# 6.2: Graphing Basic Polar Equations

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Graph polar equations.
• Graph and recognize limaçons and cardioids.
• Determine the shape of a limaçon from the polar equation.

Just as in graphing on a rectangular grid, you can also graph polar equations on a polar grid. These equations may be simple or complex. To begin, you should try something simple like \begin{align*}r = k\end{align*} or \begin{align*}\theta = k\end{align*} where \begin{align*}k\end{align*} is a constant. The solution for \begin{align*}r = 1.5\end{align*} is simply all ordered pairs such that \begin{align*}r = 1.5\end{align*} and \begin{align*}\theta\end{align*} is any real number. The same is true for the solution of \begin{align*}\theta = 30^\circ\end{align*}. The ordered pairs will be any real number for \begin{align*}r\end{align*} and \begin{align*}\theta\end{align*} will equal \begin{align*}30^\circ\end{align*}. Here are the graphs for each of these polar equations.

Example 1: On a polar plane, graph the equation \begin{align*}r = 1.5\end{align*}

Solution: The solution is all ordered pairs of \begin{align*}(r, \theta)\end{align*} such that \begin{align*}r\end{align*} is always 1.5. This means that it doesn’t matter what \begin{align*}\theta\end{align*} is, so the graph is a circle with radius 1.5 and centered at the origin.

Example 2: On a polar plane, graph the equation \begin{align*}\theta = 30^\circ\end{align*}

Solution: For this example, the \begin{align*}r\end{align*} value, or radius, is arbitrary. \begin{align*}\theta\end{align*} must equal \begin{align*}30^\circ\end{align*}, so the result is a straight line, with an angle of elevation of \begin{align*}30^\circ\end{align*}.

To begin graphing more complicated polar equations, we will make a table of values for \begin{align*}y = \sin \theta\end{align*} or in this case \begin{align*}r = \sin \theta\end{align*}. When the table has been completed, the graph will be drawn on a polar plane by using the coordinates \begin{align*}(r, \theta)\end{align*}.

Example 3: Create a table of values for \begin{align*}r = \sin \theta\end{align*} such that \begin{align*}0^\circ \le \theta \le 360^\circ\end{align*} and plot the ordered pairs. (Note: Students can be directed to use intervals of \begin{align*}30^\circ\end{align*} or allow them to create their own tables.)

\begin{align*}\theta\end{align*} \begin{align*}0^\circ\end{align*} \begin{align*}30^\circ\end{align*} \begin{align*}60^\circ\end{align*} \begin{align*}90^\circ\end{align*} \begin{align*}120^\circ\end{align*} \begin{align*}150^\circ\end{align*} \begin{align*}180^\circ\end{align*} \begin{align*}210^\circ\end{align*} \begin{align*}240^\circ\end{align*} \begin{align*}270^\circ\end{align*} \begin{align*}300^\circ\end{align*} \begin{align*}330^\circ\end{align*} \begin{align*}360^\circ\end{align*}
\begin{align*}\sin \theta\end{align*} 0 0.5 0.9 1 0.9 0.5 0 -0.5 -0.9 -1 -0.9 -0.5 0

Remember that the values of \begin{align*}\sin \theta\end{align*} are the \begin{align*}r-\end{align*}values.

This is a sinusoid curve of one revolution.

We will now repeat the process for \begin{align*}r = \cos \theta\end{align*}.

Example 4: Create a table of values for \begin{align*}r = \cos \theta\end{align*} such that \begin{align*}0^\circ \le \theta \le 360^\circ\end{align*} and plot the ordered pairs. (Note: Students can be directed to use intervals of \begin{align*}30^\circ\end{align*} or allow them to create their own tables.)

\begin{align*}\theta\end{align*} \begin{align*}0^\circ\end{align*} \begin{align*}30^\circ\end{align*} \begin{align*}60^\circ\end{align*} \begin{align*}90^\circ\end{align*} \begin{align*}120^\circ\end{align*} \begin{align*}150^\circ\end{align*} \begin{align*}180^\circ\end{align*} \begin{align*}210^\circ\end{align*} \begin{align*}240^\circ\end{align*} \begin{align*}270^\circ\end{align*} \begin{align*}300^\circ\end{align*} \begin{align*}330^\circ\end{align*} \begin{align*}360^\circ\end{align*}
\begin{align*}\cos \theta\end{align*} 1 0.9 0.5 0 -0.5 -0.9 -1 -0.9 -0.5 0 0.5 0.9 1

Remember that the values of \begin{align*}\cos \theta\end{align*} are the \begin{align*}r-\end{align*}values.

