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AS & A Level · Practice questions · Sketching polar curves

Sketch with purpose.

Six original Cambridge-style questions on sketching polar curves: tables of values, symmetry, the \(r \ge 0\) convention, tangents at the pole, greatest values of \(r\) and points of intersection.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Find the features first, then draw.

01
Recall
[2 marks]

For the curve \(r = 3(1 + \cos\theta)\), find \(r\) when \(\theta = 0, \tfrac{\pi}{3}, \tfrac{\pi}{2}, \tfrac{2\pi}{3}, \pi\), and state the line of symmetry of the curve.

  • \(\cos\theta = 1, \tfrac{1}{2}, 0, -\tfrac{1}{2}, -1\), so \(r = 6, \tfrac{9}{2}, 3, \tfrac{3}{2}, 0\). ✓
  • \(\cos(-\theta) = \cos\theta\), so the curve is symmetric about the initial line. ✓
02
Calculation
[4 marks]

The curve \(C\) has polar equation \(r = 2\cos 2\theta\), for \(-\pi \lt \theta \le \pi\).

(a) Find the set of values of \(\theta\) for which \(C\) exists.

(b) State the equations of the tangents to \(C\) at the pole and the greatest value of \(r\).

(c) Sketch \(C\).

  • (a) \(r \ge 0\) needs \(\cos 2\theta \ge 0\): \(-\tfrac{\pi}{4} \le \theta \le \tfrac{\pi}{4}\), \(\tfrac{3\pi}{4} \le \theta \le \pi\) and \(-\pi \lt \theta \le -\tfrac{3\pi}{4}\). ✓
  • (b) \(r = 0\) when \(\cos 2\theta = 0\): the tangents at the pole are \(\theta = \pm\tfrac{\pi}{4}\) and \(\theta = \pm\tfrac{3\pi}{4}\). ✓
  • Greatest \(r = 2\), at \(\theta = 0\) and \(\theta = \pi\). ✓
  • (c) Two loops, symmetric about the initial line, with tips \((2, 0)\) and \((2, \pi)\), each meeting the pole along the lines \(\theta = \pm\tfrac{\pi}{4}\), \(\theta = \pm\tfrac{3\pi}{4}\); nothing in the regions \(\tfrac{\pi}{4} \lt |\theta| \lt \tfrac{3\pi}{4}\). ✓
03
Calculation
[4 marks]

Find the polar coordinates of all the points of intersection of the curves \(r = 1 + \sin\theta\) and \(r = 3\sin\theta\), for \(0 \le \theta \lt 2\pi\).

  • \(1 + \sin\theta = 3\sin\theta\) gives \(\sin\theta = \tfrac{1}{2}\), so \(\theta = \tfrac{\pi}{6}\) or \(\tfrac{5\pi}{6}\). ✓
  • At both, \(r = 3 \times \tfrac{1}{2} = \tfrac{3}{2}\): the points \(\left(\tfrac{3}{2}, \tfrac{\pi}{6}\right)\) and \(\left(\tfrac{3}{2}, \tfrac{5\pi}{6}\right)\). ✓
  • \(r = 3\sin\theta\) passes through the pole at \(\theta = 0\) (and \(\pi\)); \(r = 1 + \sin\theta\) passes through it at \(\theta = \tfrac{3\pi}{2}\). ✓
  • So the pole is a third point of intersection, even though it is not a solution of the equation. ✓
04
Analysis
[5 marks]

The curve \(C\) has polar equation \(r = \cos\theta + \sin^{2}\theta\), for \(-\pi \lt \theta \le \pi\).

(a) Show that \(C\) is symmetric about the initial line.

(b) Find the greatest value of \(r\) on \(C\) and the values of \(\theta\) at which it occurs.

(c) Find, in the form \(\theta = \pm\cos^{-1} k\), the tangents to \(C\) at the pole.

