10.4 Polar Coordinates: Graphs (Interactive)

NoteLearning Objectives

In this section, you will:

• Test polar equations for symmetry.
• Graph polar equations by plotting points.

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 \((r,\theta).\) We interpret \(r\) as the distance from the center of the sun and \(\theta\) 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.

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

10.4.1 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 \(\theta\) on a polar grid. Recall that the coordinate pair \((r,\theta)\) indicates that we move counterclockwise from the polar axis (positive \(x\)-axis) by an angle of \(\theta,\) and extend a ray from the pole (origin) \(r\) units in the direction of \(\theta.\) 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 \(\left. r \right)\) to determine the graph of a polar equation.

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

\[\begin{array}{ll} {r = 2\sin\;\theta} & \\ {-r = 2\sin(-\theta)\begin{array}{llll} & & & \end{array}} & {\text{Replace}\;(r,\theta)\;\text{with~}(-r,-\theta).} \\ {-r = -2\sin\;\theta} & {\text{Identity:~}\sin(-\theta) = -\sin\;\theta.} \\ {r = 2\sin\;\theta} & {\text{Multiply~both~sides~by}-1.} \end{array}\]

This equation exhibits symmetry with respect to the line \(\theta = \frac{\pi}{2}.\)

In the second test, we consider symmetry with respect to the polar axis ( \(x\)-axis). We replace \((r,\theta)\) with \(\left( {r, - \theta} \right)\) or \(\left( {- r,\pi - \theta} \right)\) to determine equivalency between the tested equation and the original. For example, suppose we are given the equation \(r = 1 - 2\cos\;\theta.\)

\[\begin{array}{ll} {r = 1 - 2\cos\;\theta} & \\ {r = 1 - 2\cos(-\theta)\begin{array}{llll} & & & \end{array}} & {\text{Replace~}(r,\theta)\;\text{with}\;(r,-\theta).} \\ {r = 1 - 2\cos\;\theta} & \text{Even/Odd~identity} \end{array}\]

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 \((r,\theta)\) with \(\left( {- r,\theta} \right)\) to determine if the tested equation is equivalent to the original equation. For example, suppose we are given the equation \(r = 2\sin(3\theta\operatorname{).}\)

\[\begin{matrix} {r = 2\sin(3\theta)} \\ {- r = 2\sin(3\theta)} \end{matrix}\]

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 \(\theta = \frac{\pi}{2},\) 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.

NoteSymmetry 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.

Drag the point (𝑟,𝜃) in any graph to explore the three types of symmetry for polar graphs:

0,0

A

(𝑟,𝜃)

(−𝑟,−𝜃)

(a) 𝜃 =𝜋/2 symmetry

0,0

A

(𝑟,𝜃)

(𝑟,−𝜃)

(b) Polar axis symmetry

0,0

A

(𝑟,𝜃)

(−𝑟,𝜃)

pole

(c) Pole symmetry

𝑟 =2.5, 𝜃 =0.64

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

ImportantHow To

Given a polar equation, test for symmetry.

1. Substitute the appropriate combination of components for \(\left( {r,\theta} \right):\) \(\left( {- r, - \theta} \right)\) for \(\theta = \frac{\pi}{2}\) symmetry; \(\left( {r, - \theta} \right)\) for polar axis symmetry; and \(\left( {- r,\theta} \right)\) for symmetry with respect to the pole.
2. If the resulting equations are equivalent in one or more of the tests, the graph produces the expected symmetry.

TipExample 1 — Testing a Polar Equation for Symmetry

Test the equation \(r = 2\sin\theta\) for symmetry.

Solution (click to reveal)

Symmetry Test 1: Symmetric with respect to \(\theta = \frac{\pi}{2}\) (y-axis).

Replace \((r, \theta)\) with \((-r, -\theta)\):

\[\begin{aligned} -r &= 2\sin(-\theta) \\ -r &= -2\sin\theta \\ r &= 2\sin\theta \quad \checkmark \end{aligned}\]

This matches the original equation, so the graph is symmetric with respect to the line \(\theta = \frac{\pi}{2}\).

Symmetry Test 2: Symmetric with respect to the polar axis (x-axis).

