Polar Coordinates: Graphs

The planets move through space in elliptical, periodic orbits about the sun, as shown in Figure 1. They are in constant motion, so fixing an exact position of any planet is valid only for a moment. In other words, we can fix only a planet’s instantaneous position. This is one application of polar coordinates, represented as $(r, θ) .$ We interpret $r$ as the distance from the center of the sun and $θ$ as the planet’s angular bearing, or its direction from the center of the sun. In this section, we will focus on the polar system and the graphs that are generated directly from polar coordinates.

Illustration of the solar system with the sun at the center and orbits of the planets Mercury, Venus, Earth, and Mars shown. — 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

Just as a rectangular equation such as $y = x^{2}$ describes the relationship between $x$ and $y$ on a Cartesian grid, a polar equation describes a relationship between $r$ and $θ$ on a polar grid. Recall that the coordinate pair $(r, θ)$ indicates that we move counterclockwise from the polar axis (positive x-axis) by an angle of $θ,$ and extend a ray from the pole (origin) $r$ units in the direction of $θ .$ All points that satisfy the polar equation are on the graph.

Symmetry is a property that helps us recognize and plot the graph of any equation. If an equation has a graph that is symmetric with respect to an axis, it means that if we folded the graph in half over that axis, the portion of the graph on one side would coincide with the portion on the other side. By performing three tests, we will see how to apply the properties of symmetry to polar equations. Further, we will use symmetry (in addition to plotting key points, zeros, and maximums of $r)$ to determine the graph of a polar equation.

In the first test, we consider symmetry with respect to the line $θ = \frac{π}{2}$ (y-axis). We replace $(r, θ)$ with $(- r, - θ)$ to determine if the new equation is equivalent to the original equation. For example, suppose we are given the equation $r = 2 \sin θ;$

\begin{array}{l} r = 2 \sin θ \\ - r = 2 \sin (- θ) \begin{matrix}  \end{matrix} & Replace (r, θ) with (- r, - θ) . \\ - r = −2 \sin θ & Identity: \sin (- θ) = - \sin θ . \\ r = 2 \sin θ & Multiply both sides by −1. \end{array}

This equation exhibits symmetry with respect to the line $θ = \frac{π}{2} .$

In the second test, we consider symmetry with respect to the polar axis ( $x$ -axis). We replace $(r, θ)$ with $(r, - θ)$ or $(- r, π - θ)$ to determine equivalency between the tested equation and the original. For example, suppose we are given the equation $r = 1 - 2 \cos θ .$

\begin{array}{l} r = 1 - 2 \cos θ \\ r = 1 - 2 \cos (- θ) \begin{matrix}  \end{matrix} & Replace (r, θ) with (r, - θ) . \\ r = 1 - 2 \cos θ & Even/Odd identity \end{array}

The graph of this equation exhibits symmetry with respect to the polar axis.

In the third test, we consider symmetry with respect to the pole (origin). We replace $(r, θ)$ with $(- r, θ)$ to determine if the tested equation is equivalent to the original equation. For example, suppose we are given the equation $r = 2 \sin (3 θ).$

\begin{matrix} r = 2 \sin (3 θ) \\ - r = 2 \sin (3 θ) \end{matrix}

The equation has failed the symmetry test, but that does not mean that it is not symmetric with respect to the pole. Passing one or more of the symmetry tests verifies that symmetry will be exhibited in a graph. However, failing the symmetry tests does not necessarily indicate that a graph will not be symmetric about the line $θ = \frac{π}{2},$ the polar axis, or the pole. In these instances, we can confirm that symmetry exists by plotting reflecting points across the apparent axis of symmetry or the pole. Testing for symmetry is a technique that simplifies the graphing of polar equations, but its application is not perfect.

Example 1

Testing a Polar Equation for Symmetry

Test the equation $r = 2 \sin θ$ for symmetry.

Solution

Test for each of the three types of symmetry.

Table 1 Three rows and two columns. The first column contains the steps to test for a type of symmetry, and the second column gives an example. The first column, first row tests symmetry with respect to theta= pi/2. Test: Replacing (r, theta) with (-r, -theta) yields the same result. Thus, the graph is symmetric with respect to the line pi/2. The example is -r = 2sin(-theta). By the even-odd identity, -r = -2sin(theta). After multiplying by -1, r=2sin(theta), so it passes the test. The next test is symmetry with respect to the polar axis. Test: Replacing theta with -theta does not yield the same equation. Therefore, the graph fails the test and may or may not be symmetric with respect to the polar axis. Example: r=2sin(-theta). By the even-odd identity, r=-2sin(theta). We have then r=-2sin(theta) which does not equal 2 sin(theta), so it fails. Finally, there is symmetry with respect to the pole. Test: Replacing r with -r changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole. Example: -r = 2sin(theta). r=-2sin(theta) which does not equal 2sin(theta), so it fails the test.
1) Replacing $(r, θ)$ with $(- r, - θ)$ yields the same result. Thus, the graph is symmetric with respect to the line $θ = \frac{π}{2} .$	$\begin{array}{l} - r = 2 \sin (- θ) \\ - r = −2 \sin θ & Even-odd identity \\ r = 2 \sin θ & Multiply by −1 \\ Passed \end{array}$
2) Replacing $θ$ with $- θ$ does not yield the same equation. Therefore, the graph fails the test and may or may not be symmetric with respect to the polar axis.	$\begin{array}{l} r = 2 \sin (- θ) \\ r = −2 \sin θ & Even-odd identity \\ r = −2 \sin θ \neq 2 \sin θ \\ Failed \end{array}$
3) Replacing $r$ with $- r$ changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole.	$\begin{array}{l} - r = 2 \sin θ \\ r = −2 \sin θ \neq 2 \sin θ \\ Failed \end{array}$

