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Topos Uranos
English

Polar coordinates and polar curves

Summary

This lesson presents a second way of naming the points of the plane: by their distance to a fixed point, the pole, and by the angle they form with a fixed ray, the polar axis. It adopts the convention that admits negative radii, describes all the representations of a single point, derives the conversion formulas between polar and Cartesian coordinates, with the correction of the arctangent according to the quadrant, and obtains the distance between two points by means of the law of cosines. It then studies the polar equations of lines and circles, the symmetry tests and the sketching, by tables of values, of the cardioid, the roses, the Archimedean spiral and the limaçon, and ends with the polar equation of the conics that have a focus at the pole, from which their type, their vertices and their latus rectum are read.

Learning objectives

  1. Represent points in polar coordinates, recognize all the representations of a single point and convert from polar coordinates to Cartesian ones and vice versa.
  2. Compute distances in polar coordinates and translate equations of lines and circles from one system to the other, without losing or adding the pole.
  3. Apply the symmetry tests and tables of values to the sketching of polar curves: cardioids, roses, spirals and limaçons.
  4. Derive the polar equation of the conics with a focus at the pole and read in it their eccentricity, their vertices and their latus rectum.

Polar coordinates

Cartesian coordinates, studied in the Course I lesson on the equation of a line and Cartesian coordinate systems, name a point by its signed distances to two perpendicular lines. Now, many curves are described more naturally by the distance to a fixed point and by the direction in which they are seen from it: a circle around its center, a spiral that opens at a constant rate or the orbit of a planet around the Sun. For them a system of polar coordinates is convenient, formed by a fixed point OO, the pole, and a fixed ray starting from it, the polar axis. So that the two systems can coexist, the pole is placed at the origin and the polar axis on the positive semi-axis of abscissas.

Let rr and θ\theta be two real numbers. The point with polar coordinates (r,θ)(r, \theta) is the point of the plane with Cartesian coordinates

x=rcos⁡θ,y=rsin⁡θx = r \cos \theta, \qquad y = r \sin \theta

If r>0r > 0, it is the point at distance rr from the pole on the terminal side of the angle θ\theta in standard position, studied in the lesson on angles and their measure: it is the point (cos⁡θ,sin⁡θ)(\cos \theta, \sin \theta) of the unit circle, defined in the lesson on trigonometric ratios, with its coordinates multiplied by rr. The number rr is called the radius and the angle θ\theta, measured in radians, the polar angle.

The definition also admits r=0r = 0 and r<0r < 0, and this is what the convention of this lesson consists in. With r=0r = 0 one obtains the pole, whatever θ\theta may be. With r<0r < 0 one obtains the point located at distance −r-r from the pole in the direction opposite to that of θ\theta: by the symmetries cos⁡(θ+π)=−cos⁡θ\cos(\theta + \pi) = -\cos \theta and sin⁡(θ+π)=−sin⁡θ\sin(\theta + \pi) = -\sin \theta from the lesson on trigonometric ratios, the pairs (−r,θ)(-r, \theta) and (r,θ+π)(r, \theta + \pi) give the same point.

(−rcos⁡θ,−rsin⁡θ)=(rcos⁡(θ+π),rsin⁡(θ+π))\left( -r \cos \theta, -r \sin \theta \right) = \left( r \cos(\theta + \pi), r \sin(\theta + \pi) \right)

Many texts require r≥0r \geq 0; admitting negative radii has two advantages that this lesson will exploit: the line through the pole with direction α\alpha is described in its entirety by the equation θ=α\theta = \alpha, and curves such as the roses are traced continuously. The price is that a single point admits more representations. From this follows a rule worth fixing from now on: a point belongs to the curve of a polar equation if some representation of it satisfies the equation, even if others do not.

LemmaThe representations of a point

Let (r1,θ1)(r_1, \theta_1) be a representation of a point PP other than the pole. The pair (r2,θ2)(r_2, \theta_2) also represents PP if and only if r2=r1r_2 = r_1 and θ2=θ1+2kπ\theta_2 = \theta_1 + 2k\pi for some integer kk, or else r2=−r1r_2 = -r_1 and θ2=θ1+(2k+1)π\theta_2 = \theta_1 + (2k + 1)\pi for some integer kk. The pole is the point represented by the pairs (0,θ)(0, \theta), with θ\theta arbitrary.