This is also a sinusoid curve of one revolution.

Notice that both graphs are circles that pass through the pole and have a diameter of one unit. These graphs can be altered by adding a number to the function or by multiplying the function by a constant or by doing both. We will explore the results of these alterations by first creating a table of values and then by graphing the resulting coordinates \begin{align*}(r, \theta)\end{align*}.

Example 5: Create a table of values for \begin{align*}r = 2 + 3 \sin \theta\end{align*} such that \begin{align*}0 \le \theta \le 2 \pi\end{align*} and plot the ordered pairs. Remember that the values of \begin{align*}2 + 3 \sin \theta\end{align*} are the \begin{align*}r-\end{align*}values.

\begin{align*}\theta\end{align*} \begin{align*}\frac{\pi}{6}\end{align*} \begin{align*}\frac{\pi}{3}\end{align*} \begin{align*}\frac{\pi}{2}\end{align*} \begin{align*}\frac{2\pi}{3}\end{align*} \begin{align*}\frac{5 \pi}{6}\end{align*} \begin{align*}\pi\end{align*} \begin{align*}\frac{7\pi}{6}\end{align*} \begin{align*}\frac{4\pi}{3}\end{align*} \begin{align*}\frac{3\pi}{2}\end{align*} \begin{align*}\frac{5\pi}{3}\end{align*} \begin{align*}\frac{11\pi}{6}\end{align*} \begin{align*}2\pi\end{align*}
\begin{align*}2 + 3 \sin \theta\end{align*} 2.0 3.5 4.6 5.0 4.6 3.5 2.0 0.5 -0.6 -1.0 -0.6 0.5 2.0

This sinusoid curve is called a limaçon. It has \begin{align*}r = a \pm b \sin \theta\end{align*} or \begin{align*}r = a \pm b \cos \theta\end{align*} as its polar equation. Not all limaçons have the inner loop as a part of the shape. Some may curve to a point, have a simple indentation (known as a dimple) or curve outward. The shape of the limaçon depends upon the ratio of \begin{align*}\frac{a}{b}\end{align*} where \begin{align*}a\end{align*} is a constant and \begin{align*}b\end{align*} is the coefficient of the trigonometric function. In example 5, the ratio of \begin{align*}\frac{a}{b} = \frac{2}{3}\end{align*} which is \begin{align*}< 1\end{align*}. All limaçons that meet this criterion will have an inner loop.

Using the same format as was used in the examples above, the following limaçons were graphed. If you like, you may create the table of values for each of these functions.

i) \begin{align*}r = 4 + 3 \cos \theta\end{align*} such that \begin{align*}0 \le \theta \le 2\pi\end{align*}

\begin{align*}\frac{a}{b} = \frac{4}{3}\end{align*} which is \begin{align*}> 1\end{align*} but \begin{align*}< 2\end{align*}

This is an example of a dimpled limaçon.

ii) \begin{align*}r = 4 + 2 \sin \theta\end{align*} such that \begin{align*}0 \le \theta \le 2 \pi\end{align*}

\begin{align*}\frac{a}{b} = \frac{4}{2}\end{align*} which is \begin{align*}\ge 2\end{align*}

This is an example of a convex limaçon.