Analysis

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

Graph of the given circle on the polar coordinate grid. Center is at (0,1), and it has radius 1. — Figure 3

Graphing Polar Equations by Plotting Points

To graph in the rectangular coordinate system we construct a table of $x$ and $y$ values. To graph in the polar coordinate system we construct a table of $θ$ and $r$ values. We enter values of $θ$ into a polar equation and calculate $r .$ However, using the properties of symmetry and finding key values of $θ$ and $r$ means fewer calculations will be needed.

Finding Zeros and Maxima

To find the zeros of a polar equation, we solve for the values of $θ$ that result in $r = 0.$ Recall that, to find the zeros of polynomial functions, we set the equation equal to zero and then solve for $x .$ We use the same process for polar equations. Set $r = 0,$ and solve for $θ .$

For many of the forms we will encounter, the maximum value of a polar equation is found by substituting those values of $θ$ into the equation that result in the maximum value of the trigonometric functions. Consider $r = 5 \cos θ;$ the maximum distance between the curve and the pole is 5 units. The maximum value of the cosine function is 1 when $θ = 0,$ so our polar equation is $5 \cos θ,$ and the value $θ = 0$ will yield the maximum $| r | .$

Similarly, the maximum value of the sine function is 1 when $θ = \frac{π}{2},$ and if our polar equation is $r = 5 \sin θ,$ the value $θ = \frac{π}{2}$ will yield the maximum $| r | .$ We may find additional information by calculating values of $r$ when $θ = 0.$ These points would be polar axis intercepts, which may be helpful in drawing the graph and identifying the curve of a polar equation.

Example 2

Finding Zeros and Maximum Values for a Polar Equation

Using the equation in Example 1, find the zeros and maximum $| r |$ and, if necessary, the polar axis intercepts of $r = 2 \sin θ .$

Solution

To find the zeros, set $r$ equal to zero and solve for $θ .$

\begin{array}{l} 2 \sin θ = 0 \\ \sin θ = 0 \\ θ = \sin^{- 1} 0 \\ θ = n π & where n is an integer \end{array}

Substitute any one of the $θ$ values into the equation. We will use $0.$

\begin{array}{l} \begin{array}{l} r = 2 \sin (0) \end{array} \\ r = 0 \end{array}

The points $(0, 0)$ and $(0, \pm n π)$ are the zeros of the equation. They all coincide, so only one point is visible on the graph. This point is also the only polar axis intercept.

To find the maximum value of the equation, look at the maximum value of the trigonometric function $\sin θ,$ which occurs when $θ = \frac{π}{2} \pm 2 k π$ resulting in $\sin (\frac{π}{2}) = 1.$ Substitute $\frac{π}{2}$ for $θ.$

\begin{array}{l} r = 2 \sin (\frac{π}{2}) \\ r = 2 (1) \\ r = 2 \end{array}

Analysis

The point $(2, \frac{π}{2})$ will be the maximum value on the graph. Let’s plot a few more points to verify the graph of a circle. See Table 2 and Figure 4.

Table 2 Eight rows and 3 columns. First column is labeled theta, second column is labeled r=2sin(theta), and third column is labeled r. The table has ordered triples of these column values: (0, r=2sin(0)=0, 0), (pi/6, r=2sin(pi/6)=1, 1), (pi/3, r=2sin(pi/3) = approx. 1.73, 1.73), (pi/2, r=2sin(pi/2) = 2, 2), (2pi/3, r=2sin(2pi/3)=approx. 1.73, 1.73), (5pi/6, r=2sin(5pi/6)=1, 1), and (pi, r=2sin(pi)=0).
$θ$	$r = 2 \sin θ$	$r$
0	$r = 2 \sin (0) = 0$	$0$
$\frac{π}{6}$	$r = 2 \sin (\frac{π}{6}) = 1$	$1$
$\frac{π}{3}$	$r = 2 \sin (\frac{π}{3}) \approx 1.73$	$1.73$
$\frac{π}{2}$	$r = 2 \sin (\frac{π}{2}) = 2$	$2$
$\frac{2 π}{3}$	$r = 2 \sin (\frac{2 π}{3}) \approx 1.73$	$1.73$
$\frac{5 π}{6}$	$r = 2 \sin (\frac{5 π}{6}) = 1$	$1$
$π$	$r = 2 \sin (π) = 0$	$0$

Graph of circle on the polar coordinate grid. The center is at (0,1), and it has radius 1. Six points along the circumference are marked: (0,0), (1, pi/6), (1.3, pi/3), (2, pi/2), (1.73, 2pi/3), and (1, 5pi/6). — Figure 4

Investigating Circles

Now we have seen the equation of a circle in the polar coordinate system. In the last two examples, the same equation was used to illustrate the properties of symmetry and demonstrate how to find the zeros, maximum values, and plotted points that produced the graphs. However, the circle is only one of many shapes in the set of polar curves.

