15. Polar Coordinates

a.1. Polar Coordinate System

Rectangular coordinates (x,y)(x,y) are one way to specify a point, PP, in the plane, but they are not the only way. When we are studying circles, it is useful to use Polar Coordinates (r,θ)(r,\theta) which identify the point, PP, using

  • the radius rr and
  • the polar angle θ\theta.

Like rectangular coordinates P=(x,y)P=(x,y), the polar coordinates can be written as an ordered pair, P=(r,θ)P=(r,\theta). So you need to be careful to say whether this ordered pair is rectangular or polar.

PolarCoordSTD

Most often, the radius measures the (positive) distance from the origin, OO, to the point, PP, and the polar angle measures an angle (usually measured in radians and usually with 0θ<2π0 \le \theta \lt 2\pi) counterclockwise from the positive xx-axis to the ray OP\overrightarrow{OP}.

However, frequently the angle θ\theta is allowed to be bigger than 2π2\pi or is allowed to be negative, in which case it is measured clockwise from the positve xx-axis. Consequently, θ\theta is non-unique; we can always add or subtract an arbitrary multiple of 2π2\pi or 360360^\circ. Thus the point with polar coordinates (3,30)(3,{30^\circ}) can also be written as (3,390)(3,{390^\circ}), or (3,750)(3,{750^\circ}), or (3,330)(3,{-330^\circ}), or (3,690)(3,{-690^\circ}) all of which are shown in the plot with blue angles positive and red angles negative.

PolarMultiTheta

Further, once in a while, we allow rr to be negative, for example in the context of solving equations or graphing. Then rr is the negative of the distance from OO to PP. When rr is negative, the point (r,θ)(r,\theta) is obtained by going a distance r|r| along the ray at the angle θ±π=θ±180\theta\pm\pi=\theta\pm180^\circ. Think of this as going backwards along the ray at the angle θ\theta. Thus the point with polar coordinates (3,30)(-3,{30^\circ}) is actually the point (3,210)(3,{210^\circ}) or (3,150)(3,-150^\circ).

PolarNegRadius

Identify the point with each of the following coordinates:

(2,π3)\left(2,\dfrac{\pi}{3}\right) = AA BB CC DD
    EE FF GG HH
(2,π6)\left(2,\dfrac{\pi}{6}\right) = AA BB CC DD
    EE FF GG HH
(2,π6)\left(-2,\dfrac{\pi}{6}\right) = AA BB CC DD
    EE FF GG HH
(2,π3)\left(2,-\,\dfrac{\pi}{3}\right) = AA BB CC DD
    EE FF GG HH
(2,π3)\left(-2,-\,\dfrac{\pi}{3}\right) = AA BB CC DD
    EE FF GG HH
PolarExCoords

Notice that for the two points with negative radii, (2,π6)\left(-2,\dfrac{\pi}{6}\right) and (2,π3)\left(-2,-\,\dfrac{\pi}{3}\right), we have to go backwards along the angles π6\dfrac{\pi}{6} and π3-\,\dfrac{\pi}{3}, respectively.

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