A-Level Further Maths 9231 / Topic 1.5 / AS & A Level
Draw a curve from a few angles.
A polar sketch is not a plot of dozens of points. A short table of values at key angles, a check for symmetry, the points on the initial line, the tangents at the pole and the greatest value of \(r\) are enough to draw circles, cardioids, limaçons, spirals and roses with every significant feature in place.
To sketch \(r = f(\theta)\), make a table of values at key angles (multiples of \(\tfrac{\pi}{6}\) or \(\tfrac{\pi}{4}\)) and look for symmetry: if \(f(-\theta) = f(\theta)\) the curve is symmetric about the initial line, and if \(f(\pi - \theta) = f(\theta)\) it is symmetric about the line \(\theta = \tfrac{\pi}{2}\). With the convention \(r \ge 0\), angles where \(f(\theta) \lt 0\) give no points at all. Wherever \(f(\alpha) = 0\) the curve reaches the pole, and the line \(\theta = \alpha\) is the tangent at the pole. Mark the points on the initial line (\(\theta = 0\) and \(\theta = \pi\)) and the greatest and least values of \(r\). Two curves meet where \(f_1(\theta) = f_2(\theta)\), and also at the pole if both pass through it.
Fig. 1: the cardioid r = 2(1 + cos θ), drawn for -π < θ ≤ π. It meets the initial line at (4, 0), has r = 2 at θ = ±π/2, and reaches the pole only at θ = π, so the tangent there is the line θ = π. Since cos(-θ) = cos θ, the lower half is the mirror image of the upper half in the initial line
Section 01
A table of values and a check for symmetry.
Detailed plotting is not required; a sketch has to show the features that matter. Start with a short table: \(\theta = 0, \tfrac{\pi}{6}, \tfrac{\pi}{3}, \tfrac{\pi}{2}, \dots\) is usually enough. Then let symmetry halve the work. If replacing \(\theta\) by \(-\theta\) leaves \(r\) unchanged, the point \((r, -\theta)\) is the reflection of \((r, \theta)\) in the initial line, so the curve is symmetric about the initial line. This happens whenever \(r\) depends only on \(\cos\theta\).
The checklist for every polar sketch:
Feature
How to find it
Symmetry about the initial line
\(f(-\theta) = f(\theta)\), e.g. any function of \(\cos\theta\)
Symmetry about \(\theta = \tfrac{\pi}{2}\)
\(f(\pi - \theta) = f(\theta)\), e.g. any function of \(\sin\theta\)
Points on the initial line
put \(\theta = 0\) (and \(\theta = \pi\) for the line through the pole)
Where the curve exists
only where \(f(\theta) \ge 0\)
Tangents at the pole
solve \(f(\theta) = 0\); each root \(\alpha\) gives the tangent \(\theta = \alpha\)
Greatest and least \(r\)
from the range of \(\sin\) or \(\cos\), or solve \(\dfrac{\mathrm{d}r}{\mathrm{d}\theta} = 0\)
Worked example The cardioid in Fig. 1
Sketch the curve \(r = 2(1 + \cos\theta)\) for \(-\pi \lt \theta \le \pi\), showing its significant features.
\(\cos(-\theta) = \cos\theta\), so the curve is symmetric about the initial line: the same values hold at \(-\tfrac{\pi}{3}, -\tfrac{\pi}{2}, -\tfrac{2\pi}{3}\).
\(-1 \le \cos\theta \le 1\), so \(0 \le r \le 4\): \(r\) is never negative and the curve exists for every \(\theta\). Greatest \(r = 4\) at \(\theta = 0\); least \(r = 0\) at \(\theta = \pi\).
\(r = 0\) only at \(\theta = \pi\), so the curve reaches the pole there with tangent \(\theta = \pi\): it comes into the pole along the initial line from the left, making the pointed cusp.
Answer: A heart shape symmetric about the initial line, through \((4, 0)\), \(\left(2, \pm\tfrac{\pi}{2}\right)\), with a cusp at the pole where the tangent is \(\theta = \pi\) (Fig. 1)
Worked example A circle through the pole
Sketch \(r = 4\cos\theta\) for \(-\pi \lt \theta \le \pi\), and identify the curve.
\(r \ge 0\) requires \(\cos\theta \ge 0\), so the curve exists only for \(-\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}\). For \(\tfrac{\pi}{2} \lt |\theta| \le \pi\) there are no points.