There are five classic polar curves: cardioids, limaҫons, lemniscates, rose curves, and Archimedes’ spirals. We will briefly touch on the polar formulas for the circle before moving on to the classic curves and their variations.

Example 3

Sketching the Graph of a Polar Equation for a Circle

Sketch the graph of $r = 4 \cos θ .$

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 $| r |$ for $r = 4 \cos θ .$ First, set $r = 0,$ and solve for $θ$ . Thus, a zero occurs at $θ = \frac{π}{2} \pm k π .$ A key point to plot is $(0, \frac{π}{2}) .$

To find the maximum value of $r,$ note that the maximum value of the cosine function is 1 when $θ = 0 \pm 2 k π .$ Substitute $θ = 0$ into the equation:

\begin{matrix} r = 4 \cos θ \\ r = 4 \cos (0) \\ r = 4 (1) = 4 \end{matrix}

The maximum value of the equation is 4. A key point to plot is $(4, 0) .$

As $r = 4 \cos θ$ is symmetric with respect to the polar axis, we only need to calculate r-values for $θ$ over the interval $[0,$ $π] .$ Points in the upper quadrant can then be reflected to the lower quadrant. Make a table of values similar to Table 3. The graph is shown in Figure 6.

Table 3 Two rows and ten columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,4), (pi/6, 3.46), (pi/4, 2.83), (pi/3,2), (pi/2,0), (2pi/3,-2), (3pi/4,-2.83), (5pi/6, -3.46), and (pi,4).
$θ$	0	$\frac{π}{6}$	$\frac{π}{4}$	$\frac{π}{3}$	$\frac{π}{2}$	$\frac{2 π}{3}$	$\frac{3 π}{4}$	$\frac{5 π}{6}$	$π$
$r$	4	3.46	2.83	2	0	−2	−2.83	−3.46	−4

Graph of 4=4cos(theta) in polar coordinates. Points (0, pi/2), (-2, 2pi/3), (4,0), and (2, pi/3) are marked on the circumference. — Figure 6

Investigating Cardioids

While translating from polar coordinates to Cartesian coordinates may seem simpler in some instances, graphing the classic curves is actually less complicated in the polar system. The next curve is called a cardioid, as it resembles a heart. This shape is often included with the family of curves called limaçons, but here we will discuss the cardioid on its own.

Example 4

Sketching the Graph of a Cardioid

Sketch the graph of $r = 2 + 2 \cos θ .$

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 $r = 0,$ we have $θ = π + 2 k π .$ The zero of the equation is located at $(0, π) .$ The graph passes through this point.

The maximum value of $r = 2 + 2 \cos θ$ occurs when $\cos θ$ is a maximum, which is when $\cos θ = 1$ or when $θ = 0.$ Substitute $θ = 0$ into the equation, and solve for $r .$

\begin{array}{l} \begin{array}{l} r = 2 + 2 \cos (0) \end{array} \\ r = 2 + 2 (1) = 4 \end{array}

The point $(4, 0)$ 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 $[0, π] .$ The upper portion of the graph is then reflected over the polar axis. Next, we make a table of values, as in Table 4, and then we plot the points and draw the graph. See Figure 8.

Table 4 Two rows and six columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,4), (pi/4, 3.41), (pi/2, 2), (2pi/3, 1), and (pi, 0).
$θ$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{2 π}{3}$	$π$
$r$	4	3.41	2	1	0

Graph of r=2+2cos(theta). Cardioid extending to the right. Points on the edge (0,pi), (4,0),(3.4, pi/4), (2,pi/2), and (1, 2pi/3) are shown. — Figure 8

Investigating Limaçons

The word limaçon is Old French for “snail,” a name that describes the shape of the graph. As mentioned earlier, the cardioid is a member of the limaçon family, and we can see the similarities in the graphs. The other images in this category include the one-loop limaçon and the two-loop (or inner-loop) limaçon. One-loop limaçons are sometimes referred to as dimpled limaçons when $1 < \frac{a}{b} < 2$ and convex limaçons when $\frac{a}{b} \geq 2.$

Example 5

Sketching the Graph of a One-Loop Limaçon

Graph the equation $r = 4 - 3 \sin θ .$

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 $θ = \frac{π}{2},$ 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 $r = 0$ results in $θ$ being undefined. What does this mean? How could $θ$ be undefined? The angle $θ$ is undefined for any value of $\sin θ > 1.$ Therefore, $θ$ is undefined because there is no value of $θ$ for which $\sin θ > 1.$ 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 $r$ when $θ = 0.$

\begin{array}{l} r (0) = 4 - 3 \sin (0) \\ r = 4 - 3 \cdot 0 = 4 \end{array}

So, there is at least one polar axis intercept at $(4, 0) .$

Next, as the maximum value of the sine function is 1 when $θ = \frac{π}{2},$ we will substitute $θ = \frac{π}{2}$ into the equation and solve for $r .$ Thus, $r = 1.$

Make a table of the coordinates similar to Table 5.

