# Polar Coordinates and Equations

Examples on polar coordinates and equations are presented along with their detailed solutions.
We first review the Cartesian (or rectangualr) coordinates system. Let $A$ be a point on the $x y$ plane. In a Cartesian coordinate system, point $A$ have two coordinates: $x$ which is the signed distance from the $y$ axis to the point and $y$ which is the signed distance from the $x$ axis to the point. In the graph below are shown two points with coodinates $(x,y) = (4,2)$ and $(x,y) = (-3,4)$.

## Polar Coordinates System and Definitions

A point in polar coordinates also has two coordinates $(r , \theta)$ where $r$ is the distance from the origin called pole to the point and $\theta$ is the angle between the polar axis and the ray from the pole to the point. $r$ is called the radial coordinate and $\theta$ the angular coordinate .
As examples, the points $(r , \theta) = (5,\dfrac{\pi}{3})$ and $(r , \theta) = (4,\pi)$ are shown in the graph below. Fig.2 - Polar Coordinates By convention, the angle $\theta$ is positive if measured in counterclockwise direction and negative if measured in clockwise direction.

Example 1
Plot the given points given by their polar coordinates.
a) $(2,0)$       b) $(2,\dfrac{3\pi}{4})$       c) $(4,\dfrac{7\pi}{3})$       d) $(3,-\dfrac{5\pi}{4})$

Solution to Example 1
The points given by their polar coordinates are plotted below. Fig.3 - Plots of Points in Example 1 Note that if you keep the radial coordinate the same and add or subtract an $2 n \pi$ to the angular coordinate $\theta$ you obtain coordinates of the same point.
For example, the coordinates $(r,\theta)$, $(r,\theta + 2\pi)$, $(r,\theta - 4\pi)$... represent the same point.

## Radial Coordinate $r$ May be Negative

The above definition may be extended to use signed distances and allow the radial coordinate $r$ to be negative such that the points $(- r , \theta)$ and $(r , \theta)$ are on the same line through the pole, at the same distance $|r|$ from the pole such that point $r , \theta$ is in the same quadrant as $\theta$ and the point $- r , \theta$ is in the opposite quadrant as that of $\theta$.
The two points corresponding to $(- r , \theta)$ and $(r , \theta)$ are reflection of each other with respect to the pole (origin) and therefore we may say that the polar $(- r , \theta)$ and $(r , \theta + \pi)$ represent the same point.

Example 2
Plot the pairs of points given by their polar coordinates.
a) $(-2,0)$ and $(2,0)$       b) $(-2,\dfrac{3\pi}{4})$ and $(2,\dfrac{3\pi}{4})$       c) $(-4,\dfrac{-\pi}{3})$ and $(4,\dfrac{-\pi}{3})$

Solution to Example 2
The pair of points given by their polar coordinates with negative and positive radial distance $r$ are plotted below.

## The Same Point with Different Polar Coordinates

One of the major differences between Cartesian and polar coordinates is that polar coordinates of the same point may be written in an infinite number of ways.
For any integer $n$ and any other integer $k$, all coordinates of the form $(r,\theta)$, $(r,\theta+2n\pi)$ and $( - r,\theta+(2k+1) \pi)$ represent the same point.

Example 3
Show that all the given polar coordinates represent the same point.
a) $(2,-\dfrac{5\pi}{4})$       b) $(-2,-\dfrac{\pi}{4})$ c) $(2,\dfrac{3\pi}{4})$       d) $(2,\dfrac{19\pi}{4})$       e) $(2,-\dfrac{13\pi}{4})$

Solution to Example 3
We rewrite the given polar coordinates so that they have the equal radial coordinate $r$ and equal angular coordinate $\theta$ such that $0 \le \theta \lt 2 \pi$
a)
The angular coordinate $-\dfrac{5\pi}{4}$ is in the range $[-2\pi , 0]$, hence add $2 \pi$ to it.
$-\dfrac{5\pi}{4} + 2 \pi = \dfrac{3\pi}{4}$
Hence the given coordinates $(2,-\dfrac{5\pi}{4})$ and $(2, \dfrac{3\pi}{4})$ represent the same point.

b)
Change the polar coordinates so that it has a positive $r$ by adding $\pi$ to the angular coordinate
$(-2,-\dfrac{\pi}{4})$ and $(2,-\dfrac{\pi}{4}+\pi) = (2,\dfrac{3\pi}{4})$ represent the same point

c)
It has equal $r = 2$ and $\dfrac{3\pi}{4}$ so no change is required

d)
The angular coordinate is in the range $[4\pi , 5\pi]$; hence subtract $4 \pi$ from the angular coordinate of the given point $(2,\dfrac{19\pi}{4})$
$\dfrac{19\pi}{4} - 4\pi = \dfrac{3\pi}{4}$
Hence $(2,\dfrac{19\pi}{4})$ and $(2,\dfrac{3\pi}{4})$ represent the same point.

e)
The angular coordinate is in the range $[ -4\pi , -3\pi]$ , hence add $4 \pi$ to the angular coordinate of the given point $(2,-\dfrac{13\pi}{4})$
$-\dfrac{13\pi}{4} + 4 \pi = \dfrac{3\pi}{4}$
Hence $(2,-\dfrac{13\pi}{4})$ and $(2,\dfrac{3\pi}{4})$ represent the same point.
All the given polar coordinates represent the same point $(2,\dfrac{3\pi}{4})$.

## Relationships Between Polar and Cartesian Coordinates

The same point $P$ may be located using Cartesian or polar coordinates. The relationship between the two sets of coordinates may be found using trigonometry in a right triangle.

$x = r \cos \theta$ and $y = r \sin \theta$
$r^2 = x^2 + y^2$ and $\theta = \arctan (\dfrac{y}{x})$ Fig.5 - Relationship Between Cartesian and Polar Coordinates More on converting polar to rectangular coordinates and vice versa are included.

## Polar Equations and Curves

Polar equations are of the form $f(r,\theta) = 0$ and their graphs are all the set of points $(r,\theta)$ that are solution to the equation.

Example 4
Sketch the graph of the polar equation $r - 3 \sin \theta = 0$ .

Solution to Example 4
Solve the given equation for $r$.
$r = 3 \sin \theta$
Make a table of values

$\theta$ $r = 3 \sin \theta$
$0$ $0$
$\pi/6$ $3/2$
$\pi/4$ $3/\sqrt 2$
$\pi/3$ $3 \sqrt 3 /2$
$\pi/2$ $3$
$2\pi/3$ $3 \sqrt 3 /2$
$3\pi/4$ $3/\sqrt 2$
$5\pi/6$ $3 / 2$
$\pi$ $0$ Fig.6 - Graph of the Given Polar Equation $r - 3 \sin \theta = 0$ More on graphing of polar polar equations are included.

Note that the given polar equation can easily be converted to Cartesian (rectangular) coordinates using the relationships given above as follows
$r - 3 \sin \theta = 0$
Substitute $\sin \theta$ by $\dfrac{y}{r}$ (see relationship above) in the polar equation given above
$r - 3 \dfrac{y}{r} = 0$
Multiply all terms by $r$ and simplify
$r^2 - 3 y = 0$
Substitute $r^2$ by $x^2+y^2$ and rewrite the above equations as
$x^2+y^2 - 3 y = 0$
Complete the square and rewrite the above as
$x^2+ (y^2 - 1.5)^2 = 1.5^2$
It is a circle with center at the rectangular coordinates $(0 , 1.5)$ and has a radius equal to $1.5$ which is the graph obtained.
More on
converting equations from polar to rectangular and from rectangular to polar coordinates are included.