10.2 Polar Graphs

Graphing in Polar Coordinates

When we plot points in Cartesian coordinates, we start at the origin and move a distance right or left given by the $x$ -coordinate of the point, then move up or down according to the $y$ -coordinate. When we sketch the graph of an equation or function, we think of drawing the graph from left to right, with the "height" of the graph at each $x$ -value given by the function, as shown in figure (a).

In polar coordinates, however, the dependent variable, $r$ , gives not a height but a distance from the pole in direction $θ$ , as shown in figure (b). When graphing an equation in polar coordinates, we think of sweeping around the pole in the counterclockwise direction, and at each angle $θ$ the $r$ -value tells us how far the graph is from the pole.

Example 1

Graph the polar equation $r equals 2 sin ⁡ open parenthesis θ close parenthesis$ .

We make a table of values, choosing the special values for $θ$ . For each value of $θ$ , we evaluate $r equals 2 sin ⁡ open parenthesis θ close parenthesis$ .

$θ$	$0$	$\frac{π}{6}$	$\frac{π}{4}$	$\frac{π}{3}$	$\frac{π}{2}$	$\frac{2 π}{3}$	$\frac{3 π}{4}$	$\frac{5 π}{6}$	$π$
$r$	$0$	$1$	$the square root of 2$	$the square root of 3$	$2$	$the square root of 3$	$the square root of 2$	$1$	$0$

First we'll plot the points in the first quadrant. Observe that as $θ$ increases from $0$ to $\frac{π}{2}$ , $r$ increases from $0$ to $2$ . Starting at the pole, we connect the points in order of increasing $θ$ . Imagine a radial line sweeping around the graph through the first quadrant: as the angle increases, the length of the segment increases, so that its tip traces out the graph shown in figure (a).

Now continue plotting the points in the table as $θ$ increases from $\frac{π}{2}$ to $π$ . In the second quadrant, $r$ decreases as $θ$ increases, as shown in figure (b). The graph we obtain is, in fact, a circle, which we will prove algebraically shortly. However, we have not yet plotted points for $θ$ between $0$ and $2 π$ . Because $\sin (θ)$ is negative in the third and fourth quadrants, all the $r$ -values for these angles are negative.

$θ$	$π$	$\frac{7 π}{6}$	$\frac{5 π}{4}$	$\frac{4 π}{3}$	$\frac{3 π}{2}$	$\frac{5 π}{3}$	$\frac{7 π}{4}$	$\frac{11 π}{6}$	$π$
$r$	$0$	$minus 1$	$- \sqrt{2}$	$- \sqrt{3}$	$minus 2$	$- \sqrt{3}$	$- \sqrt{2}$	$minus 1$	$0$

When we plot the points in this table, we see that the original graph is traced out again. For example, the point $(\frac{7 π}{6}, - 1)$ is the same as the point $(\frac{π}{6}, 1)$ , the point $(\frac{5 π}{4}, - \sqrt{2})$ is the same as the point $(\frac{π}{4}, \sqrt{2})$ , and so on, around the circle. Thus, the graph of $r equals 2 sin ⁡ θ$ is a circle, traced twice for $0 less than or equal to θ less than or equal to 2 π$ .

To prove that the graph in the previous Example is really a circle, we convert the equation $r equals 2 sin ⁡ open parenthesis θ close parenthesis$ to Cartesian form. First, multiply both sides by $r$ to obtain

$r squared equals 2 r sin ⁡ open parenthesis θ close parenthesis$

Next, replace $r^{2}$ by $x^{2} + y^{2}$ and $r \sin (θ)$ by $y$ , to get

$x^{2} + y^{2} = 2 y$

This equation is quadratic in two variables, so its graph is a conic section. We put the equation in standard form by completing the square in $y$ .

$\begin{aligned} x^{2} + y^{2} - 2 y & = 0 & Add 1 to both sides. \\ x^{2} + (y^{2} - 2 y + 1) & = 1 & Write the equation in standard form. \\ (x - 0)^{2} + (y - 1)^{2} & = 1 \end{aligned}$

We have the equation of a circle with center $open parenthesis 0, 1 close parenthesis$ and radius $1$ .

Graph the polar equation $r equals 4 cos ⁡ open parenthesis θ close parenthesis$ .

Using a Graphing Calculator

You are familiar with the graphs of many equations in Cartesian coordinates, including lines, parabolas and other conic sections, and the graphs of basic functions. You should now become familiar with some standard graphs in polar coordinates. These include circles and roses, cardioids and limaçons, lemniscates, and spirals.