Table 5 Two rows and fourteen columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,4), (pi/6, 2.5), (pi/3,1.4), (pi/2, 1), (2pi/3, 1.4), (5pi/6, 2.5), (pi,4), (7pi/6, 5.5), (4pi/3,6.6), (3pi/2, 7), (5pi/3, 6.6), (11pi/6, 5.5), and (2pi, 4).
$θ$	$0$	$\frac{π}{6}$	$\frac{π}{3}$	$\frac{π}{2}$	$\frac{2 π}{3}$	$\frac{5 π}{6}$	$π$	$\frac{7 π}{6}$	$\frac{4 π}{3}$	$\frac{3 π}{2}$	$\frac{5 π}{3}$	$\frac{11 π}{6}$	$2 π$
$r$	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.

Graph of the limaçon r=4-3sin(theta). Extending down. Points on the edge are shown: (1,pi/2), (4,0), (4,pi), and (7, 3pi/2). — Figure 10 One-loop limaçon

Analysis

This is an example of a curve for which making a table of values is critical to producing an accurate graph. The symmetry tests fail; the zero is undefined. While it may be apparent that an equation involving $\sin θ$ is likely symmetric with respect to the line $θ = \frac{π}{2},$ evaluating more points helps to verify that the graph is correct.

Another type of limaçon, the inner-loop limaçon, is named for the loop formed inside the general limaçon shape. It was discovered by the German artist Albrecht Dürer(1471-1528), who revealed a method for drawing the inner-loop limaçon in his 1525 book Underweysung der Messing. A century later, the father of mathematician Blaise Pascal, Étienne Pascal(1588-1651), rediscovered it.

Example 6

Sketching the Graph of an Inner-Loop Limaçon

Sketch the graph of $r = 2 + 5 cos θ .$

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 $r = 0,$ $θ = 1.98.$ The maximum $| r |$ is found when $\cos θ = 1$ or when $θ = 0.$ 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.

See Table 6.

Table 6 Two rows and fourteen columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,7), (pi/6, 6.3.), (pi/3,4.5), (pi/2, 2), (2pi/3, -.5), (5pi/6, -2.3), (pi,-3), (7pi/6, -2.3), (4pi/3,-.5), (3pi/2, 2), (5pi/3, 4.5), (11pi/6, 6.3), and (2pi, 7).
$θ$	$0$	$\frac{π}{6}$	$\frac{π}{3}$	$\frac{π}{2}$	$\frac{2 π}{3}$	$\frac{5 π}{6}$	$π$	$\frac{7 π}{6}$	$\frac{4 π}{3}$	$\frac{3 π}{2}$	$\frac{5 π}{3}$	$\frac{11 π}{6}$	$2 π$
$r$	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 $θ = π .$ The graph is shown in Figure 12.

Graph of inner loop limaçon r=2+5cos(theta). Extends to the right. Points on edge plotted are (7,0), (4.5, pi/3), (2, pi/2), and (-3, pi). — Figure 12 Inner-loop limaçon

Investigating Lemniscates

The lemniscate is a polar curve resembling the infinity symbol $\infty$ or a figure 8. Centered at the pole, a lemniscate is symmetrical by definition.

Example 7

Sketching the Graph of a Lemniscate

Sketch the graph of $r^{2} = 4 \cos 2 θ .$

Solution

The equation exhibits symmetry with respect to the line $θ = \frac{π}{2},$ 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 $u = 2 θ .$

\begin{array}{l} 0 = 4 \cos 2 θ \\ 0 = 4 \cos u \\ 0 = \cos u \\ \cos^{- 1} 0 = \frac{π}{2} \\ u = \frac{π}{2} & Substitute 2 θ back in for u . \\ 2 θ = \frac{π}{2} \\ θ = \frac{π}{4} \end{array}

So, the point $(0, \frac{π}{4})$ is a zero of the equation.

Now let’s find the maximum value. Since the maximum of $\cos u = 1$ when $u = 0,$ the maximum $\cos 2 θ = 1$ when $2 θ = 0.$ Thus,

\begin{matrix} r^{2} = 4 \cos (0) \\ r^{2} = 4 (1) = 4 \\ r = \pm \sqrt{4} \pm 2 \end{matrix}

We have a maximum at (2, 0). Since this graph is symmetric with respect to the pole, the line $θ = \frac{π}{2},$ and the polar axis, we only need to plot points in the first quadrant.

Make a table similar to Table 7.

