FP2 June 2015 Q6

EdexcelOld spec11 marksPolar Coordinates

Figure 1: the curve C from O, through A, to the point 6a on the initial line; the region R between OA, the curve and the initial line is shaded — Figure 1

The curve \(C\), shown in Figure 1, has polar equation \[r = 3a(1 + \cos\theta), \qquad 0 \leqslant \theta \lt \pi\]

The tangent to \(C\) at the point \(A\) is parallel to the initial line.

(a) Find the polar coordinates of \(A\). (6)

The finite region \(R\), shown shaded in Figure 1, is bounded by the curve \(C\), the initial line and the line \(OA\).

(b) Use calculus to find the area of the shaded region \(R\), giving your answer in the form \(a^2\left(p\pi + q\sqrt{3}\right)\), where \(p\) and \(q\) are rational numbers. (5)

Mark scheme (a)

Scheme	Marks
\(r\sin\theta = 3a\sin\theta + 3a\sin\theta\cos\theta\) OR \(3a\sin\theta + \dfrac{3}{2}a\sin 2\theta\)	M1
\(\dfrac{\mathrm{d}(r\sin\theta)}{\mathrm{d}\theta} = 3a\cos\theta + 3a\cos^2\theta - 3a\sin^2\theta\) OR \(3a\cos\theta + 3a\cos 2\theta\)	dM1
\(2\cos^2\theta + \cos\theta - 1 = 0\) terms in any order	A1
\((2\cos\theta - 1)(\cos\theta + 1) = 0\)
\(\cos\theta = \dfrac{1}{2} \quad \theta = \dfrac{\pi}{3} \qquad (\theta = \pi\) need not be seen\()\)	ddM1A1
\(r = 3a\times\dfrac{3}{2} = \dfrac{9}{2}a\)	A1
	(6)

Notes

M1: using \(r\sin\theta\) \(r\cos\theta\) scores M0

dM1: Attempt the differentiation of \(r\sin\theta\), inc use of product rule or \(\sin 2\theta = 2\sin\theta\cos\theta\)

A1: Correct 3 term quadratic in \(\cos\theta\)

ddM1: dep on both M marks. Solve their quadratic (usual rules) giving one or two roots

A1: Correct quadratic solved to give \(\theta = \dfrac{\pi}{3}\)

A1: Correct \(r\) obtained No need to see coordinates together in brackets
Special Case: If \(r\cos\theta\) used, score M0M1A0M0A0A0to

Mark scheme (b)

Scheme	Marks
Area \(= \dfrac{1}{2}\displaystyle\int r^2\,\mathrm{d}\theta = \frac{1}{2}\int_0^{\frac{\pi}{3}} 9a^2(1 + \cos\theta)^2\,\mathrm{d}\theta\)
\(= \dfrac{9a^2}{2}\displaystyle\int_0^{\frac{\pi}{3}}\left(1 + 2\cos\theta + \cos^2\theta\right)\mathrm{d}\theta\)	M1
\(= \dfrac{9a^2}{2}\displaystyle\int_0^{\frac{\pi}{3}}\left(1 + 2\cos\theta + \frac{1}{2}(\cos 2\theta + 1)\right)\mathrm{d}\theta\)	M1
\(= \dfrac{9a^2}{2}\left[\theta + 2\sin\theta + \dfrac{1}{2}\left(\dfrac{1}{2}\sin 2\theta + \theta\right)\right]_0^{\frac{\pi}{3}}\)	dM1A1
\(\dfrac{9a^2}{2}\left[\dfrac{\pi}{3} + \sqrt{3} + \dfrac{1}{4}\times\dfrac{\sqrt{3}}{2} + \dfrac{\pi}{6}\ (-0)\right]\)
\(\dfrac{9a^2}{2}\left[\dfrac{\pi}{2} + \dfrac{9\sqrt{3}}{8}\right] = \left(\dfrac{9\pi}{4} + \dfrac{81\sqrt{3}}{16}\right)a^2\)	A1
	(5)
	(11 marks)

Notes

M1: Use of correct area formula, \(\dfrac{1}{2}\) may be seen later, inc squaring the bracket to obtain 3 terms - limits need not be shown.

M1: Use double angle formula (formula to be of form \(\cos^2\theta = \pm\dfrac{1}{2}(\cos 2\theta \pm 1)\)) to obtain an integrable function - limits need not be shown, \(\dfrac{1}{2}\) from area formula may be missing,

dM1: attempt the integration - limits not needed – dep on 2^nd M mark but not the first

A1: correct integration – substitution of limits not required

A1: correct final answer any equivalent provided in the demanded form.