Polar Coordinates: Graphs

Figure 1. Planets follow elliptical paths as they orbit around the Sun. (credit: modification of work by NASA/JPL-Caltech)

Testing Polar Equations for Symmetry

A General Note: Symmetry Tests

A polar equation describes a curve on the polar grid. The graph of a polar equation can be evaluated for three types of symmetry, as shown in Figure 2.

Figure 2. (a) A graph is symmetric with respect to the line [latex]\theta =\frac{\pi }{2}[/latex] (y-axis) if replacing [latex]\left(r,\theta \right)[/latex] with [latex]\left(-r,-\theta \right)[/latex] yields an equivalent equation. (b) A graph is symmetric with respect to the polar axis (x-axis) if replacing [latex]\left(r,\theta \right)[/latex] with [latex]\left(r,-\theta \right)[/latex] or [latex]\left(-r,\mathrm{\pi -}\theta \right)[/latex] yields an equivalent equation. (c) A graph is symmetric with respect to the pole (origin) if replacing [latex]\left(r,\theta \right)[/latex] with [latex]\left(-r,\theta \right)[/latex] yields an equivalent equation.

Analysis of the Solution

Using a graphing calculator, we can see that the equation [latex]r=2\sin \theta [/latex] is a circle centered at [latex]\left(0,1\right)[/latex] with radius [latex]r=1[/latex] and is indeed symmetric to the line [latex]\theta =\frac{\pi }{2}[/latex]. We can also see that the graph is not symmetric with the polar axis or the pole. See Figure 3.

Figure 3

 Graphing Polar Equations by Plotting Points

Finding Zeros and Maxima

Analysis of the Solution

The point [latex]\left(2,\frac{\pi }{2}\right)[/latex] will be the maximum value on the graph. Let’s plot a few more points to verify the graph of a circle.

[latex]\theta [/latex]	[latex]r=2\sin \theta [/latex]	[latex]r[/latex]
0	[latex]r=2\sin \left(0\right)=0[/latex]	[latex]0[/latex]
[latex]\frac{\pi }{6}[/latex]	[latex]r=2\sin \left(\frac{\pi }{6}\right)=1[/latex]	[latex]1[/latex]
[latex]\frac{\pi }{3}[/latex]	[latex]r=2\sin \left(\frac{\pi }{3}\right)\approx 1.73[/latex]	[latex]1.73[/latex]
[latex]\frac{\pi }{2}[/latex]	[latex]r=2\sin \left(\frac{\pi }{2}\right)=2[/latex]	[latex]2[/latex]
[latex]\frac{2\pi }{3}[/latex]	[latex]r=2\sin \left(\frac{2\pi }{3}\right)\approx 1.73[/latex]	[latex]1.73[/latex]
[latex]\frac{5\pi }{6}[/latex]	[latex]r=2\sin \left(\frac{5\pi }{6}\right)=1[/latex]	[latex]1[/latex]
[latex]\pi [/latex]	[latex]r=2\sin \left(\pi \right)=0[/latex]	[latex]0[/latex]

Figure 4

 Graphing Circles and the 5 Classic Polar Curves

Investigating Circles

A General Note: Formulas for the Equation of a Circle

Some of the formulas that produce the graph of a circle in polar coordinates are given by [latex]r=a\cos \theta [/latex] and [latex]r=a\sin \theta [/latex], where [latex]a[/latex] is the diameter of the circle or the distance from the pole to the farthest point on the circumference. The radius is [latex]\frac{|a|}{2}[/latex], or one-half the diameter. For [latex]r=a\cos \theta , [/latex] the center is [latex]\left(\frac{a}{2},0\right)[/latex]. For [latex]r=a\sin \theta [/latex], the center is [latex]\left(\frac{a}{2},\pi \right)[/latex]. Figure 5 shows the graphs of these four circles.

