Polar coordinates

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 \(k \to \infty\).

Two curves have polar equations \(r = a + 2\cos\theta\) and \(r = 2 + \cos 2\theta\), where \(r \geqslant 0\) and \(a\) is a constant.

1. Show that these curves meet when \[ 2\cos^2\theta - 2\cos\theta + 1 - a = 0. \] Hence show that these curves touch if \(a = \tfrac{1}{2}\) and find the other two values of \(a\) for which the curves touch.
2. Sketch the curves \(r = a + 2\cos\theta\) and \(r = 2 + \cos 2\theta\) on the same diagram in the case \(a = \tfrac{1}{2}\). Give the values of \(r\) and \(\theta\) at the points at which the curves touch and justify the other features you show on your sketch.
3. On two further diagrams, one for each of the other two values of \(a\), sketch both the curves \(r = a + 2\cos\theta\) and \(r = 2 + \cos 2\theta\). Give the values of \(r\) and \(\theta\) at the points at which the curves touch and justify the other features you show on your sketch.

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 \(\dfrac{\pi}{2\sqrt{2}}\).

The point \(P(a\sec \theta, b\tan \theta )\) lies on the hyperbola \[ \dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1\,, \] where \(a>b>0\,\). Show that the equation of the tangent to the hyperbola at \(P\) can be written as \[ bx- ay \sin\theta = ab \cos\theta \,. \]

1. This tangent meets the lines \(\dfrac x a = \dfrac yb\) and \(\dfrac x a =- \dfrac y b\) at \(S\) and \(T\), respectively. How is the mid-point of \(ST\) related to \(P\)?
2. The point \(Q(a\sec \phi, b\tan \phi)\) also lies on the hyperbola and the tangents to the hyperbola at \(P\) and \(Q\) are perpendicular. These two tangents intersect at \((x,y)\). Obtain expressions for \(x^2\) and \(y^2\) in terms of \(a\), \(\theta\) and \(\phi\). Hence, or otherwise, show that \(x^2+y^2 = a^2 -b^2\).

The point with cartesian coordinates \((x,y)\) lies on a curve with polar equation \(r=\f(\theta)\,\). Find an expression for \(\dfrac{\d y}{\d x}\) in terms of \(\f(\theta)\), \(\f'(\theta)\) and \(\tan\theta\,\). Two curves, with polar equations \(r=\f(\theta)\) and \(r=\g(\theta)\), meet at right angles. Show that where they meet \[ \f'(\theta) \g'(\theta) +\f(\theta)\g(\theta) = 0 \,. \] The curve \(C\) has polar equation \(r=\f(\theta)\) and passes through the point given by \(r=4\), \(\theta = - \frac12\pi\). For each positive value of \(a\), the curve with polar equation \(r= a(1+\sin\theta)\) meets~\(C\) at right angles. Find \(\f(\theta)\,\). Sketch on a single diagram the three curves with polar equations \(r= 1+\sin\theta\,\), \ \(r= 4(1+\sin\theta)\) and \(r=\f(\theta)\,\).