At the end of this section you will find a Catalog of the basic polar graphs and their properties. You can use your calculator, set in Polar mode, to experiment with these graphs.

Example 2

Graph the Archimedean spiral $r equals 2 θ$ , for $0 less than or equal to θ less than or equal to 6 π$ .

After setting the calculator in Polar mode, we enter the equation $r equals 2 θ$ and enter the window settings

$\begin{aligned} θ min & = 0 θ max = 6 π θ step = 0.1 \\ Xmin & = - 25 Xmax = 25 Xscl = 5 \\ Ymin & = - 25 Ymax = 25 Yscl = 5 \end{aligned}$

We then press Zoom 5 to set a square window. The graph is shown at right.

Studying a table of values can help us understand the shape of the graph.

$θ$	$0.25$	$0.50$	$0.75$	$1.00$	$1.25$
$r$	$0.50$	$1.00$	$1.50$	$2.00$	$2.50$

As $θ$ increases, $r$ increases also, at a constant rate. We wind our way around the pole, steadily increasing our distance from the pole as we go. We spiral outwards, tracing the graph shown above.

Use a calculator to graph the polar equation $r = e^{0.1 θ}$ . Use the window settings

$\begin{aligned} θ min & = 0 θ max = 12 π θ step = 0.5 \\ Xmin & = - 45 Xmax = 45 Xscl = 10 \\ Ymin & = - 35 Ymax = 35 Yscl = 10 \end{aligned}$

It is important to connect the points on the graph in order of increasing $θ$ .

Example 3

Graph the polar equation $r equals 2 sin ⁡ open parenthesis 3 θ close parenthesis$ .

We'll graph the equation in stages in order to see how the graph is traced out. Begin with the window settings

$\begin{aligned} θ min & = 0 θ max = \frac{π}{3} θ step = 0.5 \\ Xmin & = - 4.5 Xmax = 4.5 Xscl = 1 \\ Ymin & = - 3 Ymax = 3 Yscl = 1 \end{aligned}$

Watch as your calculator produces the graph shown in figure (a).

Observe that, as $θ$ increases from $0$ to $\frac{π}{3}$ , $r$ first increases, reaching its maximum value of 2 at $θ = \frac{π}{6}$ and then decreases back to 0. You can verify the values in the table below, which shows the points at multiples of $\frac{π}{18}$ . These points create the first loop of the graph.

$θ$	$0$	$\frac{π}{18}$	$\frac{π}{9}$	$\frac{π}{6}$	$\frac{2 π}{9}$	$\frac{5 π}{18}$	$\frac{π}{3}$
$3 θ$	$0$	$\frac{π}{6}$	$\frac{π}{3}$	$\frac{π}{2}$	$\frac{2 π}{3}$	$\frac{5 π}{6}$	$π$
$r$	$0$	$1$	$the square root of 3$	$2$	$the square root of 3$	$1$	$0$

Next, change $θ max$ to $\frac{2 π}{3}$ , and graph again. This time the calculator traces out two loops, as shown in figure (b). For $θ$ between $\frac{π}{3}$ and $\frac{2 π}{3}$ , $r$ is negative, so the second loop lies in the third and fourth quadrants.

$θ$	$\frac{π}{3}$	$\frac{7 π}{18}$	$\frac{4 π}{9}$	$\frac{π}{2}$	$\frac{5 π}{9}$	$\frac{11 π}{18}$	$\frac{2 π}{3}$
$3 θ$	$π$	$\frac{7 π}{6}$	$\frac{4 π}{3}$	$\frac{π}{2}$	$\frac{5 π}{3}$	$\frac{11 π}{6}$	$2 π$
$r$	$0$	$minus 1$	$- \sqrt{3}$	$minus 2$	$- \sqrt{3}$	$minus 1$	$0$

Finally, change $θ max$ to $π$ , and graph again. For $θ$ between $\frac{2 π}{3}$ and $π$ , the graph traces out a third loop, as shown in figure (c). For $θ$ between $π$ and $2 π$ , the entire graph is traced a second time. The finished graph is a rose with 3 petals of length 2.

Use your calculator to graph the polar equation $r equals 2 cos ⁡ open parenthesis 3 θ close parenthesis$ .

Complete the table of values for the function.

$θ$	$0$	$\frac{π}{6}$	$\frac{π}{3}$	$\frac{π}{2}$	$\frac{2 π}{3}$	$\frac{5 π}{6}$	$π$
$r$

Sketch the graph by hand on the grid at right.