Solution

First, testing the equation for symmetry, we find that the graph is symmetric about the polar axis. Next, we find the zeros and maximum [latex]|r|[/latex] for [latex]r=4\cos \theta [/latex]. First, set [latex]r=0[/latex], and solve for [latex]\theta [/latex] . Thus, a zero occurs at [latex]\theta =\frac{\pi }{2}\pm k\pi [/latex]. A key point to plot is [latex]\left(0,\text{ }\text{ }\frac{\pi }{2}\right)[/latex]. To find the maximum value of [latex]r[/latex], note that the maximum value of the cosine function is 1 when [latex]\theta =0\pm 2k\pi [/latex]. Substitute [latex]\theta =0[/latex] into the equation:

[latex]\begin{array}{c}r=4\cos \theta \\ r=4\cos \left(0\right)\\ r=4\left(1\right)=4\end{array}[/latex]

The maximum value of the equation is 4. A key point to plot is [latex]\left(4,0\right)[/latex]. As [latex]r=4\cos \theta [/latex] is symmetric with respect to the polar axis, we only need to calculate r-values for [latex]\theta [/latex] over the interval [latex]\left[0,\pi \right][/latex]. Points in the upper quadrant can then be reflected to the lower quadrant. Make a table of values similar to the table below. The graph is shown in Figure 6.

[latex]\theta [/latex]	0	[latex]\frac{\pi }{6}[/latex]	[latex]\frac{\pi }{4}[/latex]	[latex]\frac{\pi }{3}[/latex]	[latex]\frac{\pi }{2}[/latex]	[latex]\frac{2\pi }{3}[/latex]	[latex]\frac{3\pi }{4}[/latex]	[latex]\frac{5\pi }{6}[/latex]	[latex]\pi [/latex]
[latex]r[/latex]	4	3.46	2.83	2	0	−2	−2.83	−3.46	4

Figure 6

Investigating Cardioids

A General Note: Formulas for a Cardioid

The formulas that produce the graphs of a cardioid are given by [latex]r=a\pm b\cos \theta [/latex] and [latex]r=a\pm b\sin \theta [/latex] where [latex]a>0,b>0[/latex], and [latex]\frac{a}{b}=1[/latex]. The cardioid graph passes through the pole, as we can see in Figure 7.

Figure 7

Solution

First, testing the equation for symmetry, we find that the graph of this equation will be symmetric about the polar axis. Next, we find the zeros and maximums. Setting [latex]r=0[/latex], we have [latex]\theta =\pi +2k\pi [/latex]. The zero of the equation is located at [latex]\left(0,\pi \right)[/latex]. The graph passes through this point. The maximum value of [latex]r=2+2\cos \theta [/latex] occurs when [latex]\cos \theta [/latex] is a maximum, which is when [latex]\cos \theta =1[/latex] or when [latex]\theta =0[/latex]. Substitute [latex]\theta =0[/latex] into the equation, and solve for [latex]r[/latex].

[latex]\begin{array}{l}\begin{array}{l}\\ r=2+2\cos \left(0\right)\end{array}\hfill \\ r=2+2\left(1\right)=4\hfill \end{array}[/latex]

The point [latex]\left(4,0\right)[/latex] is the maximum value on the graph. We found that the polar equation is symmetric with respect to the polar axis, but as it extends to all four quadrants, we need to plot values over the interval [latex]\left[0,\pi \right][/latex]. The upper portion of the graph is then reflected over the polar axis. Next, we make a table of values, as in the table below, and then we plot the points and draw the graph. See Figure 8.

[latex]\theta [/latex]	[latex]0[/latex]	[latex]\frac{\pi }{4}[/latex]	[latex]\frac{\pi }{2}[/latex]	[latex]\frac{2\pi }{3}[/latex]	[latex]\pi [/latex]
[latex]r[/latex]	4	3.41	2	1	0

Figure 8

Investigating Limaçons

A General Note: Formulas for One-Loop Limaçons

The formulas that produce the graph of a dimpled one-loop limaçon are given by [latex]r=a\pm b\cos \theta [/latex] and [latex]r=a\pm b\sin \theta [/latex] where [latex]a>0,b>0,\text{and 1<}\frac{a}{b}<2[/latex]. All four graphs are shown in Figure 9.