Replace \(\theta\) with \(-\theta\):

\[\begin{aligned} r &= 2\sin(-\theta) \\ r &= -2\sin\theta \end{aligned}\]

This does not match the original equation \(r = 2\sin\theta\). The graph is not symmetric with respect to the polar axis.

Symmetry Test 3: Symmetric with respect to the pole (origin).

Replace \(r\) with \(-r\):

\[\begin{aligned} -r &= 2\sin\theta \\ r &= -2\sin\theta \end{aligned}\]

This does not match the original equation \(r = 2\sin\theta\). The graph is not symmetric with respect to the pole.

The graph of \(r = 2\sin\theta\) is symmetric only with respect to \(\theta = \frac{\pi}{2}\). Use the reflection buttons on Figure 3 to visualize each symmetry test.

0,0

A

(2,𝜋/2)

(0,0)

𝑟 =2⁢sin⁡(𝜃) 

Figure 3

WarningTry It #1

Test the equation for symmetry: \(r = -4\cos\theta\).

Solution (click to reveal)

Symmetry Test 1: Symmetric with respect to \(\theta = \frac{\pi}{2}\) (y-axis).

Replace \((r, \theta)\) with \((-r, -\theta)\):

\[\begin{aligned} -r &= -4\cos(-\theta) \\ -r &= -4\cos\theta \\ r &= 4\cos\theta \end{aligned}\]

This does not match \(r = -4\cos\theta\). The graph is not symmetric with respect to the line \(\theta = \frac{\pi}{2}\).

Symmetry Test 2: Symmetric with respect to the polar axis (x-axis).

Replace \(\theta\) with \(-\theta\):

\[\begin{aligned} r &= -4\cos(-\theta) \\ r &= -4\cos\theta \quad \checkmark \end{aligned}\]

This matches the original equation, so the graph is symmetric with respect to the polar axis.

Symmetry Test 3: Symmetric with respect to the pole (origin).

Replace \(r\) with \(-r\):

\[\begin{aligned} -r &= -4\cos\theta \\ r &= 4\cos\theta \end{aligned}\]

This does not match \(r = -4\cos\theta\). The graph is not symmetric with respect to the pole.

The equation \(r = -4\cos\theta\) is symmetric only with respect to the polar axis. The graph is a circle centered at \((-2, 0)\) with radius \(2\).

Graph of \(r = -4\cos\theta\)

10.4.2 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 \(\theta\) and \(r\) values. We enter values of \(\theta\) into a polar equation and calculate \(r.\) However, using the properties of symmetry and finding key values of \(\theta\) 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 \(\theta\) 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 \(\theta.\)

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

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

TipExample 2 — Finding Zeros and Maximum Values for a Polar Equation

Find the zeros and maximum |𝑟| and, if necessary, the polar axis intercepts of 𝑟 =2⁢sin⁡(𝜃).

Solution (click to open)

WarningTry It #2

Without converting to Cartesian coordinates, test the given equation for symmetry and find the zeros and maximum values of \(|r|\): \(r = 3\cos\theta\).

Solution (click to reveal)

Tests will reveal symmetry about the polar axis. Replacing \(\theta\) with \(-\theta\):

\[\begin{aligned} r &= 3\cos(-\theta) \\ r &= 3\cos\theta \quad \checkmark \end{aligned}\]

The equation is unchanged, so the graph is symmetric with respect to the polar axis.

The zero is at \(\left(0, \frac{\pi}{2}\right)\), since \(\cos\!\left(\frac{\pi}{2}\right) = 0\).

The maximum value of \(|r|\) is \(3\), occurring at \(\theta = 0\) where \(\cos\theta = 1\). The maximum point is \((3, 0)\).

Graph of \(r = 3\cos\theta\)

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.

NoteFormulas for the Equation of a Circle

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

The four circle forms in polar coordinates (radius \(= |a|/2 = 1.5\)):

\(r = 3\cos\theta\), center \((1.5, 0)\)

\(r = -3\cos\theta\), center \((-1.5, 0)\)

\(r = 3\sin\theta\), center \((0, 1.5)\)

\(r = -3\sin\theta\), center \((0, -1.5)\)

Figure 5

TipExample 3 — Sketching the Graph of a Polar Equation for a Circle

Sketch the graph of 𝑟 =4⁢cos⁡(𝜃).

