Polar coordinates and polar curvesEdexcel A-Level Further Maths: Revision notes

A point is given by its distance $r$ from the pole $O$ and the angle $\theta$ anticlockwise from the initial line (the positive $x$-axis): $(r,\theta)$. Usually $r\ge0$ and either $-\pi<\theta\le\pi$ or $0\le\theta<2\pi$. $x=r\cos\theta,\quad y=r\sin\theta,\quad r^2=x^2+y^2,\quad\tan\theta=\frac yx.$ Example: $\left(4,\frac{2\pi}{3}\right)$ has $x=4\cos\frac{2\pi}{3}=-2$ and $y=4\sin\frac{2\pi}{3}=2\sqrt3$. To convert $(-1,-1)$: $r=\sqrt2$ and $\tan\theta=1$, but the point is in the third quadrant, so $\theta=-\frac{3\pi}{4}$ (not $\frac{\pi}{4}$).

Replace $x$, $y$ and $x^2+y^2$ using the formulae above, or the reverse. Multiplying by $r$ is often the key step.

• $r=4\cos\theta\Rightarrow r^2=4r\cos\theta\Rightarrow x^2+y^2=4x$, the circle $(x-2)^2+y^2=4$.
• $r=3\sec\theta\Rightarrow r\cos\theta=3\Rightarrow x=3$.
• $r^2=9\cos2\theta\Rightarrow r^4=9r^2(\cos^2\theta-\sin^2\theta)\Rightarrow(x^2+y^2)^2=9(x^2-y^2)$.

• $r=a$ is a circle with centre the pole and radius $a$.
• $r=2a\cos\theta$ is a circle of radius $a$ through the pole, centre $(a,0)$ (for $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$). $r=2a\sin\theta$ has its centre on the $y$-axis.
• $r=p\sec(\alpha-\theta)$, i.e. $r\cos(\theta-\alpha)=p$, is a straight line at perpendicular distance $p$ from the pole, whose perpendicular from the pole makes angle $\alpha$ with the initial line. $r=3\sec\theta$ is $x=3$.
• $r=k\theta$ ($\theta\ge0$) is an Archimedean spiral: $r$ grows steadily with $\theta$, so each turn is a fixed distance $2\pi k$ further out.

$r=a(1+\cos\theta)$ is a cardioid (heart shape): greatest $r=2a$ at $\theta=0$, and $r=0$ at $\theta=\pi$, where the curve touches the pole. $r=a(1-\cos\theta)$ is the mirror image, pointing the other way. $r=a(3+2\cos\theta)$ is a limaçon: $r$ stays positive, from $5a$ at $\theta=0$ to $a$ at $\theta=\pi$, so it never reaches the pole and has no loop; it is a rounded, slightly egg-shaped curve. All of these contain $\cos\theta$, which is even, so they are symmetrical about the initial line.

$r=a\cos^2\theta$: $r\ge0$ always, with $r=a$ at $\theta=0$ and $\theta=\pi$ and $r=0$ at $\theta=\pm\frac{\pi}{2}$. The curve has two lobes, one each side of the pole along the $x$-axis, touching at the pole, with the $y$-axis as tangent there. $r^2=a^2\cos2\theta$ is a lemniscate (figure of eight). $r^2\ge0$ needs $\cos2\theta\ge0$, so it exists only for $-\frac{\pi}{4}\le\theta\le\frac{\pi}{4}$ and $\frac{3\pi}{4}\le\theta\le\frac{5\pi}{4}$, giving two loops that meet at the pole with tangents $\theta=\pm\frac{\pi}{4}$. The greatest $r$ is $a$.

1. Find the range of $r$: the greatest and least values, and where $r=0$ (curve through the pole).
2. Use symmetry: $\cos\theta$ gives symmetry about the initial line; $\sin\theta$ about $\theta=\frac{\pi}{2}$; $r(\theta+\pi)=r(\theta)$ gives symmetry about the pole.
3. Tabulate $r$ at key angles such as multiples of $\frac{\pi}{6}$ or $\frac{\pi}{4}$ and plot the points.
4. Label where the curve crosses the axes and the pole, and note any value of $\theta$ for which $r$ is undefined or negative. If $r<0$ the point is plotted in the opposite direction, at angle $\theta+\pi$.