Figure 9. Dimpled limaçons

Solution

First, testing the equation for symmetry, we find that it fails all three symmetry tests, meaning that the graph may or may not exhibit symmetry, so we cannot use the symmetry to help us graph it. However, this equation has a graph that clearly displays symmetry with respect to the line [latex]\theta =\frac{\pi }{2}[/latex], yet it fails all the three symmetry tests. A graphing calculator will immediately illustrate the graph’s reflective quality. Next, we find the zeros and maximum, and plot the reflecting points to verify any symmetry. Setting [latex]r=0[/latex] results in [latex]\theta [/latex] being undefined. What does this mean? How could [latex]\theta [/latex] be undefined? The angle [latex]\theta [/latex] is undefined for any value of [latex]\sin \theta >1[/latex]. Therefore, [latex]\theta [/latex] is undefined because there is no value of [latex]\theta [/latex] for which [latex]\sin \theta >1[/latex]. Consequently, the graph does not pass through the pole. Perhaps the graph does cross the polar axis, but not at the pole. We can investigate other intercepts by calculating [latex]r[/latex] when [latex]\theta =0[/latex].

[latex]\begin{array}{l}r\left(0\right)=4 - 3\sin \left(0\right)\hfill \\ r=4 - 3\cdot 0=4\hfill \end{array}[/latex]

So, there is at least one polar axis intercept at [latex]\left(4,0\right)[/latex]. Next, as the maximum value of the sine function is 1 when [latex]\theta =\frac{\pi }{2}[/latex], we will substitute [latex]\theta =\frac{\pi }{2}[/latex] into the equation and solve for [latex]r[/latex]. Thus, [latex]r=1[/latex]. Make a table of the coordinates similar to the table below.

[latex]\theta [/latex]	[latex]0[/latex]	[latex]\frac{\pi }{6}[/latex]	[latex]\frac{\pi }{3}[/latex]	[latex]\frac{\pi }{2}[/latex]	[latex]\frac{2\pi }{3}[/latex]	[latex]\frac{5\pi }{6}[/latex]	[latex]\pi [/latex]	[latex]\frac{7\pi }{6}[/latex]	[latex]\frac{4\pi }{3}[/latex]	[latex]\frac{3\pi }{2}[/latex]	[latex]\frac{5\pi }{3}[/latex]	[latex]\frac{11\pi }{6}[/latex]	[latex]2\pi [/latex]
[latex]r[/latex]	4	2.5	1.4	1	1.4	2.5	4	5.5	6.6	7	6.6	5.5	4

The graph is shown in Figure 10.

Figure 10. One-loop limaçon

A General Note: Formulas for Inner-Loop Limaçons

The formulas that generate the inner-loop limaçons are given by [latex]r=a\pm b\cos \theta [/latex] and [latex]r=a\pm b\sin \theta [/latex] where [latex]a>0,b>0[/latex], and [latex]a<b[/latex]. The graph of the inner-loop limaçon passes through the pole twice: once for the outer loop, and once for the inner loop. See Figure 11 for the graphs.

Figure 11

Solution

Testing for symmetry, we find that the graph of the equation is symmetric about the polar axis. Next, finding the zeros reveals that when [latex]r=0,\theta =1.98[/latex]. The maximum [latex]|r|[/latex] is found when [latex]\cos \theta =1[/latex] or when [latex]\theta =0[/latex]. Thus, the maximum is found at the point (7, 0). Even though we have found symmetry, the zero, and the maximum, plotting more points will help to define the shape, and then a pattern will emerge.

[latex]\theta [/latex]

[latex]0[/latex]

[latex]\frac{\pi }{6}[/latex]

[latex]\frac{\pi }{3}[/latex]

[latex]\frac{\pi }{2}[/latex]

[latex]\frac{2\pi }{3}[/latex]

[latex]\frac{5\pi }{6}[/latex]

[latex]\pi [/latex]

[latex]\frac{7\pi }{6}[/latex]

[latex]\frac{4\pi }{3}[/latex]

[latex]\frac{3\pi }{2}[/latex]

[latex]\frac{5\pi }{3}[/latex]

[latex]\frac{11\pi }{6}[/latex]

[latex]2\pi [/latex]

[latex]r[/latex]

7

6.3

4.5

2

−0.5

−2.3

−3

−2.3

−0.5

2

4.5

6.3

7

As expected, the values begin to repeat after [latex]\theta =\pi [/latex]. The graph is shown in Figure 12.

