac{1}{3}\pi} \tfrac{1}{2}a^2\sin^2 3\theta\,\mathrm{d}\theta\)	M1 A1	1.1 1.1
\(\displaystyle A = \int_0^{\frac{1}{3}\pi} \tfrac{1}{4}a^2(1 - \cos 6\theta)\,\mathrm{d}\theta\)	M1	3.1a
\(= \tfrac{1}{4}a^2\left[\theta - \tfrac{1}{6}\sin 6\theta\right]_0^{\frac{1}{3}\pi}\)	A1	1.1
\(= \tfrac{1}{12}\pi a^2\)	A1	1.1
	[5]

Notes

M1: \(A = \int \frac{1}{2}a^2\sin^2 3\theta\,\mathrm{d}\theta\). Condone missing \(\mathrm{d}\theta\).

A1: Limits correct. Accept alternative limits between 0 and \(\pi\) provided correct multiplication or division of integral is seen or implied at some stage.

M1: Double angle formula used correctly in their integral. Condone missing \(\mathrm{d}\theta\).

A1: \(k\left[\theta - \frac{1}{6}\sin 6\theta\right]\). Condone incorrect or missing limits

A1: SC B4 for an otherwise fully correct answer using limits outside the range 0 to \(\pi\).