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STEP II 1992 question 3 solution
From The Student RoomTSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1992 question 3 solution The circumference of the larger circle is four times the circumference of the smaller circle. Hence, the smaller circle rotates four times as quickly as OP. Let the (variable) point at which the two circles touch be Q, so that OQ and OP are collinear, with OQ = 4. Let the angle between QP and PB be θ; then θ = -4φ.
Solution by generalebriety. | 
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![\displaystyle x = OP\cos\phi + PB\cos (\phi +\theta ) = 3\cos\phi + \cos (-3\phi ) = 3\cos\phi + \cos 3\phi \\
y = OP\sin\phi + PB\sin (\phi +\theta ) = 3\sin\phi + \sin (-3\phi ) = 3\sin\phi - \sin 3\phi \\
\therefore (x, y) = (3\cos\phi + \cos 3\phi , 3\sin\phi - \sin 3\phi ).\\
A = \int_0^{2\pi} x \frac{\text{d}y}{\text{d}\phi} \text{d}\phi = \int_0^{2\pi} (3\cos\phi + \cos 3\phi )(3\cos\phi + 3\cos 3\phi ) \text{d}\phi \\
= \int_0^{2\pi} (9\cos^2 \phi - 6\cos\phi\cos 3\phi - 3\cos^2 3\phi) \text{d}\phi \\
= \frac{1}{2} \int_0^{2\pi} (9\cos 2\phi + 9 - 6\cos 2\phi - 6\cos 4\phi - 3\cos 6\phi - 3) \text{d}\phi \\
= \frac{1}{2} \left[6\phi + \tfrac{9}{2} \sin 2\phi - 3\sin 2\phi - \tfrac{3}{4} \sin 4\phi - \tfrac{1}{2} \sin 6\phi \right]_0^{2\pi} \\
= 3\left[\phi\right]_0^{2\pi} = 6\pi .](http://thestudentroom.co.uk/../latexrender/pictures/b5b3aef053b1fa902f9d4a5011e14748.png "\displaystyle x = OP\cos\phi + PB\cos (\phi +\theta ) = 3\cos\phi + \cos (-3\phi ) = 3\cos\phi + \cos 3\phi \\
y = OP\sin\phi + PB\sin (\phi +\theta ) = 3\sin\phi + \sin (-3\phi ) = 3\sin\phi - \sin 3\phi \\
\therefore (x, y) = (3\cos\phi + \cos 3\phi , 3\sin\phi - \sin 3\phi ).\\
A = \int_0^{2\pi} x \frac{\text{d}y}{\text{d}\phi} \text{d}\phi = \int_0^{2\pi} (3\cos\phi + \cos 3\phi )(3\cos\phi + 3\cos 3\phi ) \text{d}\phi \\
= \int_0^{2\pi} (9\cos^2 \phi - 6\cos\phi\cos 3\phi - 3\cos^2 3\phi) \text{d}\phi \\
= \frac{1}{2} \int_0^{2\pi} (9\cos 2\phi + 9 - 6\cos 2\phi - 6\cos 4\phi - 3\cos 6\phi - 3) \text{d}\phi \\
= \frac{1}{2} \left[6\phi + \tfrac{9}{2} \sin 2\phi - 3\sin 2\phi - \tfrac{3}{4} \sin 4\phi - \tfrac{1}{2} \sin 6\phi \right]_0^{2\pi} \\
= 3\left[\phi\right]_0^{2\pi} = 6\pi .")
