Polar Curve Sketching

Sketching a polar curve $r=f(\theta)$r=f(θ) is one of the most visually rewarding skills in AQA Further Mathematics, and it is the necessary precursor to every polar area and tangent question — you cannot integrate $\tfrac12\int r^2\,\mathrm d\theta$21​∫r2dθ correctly without first knowing the shape, the symmetry, and the interval over which the curve is traced. This lesson sets out a systematic sketching procedure and then applies it to the five families that appear in the specification: circles, cardioids, limaçons, rose curves and spirals, with the lemniscate as a stretch.

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1 Spec mapping

Polar curve sketching is compulsory pure content for Papers 1 and 2 of AQA 7367. The skill itself is mostly AO1 (carry out the standard procedure), but a good sketch demands genuine AO2 reasoning: choosing the tracing interval, exploiting symmetry, and interpreting where $r<0$r<0. It is rarely examined in isolation — almost always a sketch is part (a) of a question whose part (b) asks for an area or a tangent, so a wrong sketch cascades into lost marks downstream. Treat the sketch as load-bearing, not decorative.

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2 Core theory — a systematic sketching procedure

Given $r=f(\theta)$r=f(θ), work through the same checklist every time:

1. Tabulate $r$r at the convenient angles $\theta=0,\tfrac{\pi}{6},\tfrac{\pi}{4},\tfrac{\pi}{3},\tfrac{\pi}{2},\tfrac{2\pi}{3},\tfrac{3\pi}{4},\tfrac{5\pi}{6},\pi$θ=0,6π​,4π​,3π​,2π​,32π​,43π​,65π​,π, extending to $2\pi$2π if needed.
2. Test for symmetry (this can halve the work): $\begin{aligned} f(-\theta)&=f(\theta) &&\Rightarrow\ \text{symmetric about the initial line } (x\text{-axis}),\\ f(\pi-\theta)&=f(\theta) &&\Rightarrow\ \text{symmetric about the line } \theta=\tfrac{\pi}{2}\ (y\text{-axis}),\\ f(\theta+\pi)&=f(\theta) &&\Rightarrow\ \text{symmetric about the pole.} \end{aligned}$f(−θ)f(π−θ)f(θ+π)​=f(θ)=f(θ)=f(θ)​​⇒ symmetric about the initial line (x-axis),⇒ symmetric about the line θ=2π​ (y-axis),⇒ symmetric about the pole.​
3. Solve $r=0$r=0 — these angles are where the curve passes through the pole and give the tangents at the pole.
4. Find the extreme values of $r$r — the maximum gives the furthest point, and $\frac{\mathrm dr}{\mathrm d\theta}=0$dθdr​=0 locates turning values of the radius.
5. Identify where $r<0$r<0 and plot those points using the signed-$r$r convention $(r,\theta)=(-r,\theta+\pi)$(r,θ)=(−r,θ+π).
6. Plot and join smoothly, marking the direction of increasing $\theta$θ with an arrow.

Why symmetry first? Spotting $f(-\theta)=f(\theta)$f(−θ)=f(θ) means you only need to plot $0\le\theta\le\pi$0≤θ≤π and reflect in the $x$x-axis. In an exam that is half the table and half the chance of an arithmetic slip.

A word on the symmetry tests, because they are easy to misremember. The test $f(-\theta)=f(\theta)$f(−θ)=f(θ) is exactly the statement that replacing the angle by its negative (a reflection in the initial line) leaves the radius unchanged, so the reflected point is also on the curve — hence symmetry about the $x$x-axis. Any curve built only from $\cos$cos of $\theta$θ automatically passes this test, because cosine is even; that is why every $r=a+b\cos\theta$r=a+bcosθ limaçon and every $r=a\cos n\theta$r=acosnθ rose is symmetric about the initial line. Curves built from $\sin\theta$sinθ instead are symmetric about the vertical line $\theta=\tfrac{\pi}{2}$θ=2π​, since sine satisfies $\sin(\pi-\theta)=\sin\theta$sin(π−θ)=sinθ. Recognising this at a glance — "cosine means $x$x-axis symmetry, sine means $y$y-axis symmetry" — saves you from ever computing the test from scratch.

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3 Circles

$r=a$r=a (constant)

Every point is the same distance $a$a from the pole, so this is a circle of radius $a$a centred at $O$O.

