# Polar Coordinates, Equations and Graphs

This section covers:

So far, we’ve plotted points using rectangular (or Cartesian) coordinates, since the points since we are going back and forth $$x$$ units, and up and down $$y$$ units.

# Plotting Points Using Polar Coordinates

In the Polar Coordinate System, we go around the origin or the pole a certain distance out, and a certain angle from the positive $$x$$axis:

The ordered pairs, called polar coordinates, are in the form $$\left( {r,\theta } \right)$$, with $$r$$ being the number of units from the origin or pole (if $$r>0$$), like a radius of a circle, and $$\theta$$ being the angle (in degrees or radians) formed by the ray on the positive $$x$$axis (polar axis), going counter-clockwise. If $$r<0$$, the point is   units (like a radius) in the opposite direction (across the origin or pole) of the angle $$\theta$$. If $$\theta <0$$, you go clockwise with the angle, starting with the positive $$x$$axis.

To plot a point, you typically circle around the positive $$x$$axis $$\theta$$  degrees first, and then go out from the origin or pole $$r$$ units (if $$r$$ is negative, go the other way (180°) $$r$$ units).

Here a polar graph with some points on it. Note that we typically count in increments of 15°, or $$\displaystyle \frac{\pi }{{12}}$$.

For a point $$\left( {r,\theta } \right)$$, do you see how you always go counter-clockwise (or clockwise, if you have a negative angle) until you reach the angle you want, and then out from the center $$r$$ units, if $$r$$ is positive? If $$r$$ is negative, you go in the opposite direction from the angle $$r$$ units. If both $$r$$ and the angle $$\theta$$ are negative, you have to make sure you go clockwise to get the angle, but in the opposite direction $$r$$ units.

You may be asked to rename a point in several different ways, for example, between $$\left[ {-2\pi ,2\pi } \right)$$ or $$\left[ {-360{}^\circ ,360{}^\circ } \right)$$. For example, if we wanted to rename the point $$\left( {6,240{}^\circ } \right)$$ three other different ways between $$\left[ {-360{}^\circ ,360{}^\circ } \right)$$, by looking at the graph above, we’d get $$\left( {-6,60{}^\circ } \right)$$(make $$r$$ negative and subtract 180°), $$\left( {6,-120{}^\circ } \right)$$ (subtract 360°), and $$\left( {-6,-240{}^\circ } \right)$$ (make both negative). (Remember that 240° and –120°, and 60° and –240° are co-terminal angles). To get these, if the first number ($$r$$) is negative, you want to go in the opposite direction, and if the angle is negative, you want to go clockwise instead of counterclockwise from the positive $$x$$axis.

# Polar-Rectangular Point Conversions

You will probably be asked to convert coordinates between polar form and rectangular form.

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## Converting from Polar to Rectangular Coordinates

Let’s first convert from polar to rectangular form; to do this we use the following formulas, as we can see this from the graph:

This conversion is pretty straight-forward, and we’ll see examples below.

## Converting from Rectangular to Polar Coordinates

Converting from rectangular coordinates to polar coordinates can be a little trickier since we need to check the quadrant of the rectangular point to get the correct angle; the quadrants must match. Here are the formulas:

$$\displaystyle r\,\,=\,\,\sqrt{{{{x}^{2}}+{{y}^{2}}}}\,\,\,\,\,\,\,\,\,\text{(this will be positive)}$$

$$\displaystyle \theta \,\,=\,\,{{\tan }^{{-1}}}\left( {\frac{y}{x}} \right)\,\,\,\,\,\,\text{(check for correct quadrant)}$$

One word of caution: If $$x=0$$, we’ll get an error when trying to obtain $$\theta$$. In these cases, graph the point (it will be on the $$y$$-axis) to get the angle.

Again, we need to check quadrants when using the calculator to get $${{\tan }^{{-1}}}$$. We’ll have to add the following degrees or radians when the point is in the following quadrants (this is because the  $${{\tan }^{{-1}}}$$ function on the calculator only gives answers back in the interval $$\displaystyle \left( {-\frac{\pi }{2},\,\,\frac{\pi }{2}} \right)$$):

Note that there can be multiple “answers” when converting from rectangular to polar, since polar points can be represented in many different ways (co-terminal angles, positive or negative “$$r$$”, and so on). Thus is is typically easier to convert from polar to rectangular.

## Examples

Here are some examples; note you may be asked to convert back to polar into degrees or radians. For converting back to polar, make sure answers are either between 0 and 360° for degrees or 0 to $$2\pi$$ for radians. (And again, note that when we convert back to polar coordinates, we may not always get the same representation as the polar point we started out with.)

