Polar coordinates

A ship's radar does not report "${3}$ miles east and ${4}$ miles north." It reports "${5}$ miles away, in this direction." That is the idea of polar coordinates: a point is given by its distance from a fixed point and an angle.

The polar coordinate system

Choose a point ${O}$, called the pole (it plays the role of the origin), and a ray from it, called the polar axis (usually the positive ${x}$-axis). A point ${P}$ then has polar coordinates ${(r,\theta)}$, where

${r}$ is the distance from ${O}$ to ${P}$,
${\theta}$ is the angle from the polar axis to the ray ${OP}$, measured counterclockwise.

The point ${P}$ is a distance ${r}$ from the pole ${O}$, at angle ${\theta}$ from the polar axis.

Polar graph paper has circles around the pole (for ${r}$) and rays out from it (for ${\theta}$). To plot ${(r,\theta)}$, turn to angle ${\theta}$, then walk out a distance ${r}$.

A negative ${r}$ means walk backward: the point ${(-r,\theta)}$ is on the ray opposite to ${\theta}$, at distance ${r}$.

Three points. For ${(-2,\dfrac{\pi}{4})}$, face the direction ${\dfrac{\pi}{4}}$ (dashed) and go ${2}$ units backward.

One point, many names

In rectangular coordinates, each point has exactly one name. In polar coordinates, each point has infinitely many. Adding ${2\pi}$ to the angle gives the same point, and so does turning around and using a negative ${r}$:

${(r,\theta)}$, ${(r,\theta+2k\pi)}$, and ${(-r,\theta+\pi)}$ are all the same point

The pole is ${(0,\theta)}$ for any angle ${\theta}$.

Example 1: Give two other polar names for the point ${\left(2,\dfrac{\pi}{3}\right)}$, one with ${r<0}$.

Solution:

Add ${2\pi}$ to the angle:

$\left(2,\dfrac{\pi}{3}+2\pi\right)=\left(2,\dfrac{7\pi}{3}\right)$

Use ${-r}$ and add ${\pi}$ to the angle:

$\left(-2,\dfrac{\pi}{3}+\pi\right)=\left(-2,\dfrac{4\pi}{3}\right)$

Changing between polar and rectangular coordinates

Put the pole at the origin and the polar axis along the positive ${x}$-axis. The right triangle in the figure links the two systems:

The right triangle shows ${x=r\cos\theta}$ and ${y=r\sin\theta}$.

${x=r\cos\theta}$${y=r\sin\theta}$

${r^2=x^2+y^2}$${\tan\theta=\dfrac{y}{x}}$

The first pair turns polar into rectangular. The second pair goes back. As with direction angles of vectors, use the quadrant of ${(x,y)}$ to choose ${\theta}$.

Example 2: Find the rectangular coordinates of the point ${\left(4,\dfrac{2\pi}{3}\right)}$.

Solution:

$x=4\cos\dfrac{2\pi}{3}=4\left(-\dfrac{1}{2}\right)=-2$

$y=4\sin\dfrac{2\pi}{3}=4\cdot\dfrac{\sqrt{3}}{2}=2\sqrt{3}$

The point is ${(-2,\ 2\sqrt{3})}$.

Example 3: Find polar coordinates, with ${r>0}$ and ${0\le\theta<2\pi}$, for each point.

(a) ${(1,-1)}$ (b) ${(-\sqrt{3},1)}$

Solution:

(a) ${r=\sqrt{1+1}=\sqrt{2}}$. The point is in quadrant IV, and ${\tan\theta=-1}$, so ${\theta=\dfrac{7\pi}{4}}$. The point is ${\left(\sqrt{2},\dfrac{7\pi}{4}\right)}$.

(b) ${r=\sqrt{3+1}=2}$. The point is in quadrant II, and ${\tan\theta=\dfrac{1}{-\sqrt{3}}}$, with reference angle ${\dfrac{\pi}{6}}$. So

${\theta=\pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}}$

The point is ${\left(2,\dfrac{5\pi}{6}\right)}$.

Watch out: A calculator's ${\tan^{-1}}$ gives an angle between ${-\dfrac{\pi}{2}}$ and ${\dfrac{\pi}{2}}$ only. For points in quadrants II and III, you must add ${\pi}$.

Changing equations

The same formulas turn an equation in ${x}$ and ${y}$ into one in ${r}$ and ${\theta}$, and back. Often one form is much simpler than the other.

Example 4: Write each equation in polar form.

(a) ${x^2+y^2=9}$ (b) ${y=x}$ (c) ${x=2}$

Solution:

(a) ${x^2+y^2=r^2}$, so ${r^2=9}$, or simply ${r=3}$: the circle of radius ${3}$.

(b) Divide by ${x}$: ${\dfrac{y}{x}=1}$, so ${\tan\theta=1}$. That is the line ${\theta=\dfrac{\pi}{4}}$ (with ${r}$ allowed to be negative, it covers the whole line).

Example 5: Write ${r=4\cos\theta}$ in rectangular form, and describe its graph.

Solution:

Multiply both sides by ${r}$ to make ${r^2}$ and ${r\cos\theta}$ appear:

$\begin{align*}r^2&=4r\cos\theta\\x^2+y^2&=4x\end{align*}$

Complete the square in ${x}$ (see Coordinates, graphs, and circles):

$\begin{align*}x^2-4x+4+y^2&=4\\(x-2)^2+y^2&=4\end{align*}$

This is the circle with center ${(2,0)}$ and radius ${2}$.

The graph of ${r=4\cos\theta}$ is the circle with center ${(2,0)}$ and radius ${2}$.

Summary

A point ${(r,\theta)}$ is a distance ${r}$ from the pole, at angle ${\theta}$. Negative ${r}$ means the opposite direction.
Every point has many polar names: ${(r,\theta)}$, ${(r,\theta+2k\pi)}$, and ${(-r,\theta+\pi)}$.
${x=r\cos\theta}$, ${y=r\sin\theta}$, ${r^2=x^2+y^2}$, and ${\tan\theta=\dfrac{y}{x}}$ (check the quadrant).
To change a polar equation, it often helps to multiply by ${r}$ first.

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