FP2 June 2007 Q4

EdexcelOld spec14 marksPolar Coordinates

Loop of the curve C from O, with points P and Q on it; shaded region R between OP, OQ and C; initial line marked

The diagram above shows a sketch of the curve \(C\) with polar equation \[r = 4\sin\theta\cos^2\theta, \qquad 0 \leqslant \theta \lt \frac{\pi}{2}.\]

The tangent to \(C\) at the point \(P\) is perpendicular to the initial line.

(a) Show that \(P\) has polar coordinates \(\left(\dfrac{3}{2}, \dfrac{\pi}{6}\right)\). (6)

The point \(Q\) on \(C\) has polar coordinates \(\left(\sqrt{2}, \dfrac{\pi}{4}\right)\).

The shaded region \(R\) is bounded by \(OP\), \(OQ\) and \(C\), as shown in the diagram above.

(b) Show that the area of \(R\) is given by \[\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \left(\sin^2 2\theta\cos 2\theta + \frac{1}{2} - \frac{1}{2}\cos 4\theta\right)\mathrm{d}\theta\] (3)

(c) Hence, or otherwise, find the area of \(R\), giving your answer in the form \(a + b\pi\), where \(a\) and \(b\) are rational numbers. (5)

Mark scheme (a)

Scheme	Marks
\(x = r\cos\theta = 4\sin\theta\cos^3\theta\)	M1
\(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = 4\cos^4\theta - 12\cos^2\theta\sin^2\theta\) any correct expression	M1A1
Solving \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = 0\) \(\left[\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = 0 \Rightarrow 4\cos^2\theta(\cos^2\theta - 3\sin^2\theta) = 0\right]\)	M1
\(\sin\theta = \dfrac{1}{2}\) or \(\cos\theta = \dfrac{\sqrt{3}}{2}\) or \(\tan\theta = \dfrac{1}{\sqrt{3}} \Rightarrow \theta = \dfrac{\pi}{6}\) AG	A1 cso
\(r = 4\sin\dfrac{\pi}{6}\cos^2\dfrac{\pi}{6} = \dfrac{3}{2}\) AG	A1cso
	(6)

Notes

So many ways \(x\) may be expressing e.g. \(2\sin 2\theta\cos^2\theta,\ \sin 2\theta(1 + \cos 2\theta),\ \sin 2\theta + (1/2)\sin 4\theta\) leading to many results for \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta}\)

Some relevant equations in solving \([(1 - 4\sin^2\theta) = 0,\ (4\cos^2\theta - 3) = 0,\ (1 - 3\tan^2\theta) = 0,\ \cos 3\theta = 0]\)

Showing that \(\theta = \dfrac{\pi}{6}\) satisfies \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = 0\), allow M1 A1 providing \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta}\) correct

Starting with \(x = r\sin\theta\) can gain M0M1M1

Mark scheme (b)

Scheme	Marks
\(A = \dfrac{1}{2}\displaystyle\int_{\pi/6}^{\pi/4} r^2\,\mathrm{d}\theta = \frac{1}{2} \cdot 16\int_{\pi/6}^{\pi/4} \sin^2\theta\cos^4\theta\,\mathrm{d}\theta\)
\(8\sin^2\theta\cos^4\theta = 2\cos^2\theta(4\sin^2\theta\cos^2\theta) = 2\cos^2\theta\sin^2 2\theta\)	M1
\(= (\cos 2\theta + 1)\sin^2 2\theta\)	M1
\(= \cos 2\theta\sin^2 2\theta + \dfrac{1 - \cos 4\theta}{2}\) = Answer AG	A1 cso
	(3)

Notes

First M1 for use of double angle formula for \(\sin 2A\)
Second M1 for use of \(\cos 2A = 2\cos^2 A - 1\)
Answer given: must be intermediate step, as shown, and no incorrect work

Mark scheme (c)

Scheme	Marks
Area \(= \left[\dfrac{1}{6}\sin^3 2\theta + \dfrac{\theta}{2} - \dfrac{\sin 4\theta}{8}\right]_{\left(\frac{\pi}{6}\right)}^{\left(\frac{\pi}{4}\right)}\) ignore limits	M1A1
\(= \left(\dfrac{1}{6}\sin^3\dfrac{\pi}{2} + \dfrac{\pi}{8} - \dfrac{\sin\pi}{8}\right) - \left(\dfrac{1}{6}\sin^3\dfrac{\pi}{3} + \dfrac{\pi}{12} - \dfrac{\sin\frac{2\pi}{3}}{8}\right)\) (sub. limits)	M1
\(= \left(\dfrac{1}{6} + \dfrac{\pi}{8}\right) - \left(\dfrac{\sqrt{3}}{16} + \dfrac{\pi}{12} - \dfrac{\sqrt{3}}{16}\right) = \dfrac{1}{6},\ + \dfrac{\pi}{24}\) both cao	A1, A1
	(5)
	(14 marks)

Notes

For first M, of the form \(a\sin^3 2\theta + \dfrac{\theta}{2} \pm b\sin 4\theta\) (Allow if two of correct form)
On ePen the order of the As in answer is as written