Problems

The polar curves \(C_1\) and \(C_2\) are defined for \(0 \leqslant \theta \leqslant \pi\) by \[r = k(1 + \sin\theta)\] \[r = k + \cos\theta\] respectively, where \(k\) is a constant greater than \(1\).

1. Sketch the curves on the same diagram. Show that if \(\theta = \alpha\) at the point where the curves intersect, \(\tan\alpha = \dfrac{1}{k}\).
2. The region A is defined by the inequalities \[0 \leqslant \theta \leqslant \alpha \quad \text{and} \quad r \leqslant k(1+\sin\theta)\,.\] Show that the area of A can be written as \[\frac{k^2}{4}(3\alpha - \sin\alpha\cos\alpha) + k^2(1 - \cos\alpha)\,.\]
3. The region B is defined by the inequalities \[\alpha \leqslant \theta \leqslant \pi \quad \text{and} \quad r \leqslant k + \cos\theta\,.\] Find an expression in terms of \(k\) and \(\alpha\) for the area of B.
4. The total area of regions A and B is denoted by \(R\). The area of the region enclosed by \(C_1\) and the lines \(\theta = 0\) and \(\theta = \pi\) is denoted by \(S\). The area of the region enclosed by \(C_2\) and the lines \(\theta = 0\) and \(\theta = \pi\) is denoted by \(T\). Show that as \(k \to \infty\), \[\frac{R}{T} \to 1\] and find the limit of \[\frac{R}{S}\] as \(k \to \infty\).

1. Sketch the curve \(y = \cos x + \sqrt{\cos 2x}\) for \(-\frac{1}{4}\pi \leqslant x \leqslant \frac{1}{4}\pi\).
2. The equation of curve \(C_1\) in polar co-ordinates is \[ r = \cos\theta + \sqrt{\cos 2\theta} \qquad -\tfrac{1}{4}\pi \leqslant \theta \leqslant \tfrac{1}{4}\pi. \] Sketch the curve \(C_1\).
3. The equation of curve \(C_2\) in polar co-ordinates is \[ r^2 - 2r\cos\theta + \sin^2\theta = 0 \qquad -\tfrac{1}{4}\pi \leqslant \theta \leqslant \tfrac{1}{4}\pi. \] Find the value of \(r\) when \(\theta = \pm\frac{1}{4}\pi\). Show that, when \(r\) is small, \(r \approx \frac{1}{2}\theta^2\). Sketch the curve \(C_2\), indicating clearly the behaviour of the curve near \(r=0\) and near \(\theta = \pm\frac{1}{4}\pi\). Show that the area enclosed by curve \(C_2\) and above the line \(\theta = 0\) is \(\dfrac{\pi}{2\sqrt{2}}\).

In this question, \(r\) and \(\theta\) are polar coordinates with \(r \ge0\) and \(- \pi < \theta\le \pi\), and \(a\) and \(b\) are positive constants. Let \(L\) be a fixed line and let \(A\) be a fixed point not lying on \(L\). Then the locus of points that are a fixed distance (call it \(d\)) from \(L\) measured along lines through \(A\) is called a conchoid of Nicomedes.

1. Show that if \[ \vert r- a \sec\theta \vert = b\,, \tag{\(*\)} \] where \(a>b\), then \(\sec\theta >0\). Show that all points with coordinates satisfying (\(*\)) lie on a certain conchoid of Nicomedes (you should identify \(L\), \(d\) and \(A\)). Sketch the locus of these points.
2. In the case \(a < b\), sketch the curve (including the loop for which \(\sec\theta<0\)) given by \[ \vert r- a \sec\theta \vert = b\, . \] Find the area of the loop in the case \(a=1\) and \(b=2\). [Note: $ %\displaystyle \int \! \sec\theta \,\d \theta = \ln \vert \sec\theta + \tan\theta \vert + C \,. $]

Show that the equation (in plane polar coordinates) \(r=\cos\theta\), for $-\frac{1}{2}\pi \le \theta \le \frac{1}{2}\pi$, represents a circle. Sketch the curve \(r=\cos2\theta\) for \(0\le\theta\le 2\pi\), and describe the curves \(r=\cos2n\theta\), where \(n\) is an integer. Show that the area enclosed by such a curve is independent of \(n\). Sketch also the curve \(r=\cos3\theta\) for \(0\le\theta\le 2\pi\).

The curve \(C\) has the equation \(x^3+y^3 = 3xy\).

1. Show that there is no point of inflection on \(C\). You may assume that the origin is not a point of inflection.
2. The part of \(C\) which lies in the first quadrant is a closed loop touching the axes at the origin. By converting to polar coordinates, or otherwise, evaluate the area of this loop.

The curve \(C\) has the differential equation in polar coordinates \[ \frac{\mathrm{d}^{2}r}{\mathrm{d}\theta^{2}}+4r=5\sin3\theta,\qquad\text{for }\quad\frac{\pi}{5}\leqslant\theta\leqslant\frac{3\pi}{5}, \] and, when \(\theta=\dfrac{\pi}{2},\) \(r=1\) and \(\dfrac{\mathrm{d}r}{\mathrm{d}\theta}=-2.\) Show that \(C\) forms a closed loop and that the area of the region enclosed by \(C\) is \[ \frac{\pi}{5}+\frac{25}{48}\left[\sin\left(\frac{\pi}{5}\right)-\sin\left(\frac{2\pi}{5}\right)\right]. \]