FM June 2022 p13 q06
4244
The curve C has Cartesian equation \(x^2 + xy + y^2 = a\), where \(a\) is a positive constant.
(a) Show that the polar equation of C is \(r^2 = \frac{2a}{2 + \sin 2\theta}\).
(b) Sketch the part of C for \(0 \leq \theta \leq \frac{1}{4}\pi\).
The region R is enclosed by this part of C, the initial line and the half-line \(\theta = \frac{1}{4}\pi\).
(c) It is given that \(\sin 2\theta\) may be expressed as \(\frac{2 \tan \theta}{1 + \tan^2 \theta}\). Use this result to show that the area of R is
\(\frac{1}{2} a \int_{0}^{\frac{1}{4}\pi} \frac{1 + \tan^2 \theta}{1 + \tan \theta + \tan^2 \theta} \, d\theta\)
and use the substitution \(t = \tan \theta\) to find the exact value of this area.
Solution
(a) Start with the polar coordinates: \(x = r \cos \theta\) and \(y = r \sin \theta\). Substitute into the Cartesian equation:
\(x^2 + xy + y^2 = r^2 \cos^2 \theta + r^2 \cos \theta \sin \theta + r^2 \sin^2 \theta = a\).
This simplifies to \(r^2 (\cos^2 \theta + \sin^2 \theta + \cos \theta \sin \theta) = a\).
Using \(\cos^2 \theta + \sin^2 \theta = 1\) and \(\sin 2\theta = 2 \sin \theta \cos \theta\), we have:
\(r^2 (1 + \frac{1}{2} \sin 2\theta) = a\).
Thus, \(r^2 = \frac{2a}{2 + \sin 2\theta}\).
(b) The sketch should show a polar graph with the curve in the correct domain \(0 \leq \theta \leq \frac{1}{4}\pi\), with \(r\) strictly decreasing.
(c) Given \(\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}\), substitute \(t = \tan \theta\), then \(\frac{d\theta}{dt} = \frac{1}{1 + t^2}\).
The integral becomes:
\(R = \frac{1}{2} a \int_{0}^{1} \frac{1}{t^2 + t + 1} \, dt\).
Complete the square: \(t^2 + t + 1 = (t + \frac{1}{2})^2 + \frac{3}{4}\).
Integrate: \(R = \frac{1}{\sqrt{3}} [\arctan(\frac{2t + 1}{\sqrt{3}})]_{0}^{1}\).
Evaluate: \(R = \frac{\pi a}{6\sqrt{3}}\).
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