Polar Coordinates (Part 3)

(This page is Part 3. Click here for Part 1.  Click here for Part 2.)

Converting from Rectangular Coordinates to Polar Coordinates

Given a point with rectangular coordinates $\,(x,y)\,,$ we want formulas for polar coordinates $\,(r,\theta)\,.$ That is, we want:

• a formula for $\,r\,$ that depends on $\,x\,$ and $\,y$
• a formula for $\,\theta\,$ that depends on $\,x\,$ and $\,y$

Here's part of the punchline:  For all real numbers $\,x\,$ and $\,y\,$:

Unfortunately, $\,\theta\,$ is considerably more complicated! The details follow.

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 above suggests, there's an easy formula to get a point's distance from the origin:

The table below verifies that the expression $\,\sqrt{x^2 + y^2}\,$ returns the correct distance from the origin for every point in the plane.

Recall that for all real numbers $\,t\,$:

• $|t|\,$ gives the distance from $\,t\,$ to zero
• $\sqrt{t^2} = |t|$
• $|t|^2 = t^2\,$

Formulas for $\,\theta$

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:

• $0 \le \theta \lt 2\pi$
• $-\pi \lt \theta \le \pi$

Formula for $\,\theta\,$ in Quadrant I

Quadrant I (where $\,x \gt 0\,$ and $\,y \gt 0\,$) is no problem!

Given a point $\,(x,y)\,$ in Quadrant I, the right triangle shown above always has these properties:

• For polar coordinates, we need $\,\color{red}{\theta}\,.$
• The side OPPosite $\,\color{red}{\theta}\,$ has length $\,y\,.$
• The side ADJacent to $\,\color{red}{\theta}\,$ has length $\,x\,.$
• Thus, $\,\color{red}{\tan \theta = \frac{\text{OPP}}{\text{ADJ}} = \frac yx}\,.$

In Quadrant I, $\,\color{red}{\theta}\,$ is always (strictly) between $\,0\,$ and $\,\frac\pi 2\,.$ 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!

Formulas for $\,\theta\,$ in Other Quadrants

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: $\,x \gt 0\,$ and $\,y \gt 0\,$

In this quadrant:  $\displaystyle\theta = \arctan \frac yx$

Piecewise-Defined Function that Returns $\,\theta\in [0,2\pi)\,$ for Points $\,(x,y)$

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 $\,y = 0\,,$

$$\begin{align} \theta &= \arctan{\frac yx}\cr &= \arctan 0\cr &= 0 \end{align} $$

$\,x = 0\,$ 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 $\,y = 0\,,$

$$\begin{align} \theta &= (\arctan{\frac yx}) + \pi\cr &= (\arctan 0) + \pi\cr &= \pi \end{align} $$

$x \lt 0\,$ 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 $\,[0,2\pi)\,.$

Flowers, Spirals, and More

There are many fun/interesting/beautiful curves that can be described by simple equations involving $\,r\,$ and $\,\theta\,.$ These are sometimes called ‘polar curves’.

Spirals (and more)

Below, $\,r\,$ is a simple linear function of $\,\theta\,$:

Type in Desired Values for the Curve $\,r = m\theta + b$

Plot the curve!

$r = m\theta + b$,  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:

• The blue point is the start point:  $\,r_\text{start} := m\theta_{\text{start}} + b$
• The green point is the end point:  $\,r_\text{end} := m\theta_{\text{end}} + b$
• The red point is a glider on the curve. Drag it around!

  Rectangular coordinates are ‘attached to’ the point. Corresponding polar coordinates are displayed at the top right of the graph (with $\,\theta\,$ between $\,-180^\circ\,$ and $\,180^\circ\,$).

Note that:

• $m\,$ controls how fast $\,r\,$ changes:  When $\,\theta\,$ changes by an amount $\,\Delta\theta\,,$ then $\,r\,$ changes by $\,m\Delta\theta\,.$ That is, $\,r\,$ changes $\,m\,$ times faster than $\,\theta\,.$
• $b\,$ gives an ‘offset’:  When $\,\theta = 0\,,$ then $\,r = m(0) + b = b\,.$
• The viewing screen changes as needed, so that you can see the entire curve. If you change a value and it looks like the graph hasn't changed—check the scales on the $x$ and $y$ axes!

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 $\,\theta\in [-10,10]$

Flowers (and more)

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)

Gives a rose with $\,k\,$ petals when $\,k\,$ is an odd integer (for $\,\theta\in [0,\pi]\,$)

r = 2cos(5theta), with theta from 0 to pi

Gives a rose with $\,2k\,$ petals when $\,k\,$ is an even integer (for $\,\theta\in [0,2\pi]\,$)

r = 3cos(4theta), with theta from 0 to 2pi

Just Have Fun!

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!