Table 7 Two rows and six columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,2), (pi/6, rad2), (pi/4,0), (pi/3, rad2), (pi/2,0).
$θ$	0	$\frac{π}{6}$	$\frac{π}{4}$
$r$	$\pm 2$	$\pm \sqrt{2}$	0

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

Graph of r^2 = 4cos(2theta). Horizontal lemniscate, along x-axis. Points on edge plotted are (2,0), (rad2, pi/6), (rad2 7pi/6). — Figure 14 Lemniscate

Analysis

Making a substitution such as $u = 2 θ$ is a common practice in mathematics because it can make calculations simpler. However, we must not forget to replace the substitution term with the original term at the end, and then solve for the unknown.

Some of the points on this graph may not show up using the Trace function on the TI-84 graphing calculator, and the calculator table may show an error for these same points of $r .$ This is because there are no real square roots for these values of $θ .$ In other words, the corresponding r-values of $\sqrt{4 \cos (2 θ)}$ are complex numbers because there is a negative number under the radical.

Investigating Rose Curves

The next type of polar equation produces a petal-like shape called a rose curve. Although the graphs look complex, a simple polar equation generates the pattern.

Example 8

Sketching the Graph of a Rose Curve (n Even)

Sketch the graph of $r = 2 \cos 4 θ .$

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 $θ = \frac{π}{2}$ and the pole.

Now we will find the zeros. First make the substitution $u = 4 θ .$

\begin{matrix} 0 = 2 \cos 4 θ \\ 0 = \cos 4 θ \\ 0 = \cos u \\ \cos^{- 1} 0 = u \\ u = \frac{π}{2} \\ 4 θ = \frac{π}{2} \\ θ = \frac{π}{8} \end{matrix}

The zero is $θ = \frac{π}{8} .$ The point $(0, \frac{π}{8})$ is on the curve.

Next, we find the maximum $| r | .$ We know that the maximum value of $\cos u = 1$ when $θ = 0.$ Thus,

\begin{array}{l} r = 2 \cos (4 \cdot 0) \\ r = 2 \cos (0) \\ r = 2 (1) = 2 \end{array}

The point $(2, 0)$ is on the curve.

The graph of the rose curve has unique properties, which are revealed in Table 8.

Table 8 Two rows and eight columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,2), (pi/8, 0), (pi/4, -2), (3pi/8, 0), (pi/2, 2), (5pi/8, 0), (3pi/4, -2).
$θ$	0	$\frac{π}{8}$	$\frac{π}{4}$	$\frac{3 π}{8}$	$\frac{π}{2}$	$\frac{5 π}{8}$	$\frac{3 π}{4}$
$r$	2	0	−2	0	2	0	−2

As $r = 0$ when $θ = \frac{π}{8},$ it makes sense to divide values in the table by $\frac{π}{8}$ 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 $r = 0,$ each petal extends out a distance of $r = 2,$ and then turns back to zero $2 n$ times for a total of eight petals. See the graph in Figure 16.

Sketch of rose curve r=2*cos(4 theta). Goes out distance of 2 for each petal 2n times (here 2*4=8 times). — Figure 16 Rose curve, $n$ even

Analysis

When these curves are drawn, it is best to plot the points in order, as in the Table 8. This allows us to see how the graph hits a maximum (the tip of a petal), loops back crossing the pole, hits the opposite maximum, and loops back to the pole. The action is continuous until all the petals are drawn.

Example 9

Sketching the Graph of a Rose Curve (n Odd)

Sketch the graph of $r = 2 \sin (5 θ) .$

Solution

The graph of the equation shows symmetry with respect to the line $θ = \frac{π}{2} .$ Next, find the zeros and maximum. We will want to make the substitution $u = 5 θ .$

\begin{matrix} 0 = 2 \sin (5 θ) \\ 0 = \sin u \\ \sin^{- 1} 0 = 0 \\ u = 0 \\ 5 θ = 0 \\ θ = 0 \end{matrix}

The maximum value is calculated at the angle where $\sin θ$ is a maximum. Therefore,

\begin{array}{l} \begin{array}{l} r = 2 \sin (5 \cdot \frac{π}{2}) \end{array} \\ r = 2 (1) = 2 \end{array}

Thus, the maximum value of the polar equation is 2. This is the length of each petal. As the curve for $n$ odd yields the same number of petals as $n,$ there will be five petals on the graph. See Figure 17.

Graph of rose curve r=2sin(5theta). Five petals equally spaced around origin. Point (2, pi/2) on edge is marked. — Figure 17 Rose curve, $n$ odd

Create a table of values similar to Table 9.

Table 9 Two rows and seven columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (0,0), (pi/6, 1), (pi/3, -1.73), (pi/2, 2), (2pi/3, -1.73), (5pi/6, 1), (pi, 0).
$θ$	0	$\frac{π}{6}$	$\frac{π}{3}$	$\frac{π}{2}$	$\frac{2 π}{3}$	$\frac{5 π}{6}$	$π$
$r$	0	1	−1.73	2	−1.73	1	0

Investigating the Archimedes’ Spiral

The final polar equation we will discuss is the Archimedes’ spiral, named for its discoverer, the Greek mathematician Archimedes (c. 287 BCE-c. 212 BCE), who is credited with numerous discoveries in the fields of geometry and mechanics.