Sketching Familiar Equations

You should also be able to sketch the standard polar graphs by hand. Once you recognize an equation as a particular type of graph, say a rose or a limaçon, you can sketch the graph quickly by finding just a few well-chosen guidepoints. The next example demonstrates a technique for sketching a rose.

Function graph showing the polar curve r = a*cos(n*theta). Adjustable parameters: Petal length a (a) = 2, Petal number n (n) = 4. Viewing window: x from -5 to 5, y from -3.09 to 3.09. — The rose family r = a cos(nθ), live. The slider a sets the petal length — each petal reaches out to distance a from the pole. The slider n sets the petal count, and it follows the section's rule exactly: an odd n gives n petals, an even n gives 2n. Step n from 3 to 4 and watch the count jump from three petals to eight — with n odd the curve traces each petal twice as θ runs from 0 to 2π, so half the petals coincide, while with n even all 2n petals are distinct. At n = 1 the “rose” has a single petal: a circle of diameter a through the pole, the standard circle graph from earlier in this section.

Example 4

Graph the polar equation $r equals 3 sin ⁡ open parenthesis 2 θ close parenthesis$ .

Referring to the Catalog of Polar Graphs, we see that the graph of this equation is a rose, with petal length $a equals 3$ and four petals, because $2 n equals 4$ . If we can locate the tips of the petals, we can use them as guidepoints to sketch the graph.

Now, the points at the tips of the petals have $r equals 3$ , so we substitute $r equals 3$ into the equation of the rose to find the values of $θ$ at those points.

$\begin{aligned} 3 & = 3 \sin (2 θ) & Divide both sides by 3. \\ 1 & = \sin (2 θ) & Apply the arcsine function to both sides. \\ \frac{π}{2} & = 2 θ & Divide both sides by 2. \\ \frac{π}{4} & = θ \end{aligned}$

Thus, one of the petal tips is located at $θ = \frac{π}{4}$ . Because the 4 petals are evenly spaced around the pole, the angle between the petals is $\frac{2 π}{4} = \frac{π}{2}$ , and the other petal tips occur at

$\begin{aligned} θ & = \frac{π}{4} + \frac{π}{2} = \frac{3 π}{4} \\ θ & = \frac{π}{4} + 2 (\frac{π}{2}) = \frac{5 π}{4} \\ θ & = \frac{π}{4} + 3 (\frac{π}{2}) = \frac{7 π}{4} \end{aligned}$

We plot the tips of the petals as guidepoints, at $(3, \frac{π}{4})$ , $(3, \frac{3 π}{4})$ , $(3, \frac{5 π}{4})$ and $(3, \frac{7 π}{4})$ . Now we can sketch a rose with 4 petals of length 3, as shown at right.

Graph the polar equation $r equals 4 cos ⁡ open parenthesis 5 θ close parenthesis$ .

The limaçons, $r = a \pm \sin (θ)$ and $r = a \pm \cos (θ)$ , are another family of polar graphs. In particular, the cardioid is a special case of a limaçon with $a = b$ .

Example 5

Graph the polar equation $r = 3 + 2 \cos (θ)$ .

This is the equation for a limaçon, with $a equals 3$ and $b equals 2$ . Because $a > b$ , the limaçon will have a dent, like a lima bean, rather than a loop. As guidepoints, we locate the points at the four quadrantal angles.

$θ$	$r = 3 + 2 \cos (θ)$
$0$	$5$
$\frac{π}{2}$	$3$
$π$	$1$
$\frac{3 π}{2}$	$3$

We plot the guidepoints and connect them in order with a smooth curve, as shown above. Note that this limaçon involving cosine is symmetric about the $x$ -axis; limaçons involving sine are symmetric about the $y$ -axis.

Graph the polar equation $r = 1 - 3 \sin (θ)$ .

You should also be able to identify a polar graph and write its equation.

Example 6

Give a polar equation for each of the graphs below.

The graph is a circle centered at $open parenthesis minus 2, 0 close parenthesis$ and with radius 2, so we choose the equation $r equals 2 a cos ⁡ open parenthesis θ close parenthesis$ , with $a = - 2$ . Thus, $r = - 4 \cos (θ)$ .
The graph is a cardioid with its axis of symmetry on the $x$ -axis, and the "bottom" of the heart points in the positive $x$ direction, so its equation has the form $r = a + a \cos (θ)$ . At $θ = 0, r = 2$ , so we can solve for $a$ :
$\begin{aligned} 2 & = a + a \cos (0) \\ 2 & = a + a (1) = 2 a \end{aligned}$
Thus, $a equals 1$ and we choose the equation $r = 1 + \cos (θ)$ .