$r=2a\cos\theta$r=2acosθ

Setting $r=0$r=0 gives $\cos\theta=0$cosθ=0, i.e. $\theta=\pm\tfrac{\pi}{2}$θ=±2π​; the curve exists for $-\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}$−2π​≤θ≤2π​ (where $\cos\theta\ge0$cosθ≥0) and is a circle of radius $a$a, centre $(a,0)$(a,0), passing through the pole. (Confirm with $r=2a\cos\theta\Rightarrow r^2=2ar\cos\theta\Rightarrow x^2+y^2=2ax\Rightarrow(x-a)^2+y^2=a^2$r=2acosθ⇒r2=2arcosθ⇒x2+y2=2ax⇒(x−a)2+y2=a2.)

$r=2a\sin\theta$r=2asinθ

By the same algebra this is a circle of radius $a$a, centre $(0,a)$(0,a); $r=0$r=0 at $\theta=0$θ=0 and $\pi$π, and the curve is traced for $0\le\theta\le\pi$0≤θ≤π. A common trap: the whole circle is swept by $\theta\in[0,\pi]$θ∈[0,π], not $[0,2\pi]$[0,2π] — letting $\theta$θ run to $2\pi$2π traces it twice, which doubles a later area. The reason is that for $\pi<\theta<2\pi$π<θ<2π we have $\sin\theta<0$sinθ<0, so $r<0$r<0, and the signed-$r$r convention plots those points right back on top of the circle already drawn. This "circle through the pole traced once over a half-interval" idea recurs constantly, so internalise it now: whenever a polar equation is a single $\cos\theta$cosθ or $\sin\theta$sinθ term, the closed curve is complete after only half a revolution.

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4 Cardioids

The cardioid $r=a(1+\cos\theta)$r=a(1+cosθ) is the $a=b$a=b case of the limaçon family. Its key values are:

$\theta$θ	$0$0	$\tfrac{\pi}{3}$3π​	$\tfrac{\pi}{2}$2π​	$\tfrac{2\pi}{3}$32π​	$\pi$π
$r$r	$2a$2a	$\tfrac{3a}{2}$23a​	$a$a	$\tfrac{a}{2}$2a​	$0$0

It is heart-shaped (hence the name), symmetric about the initial line (since $\cos(-\theta)=\cos\theta$cos(−θ)=cosθ), reaches its maximum $r=2a$r=2a at $\theta=0$θ=0, and has a cusp at the pole at $\theta=\pi$θ=π. The reflections $r=a(1-\cos\theta)$r=a(1−cosθ) (cusp at $\theta=0$θ=0) and $r=a(1\pm\sin\theta)$r=a(1±sinθ) (symmetric about $\theta=\tfrac{\pi}{2}$θ=2π​) behave identically up to rotation.

Worked Example 1 — sketch $r=3(1+\cos\theta)$r=3(1+cosθ)

Build the table (here $a=3$a=3):

$\theta$θ	$0$0	$\tfrac{\pi}{6}$6π​	$\tfrac{\pi}{3}$3π​	$\tfrac{\pi}{2}$2π​	$\tfrac{2\pi}{3}$32π​	$\tfrac{5\pi}{6}$65π​	$\pi$π
$\cos\theta$cosθ	$1$1	$\tfrac{\sqrt3}{2}$23​​	$\tfrac12$21​	$0$0	$-\tfrac12$−21​	$-\tfrac{\sqrt3}{2}$−23​​	$-1$−1
$r$r	$6$6	$5.60$5.60	$4.5$4.5	$3$3	$1.5$1.5	$0.40$0.40	$0$0

Reasoning: symmetric about the $x$x-axis, so plot $0\le\theta\le\pi$0≤θ≤π and reflect. The curve leaves $(6,0)$(6,0), curls in to the pole at $\theta=\pi$θ=π forming the cusp, and the reflection completes the heart.

Mark scheme: M1 a correct table of at least five values; A1 correct shape with cusp at the pole; B1 maximum $r=6$r=6 on the axis labelled; B1 symmetry about the initial line indicated.

