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FM June 2024 p13 q07
4177
The curve C has polar equation \(r^2 = \sin 2\theta \cos \theta\), for \(0 \leq \theta \leq \pi\).
Sketch C and state the equation of the line of symmetry.
Find a Cartesian equation for C.
Find the total area enclosed by C.
Find the greatest distance of a point on C from the pole.
Solution
(a) The sketch of the curve shows two loops symmetric about the line \(\theta = \frac{1}{2} \pi\).
(b) Start with \(r^2 = \sin 2\theta \cos \theta\). Use \(\sin 2\theta = 2 \sin \theta \cos \theta\) and polar to Cartesian conversions: \(x = r \cos \theta\), \(y = r \sin \theta\). Thus, \(r^5 = 2x^2\). Therefore, \(\left( x^2 + y^2 \right)^{\frac{3}{2}} = 2x^2 y\).
(c) The total area enclosed by the curve is given by \(\frac{1}{2} \int_0^\pi \sin(2\theta) \cos \theta \, d\theta\). Simplifying, \(\int \sin \theta \cos^2 \theta \, d\theta = -\frac{1}{3} \cos^3 \theta\). Evaluating from 0 to \(\pi\), the area is \(\frac{2}{3}\).
(d) Differentiate \(r^2 = \sin 2\theta \cos \theta\) with respect to \(\theta\) to find the maximum distance. Solve \(6 \cos^3 \theta - 4 \cos \theta = 0\) to find \(\cos^2 \theta = \frac{2}{3}\). The greatest distance is \(r = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{9} \approx 0.877\).