We use the Pythagorean identity and the characterization of coterminal angles from the lesson on angles and their measure: two angles give the same point of the unit circle exactly when they differ by an integer multiple of 2π2\pi.

Proof

  1. x2+y2=r2cos⁡2θ+r2sin⁡2θ=r2→r2=r1∨r2=−r1x^2 + y^2 = \resaltar{r^2 \cos^2 \theta + r^2 \sin^2 \theta} = \resaltar{r^2} \rightarrow \dato{rr}{r_2 = r_1 \vee r_2 = -r_1}

    The distance from a point to the pole does not depend on the chosen representation: by the Pythagorean identity, x2+y2x^2 + y^2 equals r2r^2. Therefore, two representations of the same point have r22=r12r_2^2 = r_1^2, and the pole is the only point with r=0r = 0.

  2. cos⁡θ2=cos⁡θ1∧sin⁡θ2=sin⁡θ1↔θ2=θ1+2kπ\resaltar{\cos \theta_2 = \cos \theta_1 \wedge \sin \theta_2 = \sin \theta_1} \leftrightarrow \dato{co}{\theta_2 = \theta_1 + 2k\pi}

    If r2=r1≠0r_2 = r_1 \neq 0, we divide the coordinates of both points by r1r_1: the points (cos⁡θ1,sin⁡θ1)(\cos \theta_1, \sin \theta_1) and (cos⁡θ2,sin⁡θ2)(\cos \theta_2, \sin \theta_2) of the unit circle coincide, and this happens exactly when the angles are coterminal.

  3. θ2+π=θ1+2kπ↔θ2=θ1+(2k−1)π\resaltar{\theta_2 + \pi} = \theta_1 + 2k\pi \leftrightarrow \resaltar{\theta_2 = \theta_1 + (2k - 1)\pi}

    If r2=−r1r_2 = -r_1, the pair (−r1,θ2)(-r_1, \theta_2) gives the same point as (r1,θ2+π)(r_1, \theta_2 + \pi), and by the preceding step θ2+π\theta_2 + \pi is coterminal with θ1\theta_1. Since 2k−12k - 1 runs through all the odd numbers as kk runs through the integers, this is what the lemma states.

Thus, the point with polar coordinates (2,π3)\left( 2, \frac{\pi}{3} \right) also admits the representations (2,7π3)\left( 2, \frac{7\pi}{3} \right) and (2,−5π3)\left( 2, -\frac{5\pi}{3} \right), which make one turn more or one turn less, and (−2,4π3)\left( -2, \frac{4\pi}{3} \right) or (−2,−2π3)\left( -2, -\frac{2\pi}{3} \right), which look in the opposite direction and move backward.

From Cartesian to polar coordinates

The formulas x=rcos⁡θx = r \cos \theta and y=rsin⁡θy = r \sin \theta give the conversion from polar to Cartesian coordinates. The reverse conversion requires choosing one representation among all the possible ones; the one usually taken is the one with r>0r > 0 and θ\theta in the interval (−π,π]\ioc{-\pi}{\pi}. To find the angle one uses the arctangent function from the lesson on the inverse trigonometric functions, which takes values in (−π2,π2)\ioo{-\frac{\pi}{2}}{\frac{\pi}{2}} and therefore only sees the angles of the first and fourth quadrants: in the other two it must be corrected by adding or subtracting π\pi.

TheoremFrom Cartesian to polar coordinates

Every point (x,y)(x, y) other than the pole has a unique polar representation (r,θ)(r, \theta) with r>0r > 0 and θ∈(−π,π]\theta \in \ioc{-\pi}{\pi}. Its radius is r=x2+y2r = \sqrt{x^2 + y^2}, and its polar angle is given by the five cases of the following formula.

θ={arctan⁡yxx>0arctan⁡yx+πx<0∧y≥0arctan⁡yx−πx<0∧y<0π2x=0∧y>0−π2x=0∧y<0\theta = \begin{cases} \arctan \frac{y}{x} & x > 0 \\ \arctan \frac{y}{x} + \pi & x < 0 \wedge y \geq 0 \\ \arctan \frac{y}{x} - \pi & x < 0 \wedge y < 0 \\ \frac{\pi}{2} & x = 0 \wedge y > 0 \\ -\frac{\pi}{2} & x = 0 \wedge y < 0 \end{cases}

The idea is to reduce the problem to the unit circle, where the slope of the line through the pole determines two opposite points, and the sign of the abscissa decides between them.