Table: \(\theta = 0, \pm\tfrac{\pi}{3}, \pm\tfrac{\pi}{2}\) give \(r = 4, 2, 0\). Symmetric about the initial line.
\(r = 0\) at \(\theta = \pm\tfrac{\pi}{2}\), which is one line through the pole: the tangent at the pole is \(\theta = \tfrac{\pi}{2}\).
Multiply by \(r\): \(r^{2} = 4r\cos\theta\), so \(x^{2} + y^{2} = 4x\), that is \((x - 2)^{2} + y^{2} = 4\).
Answer: The circle with centre \((2, 0)\) and radius 2, traced once as \(\theta\) runs from \(-\tfrac{\pi}{2}\) to \(\tfrac{\pi}{2}\); it touches the line \(\theta = \tfrac{\pi}{2}\) at the pole
Your turn
For \(r = 2(1 - \sin\theta)\), \(-\pi \lt \theta \le \pi\), find the line of symmetry, the points on the initial line, the greatest value of \(r\) and the tangent at the pole.
\(\sin(\pi - \theta) = \sin\theta\), so the curve is symmetric about the line \(\theta = \tfrac{\pi}{2}\). At \(\theta = 0\) and \(\theta = \pi\), \(r = 2\): the points \((2, 0)\) and \((2, \pi)\). Greatest \(r = 4\) at \(\theta = -\tfrac{\pi}{2}\). \(r = 0\) at \(\theta = \tfrac{\pi}{2}\), so the tangent at the pole is \(\theta = \tfrac{\pi}{2}\). It is the cardioid of Fig. 1 turned so that its cusp points up.
Examiner trap
A symmetry test must be done on the formula, not guessed from the shape. \(r = 2(1 - \sin\theta)\) is NOT symmetric about the initial line: \(\theta = \tfrac{\pi}{2}\) gives \(r = 0\) but \(\theta = -\tfrac{\pi}{2}\) gives \(r = 4\). And remember that \(\theta = \pi\) is on the initial line extended backwards through the pole, so check it too.
Section 02
The pole and the r ≥ 0 rule.
With the convention \(r \ge 0\), an angle that makes \(f(\theta)\) negative simply contributes no point. This is what cuts curves such as \(r = a\sin 3\theta\) and \(r = a\cos 2\theta\) into separate loops, or petals. Each loop starts and ends at the pole, at the angles where \(f(\theta) = 0\), and the lines \(\theta = \alpha\) with \(f(\alpha) = 0\) are the tangents there. Draw each loop leaving the pole along these lines.
Fig. 2: the rose r = 2 sin 3θ for 0 ≤ θ < 2π. Only the three petals with sin 3θ ≥ 0 are drawn. Each petal leaves the pole along one tangent line and returns along the next, and the tips (r = 2) are at θ = π/6, 5π/6 and 3π/2
Petals of \(r = a\sin 3\theta\) occupy \(0 \le \theta \le \tfrac{\pi}{3}\), \(\tfrac{2\pi}{3} \le \theta \le \pi\) and \(\tfrac{4\pi}{3} \le \theta \le \tfrac{5\pi}{3}\). In the gaps between them \(\sin 3\theta \lt 0\).
The tip of each petal is where \(r\) is greatest, half-way between the two angles where \(r = 0\).
Without the convention \(r \ge 0\) you would plot \((-r, \theta)\) in the gaps and get the same three petals again; the 9231 convention leaves those angles empty.
Worked example The three-petal rose in Fig. 2
Sketch \(r = 2\sin 3\theta\) for \(0 \le \theta \lt 2\pi\), stating the tangents at the pole and the greatest value of \(r\).
\(r = 0\) when \(\sin 3\theta = 0\): \(\theta = 0, \tfrac{\pi}{3}, \tfrac{2\pi}{3}, \pi, \tfrac{4\pi}{3}, \tfrac{5\pi}{3}\). These are the tangents at the pole.
Greatest \(r = 2\) when \(\sin 3\theta = 1\): \(3\theta = \tfrac{\pi}{2}, \tfrac{5\pi}{2}, \tfrac{9\pi}{2}\), so \(\theta = \tfrac{\pi}{6}, \tfrac{5\pi}{6}, \tfrac{3\pi}{2}\).