Give a polar equation for each of the graphs below.

$r squared equals 4 sin ⁡ open parenthesis 2 θ close parenthesis$
$r = 1 - \sin (θ)$

Finding Intersection Points

To find the intersection points of two graphs, we solve the system made up of their equations. If the equations are $y = f (x)$ and $y = g (x)$ , we simply solve the equation $f (x) = g (x)$ . For example, we find the intersection points of $y = x^{2}$ and $y = x + 2$ by solving the equation $x^{2} - x - 2 = 0$ to get

$\begin{aligned} x^{2} - x - 2 & = 0 \\ (x - 2) (x + 1) & = 0 \\ x - 2 = 0, x + 1 & = 0 \\ x = 2, x & = - 1 \end{aligned}$

These are the $x$ -coordinates of the intersection points, and we can find the $y$ -coordinates by substituting these values into either equation.

For $x equals 2,$ we find $y = 2^{2} = 4.$
For $x = - 1,$ we find $y = (- 1)^{2} = 1.$

Thus, the intersection points are $open parenthesis 2, 4 close parenthesis$ and $open parenthesis minus 1, 1 close parenthesis$ , as shown at right.

To find the intersection points of the polar graphs $r = f (θ)$ and $r = g (θ)$ we solve the equation $f (θ) = g (θ)$ .

Example 7

Find all intersection points of the graphs of $r = 2 + 2 \sin (θ)$ and $r = 2 + 2 \cos (θ)$ , for $0 less than or equal to θ less than or equal to 2 π$ .

We equate the two expressions for $r$ and solve for $θ$ .

$\begin{aligned} r = 2 + 2 \sin (θ) & = r = 2 + 2 \cos (θ) \\ \sin (θ) & = \cos (θ) & Divide both sides by \cos (θ) . \\ \tan (θ) & = 1 & Solve for θ . \\ θ & = \frac{π}{4}, \frac{5 π}{4} \end{aligned}$

We evaluate either expression for $r$ to find the other coordinate of the intersection point.

$\begin{matrix} when θ = \frac{π}{4}, r = 2 + 2 \sin (\frac{π}{4}) = 2 + (\frac{1}{\sqrt{2}}) = 2 + \sqrt{2} \\ when θ = \frac{5 π}{4}, r = 2 + 2 \sin (\frac{5 π}{4}) = 2 + (\frac{- 1}{\sqrt{2}}) = 2 - \sqrt{2} \end{matrix}$

Thus, two of the intersection points are

$(2 + \sqrt{2}, \frac{π}{4}) and (2 - \sqrt{2}, \frac{5 π}{4})$

as shown at right.

However, you can see in the figure that the graphs also appear to intersect at the pole. To verify that the pole indeed lies on both graphs, we can solve for $θ$ in each equation when $r equals 0$ .

$\begin{aligned} 0 & = 2 + 2 \sin (θ) & 0 & = 2 + 2 \cos (θ) \\ - 1 & = \sin (θ) & - 1 & = \cos θ \\ θ & = \frac{3 π}{2} & θ & = π \end{aligned}$

Both points, $(0, \frac{3 π}{2})$ and $open parenthesis 0, π close parenthesis$ , represent the pole. Thus, the graphs intersect at three points: $(2 + \sqrt{2}, \frac{π}{4})$ , $(2 - \sqrt{2}, \frac{5 π}{4})$ , and the pole.

Find all intersection points of the graphs of $r equals 1$ and $r equals 2 cos ⁡ open parenthesis θ close parenthesis$ , for $0 less than or equal to θ less than or equal to 2 π$ .

$(1, \frac{π}{3})$ , $(1, \frac{5 π}{3})$

Review the following skills you will need for this section.

Algebra Refresher

Solve each equation. Give two solutions, $0 less than or equal to θ less than or equal to 2 π$ .

$\cos (θ) = \frac{\sqrt{3}}{2}$
$\sin (θ) = \frac{1}{\sqrt{2}}$
$\tan (θ) = \sqrt{3}$
$\tan (θ) = \frac{1}{\sqrt{3}}$
$\sin (θ) = \frac{- 1}{\sqrt{2}}$
$\cos (θ) = \frac{- 1}{2}$
$\tan (θ) = - 1$
$cos ⁡ open parenthesis θ close parenthesis equals 0$

$\underset{―}{}$

Skills Refresher Answers

$\frac{π}{6}$ , $\frac{11 π}{6}$
$\frac{π}{4}$ , $\frac{3 π}{4}$
$\frac{π}{3}$ , $\frac{4 π}{3}$
$\frac{π}{6}$ , $\frac{7 π}{6}$
$\frac{5 π}{4}$ , $\frac{7 π}{4}$
$\frac{2 π}{3}$ , $\frac{4 π}{3}$
$\frac{3 π}{4}$ , $\frac{7 π}{4}$
$\frac{π}{2}$ , $\frac{3 π}{2}$

A Catalog of Polar Curves

The Coordinate Curves

$θ = k$ ( $k$ a constant)
A line through the pole.
$k$ gives the angle of inclination of the line (in radians).
$r = k$ ( $k$ a constant)
A circle centered at the pole.
$k$ is the radius of the circle.