Example 10

Sketching the Graph of an Archimedes’ Spiral

Sketch the graph of $r = θ$ over $[0, 2 π] .$

Solution

As $r$ is equal to $θ,$ 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 Table 10.

Table 10 Two rows and seven columns. First row is labeled theta and second row is labeled r. The table has ordered pairs of each of these column values: (pi/4, 0.785), (pi/2, 1.57), (pi, 3.14), (3pi/2, 4.71), (7pi/4, 5.5), (2pi, 6.28).
$θ$	$\frac{π}{4}$	$\frac{π}{2}$	$π$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$r$	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.

Graph of Archimedes' spiral r=theta over [0,2pi]. Starts at origin and spirals out in one loop counterclockwise. Points (pi/4, pi/4), (pi/2,pi/2), (pi,pi), (5pi/4, 5pi/4), (7pi/4, pi/4), and (2pi, 2pi) are marked. — Figure 19 Archimedes’ spiral

Analysis

The domain of this polar curve is $[0, 2 π] .$ In general, however, the domain of this function is $(- \infty, \infty) .$ Graphing the equation of the Archimedes’ spiral is rather simple, although the image makes it seem like it would be complex.

Summary of Curves

We have explored a number of seemingly complex polar curves in this section. Figure 20 and Figure 21 summarize the graphs and equations for each of these curves.

Four graphs side by side - a summary. (A) is a circle: r=asin(theta) or r=acos(theta). (B) is a cardioid: r= a + or - bcos(theta), or r = a + or - b sin(theta). a>0, b>0, a/b=1. (C) is one-loop limaçons. r= a + or - bcos(theta), or r= a + or - bsin(theta). a>0, b>0, 1<a/b<2. (D) is inner-loop limaçons. R = a + or - bcos(theta), or r = a + or - bsin(theta). A>0, b>0, a<b. — Figure 20

Four graphs side by side - a summary. (A) is lemniscates. R^2 = a^2cos(2theta), or r^2=a^2sin(2theta). a is not equal to 0. (B) is a rsose curve (n even). R = acos(ntheta), or r=asin(ntheta). N is even, and there are 2n petals. (C) is a rose curve (n odd). R = acos(ntheta), or r=asin(theta). N is odd, and there are n petals. (D) is an Archimedes's spiral. R=theta, and theta >=0. — Figure 21

Key Concepts

It is easier to graph polar equations if we can test the equations for symmetry with respect to the line $θ = \frac{π}{2},$ 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. See Example 1.
Polar equations may be graphed by making a table of values for $θ$ and $r .$
The maximum value of a polar equation is found by substituting the value $θ$ that leads to the maximum value of the trigonometric expression.
The zeros of a polar equation are found by setting $r = 0$ and solving for $θ .$ See Example 2.
Some formulas that produce the graph of a circle in polar coordinates are given by $r = a \cos θ$ and $r = a \sin θ .$ See Example 3.
The formulas that produce the graphs of a cardioid are given by $r = a \pm b \cos θ$ and $r = a \pm b \sin θ,$ for $a > 0,$ $b > 0,$ and $\frac{a}{b} = 1.$ See Example 4.
The formulas that produce the graphs of a one-loop limaçon are given by $r = a \pm b \cos θ$ and $r = a \pm b \sin θ$ for $1 < \frac{a}{b} < 2.$ See Example 5.
The formulas that produce the graphs of an inner-loop limaçon are given by $r = a \pm b \cos θ$ and $r = a \pm b \sin θ$ for $a > 0,$ $b > 0,$ and $a < b .$ See Example 6.
The formulas that produce the graphs of a lemniscates are given by $r^{2} = a^{2} \cos 2 θ$ and $r^{2} = a^{2} \sin 2 θ,$ where $a \neq 0.$ See Example 7.
The formulas that produce the graphs of rose curves are given by $r = a \cos n θ$ and $r = a \sin n θ,$ where $a \neq 0;$ if $n$ is even, there are $2 n$ petals, and if $n$ is odd, there are $n$ petals. See Example 8 and Example 9.
The formula that produces the graph of an Archimedes’ spiral is given by $r = θ,$ $θ \geq 0.$ See Example 10.

Section Exercises

Verbal

Exercise 1

Describe the three types of symmetry in polar graphs, and compare them to the symmetry of the Cartesian plane.

Solution

Symmetry with respect to the polar axis is similar to symmetry about the $x$ -axis, symmetry with respect to the pole is similar to symmetry about the origin, and symmetric with respect to the line $θ = \frac{π}{2}$ is similar to symmetry about the $y$ -axis.

Exercise 2

Which of the three types of symmetries for polar graphs correspond to the symmetries with respect to the x-axis, y-axis, and origin?

Exercise 3

What are the steps to follow when graphing polar equations?