Proof

  1. r=x2+y2,(cos⁡θ,sin⁡θ)=(xr,yr)r = \sqrt{x^2 + y^2}, \qquad \dato{q}{(\cos \theta, \sin \theta) = \left( \frac{x}{r}, \frac{y}{r} \right)}

    By the preceding lemma, r2=x2+y2r^2 = x^2 + y^2, and with r>0r > 0 the radius is the square root. Dividing by rr, the angle sought is that of a point QQ of the unit circle, which exists by the covering property of the traversal axiom, admitted in the lesson on angles and their measure, and which is unique in (−π,π]\ioc{-\pi}{\pi}, because two angles of that interval, which has length 2π2\pi, cannot be coterminal.

  2. x=0→Q=(0,1)∨Q=(0,−1)x = 0 \rightarrow Q = \resaltar{(0, 1)} \vee Q = \resaltar{(0, -1)}

    If x=0x = 0, the point QQ is (0,1)(0, 1) or (0,−1)(0, -1) according to the sign of yy, and by the special values of sine and cosine the angle is π2\frac{\pi}{2} or −π2-\frac{\pi}{2}.

  3. tan⁡θ=sin⁡θcos⁡θ=yx,α=arctan⁡yx\tan \theta = \frac{\resaltar{\sin \theta}}{\resaltar{\cos \theta}} = \resaltar{\frac{y}{x}}, \qquad \dato{al}{\alpha = \arctan \frac{y}{x}}

    If x≠0x \neq 0, the tangent of θ\theta is the slope m=yxm = \frac{y}{x} of the line joining the pole to the point. The angle α=arctan⁡m\alpha = \arctan m has that tangent and lies in (−π2,π2)\ioo{-\frac{\pi}{2}}{\frac{\pi}{2}}, where the cosine is positive.

  4. u2+(mu)2=1↔u=11+m2∨u=−11+m2u^2 + \resaltar{(mu)^2} = 1 \leftrightarrow \resaltar{u = \frac{1}{\sqrt{1 + m^2}} \vee u = -\frac{1}{\sqrt{1 + m^2}}}

    On the unit circle there are exactly two points with slope mm from the pole: if (u,v)(u, v) is one of them, v=muv = mu and u2+v2=1u^2 + v^2 = 1. They are opposite, so one has positive abscissa and the other negative; by the symmetry of the angle α+π\alpha + \pi, they are the points of the angles α\alpha and α+π\alpha + \pi.

  5. x<0∧y<0→θ=α+π−2π=α−π∈(−π,−π2)x < 0 \wedge y < 0 \rightarrow \theta = \resaltar{\alpha + \pi - 2\pi} = \resaltar{\alpha - \pi} \in \ioo{-\pi}{-\frac{\pi}{2}}

    If x>0x > 0, the point QQ has positive abscissa and θ=α\theta = \alpha. If x<0x < 0, it is the point of α+π\alpha + \pi. When in addition y≥0y \geq 0, the slope is negative or zero, α∈(−π2,0]\alpha \in \ioc{-\frac{\pi}{2}}{0} and α+π\alpha + \pi already lies in (π2,π]\ioc{\frac{\pi}{2}}{\pi}; when y<0y < 0, the slope is positive and one turn must be subtracted, α+π−2π=α−π\alpha + \pi - 2\pi = \alpha - \pi.

For example, the point (−1,1)(-1, 1) lies in the second quadrant: its radius is 2\sqrt{2}, and since arctan⁡(−1)=−π4\arctan(-1) = -\frac{\pi}{4}, its polar angle is −π4+π=3π4-\frac{\pi}{4} + \pi = \frac{3\pi}{4}. A calculator that gives the arctangent of yx\frac{y}{x} without attending to the signs of xx and of yy separately would return −π4-\frac{\pi}{4}, which is the angle of the opposite point (1,−1)(1, -1).