 Polar Coordinate Convert from Polar to Rectangular Convert from Rectangular to Polar $$\displaystyle \left( {2,\frac{{4\pi }}{3}} \right)$$ $$\displaystyle x=2\cos \frac{{4\pi }}{3}=2\left( {-\frac{1}{2}} \right)=-1$$ $$\displaystyle y=2\sin \frac{{4\pi }}{3}=2\left( {-\frac{{\sqrt{3}}}{2}} \right)=-\sqrt{3}$$   $$\left( {-1,-\sqrt{3}} \right)$$ $$r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{1+3}}=2$$   $$\displaystyle \theta ={{\tan }^{{-1}}}\left( {\frac{{-\sqrt{3}}}{{-1}}} \right)=\frac{{4\pi }}{3}\text{ (3rd quadrant)}$$ $$\displaystyle \left( {2,\frac{{4\pi }}{3}} \right)$$ $$\left( {4,\pi } \right)$$ $$\begin{array}{c}x=4\cos \pi =4\left( {-1} \right)=-4\\y=4\sin \pi =4\left( 0 \right)=0\end{array}$$   $$\left( {-4,0} \right)$$ $$\displaystyle r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{16+0}}=4$$   $$\displaystyle \begin{array}{c}\theta ={{\tan }^{{-1}}}\left( {\frac{0}{{-4}}} \right)=0\text{ (between 2nd }\\\text{ and 3rd quadrants)}\\\left( {4,\pi } \right)\end{array}$$ $$\left( {-7,45{}^\circ } \right)$$ $$\displaystyle x=-7\cos \left( {45{}^\circ } \right)=-\frac{{7\sqrt{2}}}{2}$$ $$\displaystyle y=-7\sin \left( {45{}^\circ } \right)=-\frac{{7\sqrt{2}}}{2}$$   $$\displaystyle \left( {-\frac{{7\sqrt{2}}}{2},-\frac{{7\sqrt{2}}}{2}} \right)$$ $$r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{\frac{{49\left( 2 \right)}}{4}+\frac{{49\left( 2 \right)}}{4}}}=7$$   $$\theta ={{\tan }^{{-1}}}\left( 1 \right)=225{}^\circ \text{ (3rd quadrant)}$$ $$\left( {7,225{}^\circ } \right),\text{ which is the same as }\left( {-7,45{}^\circ } \right)$$ $$\left( {5,150{}^\circ } \right)$$ $$\displaystyle x=5\cos \left( {150{}^\circ } \right)=5\left( {-\frac{{\sqrt{3}}}{2}} \right)=-\frac{{5\sqrt{3}}}{2}$$ $$\displaystyle y=5\sin \left( {150{}^\circ } \right)=5\left( {\frac{1}{2}} \right)=\frac{5}{2}$$   $$\displaystyle \left( {-\frac{{5\sqrt{3}}}{2},\frac{5}{2}} \right)$$ $$\displaystyle r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{\frac{{25\left( 3 \right)}}{4}+\frac{{25}}{4}}}=5$$   $$\displaystyle \theta ={{\tan }^{{-1}}}\left( {\frac{{\frac{5}{2}}}{{-\frac{{5\sqrt{3}}}{2}}}} \right)=\text{150}{}^\circ \text{ (2nd quadrant)}$$ $$\displaystyle \left( {5,150{}^\circ } \right)$$ $$\displaystyle \left( {3,-\frac{{3\pi }}{2}} \right)$$ $$\displaystyle x=3\cos \left( {-\frac{{3\pi }}{2}} \right)=3\left( 0 \right)=0$$ $$\displaystyle y=3\sin \left( {-\frac{{3\pi }}{2}} \right)=3\left( 1 \right)=3$$   $$\displaystyle \left( {0,3} \right)$$ $$\displaystyle r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{0+9}}=3$$   $$\displaystyle \begin{array}{c}\theta ={{\tan }^{{-1}}}\left( {\frac{3}{0}} \right)=\frac{\pi }{2}\text{ (we want where tan is }\\\text{ undefined, between 1st and 2nd quadrants)}\end{array}$$ $$\displaystyle \left( {3,\frac{\pi }{2}} \right),\text{ which is the same as }\left( {3,-\frac{{3\pi }}{2}} \right)$$

Here’s one where we go from Rectangular to Polar and we can’t get the angle from the Unit Circle. Note that we had to add $$\boldsymbol {\pi }$$ to our answer since we want Quadrant II.

