Sketching polar curvesAQA A-Level Further Maths: Revision notes

A polar curve is a set of points $(r,\theta)$ with $r=f(\theta)$. To sketch one, work through these checks:

1. Range: where is $r$ real, and what range of $\theta$ is given?
2. Greatest and least $r$ (use the range of $\sin\theta$, $\cos\theta$) and where they occur.
3. Zeros of $r$: the values of $\theta$ where the curve passes through the pole; the curve is tangent to those lines there.
4. Symmetry: if $f(-\theta)=f(\theta)$ the curve is symmetrical about the initial line; if $f(\pi-\theta)=f(\theta)$ it is symmetrical about $\theta=\frac{\pi}{2}$.
5. Negative $r$: the point $(-r,\theta)$ is plotted $r$ units from the pole in the direction $\theta+\pi$.
6. Plot a few key points ($\theta=0,\frac{\pi}{2},\pi,\frac{3\pi}{2}$ and the zeros) and join smoothly.

• $r=a$ is a circle, centre the pole, radius $a$.
• $\theta=\alpha$ is a half-line through the pole at angle $\alpha$ to the initial line.
• $r=a\cos\theta$ is a circle through the pole with diameter $a$ along the initial line (centre $\left(\frac a2,0\right)$).
• $r=a\sin\theta$ is a circle through the pole with diameter $a$ along the line $\theta=\frac{\pi}{2}$ (centre $\left(0,\frac a2\right)$).
• $r=a\theta$ ($\theta\ge0$) is an Archimedean spiral: $r$ grows steadily with $\theta$, so it winds outward from the pole.

Curves of the form $r=a+b\cos\theta$ (or with $\sin\theta$) depend on the ratio of $a$ to $b$ (taking $a,b>0$):

• $a=b$: a cardioid, heart-shaped, passing through the pole once with a cusp. Example: $r=4(1+\cos\theta)$ has $r$ from $8$ (at $\theta=0$) down to $0$ (at $\theta=\pi$).
• $a>b$: a smooth limaçon with no loop (convex if $a\ge2b$, with a dimple if $b<a<2b$).
• $a<b$: a limaçon with an inner loop, because $r$ becomes negative for some $\theta$. Example: $r=2+3\cos\theta$ has $r=0$ at $\theta=2.30$ and $3.98$, $r=5$ at $\theta=0$ and $r=-1$ at $\theta=\pi$.

With $\cos\theta$ the curve is symmetrical about the initial line; with $\sin\theta$ it is symmetrical about $\theta=\frac{\pi}{2}$.

• Rose $r=a\cos n\theta$ (or $a\sin n\theta$): the petals have length $a$. There are $n$ petals when $n$ is odd and $2n$ petals when $n$ is even. Example: $r=3\cos2\theta$ has $4$ petals, with tips at $\theta=0,\frac{\pi}{2},\pi,\frac{3\pi}{2}$ and $r=0$ at $\theta=\frac{\pi}{4},\frac{3\pi}{4},\dots$. At $\theta=\frac{\pi}{2}$, $r=-3$, which is the point $(0,-3)$.
• Lemniscate $r^2=a^2\cos2\theta$: $r^2\ge0$ needs $\cos2\theta\ge0$, so it exists only for $-\frac{\pi}{4}\le\theta\le\frac{\pi}{4}$ and $\frac{3\pi}{4}\le\theta\le\frac{5\pi}{4}$. The result is a figure of eight with greatest $r=a$ at $\theta=0$ and $\pi$, passing through the pole at $\theta=\pm\frac{\pi}{4},\ \frac{3\pi}{4},\ \frac{5\pi}{4}$.

Sketch $r=2+3\cos\theta$. Greatest $r=5$ at $\theta=0$. $r=0$ when $\cos\theta=-\frac23$, i.e. $\theta=2.30$ and $3.98$. Between these $r<0$ (least value $-1$ at $\theta=\pi$), giving an inner loop. The curve is symmetrical about the initial line. Mark the intercepts on the axes, the greatest $r$ and the pole, and label the curve.

Intersections of two polar curves are found by equating the expressions for $r$ (or substituting). For $r^2=16\cos2\theta$ and $r=2\sqrt2$: $\cos2\theta=\frac12$, giving $\theta=\pm\frac{\pi}{6},\pm\frac{5\pi}{6}$ and four points at $r=2\sqrt2$.