Solution (click to open)

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.

NoteFormulas for a Cardioid

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

The four cardioid forms (\(a/b = 1\)):

\(r = 2 + 2\cos\theta\)

\(r = 2 - 2\cos\theta\)

\(r = 2 + 2\sin\theta\)

\(r = 2 - 2\sin\theta\)

Figure 7

ImportantHow To

Given the polar equation of a cardioid, sketch its graph.

1. Check equation for the three types of symmetry.
2. Find the zeros. Set \(r = 0.\)
3. Find the maximum value of the equation according to the maximum value of the trigonometric expression.
4. Make a table of values for \(r\) and \(\theta.\)
5. Plot the points and sketch the graph.

TipExample 4 — Sketching the Graph of a Cardioid

Sketch the graph of 𝑟 =2 +2⁢cos⁡(𝜃).

Solution (click to open)

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 \(\frac{a}{b} \geq 2.\)

NoteFormulas for One-Loop Limaçons

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

Dimpled (one-loop) limaçons: \(a \gt 0, b \gt 0, 1 \lt a/b \lt 2\).

\(r = 4 + 3\cos\theta\)

\(r = 4 - 3\cos\theta\)

\(r = 4 + 3\sin\theta\)

\(r = 4 - 3\sin\theta\)

Figure 9 Dimpled limaçons

ImportantHow To

Given a polar equation for a one-loop limaçon, sketch the graph.

1. Test the equation for symmetry. Remember that failing a symmetry test does not mean that the shape will not exhibit symmetry. Often the symmetry may reveal itself when the points are plotted.
2. Find the zeros.
3. Find the maximum values according to the trigonometric expression.
4. Make a table.
5. Plot the points and sketch the graph.

TipExample 5 — Sketching the Graph of a One-Loop Limaçon

Sketch the graph of 𝑟 =4 −3⁢sin⁡(𝜃).

Solution (click to open)

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 \(\sin\;\theta\) is likely symmetric with respect to the line \(\theta = \frac{\pi}{2},\) evaluating more points helps to verify that the graph is correct.

WarningTry It #3

Sketch the graph of \(r = 3 - 2\cos\theta\).

Solution (click to reveal)

The equation \(r = 3 - 2\cos\theta\) is a limaçon. Since \(\frac{3}{2} = 1.50\), which is between \(1\) and \(2\), this is a dimpled limaçon (no inner loop).

The maximum \(r\) occurs when \(\cos\theta = -1\) (at \(\theta = \pi\)): \(r = 3 + 2 = 5\). The minimum \(r\) occurs when \(\cos\theta = 1\) (at \(\theta = 0\)): \(r = 3 - 2 = 1\).

Graph of \(r = 3 - 2\cos\theta\)

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.

NoteFormulas for Inner-Loop Limaçons

The formulas that generate the inner-loop limaçons are given by \(r = a \pm b\cos\;\theta\) and \(r = a \pm b\sin\;\theta\) where \(a > 0,\) \(b > 0,\) and \(a < b.\) 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.

Inner-loop limaçons: \(a \gt 0, b \gt 0, a \lt b\).

\(r = 2 + 5\cos\theta\)

\(r = 2 - 5\cos\theta\)

\(r = 2 + 5\sin\theta\)

\(r = 2 - 5\sin\theta\)

Figure 11

TipExample 6 — Sketching the Graph of an Inner-Loop Limaçon

Sketch the graph of 𝑟 =2 +5⁢cos⁡(𝜃).

Solution (click to open)

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.

NoteFormulas for Lemniscates

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

Lemniscate forms (\(a \neq 0\)):

\(r^2 = 4\cos 2\theta\)

\(r^2 = -4\cos 2\theta\)

\(r^2 = 4\sin 2\theta\)

\(r^2 = -4\sin 2\theta\)

Figure 13

TipExample 7 — Sketching the Graph of a Lemniscate

Sketch the graph of 𝑟2 =4⁢cos⁡2⁢𝜃.

Solution (click to open)

Making a substitution such as \(u = 2\theta\) 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 \(\theta.\) In other words, the corresponding \(r\)-values of \(\sqrt{4\cos(2\theta)}\) 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.