 Rectangular Point Convert from Rectangular to Polar $$\left( {-1,5} \right)$$ $$\displaystyle r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\sqrt{{1+25}}=\sqrt{{26}}$$   $$\displaystyle \theta ={{\tan }^{{-1}}}\left( {\frac{5}{{-1}}} \right)=-1.373+\pi =1.768\text{ (2nd quadrant)}$$ $$\displaystyle \left( {\sqrt{{26}},1.768} \right)$$

Note that you can also use “2nd APPS (ANGLE)” on your graphing calculator to do these conversions, but you won’t get the answers with the roots in them (you’ll get decimals that aren’t “exact”). And you have to solve for the $$x$$ and $$y$$, or $$r$$ and $$\theta$$ separately, and use the “,” above the 7 for the comma.

Make sure you have your calculator either in DEGREES or RADIANS (in MODE), depending on what you’re working with.

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# Drawing Polar Graphs

I find that drawing polar graphs is a combination of part memorizing and part knowing how to create polar t-charts.

First, here is a table of some of the more common polar graphs. I included t-charts in both degrees and radians.

(Note that you can also put these in your graphing calculator, as an example, with radians: MODE: RADIAN, POLAR and WINDOW:   θ = [0, 2π], θstep =  π/12 or π/6, X = [–10, 10], Y = [–6, 6], and then using “Y=” to put in the equation, or just put in graph and use ZOOM ZTRIG (option 7). By putting in smaller values of θstep, such as .1, the graph is drawn more slowly and more accurately; to redraw graph, you can turn the graph off and back on by going to “=” and un-highlighting and highlighting it back again before hitting “graph”.

Here is an example:

Here are some polar equations that result in lines:

Here are graphs that we call Cardioids and Limacons. They are in the form $$r=a+b\cos \theta$$ or $$r=a+b\sin \theta$$.

Note: Unlike their rectangular equivalents, $$r=a\pm b\cos \theta$$ and $$r=-a\pm b\cos \theta$$ (same with $$r=a\pm b\sin \theta$$ and $$r=-a\pm b\sin \theta$$) are the same polar graph! Try it!

First, the Cardioids (Hearts); note that these and the Limecon “Loops” touch the pole (origin), while the Limecon “Beans” do not:

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Here are the Limacons:

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Graphs of Roses produce “petals” and are in the form $$r=a\cos \left( {b\theta } \right)$$  or  $$r=a\sin \left( {b\theta } \right)$$. Note that since we have the starting point for these graphs, and the distance between the petals, the t-chart isn’t that helpful. (In the t-charts, I made the $$\Delta$$ angle the same as the distance between petals).

And here are some sin Rose graphs:

Note: For a rose graph, you may be asked to name the order that petals are drawn. One way to do this is to use the angle measurements $$\displaystyle 0,\,\frac{\pi }{4},\,\frac{{3\pi }}{4},\,\frac{{5\pi }}{4},\,\frac{{7\pi }}{4}$$, solve for $$r$$, and observe the order of the petals. You can also use the graphing calculator as shown above, but make the θstep smaller to slow down the drawing of the graph.

Here are a couple more polar graphs (Spirals and Lemniscates) that you might see:

You might be asked to obtain the equation of a polar function from a graph:

Here’s another type of question you may get asked when studying polar graphs.

Problem:

For the rose polar graph $$5\sin \left( {10\theta } \right)$$: find the length of each petal, number of petals, spacing between each petal, and the tip of the 1st petal in Quadrant I.

Solution:
The length of each petal in the rose polar graph is $$a$$, so this length is 5. Since $$b$$ (10) is even, we need to double it to get the number of petals, so we get 20. The spacing between each petal is $$\displaystyle \frac{{360{}^\circ }}{{20}}=18{}^\circ$$, since there are 360° in a circle. The tip of the 1st petal in Quadrant I will be at $$\displaystyle \frac{{90{}^\circ }}{b}=\frac{{90{}^\circ }}{{10}}=9{}^\circ$$.

# Converting Equations from Rectangular to Polar

To convert Rectangular Equations to Polar Equations, we want to get rid of the $$x$$’s and $$y$$’s and only have $$r$$’s and/or $$\theta$$’s in the answer. We can do this with the following equations:

$$\begin{array}{l}x=r\cos \,\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}+{{y}^{2}}={{r}^{2}}\\y=r\,\sin \,\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\end{array}$$

Here are some examples; note that we want to solve for $$r$$ if we can; in the case of quadratics or higher degrees, this may involve moving everything to one side and factoring.

Note that when we also get $$r=0$$ (the pole) for the answers; this is one point only, and in these cases, the pole is included in the other part of the answers. Thus, we can discard $$r=0$$.

# Converting Equations from Polar to Rectangular

To convert Polar Equations to Rectangular Equations, we want to get rid of the $$r$$’s and $$\theta$$’s and only have $$x$$’s and/or $$y$$’s in the answer. We can do this with the following equations, depending on what we have in the polar equation:

Here are some examples. Note that sometimes we may be asked to Complete the Square to get the equation in a circle form; we learned how to do this in the Factoring Quadratics and Completing the Square section here.