Solution

Test for symmetry; find zeros, intercepts, and maxima; make a table of values. Decide the general type of graph, cardioid, limaçon, lemniscate, etc., then plot points at $θ = 0, \frac{π}{2},$ $π$ and $\frac{3 π}{2},$ and sketch the graph.

Exercise 4

Describe the shapes of the graphs of cardioids, limaçons, and lemniscates.

Exercise 5

What part of the equation determines the shape of the graph of a polar equation?

Solution

The shape of the polar graph is determined by whether or not it includes a sine, a cosine, and constants in the equation.

Graphical

For the following exercises, test the equation for symmetry.

Exercise 6

$r = 5 \cos 3 θ$

Exercise 7

$r = 3 - 3 \cos θ$

Solution

symmetric with respect to the polar axis

Exercise 8

$r = 3 + 2 \sin θ$

Exercise 9

$r = 3 \sin 2 θ$

Solution

symmetric with respect to the polar axis, symmetric with respect to the line $θ = \frac{π}{2},$ symmetric with respect to the pole

Exercise 10

$r = 4$

Exercise 11

$r = 2 θ$

Solution

symmetric with respect to the line $θ = \frac{π}{2}$

Exercise 12

$r = 4 \cos \frac{θ}{2}$

Exercise 13

$r = \frac{2}{θ}$

Solution

Symmetric with respect to line $θ = \frac{π}{2}$ (y-axis)

Exercise 14

$r = 3 \sqrt{1 - \cos^{2} θ}$

Exercise 15

$r = \sqrt{5 \sin 2 θ}$

Solution

symmetric with respect to the pole

For the following exercises, graph the polar equation. Identify the name of the shape.

Exercise 16

$r = 3 \cos θ$

Exercise 17

$r = 4 \sin θ$

Solution

circle
Graph of given circle.

Exercise 18

$r = 2 + 2 \cos θ$

Exercise 19

$r = 2 - 2 \cos θ$

Solution

cardioid
Graph of given cardioid.

Exercise 20

$r = 5 - 5 \sin θ$

Exercise 21

$r = 3 + 3 \sin θ$

Solution

cardioid
Graph of given cardioid.

Exercise 22

$r = 3 + 2 \sin θ$

Exercise 23

$r = 7 + 4 \sin θ$

Solution

one-loop/dimpled limaçon

Exercise 24

$r = 4 + 3 \cos θ$

Exercise 25

$r = 5 + 4 \cos θ$

Solution

one-loop/dimpled limaçon
Graph of given one-loop/dimpled limaçon

Exercise 26

$r = 10 + 9 \cos θ$

Exercise 27

$r = 1 + 3 \sin θ$

Solution

inner loop/two-loop limaçon

Exercise 28

$r = 2 + 5 \sin θ$

Exercise 29

$r = 5 + 7 \sin θ$

Solution

inner loop/two-loop limaçon

Exercise 30

$r = 2 + 4 \cos θ$

Exercise 31

$r = 5 + 6 \cos θ$

Solution

inner loop/two-loop limaçon
Graph of given inner loop/two-loop limaçon

Exercise 32

$r^{2} = 36 \cos (2 θ)$

Exercise 33

$r^{2} = 10 \cos (2 θ)$

Solution

lemniscate

Exercise 34

$r^{2} = 4 \sin (2 θ)$

Exercise 35

$r^{2} = 10 \sin (2 θ)$

Solution

lemniscate

Exercise 36

$r = 3 sin (2 θ)$

Exercise 37

$r = 3 cos (2 θ)$

Solution

rose curve

Exercise 38

$r = 5 sin (3 θ)$

Exercise 39

$r = 4 sin (4 θ)$

Solution

rose curve

Exercise 40

$r = 4 sin (5 θ)$

Exercise 41

$r = - θ$

Solution

Archimedes’ spiral

Exercise 42

$r = 2 θ$

Exercise 43

$r = - 3 θ$

Solution

Archimedes’ spiral

Technology

For the following exercises, use a graphing calculator to sketch the graph of the polar equation.

Exercise 44

$r = \frac{1}{θ}$

Exercise 45

$r = \frac{1}{\sqrt{θ}}$

Solution

Exercise 46

$r = 2 \sin θ \tan θ,$ a cissoid

Exercise 47

$r = 2 \sqrt{1 - \sin^{2} θ}$ , a hippopede

Solution

Graph of given hippopede (two circles that are centered along the x-axis and meet at the origin)

Exercise 48

$r = 5 + \cos (4 θ)$

Exercise 49

$r = 2 - \sin (2 θ)$

Solution

Exercise 50

$r = θ^{2}$

Exercise 51

$r = θ + 1$

Solution

Graph of given equation. Similar to original Archimedes' spiral.

Exercise 52

$r = θ \sin θ$

Exercise 53

$r = θ \cos θ$

Solution

For the following exercises, use a graphing utility to graph each pair of polar equations on a domain of $[0, 4 π]$ and then explain the differences shown in the graphs.

Exercise 54

$r = θ, r = - θ$

Exercise 55

$r = θ, r = θ + \sin θ$

Solution

They are both spirals, but not quite the same.