Example 6: Create a table of values for \begin{align*}r = 2 + 2 \cos \theta\end{align*} such that \begin{align*}0 \le \theta \le 2 \pi\end{align*} and plot the ordered pairs. Remember that the values of \begin{align*}2 + 2 \cos \theta\end{align*} are the \begin{align*}r-\end{align*}values.

\begin{align*}\theta\end{align*} \begin{align*}\frac{\pi}{6}\end{align*} \begin{align*}\frac{\pi}{3}\end{align*} \begin{align*}\frac{\pi}{2}\end{align*} \begin{align*}\frac{2\pi}{3}\end{align*} \begin{align*}\frac{5 \pi}{6}\end{align*} \begin{align*}\pi\end{align*} \begin{align*}\frac{7\pi}{6}\end{align*} \begin{align*}\frac{4\pi}{3}\end{align*} \begin{align*}\frac{3\pi}{2}\end{align*} \begin{align*}\frac{5\pi}{3}\end{align*} \begin{align*}\frac{11\pi}{6}\end{align*} \begin{align*}2\pi\end{align*}
\begin{align*}2 + 2 \cos \end{align*} 4.0 3.7 3.0 2.0 1.0 0.27 0 .27 1.0 2.0 3.0 3.7 4.0

This type of curve is called a cardioid. It is a special type of limaçon that has \begin{align*}r = a + a \cos \theta\end{align*} or \begin{align*}r = a + a \sin \theta\end{align*} as its polar equation. The ratio of \begin{align*}\frac{a}{b} = \frac{2}{2}\end{align*} which is equal to 1.

Examples 3 and 4 were shown with \begin{align*}\theta\end{align*} measured in degrees while examples 5 and 6 were shown with \begin{align*}\theta\end{align*} measured in radians. The results in the tables and the resulting graphs will be the same in both units.

Now that you are familiar with the limaçon and the cardioid, also called classical curves, it is time to examine the polar pattern of the cardioid microphone. The polar pattern is modeled by the polar equation \begin{align*}r = 2.5 + 2.5 \cos \theta\end{align*}. The values of \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are equal which means that the ratio \begin{align*}\frac{a}{b} = 1\end{align*}. Therefore the limaçon will be a cardioid.

Create a table of values for \begin{align*}r = 2.5 + 2.5 \cos \theta\end{align*} such that \begin{align*}0^\circ \le \theta \le 360^\circ\end{align*} and graph the results.

\begin{align*}\theta\end{align*} \begin{align*}0^\circ\end{align*} \begin{align*}30^\circ\end{align*} \begin{align*}60^\circ\end{align*} \begin{align*}90^\circ\end{align*} \begin{align*}120^\circ\end{align*} \begin{align*}150^\circ\end{align*} \begin{align*}180^\circ\end{align*} \begin{align*}210^\circ\end{align*} \begin{align*}240^\circ\end{align*} \begin{align*}270^\circ\end{align*} \begin{align*}300^\circ\end{align*} \begin{align*}330^\circ\end{align*} \begin{align*}360^\circ\end{align*}
\begin{align*}2.5 + 2.5 \cos \theta\end{align*} 5.0 4.7 3.8 2.5 1.3 0.3 0 0.3 1.3 2.5 3.8 4.7 5.0

## Transformations of Polar Graphs

Equations of limaçons have two general forms:

\begin{align*}r = a \pm b \sin \theta \qquad \text{and} \qquad r = a \pm b \cos \theta\end{align*}

The values of \begin{align*}“a”\end{align*} and \begin{align*}“b”\end{align*} will determine the shape of the graph and whether or not it passes through the origin. When the values of \begin{align*}“a”\end{align*} and \begin{align*}“b”\end{align*} are equal, the graph will be a rounded heart-shape called a cardioid. The general polar equation of a cardioid can be written as \begin{align*}r = a(1 \pm \sin \theta)\end{align*} and \begin{align*}r = a (1 \pm \cos \theta)\end{align*}. Note: The general polar equation of a cardioid can also be written as \begin{align*}r = a(-1 \pm \sin \theta)\end{align*} and \begin{align*}r = a (-1 \pm \cos \theta)\end{align*}. This will be discussed later in the chapter.

Example 7: Graph the following polar equations on the same polar grid and compare the graphs.

\begin{align*}r & = 5 + 5 \sin \theta && r = 5 - 5 \sin \theta \\ r & = 5(1 + \sin \theta) && r = 5(1 - \sin \theta) \end{align*}

Solution:

The cardioid is symmetrical about the positive \begin{align*}y-\end{align*}axis and the point of indentation is at the pole. The result of changing + to - is a reflection in the \begin{align*}x-\end{align*}axis. The cardioid is symmetrical about the negative \begin{align*}y-\end{align*}axis and the point of indentation is at the pole.