Circles

$r equals 2 a sin ⁡ open parenthesis θ close parenthesis$
A circle with center $open parenthesis 0, a close parenthesis$ on the $y$ -axis, and radius $| a |$ .
$r equals 2 a cos ⁡ open parenthesis θ close parenthesis$
A circle with center $open parenthesis a, 0 close parenthesis$ on the $x$ -axis, and radius $| a |$ .

Roses

$r = a \sin (n θ)$
A rose with petal length $a$ .
$n$ petals if $n$ is odd; $2 n$ petals if $n$ is even.
$r = a \cos (n θ)$
A rose with petal length $a$ .
$n$ petals if $n$ is odd; $2 n$ petals if $n$ is even.

Limaçons

$r = a \pm b \sin (θ)$ or $r = a \pm b \cos (θ)$

$a > b$ , with a dent.
$a < b$ , with a loop.
$a > b$ , a cardioid.

Lemniscates

$r squared equals a squared cos ⁡ open parenthesis 2 θ close parenthesis$
$r squared equals a squared sin ⁡ open parenthesis 2 θ close parenthesis$

Spirals

Archimedean spiral
$r = a θ$
Logarithmic spiral
$r = e^{a θ}$

Section 10.2 Summary

Vocabulary

Rose
Limaçon
Cardioid
Lemniscate

Concepts

When graphing an equation in polar coordinates, we think of sweeping around the pole in the counterclockwise direction, and at each angle $θ$ the $r$ -value tells us how far the graph is from the pole.
Standard graphs in polar coordinates include circles and roses, cardioids and limaçons, lemniscates, and spirals.
To find the intersection points of the polar graphs $r = f (θ)$ and $r = g (θ)$ we solve the equation $f (θ) = g (θ)$ . In addition, we should always check whether the pole is a point on both graphs.

Study Questions

Delbert says that the graph of $r equals 2 sin ⁡ open parenthesis θ close parenthesis$ in the first example cannot be correct, because there are no points on the graph for angles between $π$ and $2 π$ . How do you respond?
Is it possible to have a rose with only two petals? What would its equation be?
Francine says that a circle of the form $r equals 2 a cos ⁡ open parenthesis θ close parenthesis$ is just a special case of a limaçon. Support or refute her statement.
There are no points on the graph of $r squared equals 4 cos ⁡ open parenthesis 2 θ close parenthesis$ for angles between $\frac{π}{4}$ and $\frac{3 π}{4}$ Why is that?

Skills

Describe the effect of parameters in polar curves #1–16, 83–84
Compare polar and Cartesian graphs #21–24
Sketch standard polar graphs #17–20, 25–42, 75–82
Identify standard polar graphs #43–58
Write equations for standard polar graphs #59–66
Find intersection points of polar graphs #67–74

Homework 10-2

In Problems 1-4, use graphing technology to graph the equations.

Graph $r = k$ , for $k equals 1, 2, 3$ . How does the graph change for different values of $k$ ?
Write a Cartesian equation for each graph in part (a).

$k$ is the radius
$x^{2} + y^{2} = 1, x^{2} + y^{2} = 4, x^{2} + y^{2} = 9$

Graph $r = k$ , for $k = - 1, - 2, - 3$ . How does these graphs compare to the graphs in Problem 1?
Write a Cartesian equation for each graph in part (a).

Graph $θ = k$ , for $k = \frac{π}{6}, \frac{π}{3}, \frac{2 π}{3}, \frac{5 π}{6}$ . How does the graph change for different values of $k$ ?
Write a Cartesian equation for each graph in part (a).

$\tan k$ is the slope
$y = \frac{x}{\sqrt{3}}, y = \sqrt{3} x, y = - \sqrt{3} x, y = \frac{x}{\sqrt{3}}$

Graph $θ = k$ , for $k = \frac{7 π}{6}, \frac{4 π}{3}, \frac{5 π}{3}, \frac{11 π}{6}$ . How does the graph change for different values of $k$ ?
Write a Cartesian equation for each graph in part (a).

