A-Level Further Maths 9231 / Topic 1.5 / AS & A Level
A distance and a direction.
Polar coordinates fix a point by how far it is from the pole and which way it lies from the initial line. Four short formulas link \((r, \theta)\) to \((x, y)\), so you can move points and whole curves between the two systems, and the convention \(r \ge 0\) decides which parts of a polar curve exist at all.
A point \(P\) has polar coordinates \((r, \theta)\): \(r \ge 0\) is the distance \(OP\) from the pole and \(\theta\) is the angle from the initial line, measured anticlockwise. Then \(x = r\cos\theta\), \(y = r\sin\theta\), \(r^{2} = x^{2} + y^{2}\) and \(\tan\theta = \dfrac{y}{x}\), where the quadrant of \((x, y)\) decides which angle with that tangent is correct.
Fig. 1: the polar grid. Circles are curves of constant r and the dashed rays are lines of constant θ. P(r, θ) is r from the pole O at angle θ anticlockwise from the initial line; dropping a perpendicular to the initial line shows x = r cos θ and y = r sin θ
Section 01
Points in polar form.
Fix a point \(O\), the pole, and a half-line from it, the initial line (it plays the part of the positive \(x\)-axis). Any point \(P\) is then described by \((r, \theta)\): its distance \(r = OP\) and the angle \(\theta\) turned anticlockwise from the initial line to \(OP\). In this course \(r \ge 0\), and \(\theta\) is usually given in radians in the range \(0 \le \theta < 2\pi\) or \(-\pi < \theta \le \pi\); the question tells you which. Fig. 1 shows the right-angled triangle that links the two systems.
Polar to Cartesian
Cartesian to polar
\(x = r\cos\theta\)
\(r = \sqrt{x^{2} + y^{2}}\)
\(y = r\sin\theta\)
\(\tan\theta = \dfrac{y}{x}\), then choose \(\theta\) in the quadrant of \((x, y)\)
Worked example Polar to Cartesian
Find the Cartesian coordinates of the point with polar coordinates \(\left(4, \dfrac{2\pi}{3}\right)\).
\(\tan\theta = \dfrac{-3\sqrt{3}}{-3} = \sqrt{3}\). A calculator gives \(\dfrac{\pi}{3}\), which is in the first quadrant.
The point has \(x < 0\) and \(y < 0\), so it is in the third quadrant: \(\theta = \pi + \dfrac{\pi}{3} = \dfrac{4\pi}{3}\).
Answer: \(\left(6, \dfrac{4\pi}{3}\right)\); in the range \(-\pi < \theta \le \pi\) it would be \(\left(6, -\dfrac{2\pi}{3}\right)\)
Your turn
Find polar coordinates of \((2, -2\sqrt{3})\) with \(0 \le \theta < 2\pi\).
\(r = \sqrt{4 + 12} = 4\); \(\tan\theta = -\sqrt{3}\) with the point in the fourth quadrant, so \(\theta = 2\pi - \dfrac{\pi}{3} = \dfrac{5\pi}{3}\). Answer \(\left(4, \dfrac{5\pi}{3}\right)\).
Examiner trap
\(\tan^{-1}\) only returns angles between \(-\dfrac{\pi}{2}\) and \(\dfrac{\pi}{2}\), so it is wrong for every point with \(x < 0\). Sketch the point first, find the acute angle from \(\tan^{-1}\left|\dfrac{y}{x}\right|\), then place it in the right quadrant.
Section 02
Curves from Cartesian to polar.
To turn a Cartesian equation into a polar one, replace \(x\) by \(r\cos\theta\) and \(y\) by \(r\sin\theta\) (and \(x^{2} + y^{2}\) by \(r^{2}\) straight away), simplify with trigonometric identities, and make \(r\) the subject where you can. Circles through or centred at the pole and straight lines give the neatest results.
Circle centre \(O\), radius \(a\): \(x^{2} + y^{2} = a^{2}\) becomes \(r = a\).
Line \(x = a\) (\(a > 0\)): \(r\cos\theta = a\), so \(r = a\sec\theta\).
Line \(y = a\) (\(a > 0\)): \(r\sin\theta = a\), so \(r = a\operatorname{cosec}\theta\).
Half-line from \(O\) at angle \(\alpha\): \(\theta = \alpha\).
Worked example A circle and a line
Find polar equations of (a) \(x^{2} + y^{2} = 9\) and (b) the line \(x = 4\).
(a) \(x^{2} + y^{2} = r^{2}\), so \(r^{2} = 9\). With \(r \ge 0\), \(r = 3\).
(b) \(r\cos\theta = 4\), so \(r = \dfrac{4}{\cos\theta} = 4\sec\theta\).
\(r \ge 0\) needs \(\cos\theta > 0\), so the line is traced for \(-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}\).
