Coordinate systems: What are they good for?

By now you're used to the Cartesian coordinate system, the $x-y$ grid that we generally use to graph functions and plot points. Each point has a set of coordinates $(x, \; y)$, a recipe for finding the point starting from the origin.

Aside from its convenience—axes at 90˚ angles, evenly-spaced grid points—there's nothing unique about the Cartesian plane. In fact, there are an infinite number of different coordinate systems, mostly silly ones, that we could use. Though less convenient, any of them will work to represent any mathematical problem or function in a plane.

For example, if you're reading this section, you've probably dealt with balls rolling down ramps in physics. Recall that in those problems you redefined your coordinate system. You tilted it to make the x-axis parallel to the ramp and the y-axis perpendicular or normal to the ramp.

Examples of coordinate systems

Drawn below are a Cartesian grid (left) and a couple of goofy coordinate systems no one uses, but they can locate a point just the same. The same point on each square is represented with an ordered pair of coordinates in each system.

Polar coordinates

Polar coordinates, defined below, come in handy when we're describing things that are centrosymmetric (have a center of symmetry, like a circle) or that rotate in a circle, like a wheel or a spinning molecule.

In the Cartesian coordinate system, we move over (left-right) $x$ units, and $y$ units in the up-down direction to find our point.

Here is an example of a polar graph →

The point $(r, \; \theta) = (3, \; 60^{\circ})$ is plotted by moving a distance 3 to the right along the zero-degree line, then rotating that line segment by 60˚ counterclockwise to reach the point.

Any point on a plane can be located in this manner, just like with Cartesian $(x, \; y)$ coordinates.

The point in the center is usually called the "pole."

Practice problems

A. Convert the following cartesian coordinates $(x, \; y)$ to polar coordinates $(r, \; \theta)$:

1. $(1, \; 1)$

   Solution

   First make a sketch of the vector. For this onee its easy to see that the angle that the vector makes with the x-axis if 45˚.

   

   Now calculate the vector length using a 45-45-90 triangle and the Pythagorean theorem: $\sqrt{1^2+1^2}=\sqrt{2}$.

   

   In polar coordinates the vector is

   $$(r, \theta) = (\sqrt{2}, 45^{\circ}) = (\sqrt{2}, \frac{\pi}{4})$$




2. $(-1, \; 1)$

   Solution

   First make a sketch of the vector. For this one we notice that we form a 45-45-90 triangle in the 2^nd quadrant. The rotation of the vector from 0˚ is 135˚.

   

   Now calculate the vector length using a 45-45-90 triangle and the Pythagorean theorem: $\sqrt{1^2+1^2}=\sqrt{2}$

   

   In polar coordinates, the vector is $(r, \theta) = (\sqrt{2}, 135^{\circ}) = (\sqrt(2), 3\pi/4)$.




3. $(1, \; 2)$

   Solution

   First make a sketch of the vector. For this one, the angle isn't obvious so we'll have to calculate it.

   

   $$ \begin{align} \theta &= \text{tan}^{-1} \left( \frac{y}{x}\right) \\[5pt] &= \text{tan}^{-1} \left( \frac{2}{1}\right) \\[5pt] &= 63.43^{\circ} \end{align}$$

   Now calculate the vector length using the Pythagoream theorem:

   

   In polar coordinates, the vector is $(r,\theta) = (\sqrt{5}, 63.43^{\circ})$




4. $(2, \; -3)$

   Solution

   First make a sketch of the vector. Calculate the angle between the x-axis and the vector using the inverse tangent. The polar angle is the counterclockwise angle to the vector from 0˚, or its negative in the clockwise direction:

   

   $$\theta = \text{tan}^{-1} \left( \frac{3}{2} \right) = -56.31^{\circ}$$

   Now calculate the length of the vector using a right triangle and the Pythagorean theorem:

   

   In polar coordinates, the vector is $(r,\theta) = (\sqrt{13}, 303.69^{\circ})$ or $(\sqrt{13}, -56.31^{\circ})$.