Complete the table of values for each equation. Plot the points in order of increasing $θ$ . What is different about the two graphs?

Equation 1: $r equals 2$

$θ$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$
$r equals 2$

Equation 2: $r = - 2$

$θ$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$
$r = - 2$

$θ$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$
$r equals 2$	$2$	$2$	$2$	$2$	$2$	$2$	$2$	$2$

$θ$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$
$r = - 2$	$minus 2$	$minus 2$	$minus 2$	$minus 2$	$minus 2$	$minus 2$	$minus 2$	$minus 2$

The graph of $r equals 2$ begins at the right-most point (and proceeds counter-clockwise); the graph of $r = - 2$ begins at the left-most point.

Graph each line, and label the points with their coordinates. How are the points on the two lines related?

Equation 1: $θ = \frac{π}{4}$

$θ = \frac{π}{4}$	$\frac{π}{4}$	$\frac{π}{4}$	$\frac{π}{4}$	$\frac{π}{4}$	$\frac{π}{4}$
$r$	$minus 2$	$minus 1$	$0$	$1$	$2$

Equation 1: $θ = \frac{5 π}{4}$

$θ = \frac{5 π}{4}$	$\frac{5 π}{4}$	$\frac{5 π}{4}$	$\frac{5 π}{4}$	$\frac{5 π}{4}$	$\frac{5 π}{4}$
$r$	$minus 2$	$minus 1$	$0$	$1$	$2$

Graph the circle $r equals 4 cos ⁡ open parenthesis θ close parenthesis$ . Label the points corresponding to $θ = 0, \frac{π}{4}, \frac{π}{2}, \frac{3 π}{4},$ and $π$ .

Complete the table of values. What happens to the graph as

θ

increases from

π

2 π

$θ$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$r$

Find the center and radius of the circle.
Give the Cartesian equation of the circle.

$θ$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$r$	$minus 4$	$- 2 \sqrt{2}$	$0$	$2 the square root of 2$	$4$

The graph is traced again.

center: $open parenthesis 2, 0 close parenthesis$ , radius: $2$
$(x - 2)^{2} + y^{2} = 4$

Graph the circle $r equals 4 sin ⁡ open parenthesis θ close parenthesis$ . Label the points corresponding to $θ = 0, \frac{π}{4}, \frac{π}{2}, \frac{3 π}{4},$ and $π$ .

Complete the table of values. What happens to the graph as

θ

increases from

π

2 π

$θ$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$r$

Find the center and radius of the circle.
Give the Cartesian equation of the circle.

Graph $r equals 2 a sin ⁡ open parenthesis θ close parenthesis$ for $a = - 2, - 1, 1, 2$ .
How do the graphs change for different values of $a$ ?

For $a greater than 0$ , $a$ is the radius of a circle centered on the positive $y$ -axis; for $a less than 0$ , $| a |$ is the radius of a circle centerd on the negative $y$ -axis.

Graph $r equals 2 a cos ⁡ open parenthesis θ close parenthesis$ for $a = - 2, - 1, 1, 2$ .
How do the graphs change for different values of $a$ ?

Complete the table of values for each cardioid, and graph the equation.

$θ$	$0$	$\frac{π}{2}$	$π$	$\frac{3 π}{2}$	$2 π$
$r$

$r = 1 + \sin (θ)$
$r = - 1 + \sin (θ)$
$r = 1 - \sin (θ)$
$r = - 1 - \sin (θ)$

$θ$ $0$ $\frac{π}{2}$ $π$ $\frac{3 π}{2}$ $2 π$
$r$ $1$ $2$ $1$ $0$ $1$
$θ$ $0$ $\frac{π}{2}$ $π$ $\frac{3 π}{2}$ $2 π$
$r$ $minus 1$ $0$ $minus 1$ $minus 2$ $minus 1$
$θ$ $0$ $\frac{π}{2}$ $π$ $\frac{3 π}{2}$ $2 π$
$r$ $1$ $0$ $1$ $2$ $1$
$θ$ $0$ $\frac{π}{2}$ $π$ $\frac{3 π}{2}$ $2 π$
$r$ $minus 1$ $minus 2$ $minus 1$ $0$ $minus 1$

$θ$	$0$	$\frac{π}{2}$	$π$	$\frac{3 π}{2}$	$2 π$
$r$	$1$	$2$	$1$	$0$	$1$

$θ$	$0$	$\frac{π}{2}$	$π$	$\frac{3 π}{2}$	$2 π$
$r$	$minus 1$	$0$	$minus 1$	$minus 2$	$minus 1$

$θ$	$0$	$\frac{π}{2}$	$π$	$\frac{3 π}{2}$	$2 π$
$r$	$1$	$0$	$1$	$2$	$1$

$θ$	$0$	$\frac{π}{2}$	$π$	$\frac{3 π}{2}$	$2 π$
$r$	$minus 1$	$minus 2$	$minus 1$	$0$	$minus 1$

Complete the table of values for each cardioid, and graph the equation.

