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:
Depending on what you're doing, one type of coordinate system might be much simpler to use than another! 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 widely-used) rectangular coordinates. |
![]() radar systems are well-suited 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:
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![]() To plot a point:
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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 $\,(r,\theta)\,$:
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![]() starting set-up 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:
![]() $r = 2$ $\theta = 90^\circ$
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![]() $r = 2$ $\theta = -90^\circ$
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![]() $r = -2$ $\theta = 90^\circ$
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![]() $r = -2$ $\theta = -90^\circ$
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![]() $r = 2$ $\theta = \pi/4$
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![]() $r = 2$ $\theta = 90^\circ$
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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 |
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$$\cssId{sb11}{r = 3}$$
circle with radius $\,3\,,$ centered at the pole |
This is an equation in two variables:
$$\cssId{sb15}{r + 0\cdot\theta = 3}\,$$
The graph is (using set-builder notation): $$\cssId{sb17}{\{\,(3,\theta) \ \ |\ \ \theta\in\Bbb R\,\}}$$ |
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$$\cssId{sb18}{\theta = \frac{\pi}4}$$ or $$\cssId{sb20}{\theta = 45^\circ}$$ line through the pole |
This is an equation in two variables:
$$\cssId{sb23}{0\cdot r + \theta = \frac{\pi}4}\,$$
The graph is: $$\cssId{sb25}{\{\,(r,\frac{\pi}4) \ \ |\ \ r\in\Bbb R\,\}}$$ |
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:
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![]() 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:
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![]() 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
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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 : $$\cssId{sb107}{r = \sqrt{x^2 + y^2}}$$ |
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point(s) $\,(x,y)\,$ | distance from $\,(x,y)\,$ to origin | the expression $\,\sqrt{x^2 + y^2}\,$ gives: |
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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:
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$\color{red}{\sqrt{x^2 + y^2}}$ |
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! $$ \cssId{sb149}{\color{red}{\theta \ = \ \arctan \frac yx \ = \ \overbrace{\tan^{-1}\frac yx}^{\text{alternate notation}}}} $$ 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 piecewise-defined 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 piecewise-defined 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 piecewise-defined description of $\,\theta\,$ is given below. Similarly, you can find formulas 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 $\,[0,2\pi)\,.$ |
$$ \cssId{sb202}{\theta = \cases{\cr 0 &if $\,x = 0\,$ and $\,y = 0\,$\cr\cr \arctan\frac yx &if $\,x \gt 0\,$ and $\,y \ge 0\,$\cr\cr \frac\pi 2 &if $\,x = 0\,$ and $\,y \gt 0\,$\cr\cr \pi +\arctan\frac yx &of $\,x \lt 0\,$\cr\cr \frac{3\pi} 2 &if $\,x = 0\,$ and $\,y \lt 0\,$\cr\cr 2\pi +\arctan\frac yx &if $\,x \gt 0\,$ and $\,y \lt 0\,$ }} $$ |
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 $\,\theta\,$:
$$\cssId{sb208}{r = m\theta + b\,, \qquad \text{ for } \theta\in [\theta_{\text{start}},\theta_{\text{end}}]}$$
Change parameters by setting values and clicking the ‘Plot the curve!’ button. Three points are given:
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Here are some for you to try:
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$r = \theta\,,$ 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 cut-and-paste 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 $\,\theta\in [0,2\pi]\,$) |
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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 cut-and-paste 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 |