Figure 12. Inner-loop limaçon

Investigating Lemniscates

A General Note: Formulas for Lemniscates

The formulas that generate the graph of a lemniscate are given by [latex]{r}^{2}={a}^{2}\cos 2\theta [/latex] and [latex]{r}^{2}={a}^{2}\sin 2\theta [/latex] where [latex]a\ne 0[/latex]. The formula [latex]{r}^{2}={a}^{2}\sin 2\theta [/latex] is symmetric with respect to the pole. The formula [latex]{r}^{2}={a}^{2}\cos 2\theta [/latex] is symmetric with respect to the pole, the line [latex]\theta =\frac{\pi }{2}[/latex], and the polar axis. See Figure 13 for the graphs.

Figure 13

Solution

The equation exhibits symmetry with respect to the line [latex]\theta =\frac{\pi }{2}[/latex], the polar axis, and the pole. Let’s find the zeros. It should be routine by now, but we will approach this equation a little differently by making the substitution [latex]u=2\theta [/latex].

[latex]\begin{array}{ll}0=4\cos 2\theta \hfill & \hfill \\ 0=4\cos u\hfill & \hfill \\ 0=\cos u\hfill & \hfill \\ {\cos }^{-1}0=\frac{\pi }{2}\hfill & \hfill \\ u=\frac{\pi }{2}\hfill & \text{Substitute }2\theta \text{ back in for }u.\hfill \\ 2\theta =\frac{\pi }{2}\hfill & \hfill \\ \theta =\frac{\pi }{4}\hfill & \hfill \end{array}[/latex]

So, the point [latex]\left(0,\frac{\pi }{4}\right)[/latex] is a zero of the equation. Now let’s find the maximum value. Since the maximum of [latex]\cos u=1[/latex] when [latex]u=0[/latex], the maximum [latex]\cos 2\theta =1[/latex] when [latex]2\theta =0[/latex]. Thus,

[latex]\begin{array}{c}{r}^{2}=4\cos \left(0\right)\\ {r}^{2}=4\left(1\right)=4\\ r=\pm \sqrt{4}=2\end{array}[/latex]

We have a maximum at (2, 0). Since this graph is symmetric with respect to the pole, the line [latex]\theta =\frac{\pi }{2}[/latex], and the polar axis, we only need to plot points in the first quadrant. Make a table similar to the table below.

[latex]\theta [/latex]	0	[latex]\frac{\pi }{6}[/latex]	[latex]\frac{\pi }{4}[/latex]	[latex]\frac{\pi }{3}[/latex]	[latex]\frac{\pi }{2}[/latex]
[latex]r[/latex]	2	[latex]\sqrt{2}[/latex]	0	[latex]\sqrt{2}[/latex]	0

Plot the points on the graph, such as the one shown in Figure 14.

Figure 14. Lemniscate

Investigating Rose Curves

A General Note: Rose Curves

The formulas that generate the graph of a rose curve are given by [latex]r=a\cos n\theta [/latex] and [latex]r=a\sin n\theta [/latex] where [latex]a\ne 0[/latex]. If [latex]n[/latex] is even, the curve has [latex]2n[/latex] petals. If [latex]n[/latex] is odd, the curve has [latex]n[/latex] petals.

Figure 15

Solution

Testing for symmetry, we find again that the symmetry tests do not tell the whole story. The graph is not only symmetric with respect to the polar axis, but also with respect to the line [latex]\theta =\frac{\pi }{2}[/latex] and the pole. Now we will find the zeros. First make the substitution [latex]u=4\theta [/latex].

[latex]\begin{array}{c}0=2\cos 4\theta \\ 0=\cos 4\theta \\ 0=\cos u\\ {\cos }^{-1}0=u\\ u=\frac{\pi }{2}\\ 4\theta =\frac{\pi }{2}\\ \theta =\frac{\pi }{8}\end{array}[/latex]

The zero is [latex]\theta =\frac{\pi }{8}[/latex]. The point [latex]\left(0,\frac{\pi }{8}\right)[/latex] is on the curve. Next, we find the maximum [latex]|r|[/latex]. We know that the maximum value of [latex]\cos u=1[/latex] when [latex]\theta =0[/latex]. Thus,

[latex]\begin{array}{l}\\ \begin{array}{l}r=2\cos \left(4\cdot 0\right)\hfill \\ r=2\cos \left(0\right)\hfill \\ r=2\left(1\right)=2\hfill \end{array}\end{array}[/latex]

The point [latex]\left(2,0\right)[/latex] is on the curve. The graph of the rose curve has unique properties, which are revealed in the table below.