$θ$	$0$	$\frac{π}{2}$	$π$	$\frac{3 π}{2}$	$2 π$
$r$

$r = 1 + \cos (θ)$
$r = - 1 + \cos (θ)$
$r = 1 - \cos (θ)$
$r = - 1 - \cos (θ)$

Complete the table of values for each limaçon, and graph the equation.

$θ$	$0$	$\frac{π}{2}$	$π$	$\frac{3 π}{2}$	$2 π$
$r$

$r = 2 + \cos (θ)$
$r = 2 - \cos (θ)$
$r = 1 + 2 \cos (θ)$
$r = 1 - 2 \cos (θ)$

$θ$ $0$ $\frac{π}{2}$ $π$ $\frac{3 π}{2}$ $2 π$
$r$ $3$ $2$ $1$ $2$ $3$
$θ$ $0$ $\frac{π}{2}$ $π$ $\frac{3 π}{2}$ $2 π$
$r$ $1$ $2$ $3$ $2$ $1$
$θ$ $0$ $\frac{π}{2}$ $π$ $\frac{3 π}{2}$ $2 π$
$r$ $3$ $1$ $minus 1$ $1$ $3$
$θ$ $0$ $\frac{π}{2}$ $π$ $\frac{3 π}{2}$ $2 π$
$r$ $minus 1$ $1$ $3$ $1$ $minus 1$

$θ$	$0$	$\frac{π}{2}$	$π$	$\frac{3 π}{2}$	$2 π$
$r$	$3$	$2$	$1$	$2$	$3$

$θ$	$0$	$\frac{π}{2}$	$π$	$\frac{3 π}{2}$	$2 π$
$r$	$1$	$2$	$3$	$2$	$1$

$θ$	$0$	$\frac{π}{2}$	$π$	$\frac{3 π}{2}$	$2 π$
$r$	$3$	$1$	$minus 1$	$1$	$3$

$θ$	$0$	$\frac{π}{2}$	$π$	$\frac{3 π}{2}$	$2 π$
$r$	$minus 1$	$1$	$3$	$1$	$minus 1$

Complete the table of values for each limaçon, and graph the equation.

$θ$	$0$	$\frac{π}{2}$	$π$	$\frac{3 π}{2}$	$2 π$
$r$

$r = 2 + \sin (θ)$
$r = 2 - \sin (θ)$
$r = 1 + 2 \sin (θ)$
$r = 1 - 2 \sin (θ)$

Graph the following roses and compare. How is the number of petals related to the value of $n$ in the equation $r = a \sin (n θ)$ ?
$r = \sin (2 θ), r = \sin (3 θ), r = \sin (4 θ), r = \sin (5 θ)$
For each graph above, list the values of $θ$ where the tips of the petals occur.
Graph $r equals a sin ⁡ open parenthesis 3 θ close parenthesis$ for $a equals 1, 2,$ and $3$ . How does the value of $a$ affect the graph?

There are $n$ petals if $n$ is odd, and $2 n$ petals if $n$ is even.
$n = 2 : \frac{π}{4},$ $\frac{3 π}{4},$ $\frac{5 π}{4},$ $\frac{7 π}{4};$ $n = 3 : \frac{π}{6},$ $\frac{5 π}{6},$ $\frac{3 π}{2};$ $n = 4 : \frac{π}{8},$ $\frac{3 π}{8},$ $\frac{5 π}{8},$ $\frac{7 π}{8},$ $\frac{9 π}{8},$ $\frac{11 π}{8},$ $\frac{13 π}{8},$ $\frac{15 π}{8};$ $n = 5 : \frac{π}{10},$ $\frac{π}{2},$ $\frac{9 π}{10},$ $\frac{13 π}{10},$ $\frac{17 π}{10}$
$a$ is the length of the petal.

