Roughly, polar coordinates determine the position of a point in a plane
by specifying a distance from a fixed point in a given direction.
Pick a point in a plane. (Think of a dot on a piece of paper.)
A coordinate system is used to answer these questions:
A coordinate system always involves:
As you'll see, applications involving concentric circles (circles with a common center) are probably going to be much better suited to polar coordinates (this section) than to the (more familiar and widelyused) rectangular coordinates. 
Depending on what you're doing, one type of coordinate system might be much simpler to use than another!
radar systems are wellsuited to polar coordinates 
It's fun (and instructive) to compare the ways the plane is ‘divided up’ for rectangular versus polar coordinates.
A more precise discussion of polar coordinates follows.
To plot a point:

To plot a point:

DEFINITION
polar coordinates
NOTES ON POLAR COORDINATES:
A polar coordinate system is determined by:
If $\,\theta\,$ has no units, then it is the radian measure of an angle. Other units (like degrees) can also be used for $\,\theta\,$. To determine the position of the point with polar coordinates $\,(r,\theta)\,$:

starting setup for a polar coordinate system rotate the polar axis about the pole by an amount $\,\theta\,$ move the point from the pole a distance $\,r\,$ 
In all these examples:
$r = 2$ $\theta = 90^\circ$

$r = 2$ $\theta = 90^\circ$

$r = 2$ $\theta = 90^\circ$

$r = 2$ $\theta = 90^\circ$

$r = 2$ $\theta = \pi/4$

$r = 2$ $\theta = 90^\circ$

In both rectangular and polar coordinates, a given coordinate pair specifies a unique point in the plane:
However, when going from a point in a plane to a coordinate pair representing it—rectangular coordinates are simpler!
There are different ways that you can restrict $\,r\,$ and $\,\theta\,$ to get a unique representation for
each point in polar coordinates.
Here are two of the most common ways:
Different coordinate systems make different curves extremely easy to plot:
GRAPH  EQUATION/DESCRIPTION  COMMENTS 
$$r = 3$$
circle with radius $\,3\,$, centered at the pole 
This is an equation in two variables:
$$r + 0\cdot\theta = 3\,$$
The graph is (using setbuilder notation): $$\{\,(3,\theta) \ \ \ \ \theta\in\Bbb R\,\}$$ 

$$\theta = \frac{\pi}4$$ or $$\theta = 45^\circ$$ line through the pole 
This is an equation in two variables:
$$0\cdot r + \theta = \frac{\pi}4\,$$
The graph is: $$\{\,(r,\frac{\pi}4) \ \ \ \ r\in\Bbb R\,\}$$ 
Given a point with polar coordinates $\,(r,\theta)\,$, we want formulas for the rectangular coordinates $\,(x,y)\,$.
That is, we want:
To convert from polar to rectangular coordinates, first introduce a rectangular coordinate system:
NOTE: 
Use the polar axis to introduce a rectangular coordinate system. 
Plot a point in polar coordinates ($\,r \ge 0\,$) and find its rectangular coordinates:

getting rectangular coordinates for a point with polar coordinates $\,(r,\theta)\,$: $\,r\ge 0\,$ 
Plot a point in polar coordinates ($\,r \lt 0\,$) and find its rectangular coordinates :

getting rectangular coordinates for a point with polar coordinates $\,(r,\theta)\,$: $\,r\lt 0\,$ 
Given a point with rectangular coordinates $\,(x,y)\,$, we want formulas for polar coordinates $\,(r,\theta)\,$.
That is, we want:
To convert from rectangular to polar coordinates,
first introduce a polar axis
.
The formulas are simplest by placing the pole at the origin, with the ray pointing in the direction of
the positive $x$axis.
Since every point has infinitely many names in polar coordinates,
decisions need to be made about
which names to return!
Formula for $\,r\,$As the sketch at right suggests, there's an easy formula to get a point's distance from the origin : $$r = \sqrt{x^2 + y^2}$$ 

point(s) $\,(x,y)\,$  distance from $\,(x,y)\,$ to origin  the expression $\,\sqrt{x^2 + y^2}\,$ gives: 

the origin: $\,(0,0)\,$  $\color{red}{0}$  $\sqrt{x^2 +y^2} = \sqrt{0^2 + 0^2} = \color{red}{0}$  
$x$axis: $\,(x,0)\,$  $\color{red}{x}$  $\sqrt{x^2 +y^2} = \sqrt{x^2 + 0^2} = \sqrt{x^2} = \color{red}{x}$  
$y$axis: $\,(0,y)\,$  $\color{red}{y}$  $\sqrt{x^2 +y^2} = \sqrt{0^2 + y^2} = \sqrt{y^2} = \color{red}{y}$  
four quadrants: $\,(x,y)\,$ for $\,x\ne 0\,$ and $\,y\ne 0\,$ 
See sketch at right:

$\color{red}{\sqrt{x^2 + y^2}}$ 
Using the formula $\,r = \sqrt{x^2 + y^2}\,$ guarantees that $\,r\,$ is nonnegative.
With nonnegative $\,r\,$, there are two conventional choices for $\,\theta\,$ to get unique names for all points in the plane:
Quadrant I (where $\,x > 0\,$ and $\,y > 0\,$) is no problem! Given a point $\,(x,y)\,$ in Quadrant I, the right triangle shown at right always has these properties:
So, we want the angle between $\,0\,$ and $\,\frac\pi 2\,$ whose tangent is $\,\frac yx\,$. In this case, the arctangent gives us precisely the angle we want! $$ \color{red}{\theta \ = \ \arctan \frac yx \ = \ \overbrace{\tan^{1}\frac yx}^{\text{alternate notation}}} $$ Since (by definition) the arctangent function always returns angles in the interval $\,(\frac{\pi}2\,,\,\frac{\pi}2)\,$, once we leave Quadrant I, the formulas for $\,\theta\,$ may require some ‘tweaking’. As you'll see next, the formula for $\,\theta\,$ is a piecewisedefined function—with quite a few pieces! 
Quadrant I: $\,x > 0\,$ and $\,y > 0\,$ In this quadrant: $\displaystyle\theta = \arctan \frac yx$ 
These sketches are used to get the different ‘pieces’ for the piecewisedefined function.
In all sketches, the black point has rectangular coordinates $\,(x,y)\,$.
$\,x \gt 0\,$ and $\,y\ge 0\,$ Here: $\displaystyle\,\frac yx \ge 0\,$ $\displaystyle\,\color{red}{\theta = \arctan{\frac yx}}\,$ Note: $\displaystyle\,\theta\in [0,\frac\pi 2)\,$ Note: when $\,y = 0\,$, $\,\theta = \arctan{\frac yx} = \arctan 0 = 0\,$ 
$\,x = 0\,$ and $\,y\gt 0\,$ Here: $\displaystyle\,\frac yx\,$ is not defined $\displaystyle\,\color{red}{\theta = \frac\pi 2}\,$ 
$\,x \lt 0\,$ and $\,y\ge 0\,$ Here: $\displaystyle\,\frac yx \le 0\,$ $\displaystyle\,\color{red}{\theta = (\arctan{\frac yx}) + \pi}\,$ Note: $\displaystyle\,\theta\in (\frac\pi 2,\pi]\,$ Note: when $\,y = 0\,$, $\,\theta = (\arctan{\frac yx}) + \pi = (\arctan 0) + \pi = \pi\,$ 
$\,x \lt 0\,$ and $\,y\lt 0\,$ Here: $\displaystyle\,\frac yx \gt 0\,$ $\displaystyle\,\color{red}{\theta = (\arctan{\frac yx}) + \pi}\,$ Note: $\displaystyle\,\theta\in (\pi,\frac{3\pi}2)\,$ 
$\,x = 0\,$ and $\,y\lt 0\,$ Here: $\displaystyle\,\frac yx\,$ is not defined $\displaystyle\,\color{red}{\theta = \frac{3\pi}{2}}\,$ 
$\,x \gt 0\,$ and $\,y\lt 0\,$ Here: $\displaystyle\,\frac yx \lt 0\,$ $\displaystyle\,\color{red}{\theta = (\arctan{\frac yx}) + 2\pi}\,$ Note: $\displaystyle\,\theta\in (\frac{3\pi}2,2\pi)\,$ 
$\,x = 0\,$ and $\,y = 0\,$ Here: $\displaystyle\,\frac yx\,$ is not defined $\displaystyle\,\color{red}{\theta = 0}\,$ Note: For uniqueness, we can agree that the point with rectangular coordinates $\,(0,0)\,$ gets represented only by $\,r = 0\,$ and $\,\theta = 0\,$. 
The piecewisedefined description of $\,\theta\,$ is given below. In the exercises, you will find the formula for $\,\theta\,$ when the desired angles to be returned are in the interval $\,(\pi,\pi]\,$. 
This formula for $\,\theta\,$, together with $\,r = \sqrt{x^2 + y^2}\,$, gives a unique representation $\,(r,\theta)\,$ for each point with rectangular coordinates $\,(x,y)\,$. The angles $\,\theta\,$ are in the interval $\,[0,2\pi)\,$. 
$$ \theta = \cases{ 0 & \text{if }\ x = 0 \text{ and } y = 0\cr\cr \arctan\frac yx & \text{if }\ x \gt 0 \text{ and } y \ge 0\cr\cr \frac\pi 2 & \text{if }\ x = 0 \text{ and } y \gt 0\cr\cr \pi +\arctan\frac yx & \text{if }\ x \lt 0\cr\cr \frac{3\pi} 2 & \text{if }\ x = 0 \text{ and } y \lt 0\cr\cr 2\pi +\arctan\frac yx & \text{if }\ x \gt 0 \text{ and } y \lt 0 } $$ 
There are many fun/interesting/beautiful curves that can be described by simple equations involving $\,r\,$ and $\,\theta\,$.
These are sometimes called ‘polar curves’.
At right, $\,r\,$ is a simple linear function of $\,\theta\,$:
$$r = m\theta + b\,, \qquad \text{ for } \theta\in [\theta_{\text{start}},\theta_{\text{end}}]$$
Change parameters by setting values and clicking the ‘Plot the curve!’ button. Three points are given:


Here are some for you to try:




$r = \theta\,$, for $\,\theta\in [0,4\pi]$  $r = \theta\,$, for $\,\theta\in [0,2\pi]$  $r = 3\theta  10\,$, for $\,\theta\in [0,5\pi]$  $r = 0.3\,\theta\,$, for $\,\theta\in [10,10]$ 
Of course, WolframAlpha can plot polar curves!
Here are some for you to try—you can cutandpaste the text below each image (if desired).
Polar Rose:
$r = a\cos(k\theta)\,$ $a$ gives the length of the petals 
Just have fun!  
Gives a rose with $\,k\,$ petals when $\,k\,$ is an odd integer (for $\,\theta\in [0,\pi]\,$) 
Gives a rose with $\,2k\,$ petals when $\,k\,$ is an even integer (for $\,\theta\in [0,2\pi]\,$) 





r = 2cos(5theta), with theta from 0 to pi 
r = 3cos(4theta), with theta from 0 to 2pi 
r = 10 + sin(2*pi*theta), for theta from 0 to 14pi 
r = theta * cos(theta), for theta from 4pi to 4pi 
You can cutandpaste the text above into WolframAlpha, and then play around! 
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
IN PROGRESS 