[latex]\theta [/latex]	0	[latex]\frac{\pi }{8}[/latex]	[latex]\frac{\pi }{4}[/latex]	[latex]\frac{3\pi }{8}[/latex]	[latex]\frac{\pi }{2}[/latex]	[latex]\frac{5\pi }{8}[/latex]	[latex]\frac{3\pi }{4}[/latex]
[latex]r[/latex]	2	0	−2	0	2	0	−2

As [latex]r=0[/latex] when [latex]\theta =\frac{\pi }{8}[/latex], it makes sense to divide values in the table by [latex]\frac{\pi }{8}[/latex] units. A definite pattern emerges. Look at the range of r-values: 2, 0, −2, 0, 2, 0, −2, and so on. This represents the development of the curve one petal at a time. Starting at [latex]r=0[/latex], each petal extends out a distance of [latex]r=2[/latex], and then turns back to zero [latex]2n[/latex] times for a total of eight petals. See the graph in Figure 16.

Figure 16. Rose curve, [latex]n[/latex] even

Solution

The graph of the equation shows symmetry with respect to the line [latex]\theta =\frac{\pi }{2}[/latex]. Next, find the zeros and maximum. We will want to make the substitution [latex]u=5\theta [/latex].

[latex]\begin{array}{c}0=2\sin \left(5\theta \right)\\ 0=\sin u\\ {\sin }^{-1}0=0\\ u=0\\ 5\theta =0\\ \theta =0\end{array}[/latex]

The maximum value is calculated at the angle where [latex]\sin \theta [/latex] is a maximum. Therefore,

[latex]\begin{array}{l}\begin{array}{l}\\ r=2\sin \left(5\cdot \frac{\pi }{2}\right)\end{array}\hfill \\ r=2\left(1\right)=2\hfill \end{array}[/latex]

Thus, the maximum value of the polar equation is 2. This is the length of each petal. As the curve for [latex]n[/latex] odd yields the same number of petals as [latex]n[/latex], there will be five petals on the graph. Create a table of values similar to the table below.

[latex]\theta [/latex]	0	[latex]\frac{\pi }{6}[/latex]	[latex]\frac{\pi }{3}[/latex]	[latex]\frac{\pi }{2}[/latex]	[latex]\frac{2\pi }{3}[/latex]	[latex]\frac{5\pi }{6}[/latex]	[latex]\pi [/latex]
[latex]r[/latex]	0	1	−1.73	2	−1.73	1	0

Figure 17. Rose curve, [latex]n[/latex] odd

Investigating the Archimedes’ Spiral

A General Note: Archimedes’ Spiral

The formula that generates the graph of the Archimedes’ spiral is given by [latex]r=\theta [/latex] for [latex]\theta \ge 0[/latex]. As [latex]\theta [/latex] increases, [latex]r[/latex] increases at a constant rate in an ever-widening, never-ending, spiraling path.

Figure 18

Solution

As [latex]r[/latex] is equal to [latex]\theta [/latex], the plot of the Archimedes’ spiral begins at the pole at the point (0, 0). While the graph hints of symmetry, there is no formal symmetry with regard to passing the symmetry tests. Further, there is no maximum value, unless the domain is restricted. Create a table such as the one below.

[latex]\theta [/latex]	[latex]\frac{\pi }{4}[/latex]	[latex]\frac{\pi }{2}[/latex]	[latex]\pi [/latex]	[latex]\frac{3\pi }{2}[/latex]	[latex]\frac{7\pi }{4}[/latex]	[latex]2\pi [/latex]
[latex]r[/latex]	0.785	1.57	3.14	4.71	5.50	6.28

Notice that the r-values are just the decimal form of the angle measured in radians. We can see them on a graph in Figure 19.