Changing the value of \begin{align*}“a”\end{align*} to a negative did not change the graph of the cardioid.

Example 8: What effect will changing the values of \begin{align*}a\end{align*} and \begin{align*}b\end{align*} have on the cardioid if \begin{align*}a > b\end{align*}? We can discover the answer to this question by plotting the graph of \begin{align*}r = 5 + 3 \sin \theta\end{align*}.

Solution:

The cardioid is symmetrical about the positive \begin{align*}y-\end{align*}axis and the point of indentation is pulled away from the pole.

Example 9: What effect will changing the values of \begin{align*}a\end{align*} and \begin{align*}b\end{align*} or changing the function have on the cardioid if \begin{align*}a < b\end{align*}? We can discover the answer to this question by plotting the graph of \begin{align*}r = 2 + 3 \sin \theta\end{align*}.

Solution:

The cardioid is now a looped limaçon symmetrical about the positive \begin{align*}y-\end{align*}axis. The loop crosses the pole.

\begin{align*}r = 2 + 3 \cos \theta\end{align*}

The cardioid is now a looped limaçon symmetrical about the positive \begin{align*}x-\end{align*}axis. The loop crosses the pole. Changing the function to cosine rotated the limaçon \begin{align*}90^\circ\end{align*} clockwise.

As you have seen from all of the graphs, transformations can be performed by making changes in the constants and/or the functions of the polar equations.

## Applications

In this subsection we will explore examples of real-world problems that use polar coordinates and polar equations as their solutions.

Example 10: A local charity is sponsoring an outdoor concert to raise money for the children’s hospital. To accommodate as many patrons as possible, they are importing bleachers so that all the fans will be seated during the performance. The seats will be placed in an area such that \begin{align*}\frac{- \pi}{3} \le \theta \le \frac{\pi}{3}\end{align*} and \begin{align*}0 \le r \le 4\end{align*}, where \begin{align*}r\end{align*} is measured in hundreds of feet. The stage will be placed at the origin (pole) and the performer will face the audience in the direction of the polar axis \begin{align*}(r)\end{align*}.

a. Create a polar graph of this area.

b. If all the seats are occupied and each seat takes up 5 square feet of space, how many people will be seated in the bleachers?

Solution: Now that the region has been graphed, the next step is to calculate the area of this sector. To do this, use the formula \begin{align*}A = \frac{1}{2}r^2 \theta\end{align*}.

\begin{align*}A & = \frac{1}{2}r^2 \theta \\ A & = \frac{1}{2} (400)^2 \left (\frac{2 \pi}{3} \right ) \\ A & \approx 167552\ ft^2.\end{align*}

\begin{align*}167552\ ft^2 . \div 5\ ft^2 . \approx 33510\end{align*}

The number of people in the bleachers is 33510.

Example 11: When Valentine’s Day arrives, hearts can be seen everywhere. As an alternative to purchasing a greeting card, use a computer to create a heart shape. Write an equation that could be used to create this heart and be careful to ensure that it is displayed in the correct position.

Solution: The classical curve that resembles a heart is a cordioid. You may have to experiment with the equation to create a heart shape that is displayed in the correct direction. One example of an equation that produces a proper heart shape is \begin{align*}r = -2 -2 \sin \theta\end{align*}.

You can create other hearts by replacing the number 2 in the equation. Another equation is \begin{align*}r = -3 - 3 \sin \theta\end{align*}.

Example 12: For centuries, people have been making quilts. These are frequently created by sewing a uniform fabric pattern onto designated locations on the quilt. Using the equation that models a rose curve, create three patterns that could be used for a quilt. Write the equation for each rose and sketch its graph. Explain why the patterns have different numbers of petals. Can you create a sample quilt?