  • (a) \(\cos(-\theta) = \cos\theta\) and \(\sin^{2}(-\theta) = \sin^{2}\theta\), so \(r(-\theta) = r(\theta)\). ✓
  • (b) \(\dfrac{\mathrm{d}r}{\mathrm{d}\theta} = -\sin\theta + 2\sin\theta\cos\theta = \sin\theta(2\cos\theta - 1)\), which is zero at \(\theta = 0, \pi\) or \(\cos\theta = \tfrac{1}{2}\). ✓
  • \(r(0) = 1\); \(r(\pi) = -1\), not a point of \(C\); \(r\!\left(\pm\tfrac{\pi}{3}\right) = \tfrac{1}{2} + \tfrac{3}{4} = \tfrac{5}{4}\). ✓
  • So the greatest value is \(r = \tfrac{5}{4}\), at \(\theta = \pm\tfrac{\pi}{3}\) (the point on the initial line, \(r = 1\), is a local minimum). ✓
  • (c) With \(c = \cos\theta\): \(r = c + 1 - c^{2} = 0\) gives \(c^{2} - c - 1 = 0\), \(c = \dfrac{1 \pm \sqrt{5}}{2}\). Only \(c = \dfrac{1 - \sqrt{5}}{2}\) lies in \([-1, 1]\), so the tangents are \(\theta = \pm\cos^{-1}\!\left(\dfrac{1 - \sqrt{5}}{2}\right)\), about \(\theta = \pm 2.24\). ✓
05
Analysis
[4 marks]

The spiral \(S\) has polar equation \(r = \theta\), for \(0 \le \theta \lt 2\pi\).

(a) State the tangent to \(S\) at the pole.

(b) Find the point where \(S\) meets the line \(\theta = \pi\), and explain why \(S\) meets the initial line only at the pole.

(c) Find the point where \(S\) meets the circle \(r = 3\), and explain why there is only one such point.

  • (a) \(r = 0\) only at \(\theta = 0\), so the tangent at the pole is the initial line \(\theta = 0\). ✓
  • (b) At \(\theta = \pi\), \(r = \pi\): the point \((\pi, \pi)\), at distance \(\pi\) from the pole on the initial line extended backwards. ✓
  • The initial line is \(\theta = 0\) (or \(2\pi\)); \(\theta = 0\) gives the pole and \(2\pi\) is outside the interval, so the spiral returns towards the initial line but does not reach it again. ✓
  • (c) \(\theta = 3\), which is in \([0, 2\pi)\): the point \((3, 3)\). Since \(r = \theta\) is strictly increasing, \(r\) takes the value 3 exactly once. ✓
06
Proof
[5 marks]

The curve \(R\) has polar equation \(r = a\sin 3\theta\), where \(a \gt 0\), for \(0 \le \theta \lt 2\pi\).

(a) Show that \(R\) is symmetric about the line \(\theta = \tfrac{\pi}{2}\).

(b) Show that \(R\) is unchanged by a rotation through \(\tfrac{2\pi}{3}\) about the pole.

(c) Hence explain why \(R\) consists of three identical petals, and state the tangents at the pole of the petal in \(0 \le \theta \le \tfrac{\pi}{3}\).

  • (a) The reflection of \((r, \theta)\) in the line \(\theta = \tfrac{\pi}{2}\) is \((r, \pi - \theta)\). \(a\sin(3\pi - 3\theta) = a\sin 3\theta\), since \(\sin(3\pi - x) = \sin x\). So the reflected point satisfies the equation too. ✓
  • (b) Rotating \((r, \theta)\) through \(\tfrac{2\pi}{3}\) gives \(\left(r, \theta + \tfrac{2\pi}{3}\right)\), and \(a\sin(3\theta + 2\pi) = a\sin 3\theta\). So the rotated point is on \(R\). ✓
  • (c) \(r \ge 0\) on \(0 \le \theta \le \tfrac{\pi}{3}\), with \(r = 0\) at both ends and \(r \gt 0\) between: one petal. ✓
  • In \(\tfrac{\pi}{3} \lt \theta \lt \tfrac{2\pi}{3}\), \(r \lt 0\), so there are no points. By (b), rotating this petal through \(\tfrac{2\pi}{3}\) and \(\tfrac{4\pi}{3}\) gives the whole curve: three identical petals. ✓
  • The first petal leaves and returns to the pole along \(\theta = 0\) and \(\theta = \tfrac{\pi}{3}\), so these are its tangents at the pole. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.