NoteRose Curves

The formulas that generate the graph of a rose curve are given by \(r = a\cos\; n\theta\) and \(r = a\sin\; n\theta\) where \(a \neq 0.\) If \(n\) is even, the curve has \(2n\) petals. If \(n\) is odd, the curve has \(n\) petals. See Figure 15.

Rose curves: if \(n\) is even the curve has \(2n\) petals; if \(n\) is odd it has \(n\) petals.

\(r = 2\cos(4\theta)\) — \(n\) even, \(2n = 8\) petals

\(r = 2\sin(3\theta)\) — \(n\) odd, \(n = 3\) petals

Figure 15

TipExample 8 — Sketching the Graph of a Rose Curve (\(n\) Even)

Sketch the graph of 𝑟 =2⁢cos⁡(4⁢𝜃).

Amplitude 𝑎 =2 = 2

Frequency 𝑛 =4 = 4

Solution (click to open)

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.

WarningTry It #4

Sketch the graph of \(r = 4\sin(2\theta)\).

Solution (click to reveal)

The graph is a rose curve with \(n = 2\) (even), so it has \(4\) petals and a maximum \(|r| = 4\). The equation \(r = 4\sin(2\theta)\) is symmetric with respect to the line \(\theta = \frac{\pi}{2}\) and the pole.

Graph of \(r = 4\sin(2\theta)\)

TipExample 9 — Sketching the Graph of a Rose Curve (\(n\) Odd)

Sketch the graph of 𝑟 =2⁢sin⁡(5⁢𝜃).

Amplitude 𝑎 =2 = 2

Frequency 𝑛 =5 = 5

Solution (click to open)

WarningTry It #5

Sketch the graph of \(r = 3\cos(3\theta)\).

Solution (click to reveal)

The graph is a rose curve with \(n = 3\) (odd), so it has \(3\) petals and a maximum \(|r| = 3\). The equation \(r = 3\cos(3\theta)\) is symmetric with respect to the polar axis.

Graph of \(r = 3\cos(3\theta)\)

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.

NoteArchimedes’ Spiral

The formula that generates the graph of the Archimedes’ spiral is given by \(r = \theta\) for \(\theta \geq 0.\) As \(\theta\) increases, \(r\) increases at a constant rate in an ever-widening, never-ending, spiraling path. See Figure 18.

Archimedes' spirals: \(r = \theta\) for \(\theta \geq 0\).

\(r = \theta\), \([0, 2\pi]\)

\(r = \theta\), \([0, 4\pi]\)

Figure 18

ImportantHow To

Given an Archimedes’ spiral over \(\left\lbrack {0,2\pi} \right\rbrack,\) sketch the graph.

1. Make a table of values for \(r\) and \(\theta\) over the given domain.
2. Plot the points and sketch the graph.

TipExample 10 — Sketching the Graph of an Archimedes’ Spiral

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

Max angle =2⁢𝜋 = 2

Solution (click to open)

The domain of this polar curve is \(\left\lbrack {0,2\pi} \right\rbrack.\) In general, however, the domain of this function is \(\left( {- \infty,\infty} \right).\) Graphing the equation of the Archimedes’ spiral is rather simple, although the image makes it seem like it would be complex.

WarningTry It #6

Sketch the graph of \(r = -\theta\) over \([0,\; 4\pi]\).

Solution (click to reveal)

The equation \(r = -\theta\) is an Archimedes’ spiral reflected through the pole. The negative sign means the curve spirals in the opposite direction compared to \(r = \theta\). The maximum \(|r|\) over \([0,\; 4\pi]\) is \(4\pi \approx 12.57\).

Graph of \(r = -\theta\)

10.4.3 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.

Figure 20

Figure 21

NoteMedia

Access these online resources for additional instruction and practice with graphs of polar coordinates.

• Graphing Polar Equations Part 1
• Graphing Polar Equations Part 2
• Animation: The Graphs of Polar Equations
• Graphing Polar Equations on the TI-84

Section Exercises

Verbal

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

Solution (click to reveal)

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 \(\theta = \frac{\pi}{2}\) is similar to symmetry about the \(y\)-axis.

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?

3. What are the steps to follow when graphing polar equations?

Solution (click to reveal)

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 \(\theta = 0,\;\frac{\pi}{2},\) \(\pi\) and  \(\frac{3\pi}{2},\) and sketch the graph.