B. Convert the following polar coordinates, $(r, \; \theta)$ to cartesian coordinates $(x, \; y)$:

1. $(1, \; 45^{\circ})$

   Solution

   First make a sketch of the polar vector. Then it's a matter of finding the lengths of the sides of the resulting right triangle using trigonometry:

   

   To find x and y, use:

   $$ \begin{align} x &= r \cdot \text{cos}(\theta) \\[5pt] &= 1 \cdot \text{cos}(45^{\circ}) = \frac{\sqrt{2}}{2} \\[5pt] y &= r \cdot \text{sin}(\theta) \\[5pt] &= 1 \cdot \text{sin}(45^{\circ}) = \frac{\sqrt{2}}{2} \end{align}$$

   In Cartesian coordinates, the location of our point is

   $$(x,y) = \left( \frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right)$$




2. $(3, \; 4\pi/3)$

   Solution

   First make a sketch of the polar vector. The polar angle is $4\pi/3$, but we'll want to use the smallest angle to the x-axis in order to find the x- and y-coordinates, thus the angle $\pi/3$ below:

   

   To find x and y, use:

   $$ \begin{align} x &= r \cdot \text{cos}(\theta) \\[5pt] &= 3 \cdot \text{cos}(\pi/3) = \frac{1}{2} \\[5pt] y &= r \cdot \text{sin}(\theta) \\[5pt] &= 3 \cdot \text{sin}(\pi/3) = \frac{\sqrt{3}}{2} \end{align}$$

   

   In Cartesian coordinates, the location of our point is

   $$(x,y) = \left( -\frac{1}{2},-\frac{\sqrt{3}}{2} \right)$$




3. $(2, \; 30^{\circ})$

   Solution

   First make a sketch of the polar vector:

   

   $$ \begin{align} x &= r \cdot \text{cos}(\theta) \\[5pt] &= 2 \cdot \text{cos}(30^{\circ}) = \frac{2\sqrt{3}}{2} = \sqrt{3} \\[5pt] y &= r \cdot \text{sin}(\theta) \\[5pt] &= 2 \cdot \text{sin}(30^{\circ}) = \frac{2}{2} = 1 \end{align}$$

   The Cartesian coordinates are $(x,y)=(\sqrt{3}, 1)$.




4. $(-2, \; 120^{\circ})$

   Solution

   First make a sketch of the polar vector. This one is a little different because we begin with a negative $r$ component. The triangle we'll use on the right (in the 4^th quadrant) is a 30-60-90 triangle.

   

   

   $$ \begin{align} x &= r \cdot \text{cos}(\theta) \\[5pt] &= 2 \cdot \text{cos}(60^{\circ}) = \frac{2\sqrt{3}}{2} = \sqrt{3}1 \\[5pt] y &= r \cdot \text{sin}(\theta) \\[5pt] &= 2 \cdot \text{sin}(60^{\circ}) = \sqrt{3} \end{align}$$

   The Cartesian coordinates are $(x,y)=(1,-\sqrt{3})$, where we've adjusted the sign of the y-component to match the picture.

Functions in polar coordinates

We can express all kinds of functions in polar coordinates, but some are more suitable for them than others. For example, the vertical and horizontal lines in the first graph below are what we could call "native" to the rectilinear (made up of lines and 90˚ angles) Cartesian coordinate system. That's why their equations are so simple. The equation of a circle of radius R, centered at the origin, however, is $x^2 + y^2 = R^2$ in Cartesian coordinates, but just $r = R$ in polar coordinates. The circle is a native figure in polar coordinates.

Example 1

Convert the Cartesian equation $2x - 3y = 7$ to polar form

To convert this function to polar coordinates, we begin with the conversions $x = r \, \text{sin}(\theta)$ and $y = r \, sin(\theta)$:

$$2x - 3y = 7$$

Plugging those into our linear equation gives:

$$2r \, cos(\theta) - 3r \, sin(\theta) = 7$$

Example 2

Convert the Cartesian circle $x^2 + y^2 = 16$ to polar form

To convert this function to polar coordinates, we begin with the squares of the conversions $x^2 = r^2 \, \text{cos}^2(\theta)$ and $y^2 = r^2 \, \text{sin}^2(\theta)$

Plugging those into our circle equation gives:

$$r^2 \, \text{cos}^2 (\theta) + r^2 \, \text{sin}^2 (\theta) = 16$$

If we factor out the $r^2$, and use the Pythagorean identity on what remains

Example 3

Now look at a circle not centered at the origin: $(x - 2)^2 + y^2 = 4$

Convert this circle to polar coordinates.

The function is

$$(x - 1)^2 + y^2 = 4$$

The first step is just to substitute $x = r cos(\theta)$ and $y = r sin(\theta)$ for $x$ and $y$:

$$[r \, \text{cos}(\theta) - 1]^2 + r^2 \, \text{sin}^2(\theta) = 4$$

Now do the squaring, remembering that the binomial-squared has three terms,

$$r^2 \, \text{cos}^2 (\theta) - 2 r \, \text{cos}(\theta) + 1 + r^2 \, \text{sin}^2 (\theta) = 4$$

Gathering terms with $r^2$ gives us a Pythagorean relationship that we can evaluate to 1

Practice problems

Convert these polar functions in the form of $r(\theta)$ to Cartesian equations in the form of $y(x)$.