Show Solution

\((x, y) = (f(\theta)\cos(\theta), f(\theta)\sin(\theta))\) so \begin{align*} \frac{dy}{d\theta} &= -f(\theta)\sin(\theta) + f'(\theta)\cos(\theta) \\ \frac{dx}{d\theta} &= f(\theta)\cos(\theta) + f'(\theta)\sin(\theta) \\ \frac{dy}{dx} &= \frac{-f(\theta)\sin(\theta) + f'(\theta)\cos(\theta)}{f(\theta)\cos(\theta) + f'(\theta)\sin(\theta) } \\ &= \frac{-f(\theta)\tan(\theta) + f'(\theta)}{f(\theta) + f'(\theta)\tan(\theta) } \end{align*} If the curves meet at right angles then the product of their gradients is \(-1\), ie \begin{align*} \frac{-f(\theta)\tan(\theta) + f'(\theta)}{f(\theta) + f'(\theta)\tan(\theta) } \cdot \frac{-g(\theta)\tan(\theta) + g'(\theta)}{g(\theta) + g'(\theta)\tan(\theta) } &= -1 \\ f(\theta)g(\theta)\tan^2 \theta - f(\theta)g'(\theta)\tan \theta - f'(\theta)g(\theta)\tan \theta + f'(\theta)g'(\theta) &= \\ \quad - \l f(\theta)g(\theta) + f(\theta)g'(\theta)\tan(\theta) + f'(\theta)g(\theta)\tan(\theta) + f'(\theta)g'(\theta)\tan^2 \theta \r \\ \tan^2\theta \l f(\theta)g(\theta) + f'(\theta)g'(\theta) \r + f'(\theta)g'(\theta) + f(\theta)g(\theta) &= 0 \\ (\tan^2\theta + 1) \l f(\theta)g(\theta) + f'(\theta)g'(\theta) \r &= 0 \\ f(\theta)g(\theta) + f'(\theta)g'(\theta) &= 0 \end{align*} \(g(\theta) = a(1+\sin\theta), g'(\theta) = a\cos\theta\) Therefore \(f'(\theta)a\cos \theta+f(\theta)a(1+\sin(\theta)) = 0\) \begin{align*} && \frac{f'(\theta)}{f(\theta)} &= -\sec(\theta) - \tan(\theta) \\ \Rightarrow && \ln(f(\theta)) &= -\ln |\tan(\theta) + \sec(\theta)| + \ln |\cos(\theta)| + C \\ \Rightarrow && f(\theta) &= A \frac{\cos \theta}{\tan \theta + \sec \theta} \\ &&&= A \frac{\cos^2 \theta}{\sin \theta + 1} \\ &&&= A \frac{1-\sin^2 \theta}{\sin \theta + 1} \\ &&&= A (1-\sin \theta) \end{align*} When \(\theta = -\frac12 \pi, r = 4\), so \(A = 2\).

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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 \,. $]

1. Show that under the changes of variable \(x= r\cos\theta\) and \(y = r\sin\theta\), where \(r\) is a function of \(\theta\) with \(r>0\), the differential equation \[ (y+x)\frac{\d y}{\d x} = y-x \] becomes \[ \frac{\d r}{\d\theta} + r=0 \,. \] Sketch a solution in the \(x\)-\(y\) plane.
2. Show that the solutions of \[ \left( y+x -x(x^2+y^2) \right) \, \frac{\d y }{\d x} = y-x - y(x^2+y^2) \] can be written in the form \\ \[ r^2 = \dfrac 1 {1+A\e^{2\theta}}\, \]\\ and sketch the different forms of solution that arise according to the value of \(A\).

A movable point \(P\) has cartesian coordinates \((x,y)\), where \(x\) and \(y\) are functions of \(t\). The polar coordinates of \(P\) with respect to the origin \(O\) are \(r\) and \(\theta\). Starting with the expression \[ \tfrac12 \int r^2 \, \d \theta \] for the area swept out by \(OP\), obtain the equivalent expression \[ \tfrac12 \int \left( x\frac{\d y}{\d t} - y \frac{\d x}{\d t}\right)\d t \,. \tag{\(*\)} \] The ends of a thin straight rod \(AB\) lie on a closed convex curve \(\cal C\). The point \(P\) on the rod is a fixed distance \(a\) from \(A\) and a fixed distance \(b\) from \(B\). The angle between \(AB\) and the positive \(x\) direction is \(t\). As \(A\) and \(B\) move anticlockwise round \(\cal C\), the angle \(t\) increases from \(0\) to \(2\pi\) and \(P\) traces a closed convex curve \(\cal D\) inside \(\cal C\), with the origin \(O\) lying inside \(\cal D\), as shown in the diagram.

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Show that in polar coordinates the gradient of any curve at the point \((r,\theta)\) is \[ \frac{ \ \ \dfrac{\d r }{\d\theta} \tan\theta + r \ \ } { \dfrac{\d r }{\d\theta} -r\tan\theta}\,. \] \noindent

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 \(0\le\theta\le 2\pi\).

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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.

Sketch the curve \(C\) whose polar equation is \[ r=4a\cos2\theta\qquad\mbox{ for }-\tfrac{1}{4}\pi<\theta<\tfrac{1}{4}\pi. \] The ellipse \(E\) has parametric equations \[ x=2a\cos\phi,\qquad y=a\sin\phi. \] Show, without evaluating the integrals, that the perimeters of \(C\) and \(E\) are equal. Show also that the areas of the regions enclosed by \(C\) and \(E\) are equal.