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

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

Solution (click to reveal)

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.

6. \(r = 5\cos\; 3\theta\)

7. \(r = 3 - 3\cos\;\theta\)

Solution (click to reveal)

symmetric with respect to the polar axis

8. \(r = 3 + 2\sin\;\theta\)

9. \(r = 3\sin\; 2\theta\)

Solution (click to reveal)

symmetric with respect to the polar axis, symmetric with respect to the line \(\theta = \frac{\pi}{2},\) symmetric with respect to the pole

10. \(r = 4\)

11. \(r = 2\theta\)

Solution (click to reveal)

symmetric with respect to the line \(\theta = \frac{\pi}{2}\)

12. \(r = 4\cos\;\frac{\theta}{2}\)

13. \(r = \frac{2}{\theta}\)

Solution (click to reveal)

Symmetric with respect to line \(\theta = \frac{\pi}{2}\) (y-axis)

14. \(r = 3\sqrt{1 - \cos^{2}\theta}\)

15. \(r = \sqrt{5\sin\; 2\theta}\)

Solution (click to reveal)

symmetric with respect to the pole

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

16. \(r = 3\cos\;\theta\)

17. \(r = 4\sin\;\theta\)

Solution (click to reveal)

circle

18. \(r = 2 + 2\cos\;\theta\)

19. \(r = 2 - 2\cos\;\theta\)

Solution (click to reveal)

cardioid

20. \(r = 5 - 5\sin\;\theta\)

21. \(r = 3 + 3\sin\;\theta\)

Solution (click to reveal)

cardioid

22. \(r = 3 + 2\sin\;\theta\)

23. \(r = 7 + 4\sin\;\theta\)

Solution (click to reveal)

one-loop/dimpled limaçon

24. \(r = 4 + 3\cos\;\theta\)

25. \(r = 5 + 4\cos\;\theta\)

Solution (click to reveal)

one-loop/dimpled limaçon

26. \(r = 10 + 9\cos\;\theta\)

27. \(r = 1 + 3\sin\;\theta\)

Solution (click to reveal)

inner loop/two-loop limaçon

28. \(r = 2 + 5\sin\;\theta\)

29. \(r = 5 + 7\sin\;\theta\)

Solution (click to reveal)

inner loop/two-loop limaçon

30. \(r = 2 + 4\cos\;\theta\)

31. \(r = 5 + 6\cos\;\theta\)

Solution (click to reveal)

inner loop/two-loop limaçon

32. \(r^{2} = 36\cos\left( {2\theta} \right)\)

33. \(r^{2} = 10\cos\left( {2\theta} \right)\)

Solution (click to reveal)

lemniscate

34. \(r^{2} = 4\sin\left( {2\theta} \right)\)

35. \(r^{2} = 10\sin\left( {2\theta} \right)\)

Solution (click to reveal)

lemniscate

36. \(r = 3\text{sin}(2\theta)\)

37. \(r = 3\text{cos}(2\theta)\)

Solution (click to reveal)

rose curve

38. \(r = 5\text{sin}(3\theta)\)

39. \(r = 4\text{sin}(4\theta)\)

Solution (click to reveal)

rose curve

40. \(r = 4\text{sin}(5\theta)\)

41. \(r = -\theta\)

Solution (click to reveal)

Archimedes’ spiral

42. \(r = 2\theta\)

43. \(r = - 3\theta\)

Solution (click to reveal)

Archimedes’ spiral

Technology

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

44. \(r = \frac{1}{\theta}\)

45. \(r = \frac{1}{\sqrt{\theta}}\)

Solution (click to reveal)

46. \(r = 2\sin\;\theta\tan\;\theta,\) a cissoid

47. \(r = 2\sqrt{1 - \sin^{2}\theta}\), a hippopede

Solution (click to reveal)

48. \(r = 5 + \cos\left( {4\theta} \right)\)

49. \(r = 2 - \sin\left( {2\theta} \right)\)

Solution (click to reveal)

50. \(r = \theta^{2}\)

51. \(r = \theta + 1\)

Solution (click to reveal)

52. \(r = \theta\sin\;\theta\)

53. \(r = \theta\cos\;\theta\)

Solution (click to reveal)

For the following exercises, use a graphing utility to graph each pair of polar equations on a domain of \(\left\lbrack {0,4\pi} \right\rbrack\) and then explain the differences shown in the graphs.