Solution: The rose curve is a graph of the polar equation of the form \begin{align*}r = a \cos n \theta\end{align*} or \begin{align*}r = a \sin n \theta\end{align*}. If \begin{align*}n\end{align*} is odd, then the number of petals will be equal to \begin{align*}n\end{align*}. If \begin{align*}n\end{align*} is even, then the number of petals will be equal to \begin{align*}2 n\end{align*}.

A Sample Quilt:

## Graphing Polar Equations on the Calculator

You can use technology, the TI graphing calculator, to create these graphs. However, there are steps that must be followed in order to graph polar equations correctly on the graphing calculator. We will go through the step by step process to plot the polar equation \begin{align*}r = 3 \cos \theta\end{align*}.

Example 13: Graph \begin{align*}r = 3 \cos \theta\end{align*} using the TI-83 graphing calculator.

Solution: Press the \begin{align*}\fbox{\text{MODE}}\end{align*} button. Scroll down to Func and over to highlight Pol. Also, while on this screen, make sure that Radian is highlighted. Now you must edit the axes for the graph. Press \begin{align*}\fbox{\text{WINDOW}}\ 0\ \fbox{\text{ENTER}} \begin{array} {|c|} \hline 2^{\text{nd}}\\ \hline \end{array}\ [\pi]\ \fbox{\text{ENTER}}\ .05\ \fbox{\text{ENTER}}\ (-)\ 4\ \fbox{\text{ENTER}}\ 4\end{align*}

\begin{align*}\fbox{\text{ENTER}}\ 1\ \fbox{\text{ENTER}}\ (-)\ 3\ \fbox{\text{ENTER}}\ 3\ \fbox{\text{ENTER}}\ 1\ \fbox{\text{ENTER}}\end{align*}.

When you have completed these steps, the screen should look like this:

The second WINDOW shows part of the first screen since you had to scroll down to access the remaining items.

Enter the equation. Press \begin{align*} \begin{array} {|c|} \hline Y = 3 \cos X, T, \theta, n)\\ \hline \end{array}\end{align*} Press \begin{align*}\fbox{GRAPH}\end{align*}.

Sometimes the polar equation you graph will look more like an ellipse than a circle. If this happens, press \begin{align*}\fbox{ZOOM}\end{align*} 5 to set a square viewing window. This will make the graph appear like a circle.

Note: If you want the calculator to graph complete rose petals when \begin{align*}n\end{align*} is even, you must set \begin{align*}\theta \text{max} = 2 \pi\end{align*}.

## Review Questions

1. Name the classical curve in each of the following diagrams and explain why you feel you’re your answer is correct. Also, find the equation of each curve.
2. Graph each curve below. Comparing your answers from part one, determine if you can find a pattern for how to find the equation of a classical curve from its graph.
1. \begin{align*}r = -3 - 3\cos \theta\end{align*}
2. \begin{align*}r = 2 + 4 \sin \theta\end{align*}
3. \begin{align*}r = 4\end{align*}
4. \begin{align*}\theta = \frac{\pi}{2}\end{align*}
5. \begin{align*}r = 5 + 3 \cos \theta\end{align*}
6. \begin{align*}r = -6 - 5 \sin \theta\end{align*}
3. Another classical curve we saw is called a rose and it is modeled by the function \begin{align*}r = a \cos n \theta\end{align*} or \begin{align*}r = a \sin n \theta\end{align*} where \begin{align*}n\end{align*} is any positive integer. Graph \begin{align*}r = 4 \cos 2 \theta\end{align*} and \begin{align*}r = 4 \cos 3 \theta\end{align*}. Is there a difference in the curves? Explain.
4. Graph the roses below. Determine if you can find a pattern for how to find the equation of a rose from its graph.
1. \begin{align*}r = 3 \sin 4 \theta\end{align*}
2. \begin{align*}r = 2 \sin 5 \theta\end{align*}
3. \begin{align*}r = 3 \cos 3 \theta\end{align*}
4. \begin{align*}r = -4 \sin 2 \theta\end{align*}
5. \begin{align*}r = 5 \cos 4 \theta\end{align*}
6. \begin{align*}r = -2 \cos 6 \theta\end{align*}