# Polar Graph Points of Intersection

To find the intersection points for sets of polar curves, it’s helpful to draw the curves and also to solve algebraically. To solve algebraically, we just set the $$r$$’s together and solve for the $$\theta$$.

Note also that after we solve for one variable (like $$\theta$$), we have to plug it back in either equation to get the other coordinate (like $$r$$).

We also have to be careful since there are “phantom” or “elusive” points, typically at the pole. The reason these points are “phantom” is because, although we don’t necessarily get them algebraically, we can see them on a graph. This is because, with an “$$r$$” of 0, the $$\theta$$ could really be anything, since we aren’t going out any distance.

We will also see phantom points when one of the equations is “$$r=$$ constant”, since another way to write this is “$$r=$$ the negative of that constant”.

Note that with “phantom” points, both equations do not have to work; I know, it’s weird. To get all these elusive points, you put in the r value in both curves to see what additional points you get.

Find the intersection points for the following sets of polar curves (algebraically) and also draw a sketch. Find the intersections when $$\theta$$ is between 0 and 2π.

It also might be good to know the sequence in which the polar graphs are drawn; in other words, from 0 to 2π, which parts of the graphs are drawn before the other graphs. (Check it out on a graphing calculator, where you can see it!)

You can use a t-chart, or set the polar equation to 0 if the graph crosses the pole, and test points in between. Here are some examples:

Problem and Solution T-Chart Graph
The inner loop for $$2+4\cos \theta$$ is formed between what two values of $$\theta$$?

Solution:

Find out the two cases when $$r=0$$, since that’s before and after the graph draws its inner loop:

$$\displaystyle 0=2+4\cos \theta ;\,\,\,\,\,\cos \theta =-\frac{1}{2}$$

$$\displaystyle \theta =\frac{{2\pi }}{3};\,\,\,\,\,\theta =\frac{{4\pi }}{3}$$

Since the end of the inner loop is at $$\left( {-2,\pi } \right)$$ (same as $$\displaystyle \left( {2,0{}^\circ } \right)$$), and this is between $$\displaystyle \frac{{2\pi }}{3}$$ and $$\displaystyle \frac{{4\pi }}{3}$$, the inner loop is formed when

$$\displaystyle \frac{{2\pi }}{3}<\theta <\frac{{4\pi }}{3}$$

 r θ        ° 6 0         0 2 $$\displaystyle \frac{\pi }{2}$$       90 0 $$\displaystyle \frac{{2\pi }}{3}$$    120 –2 $$\pi$$      180 0 $$\displaystyle \frac{{4\pi }}{3}$$    240 2 $$\displaystyle \frac{{3\pi }}{2}$$    270
The leftmost petal for $$\cos 2\theta$$ is drawn between what two values of $$\theta$$? In what order are the petals drawn?

Solution:

With this problem, we can create the following t-chart and see the sequence of petals being drawn. To determine which values to use in the t-chart, set $$\cos 2\theta$$ to 0 to see what $$\theta$$ values are between each petal:

$$\displaystyle \begin{array}{c}0=\cos 2\theta \\0=2{{\cos }^{2}}\theta -1\,\,\,\text{(identity)}\,\,\,\,\,\end{array}$$

$$\displaystyle \text{cos}\theta \,\,\text{=}\,\,\pm \sqrt{{\frac{1}{2}}}=\pm \frac{{\sqrt{2}}}{2}$$

$$\displaystyle \theta =\frac{\pi }{4};\,\,\,\,\,\theta =\frac{{3\pi }}{4};\,\,\,\,\,\theta =\frac{{5\pi }}{4};\,\,\,\,\,\theta =\frac{{7\pi }}{4}$$

Thus, the leftmost (3rd) petal is formed when

$$\displaystyle \frac{{3\pi }}{4}<\theta <\frac{{5\pi }}{4}$$

 r θ        ° 1 0         0 0 $$\displaystyle \frac{\pi }{4}$$        45 –1 $$\displaystyle \frac{\pi }{2}$$        90 0 $$\displaystyle \frac{{3\pi }}{4}$$     135 1 $$\displaystyle \pi$$        180 0 $$\displaystyle \frac{{5\pi }}{4}$$     225 –1 $$\displaystyle \frac{{3\pi }}{2}$$     270 0 $$\displaystyle \frac{{7\pi }}{4}$$     315

Learn these rules, and practice, practice, practice!

Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

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You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions.  There is even a Mathway App for your mobile device.  Enjoy!

On to Trigonometry and the Complex Plane  – you are ready!

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