54. \(r = \theta,r = - \theta\)

55. \(r = \theta,r = \theta + \sin\;\theta\)

Solution (click to reveal)

They are both spirals, but not quite the same.

56. \(r = \sin\;\theta + \theta,r = \sin\;\theta - \theta\)

57. \(r = 2\sin\left( \frac{\theta}{2} \right),r = \theta\sin\left( \frac{\theta}{2} \right)\)

Solution (click to reveal)

Both graphs are curves with 2 loops. The equation with a coefficient of \(\theta\) has two loops on the left, the equation with a coefficient of 2 has two loops side by side. Graph these from 0 to \(4\pi\) to get a better picture.

58. \(r = \sin\left( {\cos(3\theta)} \right)\; r = \sin(3\theta)\)

59. On a graphing utility, graph \(r = \sin\left( {\frac{16}{5}\theta} \right)\) on \(\left\lbrack 0 \right.\), \(\left. 4\pi \right\rbrack\), \(\left\lbrack 0 \right.\), \(\left. 8\pi \right\rbrack\), \(\left\lbrack 0 \right.\), \(\left. 12\pi \right\rbrack\), and \(\left\lbrack 0 \right.\), \(\left. 16\pi \right\rbrack.\) Describe the effect of increasing the width of the domain.

Solution (click to reveal)

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

60. On a graphing utility, graph and sketch \(r = \sin\;\theta + \left( {\sin\left( {\frac{5}{2}\theta} \right)} \right)^{3}\) on \(\left\lbrack {0,4\pi} \right\rbrack.\)

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

\[\begin{array}{l} {r_{1} = 3\sin(3\theta)} \\ {r_{2} = 2\sin(3\theta)} \\ {r_{3} = \sin(3\theta)} \end{array}\]

Solution (click to reveal)

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

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

\[\begin{array}{l} {r_{1} = 3 + 3\cos\;\theta} \\ {r_{2} = 2 + 2\cos\;\theta} \\ {r_{3} = 1 + \cos\;\theta} \end{array}\]

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

\[\begin{array}{l} {r_{1} = 3\theta} \\ {r_{2} = 2\theta} \\ {r_{3} = \theta} \end{array}\]

Solution (click to reveal)

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.

64. \(r_{1} = 3 + 2\sin\;\theta,\; r_{2} = 2\)

65. \(r_{1} = 6 - 4\cos\;\theta,\; r_{2} = 4\)

Solution (click to reveal)

\(\left( {4,\frac{\pi}{3}} \right),\left( {4,\frac{5\pi}{3}} \right)\)

66. \(r_{1} = 1 + \sin\;\theta,\; r_{2} = 3\sin\;\theta\)

67. \(r_{1} = 1 + \cos\;\theta,\; r_{2} = 3\cos\;\theta\)

Solution (click to reveal)

\(\left( {\frac{3}{2},\frac{\pi}{3}} \right),\left( {\frac{3}{2},\frac{5\pi}{3}} \right)\)

68. \(r_{1} = \cos\left( {2\theta} \right),\; r_{2} = \sin\left( {2\theta} \right)\)

69. \(r_{1} = \sin^{2}\left( {2\theta} \right)\), \(r_{2} = 1 - \cos\left( {4\theta} \right)\)

Solution (click to reveal)

\(\left( {0,\frac{\pi}{2}} \right),\;\left( {0,\pi} \right),\;\left( {0,\frac{3\pi}{2}} \right),\;\left( {0,2\pi} \right)\)

70. \(r_{1} = \sqrt{3},\; r_{2} = 2\sin(\theta)\)

71. \(r_{1}{}^{2} = \sin\;\theta,r_{2}{}^{2} = \cos\;\theta\)

Solution (click to reveal)

\(= \frac{\pi}{4}\) and \(\sqrt{\text{sin}\frac{\pi}{4}} = \sqrt{\frac{\sqrt{2}}{2}} = \frac{\sqrt[4]{8}}{2}\)

72. \(r_{1} = 1 + \cos\;\theta\), \(r_{2} = 1 - \sin\;\theta\)