Table for the first petal: \(\theta = 0, \tfrac{\pi}{12}, \tfrac{\pi}{6}, \tfrac{\pi}{4}, \tfrac{\pi}{3}\) give \(r = 0, \sqrt{2}, 2, \sqrt{2}, 0\). The other two petals are copies of it turned through \(\tfrac{2\pi}{3}\) and \(\tfrac{4\pi}{3}\).
Answer: Three equal petals with tips \(\left(2, \tfrac{\pi}{6}\right)\), \(\left(2, \tfrac{5\pi}{6}\right)\), \(\left(2, \tfrac{3\pi}{2}\right)\); tangents at the pole \(\theta = 0, \tfrac{\pi}{3}, \tfrac{2\pi}{3}, \pi, \tfrac{4\pi}{3}, \tfrac{5\pi}{3}\) (Fig. 2)
\(r \ge 0\) needs \(\cos 2\theta \ge 0\): \(-\tfrac{\pi}{4} \le \theta \le \tfrac{\pi}{4}\), and \(\tfrac{3\pi}{4} \le \theta \le \pi\) together with \(-\pi \lt \theta \le -\tfrac{3\pi}{4}\).
So there are two loops: one round \(\theta = 0\) and one round \(\theta = \pi\). For \(\tfrac{\pi}{4} \lt |\theta| \lt \tfrac{3\pi}{4}\) there are no points.
\(\cos(-2\theta) = \cos 2\theta\): symmetric about the initial line. Greatest \(r = 3\) at \(\theta = 0\) and \(\theta = \pi\), the points \((3, 0)\) and \((3, \pi)\).
\(r = 0\) at \(\theta = \pm\tfrac{\pi}{4}, \pm\tfrac{3\pi}{4}\): these are the tangents at the pole, the lines \(y = x\) and \(y = -x\).
Answer: Two loops along the initial line, tips \((3, 0)\) and \((3, \pi)\), each touching the lines \(\theta = \pm\tfrac{\pi}{4}\) or \(\theta = \pm\tfrac{3\pi}{4}\) at the pole
Your turn
For \(r = 3\sin 2\theta\), \(0 \le \theta \lt 2\pi\), find the intervals where the curve exists, the tangents at the pole and the tips of the loops.
\(\sin 2\theta \ge 0\) for \(0 \le \theta \le \tfrac{\pi}{2}\) and \(\pi \le \theta \le \tfrac{3\pi}{2}\): two loops. \(r = 0\) at \(\theta = 0, \tfrac{\pi}{2}, \pi, \tfrac{3\pi}{2}\), so the tangents at the pole are the initial line and the line \(\theta = \tfrac{\pi}{2}\). The tips are \(\left(3, \tfrac{\pi}{4}\right)\) and \(\left(3, \tfrac{5\pi}{4}\right)\).
Examiner trap
Drawing four petals for \(r = a\cos 2\theta\) (or six for \(r = a\sin 3\theta\)) is the classic error: it plots the angles where \(r \lt 0\). With \(r \ge 0\), first find where the curve exists, then sketch. Also draw each loop meeting the pole along its tangent lines, not with a rounded end.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
The curve \(r = 3 + 2\cos\theta\) is symmetric about
Quick check+10 XP
Using the convention \(r \ge 0\), how many loops does \(r = 2\cos 2\theta\), \(-\pi \lt \theta \le \pi\), have?
Quick check+10 XP
The curve \(r = 1 - \cos\theta\) reaches the pole. The tangent to the curve at the pole is
Quick check+10 XP
For \(r = 5 - 3\sin\theta\), \(-\pi \lt \theta \le \pi\), the greatest value of \(r\) is
Section 03
Standard shapes and where two curves meet.
A handful of curves come up again and again, and knowing their shapes lets you check a sketch at a glance. Here \(a\) and \(b\) are positive constants. To find where two polar curves meet, solve \(f_1(\theta) = f_2(\theta)\) in the given interval; then ask separately whether both curves pass through the pole, because they can reach it at different values of \(\theta\).