A schematic of the cardioid (not to scale):

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5 Limaçons

The general limaçon is $r=a+b\cos\theta$r=a+bcosθ (or with $\sin\theta$sinθ). The ratio $\tfrac{a}{b}$ba​ decides the shape:

Condition	Shape
$a=b$a=b	cardioid (cusp at the pole)
$a>b>0$a>b>0	dimpled limaçon (a dent, no inner loop)
$0<a<b$0<a<b	limaçon with an inner loop ($r$r goes negative)
$a\ge 2b$a≥2b	convex limaçon (no dimple at all)

The inner loop appears precisely when $r$r becomes negative for some $\theta$θ, which needs $a<b$a<b so that $a+b\cos\theta$a+bcosθ can dip below zero. The dimple (the gentle dent opposite the maximum) appears in the intermediate band $b<a<2b$b<a<2b: here $r$r never becomes negative, but its rate of change is large enough near $\theta=\pi$θ=π to pull the curve inward into a concave dent. Once $a\ge2b$a≥2b even that dent flattens out and the curve is everywhere convex. Knowing which of the four cases you are in before you tabulate tells you what shape to expect, so you can sanity-check your plotted points against it — a plotted "loop" on a curve with $a>b$a>b is a sure sign of an arithmetic slip.

Worked Example 2 — limaçon with an inner loop

Describe and sketch $r=2+3\cos\theta$r=2+3cosθ.

Here $a=2,\ b=3$a=2, b=3, and since $a<b$a<b there is an inner loop. Solve $r=0$r=0:

$2+3\cos\theta=0 \ \implies\ \cos\theta=-\tfrac23 \ \implies\ \theta=\arccos\!\left(-\tfrac23\right)\approx 2.30 \ \text{ or }\ \theta\approx 3.98\ (\approx 2\pi-2.30).$2+3cosθ=0 ⟹ cosθ=−32​ ⟹ θ=arccos(−32​)≈2.30  or  θ≈3.98 (≈2π−2.30).

For $\theta$θ between these values $\cos\theta<-\tfrac23$cosθ<−32​, so $r<0$r<0: those points are plotted backwards and trace the inner loop. The outer curve has maximum $r=5$r=5 at $\theta=0$θ=0; the minimum value $r=-1$r=−1 at $\theta=\pi$θ=π corresponds (by the signed-$r$r rule) to the Cartesian point $(1,0)$(1,0), the tip of the inner loop. It helps to picture the trace dynamically: starting at $(5,0)$(5,0), the radius shrinks as $\theta$θ grows, reaching the pole at $\theta\approx2.30$θ≈2.30; then $r$r goes negative and the curve, plotted backwards, swings inside to carve the small loop; $r$r returns to zero at $\theta\approx3.98$θ≈3.98 and the outer curve resumes, closing back at $(5,0)$(5,0). The inner loop is entirely a consequence of $a<b$a<b, and its presence is exactly what you must show in the sketch to earn the final accuracy mark.

Mark scheme: M1 solve $r=0$r=0; A1 the pole angles $\theta=\arccos(-\tfrac23)$θ=arccos(−32​); M1 identify the negative-$r$r interval as the inner loop; A1 correct sketch showing both loops and $r_{\max}=5$rmax​=5.

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6 Rose curves

For $r=a\cos(n\theta)$r=acos(nθ) or $r=a\sin(n\theta)$r=asin(nθ) the curve is a rose:

$n\ \text{odd} \ \Rightarrow\ n\ \text{petals}, \qquad n\ \text{even} \ \Rightarrow\ 2n\ \text{petals},$n odd ⇒ n petals,n even ⇒ 2n petals,

each petal of length $a$a. The parity rule is subtle: when $n$n is odd the negative-$r$r arcs retrace petals you already have, so you get only $n$n; when $n$n is even the negative-$r$r arcs fall in fresh directions and double the count to $2n$2n. The single most common slip in the exam is to count petals straight off "$n$n" without checking the parity — a candidate who writes "$r=\sin 4\theta$r=sin4θ has 4 petals" has thrown away marks, because the answer is 8. A reliable mental check: a rose with an even number of petals is symmetric about both axes, whereas a rose with an odd number is not, so if your sketch has the wrong symmetry you have miscounted. Each petal spans an angular width of $\tfrac{\pi}{n}$nπ​, bounded by two consecutive directions where $r=0$r=0.

Worked Example 3 — three-petalled rose $r=2\cos 3\theta$r=2cos3θ