1. $r = 3 \, \text{sec}(\theta)$

   Solution

   $$ \begin{align} r &= 3 \, \text{sec}(\theta) \\[5pt] r &= \frac{3}{\text{cos}(\theta)} \\[5pt] r \, \text{cos}(\theta) &= 3 \\[5pt] x &= 3 \end{align}$$

   A vertical line at x = 3.




2. $r = -2 \, \text{csc}(\theta)$

   Solution

   $$ \begin{align} r &= -2 \, \text{csc}(\theta) \\[5pt] r &= \frac{-2}{\text{sin}(\theta)} \\[5pt] r \, \text{sin}(\theta) &= -2 \\[5pt] y &= -2 \end{align}$$

   A horizontal line at y = -2.




3. $2 \, \text{sin}(\theta) - 4 \, \text{cos}(\theta) = r$

   Solution

   First recall that

   $$\text{sin}(\theta) = \frac{y}{r} \: \text{and} \: \text{cos}(\theta) = \frac{x}{r}$$

   then

   $$ \begin{align} 2 \text{sin}(\theta) - 4 \text{cos}(\theta) &= r \\[4pt] 2 \frac{y}{r} - 4 \frac{x}{r} &= r \\[4pt] 2y - 4x &= r^2 \\[4pt] 2y - 4x &= x^2 + y^2 \\[4pt] x^2 - 4x + y^2 + 2y &= 0 \\[4pt] x^2 - 4x + 2^2 + y^2 + 2y + 1^2 &= 5 \\[4pt] (x - 2)^2 + (y + 1)^2 &= 5 \end{align}$$

   A circle of radius $\sqrt{5}$ centered at (2, -1). (Note: I completed the square on x and y two lines up.)

Convert these Cartesian functions in the form of $y(x)$ to polar coordinates in the form of $r(\theta)$. You might want to plot these functions on a calculator to make sure they meet your expectations.

1. $x^2 + y^2 = 19$

   Solution

   $$r = \sqrt{19}$$

   (because $r^2 = x^2 + y^2$)




2. $-2x + 3y = 5$

   Solution

   $$ \begin{align} -2x + 3y &= 5 \\[4pt] -2r \text{cos}(\theta) + 3r \text{sin}(\theta) &= 5 \\[4pt] r(3 \text{sin}(\theta) - 2 \text{cos}(\theta)) &= 5 \\[4pt] \frac{5}{3 \text{sin}(\theta)- 2 \text{cos}(\theta)} &= r \end{align}$$

   The equation of a line is awkward in polar coordinates.




3. $(x - 3)^2 + y^2 = 9$

   Solution

   $$ \begin{align} (x - 3)^2 + y^2 &= 9 \\[4pt] x^2 - 6x + 9 + y^2 &= 9 \\[4pt] x^2 + 6x + y^2 &= 0 \\[4pt] x^2 + y^2 + 6x &= 0 \\[4pt] r^2 - 6 r \text{cos}(\theta) &= 0 \\[4pt] r &= 6 \text{cos} (\theta) \end{align}$$

Polar graphs – some examples

Here are just a few examples of the graphs of functions in polar coordinates. Notice how some functions are much simpler in polar than in Cartesian coordinates. In the exercises above, you might have noticed that some polar functions are very complicated in the Cartesian world.

Trigonometric functions

The graphs of the sine and cosine functions in polar coordinates are shown here. They are simply circles.

The black curve is the sine function, the red $r = \text{cos}(\theta)$.

What you don't see in these graphs is that as the polar angle, $\theta$, is traced around through 360˚, the sin or cos circle is traced twice. If you plot the polar graph of the function $r(\theta) = \text{sin}^2(\theta),$ you'll see a figure-8 shape, the upper loop traced as θ runs from 0˚ to 180˚, and the lower loop as the lower half between 180˚ and 360˚.

The polar graphs of $r = \text{sin}^2(\theta)$ and $r = \text{cos}^2(\theta)$ are shown below in blue and black, respectively.

Cardioids

Flowers

Some functions are quite fanciful in polar coordinates. Flower functions like these are formed by increasing the frequency factor (4 in this case). Just as it increases the number of periods per unit length in a Cartesian graph of a trig. function, it increases the number of loops in the polar graph.

You might want to try to graph some of these polar functions on your calculator:

You can graph polar functions on your TI-84 (or similar) calculator. Put the calculator in polar mode, then enter r(θ) (the variable button will automatically switch to θ), then enter the window values, x, y and the range of θ, and graph. Sometimes a small θ step size will cause your calculator to graph a complicated function very slowly, so be careful with that.