Equation
Shape
Key features
\(r = a\)
circle, centre the pole
radius \(a\)
\(r = a\cos\theta\)
circle through the pole
diameter \(a\) along the initial line; exists for \(|\theta| \le \tfrac{\pi}{2}\); tangent at pole \(\theta = \tfrac{\pi}{2}\)
\(r = a(1 + \cos\theta)\)
cardioid
greatest \(r = 2a\) at \(\theta = 0\); cusp at the pole, tangent \(\theta = \pi\)
\(r = a(1 - \sin\theta)\)
cardioid, cusp upward
symmetric about \(\theta = \tfrac{\pi}{2}\); greatest \(r = 2a\) at \(\theta = -\tfrac{\pi}{2}\); tangent at pole \(\theta = \tfrac{\pi}{2}\)
\(r = a + b\cos\theta\), \(a \gt b\)
limaçon
\(a - b \le r \le a + b\), never reaches the pole; a dimple near \(\theta = \pi\) if \(a \lt 2b\), convex (no dimple) if \(a \ge 2b\)
\(r = a\theta\), \(0 \le \theta \lt 2\pi\)
spiral
starts at the pole with tangent \(\theta = 0\); \(r\) increases steadily; crosses \(\theta = \pi\) at \(r = a\pi\)
\(r = a\sin 3\theta\)
rose, three petals
petals of length \(a\) at \(\theta = \tfrac{\pi}{6}, \tfrac{5\pi}{6}, \tfrac{3\pi}{2}\) (Fig. 2)
Worked example A cardioid and a circle
Find the polar coordinates of the points where \(r = 2(1 + \cos\theta)\) meets the circle \(r = 3\), for \(-\pi \lt \theta \le \pi\).
Set the two expressions for \(r\) equal: \(2(1 + \cos\theta) = 3\).
\(\cos\theta = \tfrac{1}{2}\), so \(\theta = \tfrac{\pi}{3}\) or \(\theta = -\tfrac{\pi}{3}\) in the interval.
At both, \(r = 3\). The circle \(r = 3\) does not pass through the pole, so there are no other points.
Check with symmetry: both curves are symmetric about the initial line, so the points come as a mirror pair. One of them is marked in Fig. 1.
Answer: \(\left(3, \tfrac{\pi}{3}\right)\) and \(\left(3, -\tfrac{\pi}{3}\right)\)
Worked example Do not forget the pole
Find all the points of intersection of \(r = 3\cos\theta\) and \(r = 1 + \cos\theta\), for \(-\pi \lt \theta \le \pi\).
\(3\cos\theta = 1 + \cos\theta\) gives \(\cos\theta = \tfrac{1}{2}\), so \(\theta = \pm\tfrac{\pi}{3}\) and \(r = \tfrac{3}{2}\).
\(r = 3\cos\theta\) passes through the pole at \(\theta = \tfrac{\pi}{2}\); \(r = 1 + \cos\theta\) passes through it at \(\theta = \pi\).
These are different values of \(\theta\), so solving \(f_1 = f_2\) cannot find this meeting point, but the pole is on both curves.
So the curves meet in three points.
Answer: \(\left(\tfrac{3}{2}, \tfrac{\pi}{3}\right)\), \(\left(\tfrac{3}{2}, -\tfrac{\pi}{3}\right)\) and the pole
Your turn
Describe the curve \(r = 2 + \cos\theta\), \(-\pi \lt \theta \le \pi\): its symmetry, greatest and least \(r\), whether it reaches the pole and whether it has a dimple.
Symmetric about the initial line. Greatest \(r = 3\) at \(\theta = 0\), least \(r = 1\) at \(\theta = \pi\), so it never reaches the pole. Here \(a = 2\), \(b = 1\) and \(a \ge 2b\), so it is a convex limaçon with no dimple: an oval slightly fatter on the right, through \((3, 0)\), \(\left(2, \pm\tfrac{\pi}{2}\right)\) and \((1, \pi)\).
Examiner trap
Solving \(f_1(\theta) = f_2(\theta)\) finds only the points where both curves have the same \(r\) at the same \(\theta\). The pole has many polar representations, \((0, \theta)\) for any \(\theta\), so two curves can share it without a common root. Always check whether each curve has \(r = 0\) somewhere in the interval.
Stage 2 · Exam
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Exam style+20 XP
Find the points where the curve \(r = 2(1 + \cos\theta)\) meets the circle \(r = 3\), for \(-\pi \lt \theta \le \pi\).
Exam style+20 XP
For the curve \(r = a\sin 3\theta\), \(0 \le \theta \lt 2\pi\), where \(a \gt 0\), at which angles are the tips of the petals?
Exam style+20 XP
Which statement about the curve \(r = 3 + 2\cos\theta\) is correct?
Exam style+20 XP
Find all the points of intersection of \(r = 3\cos\theta\) and \(r = 1 + \cos\theta\), for \(-\pi \lt \theta \le \pi\).
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