1. a limaçon with an innerloop. \begin{align*}r = 2 - 3 \sin \theta\end{align*}
2. a cardioid \begin{align*}r = 2 + 2 \sin \theta\end{align*}
3. a dimpled limaçon \begin{align*}r = 5.5 + 4.5 \sin \theta\end{align*}
1. \begin{align*}r = -3 - 3 \cos \theta\end{align*}
2. \begin{align*}r = 2 + 4 \sin \theta\end{align*}
3. \begin{align*}r = 4\end{align*}
4. \begin{align*}\theta = \frac{\pi}{2}\end{align*}
5. \begin{align*}r = 5 + 3 \cos \theta\end{align*}
6. \begin{align*}r = -6 - 5 \sin \theta\end{align*}

To determine the equation of these curves, first notice that cosine curves are along the horizontal axis and sine curves are along the vertical axis. Second, where the curve passes through the axis on the non-dimpled side is \begin{align*}a\end{align*} in \begin{align*}r = a+b \sin \theta\end{align*}. \begin{align*}b\end{align*} is a little harder to see, but it is the average of the two intercepts where the curve crosses the axis on the dimpled axis. If there is an inner loop, use the innermost value of the loop.

\begin{align*}\theta\end{align*} \begin{align*}0^\circ\end{align*} \begin{align*}30^\circ\end{align*} \begin{align*}60^\circ\end{align*} \begin{align*}90^\circ\end{align*} \begin{align*}120^\circ\end{align*} \begin{align*}150^\circ\end{align*} \begin{align*}180^\circ\end{align*} \begin{align*}210^\circ\end{align*} \begin{align*}240^\circ\end{align*} \begin{align*}270^\circ\end{align*} \begin{align*}300^\circ\end{align*} \begin{align*}330^\circ\end{align*} \begin{align*}360^\circ\end{align*}
\begin{align*}4 \cos 2 \theta\end{align*} 4 2 -2 -4 -2 2 4 2 -2 -4 -2 2 4

\begin{align*}\theta\end{align*} \begin{align*}0^\circ\end{align*} \begin{align*}30^\circ\end{align*} \begin{align*}60^\circ\end{align*} \begin{align*}90^\circ\end{align*} \begin{align*}120^\circ\end{align*} \begin{align*}150^\circ\end{align*} \begin{align*}180^\circ\end{align*} \begin{align*}210^\circ\end{align*} \begin{align*}240^\circ\end{align*} \begin{align*}270^\circ\end{align*} \begin{align*}300^\circ\end{align*} \begin{align*}330^\circ\end{align*} \begin{align*}360^\circ\end{align*}
\begin{align*}4 \cos 3 \theta\end{align*} 4 0 -4 0 4 0 -4 0 4 0 -4 0 4

In the graph of \begin{align*}r = 4 \cos 2 \theta\end{align*}, the rose has four petals on it but the graph of \begin{align*}r = 4 \cos 3 \theta\end{align*} has only three petals. It appears, that if \begin{align*}n\end{align*} is an even positive integer, the rose will have an even number of petals and if \begin{align*}n\end{align*} is an odd positive integer, the rose will have an odd number of petals.

1. \begin{align*}r = 3 \sin 4 \theta\end{align*}
2. \begin{align*}r = 2 \sin 5 \theta\end{align*}
3. \begin{align*}r = 3 \cos 3 \theta\end{align*}
4. \begin{align*}r = -4 \sin 2 \theta\end{align*}
5. \begin{align*}r = 5 \cos 4 \theta\end{align*}
6. \begin{align*}r = -2 \cos 6 \theta\end{align*}

For roses, the general equation is \begin{align*}r = a \sin n \theta\end{align*} or \begin{align*}r = a \cos n \theta\end{align*}. \begin{align*}a\end{align*} indicates how long each petal is, and depending on if \begin{align*}n\end{align*} is even or odd indicates the number of pedals. If \begin{align*}n\end{align*} is ood, there are \begin{align*}n\end{align*} pedals and if \begin{align*}n\end{align*} is even there are \begin{align*}2n\end{align*} pedals.

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