Divide by \(r\): \(r = 4\cos\theta\). The pole is not lost, because \(r = 0\) when \(\theta = \dfrac{\pi}{2}\).
\(r \ge 0\) needs \(\cos\theta \ge 0\): the whole circle is traced once for \(-\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}\).
Answer: \(r = 4\cos\theta\)
Your turn
Find a polar equation of the parabola \(y = x^{2}\), giving \(r\) in terms of \(\theta\).
\(r\sin\theta = r^{2}\cos^{2}\theta\), so (dividing by \(r\)) \(r = \dfrac{\sin\theta}{\cos^{2}\theta} = \tan\theta\sec\theta\), for \(0 \le \theta \le \pi\), \(\theta \ne \dfrac{\pi}{2}\) (where \(\sin\theta \ge 0\), so \(r \ge 0\)).
Examiner trap
With \(r \ge 0\), \(\theta = \dfrac{\pi}{4}\) is only the HALF of \(y = x\) in the first quadrant. The whole line needs \(\theta = \dfrac{\pi}{4}\) and \(\theta = \dfrac{5\pi}{4}\).
Stage 1 · Learn
Check your understanding
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What are the Cartesian coordinates of the point with polar coordinates \(\left(2, \dfrac{\pi}{6}\right)\)?
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Polar coordinates of \((-1, -1)\) with \(0 \le \theta < 2\pi\) are
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Which is a polar equation of \(x^{2} + y^{2} = 16\)?
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What is the Cartesian equation of \(r = 2\cos\theta\)?
Section 03
Curves from polar to Cartesian.
Going back, you want to create the combinations \(r\cos\theta\), \(r\sin\theta\) and \(r^{2}\). The standard trick is to multiply the whole equation by \(r\) (or by \(r^{2}\)) so that every \(\cos\theta\) and \(\sin\theta\) has an \(r\) beside it. Whatever \(r\) is left over is replaced by \(\sqrt{x^{2} + y^{2}}\), or isolated and squared.
The convention \(r \ge 0\) also means that a value of \(\theta\) which makes the formula for \(r\) negative gives no point at all. For example \(r = 1 + 2\cos\theta\) has no points for \(\dfrac{2\pi}{3} < \theta < \dfrac{4\pi}{3}\), where \(\cos\theta < -\dfrac{1}{2}\). The next lesson uses this when sketching.
Worked example Multiply through by r
Find the Cartesian equation of \(r = 6\sin\theta\) and describe the curve.
Multiply by \(r\): \(r^{2} = 6r\sin\theta\).
Substitute: \(x^{2} + y^{2} = 6y\).
Complete the square: \(x^{2} + (y - 3)^{2} = 9\).
Answer: \(x^{2} + (y - 3)^{2} = 9\), a circle with centre \((0, 3)\) and radius 3, touching the initial line at the pole
Worked example Isolate r, then square
Find the Cartesian equation of \(r = \dfrac{2}{1 - \cos\theta}\), \(0 < \theta < 2\pi\).
Multiply out: \(r - r\cos\theta = 2\), so \(r - x = 2\).
So \(y^{2} = 4x + 4\). On this curve \(x \ge -1\), so \(x + 2 > 0\) and squaring added no extra points.
Answer: \(y^{2} = 4(x + 1)\), a parabola with vertex \((-1, 0)\)
Your turn
Find the Cartesian equation of \(r^{2} = \sin 2\theta\), and say for which \(\theta\) in \(0 \le \theta < 2\pi\) there are points.
\(r^{2} = 2\sin\theta\cos\theta\); multiply by \(r^{2}\): \(r^{4} = 2(r\sin\theta)(r\cos\theta)\), so \((x^{2} + y^{2})^{2} = 2xy\). Points need \(\sin 2\theta \ge 0\): \(0 \le \theta \le \dfrac{\pi}{2}\) or \(\pi \le \theta \le \dfrac{3\pi}{2}\).
Examiner trap
Do not plot a negative \(r\) 'backwards' through the pole. Under the \(r \ge 0\) convention those \(\theta\) values simply give nothing: \(r^{2} = \sin 2\theta\) has no points in the second and fourth quadrants.
Stage 2 · Exam
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Find a polar equation of the circle \(x^{2} + y^{2} = 6x + 8y\).
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Find the Cartesian equation of the curve \(r = \dfrac{4}{2 - \cos\theta}\).
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The curve \(r = 1 + 2\cos\theta\) is considered for \(0 \le \theta < 2\pi\) with the convention \(r \ge 0\). For which values of \(\theta\) are there no points on the curve?
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The point \(A\) has Cartesian coordinates \((-2\sqrt{3}, 2)\). Find polar coordinates of \(A\) with \(-\pi < \theta \le \pi\).
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