Figure 19. Archimedes’ spiral

Key Concepts

• It is easier to graph polar equations if we can test the equations for symmetry with respect to the line [latex]\theta =\frac{\pi }{2}[/latex], the polar axis, or the pole.
• There are three symmetry tests that indicate whether the graph of a polar equation will exhibit symmetry. If an equation fails a symmetry test, the graph may or may not exhibit symmetry.
• Polar equations may be graphed by making a table of values for [latex]\theta [/latex] and [latex]r[/latex].
• The maximum value of a polar equation is found by substituting the value [latex]\theta [/latex] that leads to the maximum value of the trigonometric expression.
• The zeros of a polar equation are found by setting [latex]r=0[/latex] and solving for [latex]\theta [/latex].
• Some formulas that produce the graph of a circle in polar coordinates are given by [latex]r=a\cos \theta [/latex] and [latex]r=a\sin \theta [/latex].
• The formulas that produce the graphs of a cardioid are given by [latex]r=a\pm b\cos \theta [/latex] and [latex]r=a\pm b\sin \theta [/latex], for [latex]a>0,b>0[/latex], and [latex]\frac{a}{b}=1[/latex].
• The formulas that produce the graphs of a one-loop limaçon are given by [latex]r=a\pm b\cos \theta [/latex] and [latex]r=a\pm b\sin \theta [/latex] for [latex]1<\frac{a}{b}<2[/latex].
• The formulas that produce the graphs of an inner-loop limaçon are given by [latex]r=a\pm b\cos \theta [/latex] and [latex]r=a\pm b\sin \theta [/latex] for [latex]a>0,b>0[/latex], and [latex]a<b[/latex].
• The formulas that produce the graphs of a lemniscates are given by [latex]{r}^{2}={a}^{2}\cos 2\theta [/latex] and [latex]{r}^{2}={a}^{2}\sin 2\theta [/latex], where [latex]a\ne 0[/latex].
• The formulas that produce the graphs of rose curves are given by [latex]r=a\cos n\theta [/latex] and [latex]r=a\sin n\theta [/latex], where [latex]a\ne 0[/latex]; if [latex]n[/latex] is even, there are [latex]2n[/latex] petals, and if [latex]n[/latex] is odd, there are [latex]n[/latex] petals.
• The formula that produces the graph of an Archimedes’ spiral is given by [latex]r=\theta ,\theta \ge 0[/latex].

Summary of Curves

Glossary

Archimedes’ spiral

a polar curve given by [latex]r=\theta [/latex]. When multiplied by a constant, the equation appears as [latex]r=a\theta [/latex]. As [latex]r=\theta [/latex], the curve continues to widen in a spiral path over the domain.

cardioid

a member of the limaçon family of curves, named for its resemblance to a heart; its equation is given as [latex]r=a\pm b\cos \theta [/latex] and [latex]r=a\pm b\sin \theta [/latex], where [latex]\frac{a}{b}=1[/latex]

convex limaҫon

a type of one-loop limaçon represented by [latex]r=a\pm b\cos \theta [/latex] and [latex]r=a\pm b\sin \theta [/latex] such that [latex]\frac{a}{b}\ge 2[/latex]

dimpled limaҫon

a type of one-loop limaçon represented by [latex]r=a\pm b\cos \theta [/latex] and [latex]r=a\pm b\sin \theta [/latex] such that [latex]1<\frac{a}{b}<2[/latex]

inner-loop limaçon

a polar curve similar to the cardioid, but with an inner loop; passes through the pole twice; represented by [latex]r=a\pm b\cos \theta [/latex] and [latex]\text{ }r=a\pm b\sin \theta \text{ }[/latex] where [latex]a<b[/latex]

lemniscate

a polar curve resembling a figure 8 and given by the equation [latex]{r}^{2}={a}^{2}\cos 2\theta [/latex] and [latex]{r}^{2}={a}^{2}\sin 2\theta ,a\ne 0[/latex]

one-loop limaҫon

a polar curve represented by [latex]r=a\pm b\cos \theta [/latex] and [latex]r=a\pm b\sin \theta [/latex] such that [latex]a>0,b>0[/latex], and [latex]\frac{a}{b}>1[/latex]; may be dimpled or convex; does not pass through the pole

polar equation

an equation describing a curve on the polar grid.

rose curve

a polar equation resembling a flower, given by the equations [latex]r=a\cos n\theta [/latex] and [latex]r=a\sin n\theta [/latex]; when [latex]n[/latex] is even there are [latex]2n[/latex] petals, and the curve is highly symmetrical; when [latex]n[/latex] is odd there are [latex]n[/latex] petals.

Section Exercises

Licenses & Attributions