Show Solution

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\begin{align*} && \text{Perimeter}(C) &= \int_{-\pi/4}^{\pi/4} \sqrt{r^2 + \left ( \frac{\d r}{\d \theta} \right)^2} \d \theta \\ &&&= \int_{-\pi/4}^{\pi/4} \sqrt{16a^2 \cos^2 2 \theta + 64a^2 \sin^2 2 \theta } \d \theta \\ &&&= \int_{-\pi/4}^{\pi/4} 4a\sqrt{1 + 3 \sin^2 2 \theta } \d \theta \\ \\ \\ && \text{Perimeter}(D) &= \int_0^{2 \pi} \sqrt{\left ( \frac{\d x}{\d \phi}\right)^2+\left ( \frac{\d y}{\d \phi}\right)^2} \d \phi \\ &&&= \int_0^{2 \pi} \sqrt{ 4a^2 \sin^2 \phi+a^2 \cos^2 \phi} \d \phi \\ &&&= a^2\int_0^{2 \pi} \sqrt{ 1+3 \sin^2 \phi} \d \phi \\ \end{align*} But clearly these two integrals are equal. \begin{align*} && \text{A}(C) &= \frac12 \int_{-\pi/4}^{\pi/4} r^2 \d \theta \\ &&&= \frac12 \int_{-\pi/4}^{\pi/4} 16a^2 \cos^2 2 \theta \d \theta \\ &&&= 8a^2\int_{-\pi/4}^{\pi/4} \cos^2 2 \theta \d \theta \\ &&&= 8a^2 \frac{\pi}{4} = 2\pi a^2 \\ && \text{A}(D) &= 2\pi a^2 \end{align*}

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 \[ \frac{\pi}{5}+\frac{25}{48}\left[\sin\left(\frac{\pi}{5}\right)-\sin\left(\frac{2\pi}{5}\right)\right]. \]

The parametric equations \(E_{1}\) and \(E_{2}\) define the same ellipse, in terms of the parameters \(\theta_{1}\) and \(\theta_{2}\), (though not referred to the same coordinate axes). \begin{alignat*}{2} E_{1}:\qquad & x=a\cos\theta_{1}, & \quad & y=b\sin\theta_{1},\\ E_{2}:\qquad & x=\dfrac{k\cos\theta_{2}}{1+e\cos\theta_{2}}, & \quad & y=\dfrac{k\sin\theta_{2}}{1+e\cos\theta_{2}}, \end{alignat*} where \(0< b< a,\) \(0< e< 1\) and \(0< k\). Find the position of the axes for \(E_{2}\) relative to the axes for \(E_{1}\) and show that \(k=a(1-e^{2})\) and \(b^{2}=a^{2}(1-e^{2}).\) {[}The standard polar equation of an ellipse is \(r=\dfrac{\ell}{1+e\cos\theta}.]\) By considering expressions for the length of the perimeter of the ellipse, or otherwise, prove that \[ \int_{0}^{\pi}\sqrt{1-e^{2}\cos^{2}\theta}\,\mathrm{d}\theta=\int_{0}^{\pi}\frac{1-e^{2}}{(1+e\cos\theta)^{2}}\sqrt{1+e^{2}+2e\cos\theta}\,\mathrm{d}\theta. \] Given that \(e\) is so small that \(e^{6}\) may be neglected, show that the value of either integral is \[ \tfrac{1}{64}\pi(64-16e^{2}-3e^{4}). \]

Show by means of a sketch that the parabola \(r(1+\cos\theta)=1\) cuts the interior of the cardioid \(r=4(1+\cos\theta)\) into two parts. Show that the total length of the boundary of the part that includes the point \(r=1,\theta=0\) is \(18\sqrt{3}+\ln(2+\sqrt{3}).\)