Exercise 56

$r = \sin θ + θ, r = \sin θ - θ$

Exercise 57

$r = 2 \sin (\frac{θ}{2}), r = θ \sin (\frac{θ}{2})$

Solution

Both graphs are curves with 2 loops. The equation with a coefficient of $θ$ has two loops on the left, the equation with a coefficient of 2 has two loops side by side. Graph these from 0 to $4 π$ to get a better picture.

Exercise 58

$r = \sin (\cos (3 θ)) r = \sin (3 θ)$

Exercise 59

On a graphing utility, graph $r = \sin (\frac{16}{5} θ)$ on $[0$ , $4 π]$ , $[0$ , $8 π]$ , $[0$ , $12 π]$ , and $[0$ , $16 π] .$ Describe the effect of increasing the width of the domain.

Solution

When the width of the domain is increased, more petals of the flower are visible.

Exercise 60

On a graphing utility, graph and sketch $r = \sin θ + {(\sin (\frac{5}{2} θ))}^{3}$ on $[0, 4 π] .$

Exercise 61

On a graphing utility, graph each polar equation. Explain the similarities and differences you observe in the graphs.

\begin{array}{l} r_{1} = 3 \sin (3 θ) \\ r_{2} = 2 \sin (3 θ) \\ r_{3} = \sin (3 θ) \end{array}

Solution

The graphs are three-petal, rose curves. The larger the coefficient, the greater the curve’s distance from the pole.

Exercise 62

On a graphing utility, graph each polar equation. Explain the similarities and differences you observe in the graphs.

\begin{array}{l} r_{1} = 3 + 3 \cos θ \\ r_{2} = 2 + 2 \cos θ \\ r_{3} = 1 + \cos θ \end{array}

Exercise 63

On a graphing utility, graph each polar equation. Explain the similarities and differences you observe in the graphs.

\begin{array}{l} r_{1} = 3 θ \\ r_{2} = 2 θ \\ r_{3} = θ \end{array}

Solution

The graphs are spirals. The smaller the coefficient, the tighter the spiral.

Extensions

For the following exercises, draw each polar equation on the same set of polar axes, and find the points of intersection.

Exercise 64

$r_{1} = 3 + 2 \sin θ, r_{2} = 2$

Exercise 65

$r_{1} = 6 - 4 \cos θ, r_{2} = 4$

Solution

$(4, \frac{π}{3}), (4, \frac{5 π}{3})$

Exercise 66

$r_{1} = 1 + \sin θ, r_{2} = 3 \sin θ$

Exercise 67

$r_{1} = 1 + \cos θ, r_{2} = 3 \cos θ$

Solution

$(\frac{3}{2}, \frac{π}{3}), (\frac{3}{2}, \frac{5 π}{3})$

Exercise 68

$r_{1} = \cos (2 θ), r_{2} = \sin (2 θ)$

Exercise 69

$r_{1} = \sin^{2} (2 θ)$ , $r_{2} = 1 - \cos (4 θ)$

Solution

$(0, \frac{π}{2}), (0, π), (0, \frac{3 π}{2}), (0, 2 π)$

Exercise 70

$r_{1} = \sqrt{3}, r_{2} = 2 \sin (θ)$

Exercise 71

$r_{1}^{2} = \sin θ, r_{2}^{2} = \cos θ$

Solution

$= \frac{π}{4}$ and $\sqrt{sin \frac{π}{4}} = \sqrt{\frac{\sqrt{2}}{2}} = \frac{\sqrt[4]{8}}{2}$

Exercise 72

$r_{1} = 1 + \cos θ$ , $r_{2} = 1 - \sin θ$

Archimedes’ spiral

a polar curve given by

r = θ .

When multiplied by a constant, the equation appears as

r = a θ .

r = θ,

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

r = a \pm b \cos θ

and

r = a \pm b \sin θ,

where

\frac{a}{b} = 1

convex limaҫon

a type of one-loop limaçon represented by

r = a \pm b \cos θ

and

r = a \pm b \sin θ

such that

\frac{a}{b} \geq 2

dimpled limaҫon

a type of one-loop limaçon represented by

r = a \pm b \cos θ

and

r = a \pm b \sin θ

such that

1 < \frac{a}{b} < 2

inner-loop limaçon

a polar curve similar to the cardioid, but with an inner loop; passes through the pole twice; represented by

r = a \pm b \cos θ

and

r = a \pm b \sin θ

where

a < b

lemniscate

a polar curve resembling a figure 8 and given by the equation

r^{2} = a^{2} \cos 2 θ

and

r^{2} = a^{2} \sin 2 θ,

a \neq 0

one-loop limaҫon

a polar curve represented by

r = a \pm b \cos θ

and

r = a \pm b \sin θ

such that

a > 0, b > 0,

and

\frac{a}{b} > 1;

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

r = a \cos n θ

and

r = a \sin n θ;

when

n

is even there are

2 n

petals, and the curve is highly symmetrical; when

n

is odd there are

n

petals.