Show Solution

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The curves will intersect when: \begin{align*} && \frac{1}{1+\cos \theta} &= 4 (1 + \cos \theta) \\ \Rightarrow && 1 + \cos \theta &= \pm \frac{1}{2} \\ \Rightarrow && \cos \theta &= -\frac12 \\ \Rightarrow && \theta &= \pm \frac{2\pi}{3}, \end{align*} Therefore we can measure the two sides of the boundaries. For the cardioid it will be: \begin{align*} s &= \int_{-2\pi/3}^{2 \pi /3} \sqrt{r^2 + \left ( \frac{\d r}{\d \theta} \right)^2} \d \theta \\ &= \int_{-2\pi/3}^{2 \pi /3}\sqrt{r^2 + \left ( \frac{\d r}{\d \theta} \right)^2} \d \theta \\ &= \int_{-2\pi/3}^{2 \pi /3}\sqrt{16(1 + \cos \theta)^2 + 16 \sin^2 \theta} \d \theta \\ &= 4\int_{-2\pi/3}^{2 \pi /3}\sqrt{2 + 2 \cos \theta} \d \theta \\ &= 8\int_{-2\pi/3}^{2 \pi /3}\sqrt{\cos^2 \frac{\theta}{2}} \d \theta \\ &= 8\int_{-2\pi/3}^{2 \pi /3}|\cos \frac{\theta}{2}| \d \theta \\ &= 16\int_{\pi}^{2\pi/3}(-\cos \frac{\theta}{2}) \d \theta + 8\int_{-\pi}^{\pi}\cos \frac{\theta}{2} \d \theta \\ &= 16 \cdot \left [ 2\sin \frac{\theta}{2}\right]_{\pi}^{2\pi/3}+ 8 \cdot 4 \\ &= 16 \cdot (\sqrt{3}-2) + 8 \cdot 4 \\ &= 16\sqrt{3} \end{align*} For the parabola we have that \(\sqrt{x^2+y^2} + x = 1 \Rightarrow x^2 + y^2 = 1 - 2x + x^2 \Rightarrow y^2 = 1-2x\). So we can parameterise our parabola as \(y = t, x = \frac{1-t^2}2\). And we are interested in the points \(t = -\sqrt{3}\) and \(t =\sqrt3\) \begin{align*} &&s &= \int_{-\sqrt3}^\sqrt3 \sqrt{\left ( \frac{\d x}{\d t} \right)^2 + \left ( \frac{\d y}{\d t} \right)^2 } \d t \\ &&&= \int_{-\sqrt3}^\sqrt3 \sqrt{t^2+1^2} \d t \\ \sinh u = t, \frac{\d t}{\d u} = \cosh u&&&= \int_{-\sinh^{-1} \sqrt3}^{\sinh^{-1}\sqrt3} \cosh^2 u \d u \\ &&&= \left [\frac12 u + \frac14 \sinh(2u) \right ]_{-\sinh^{-1} \sqrt3}^{\sinh^{-1}\sqrt3} \\ &&&= \sinh^{-1} \sqrt{3} + 2\sqrt{3} \\ &&&= \ln(2 + \sqrt{3}) + 2\sqrt{3} \end{align*} Therefore the total distance is as required.

1. Show that in polar coordinates, the gradient of any curve at the point \((r,\theta)\) is \[ \left.\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}\tan\theta+r\right)\right/\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}-r\tan\theta\right). \]
   

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2. A mirror is designed so that any ray of light which hits one side of the mirror and which is parallel to a certain fixed line \(L\) is reflected through a fixed point \(O\) on \(L\). For any ray hitting the mirror, the normal to the mirror at the point of reflection bisects the angle between the incident ray and the reflected ray, as shown in the figure. Prove that the mirror intersects any plane containing \(L\) in a parabola.

Show that, for a given constant \(\gamma\) \((\sin\gamma\neq0)\) and with suitable choice of the constants \(A\) and \(B\), the line with cartesian equation \(lx+my=1\) has polar equations \[ \frac{1}{r}=A\cos\theta+B\cos(\theta-\gamma). \] The distinct points \(P\) and \(Q\) on the conic with polar equations \[ \frac{a}{r}=1+e\cos\theta \] correspond to \(\theta=\gamma-\delta\) and \(\theta=\gamma+\delta\) respectively, and \(\cos\delta\neq0.\) Obtain the polar equation of the chord \(PQ.\) Hence, or otherwise, obtain the equation of the tangent at the point where \(\theta=\gamma.\) The tangents at \(L\) and \(M\) to a conic with focus \(S\) meet at \(T.\) Show that \(ST\) bisects the angle \(LSM\) and find the position of the intersection of \(ST\) and \(LM\) in terms of your chosen parameters for \(L\) and \(M.\)