Graph the following roses and compare. How is the number of petals related to the value of $n$ in the equation $r = a \cos (n θ)$ ?
$r = \cos (2 θ), r = \cos (3 θ), r = \cos (4 θ), r = \cos (5 θ)$
For each graph above, list the values of $θ$ where the tips of the petals occur.
Graph $r equals a cos ⁡ open parenthesis 3 θ close parenthesis$ for $a equals 1, 2,$ and $3$ . How does the value of $a$ affect the graph?

Solve $r squared equals 9 cos ⁡ open parenthesis 2 θ close parenthesis$ for $r$ . (You should get two equations for $r$ .)
Graph both equations together. Change $θ$ step to 0.02 to see the whole graph.
How does the value of $a$ affect the graph of $r squared equals a squared cos ⁡ open parenthesis 2 θ close parenthesis$ ?

$r = \pm 3 \sqrt{\cos 2 θ}$
$a$ is the length of the loop.

Solve $r squared equals 9 sin ⁡ open parenthesis 2 θ close parenthesis$ for $r$ . (You should get two equations for $r$ .)
Graph both equations together. Change $θ$ step to 0.02 to see the whole graph.
How does this graph differ from the graph in Problem 17?

Graph the Archimedean spiral $r = θ$ . Set your window to

$\begin{aligned} θ min & = 0 θ max = 8 π \\ Xmin & = - 20 Xmax = 20 \\ Ymin & = - 20 Ymax = 20 \end{aligned}$

Then graph by pressing Zoom 5.

Graph the logarithmic spiral $r = e^{0.2 θ}$ . Set your window to

$\begin{aligned} θ min & = 0 θ max = 8 π \\ Xmin & = - 100 Xmax = 100 \\ Ymin & = - 100 Ymax = 100 \end{aligned}$

Then graph by pressing Zoom 5.

Complete the table and graph the equation

y equals sin ⁡ open parenthesis 3 θ close parenthesis

in Cartesian coordinates, for

0 less than or equal to θ less than or equal to 2 π

$θ$
$3 θ$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$y$

Complete the table and graph the equation

r equals sin ⁡ open parenthesis 3 θ close parenthesis

in polar coordinates, for

0 less than or equal to θ less than or equal to 2 π

$θ$
$3 θ$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$r$

$θ$	$0$	$\frac{π}{12}$	$\frac{π}{6}$	$\frac{π}{4}$	$\frac{π}{3}$	$\frac{5 π}{12}$	$\frac{π}{2}$	$\frac{7 π}{12}$	$\frac{2 π}{3}$
$3 θ$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$r$	$0$	$\frac{\sqrt{2}}{2}$	$1$	$\frac{\sqrt{2}}{2}$	$0$	$\frac{- \sqrt{2}}{2}$	$minus 1$	$\frac{- \sqrt{2}}{2}$	$0$

$θ$	$0$	$\frac{π}{12}$	$\frac{π}{6}$	$\frac{π}{4}$	$\frac{π}{3}$	$\frac{5 π}{12}$	$\frac{π}{2}$	$\frac{7 π}{12}$	$\frac{2 π}{3}$
$3 θ$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$r$	$0$	$\frac{\sqrt{2}}{2}$	$1$	$\frac{\sqrt{2}}{2}$	$0$	$\frac{- \sqrt{2}}{2}$	$minus 1$	$\frac{- \sqrt{2}}{2}$	$0$

Complete the table and graph the equation

y equals cos ⁡ open parenthesis 2 θ close parenthesis

in Cartesian coordinates, for

0 less than or equal to θ less than or equal to 2 π

$θ$
$2 θ$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$y$

Complete the table and graph the equation

r equals cos ⁡ open parenthesis 2 θ close parenthesis

in polar coordinates, for

0 less than or equal to θ less than or equal to 2 π

$θ$
$2 θ$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$r$

Complete the table and graph the equation

y = 2 + 2 \cos (θ)

in Cartesian coordinates, for

0 less than or equal to θ less than or equal to 2 π

$θ$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$y$

Complete the table and graph the equation

r = 2 + 2 \cos (θ)

in polar coordinates, for

0 less than or equal to θ less than or equal to 2 π

$θ$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$r$

$θ$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$y$	$4$	$2 + \sqrt{2}$	$2$	$2 - \sqrt{2}$	$0$	$2 - \sqrt{2}$	$2$	$2 + \sqrt{2}$	$4$

$θ$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$y$	$4$	$2 + \sqrt{2}$	$2$	$2 - \sqrt{2}$	$0$	$2 - \sqrt{2}$	$2$	$2 + \sqrt{2}$	$4$

Complete the table and graph the equation

y = 1 - \sin (θ)

in Cartesian coordinates, for

0 less than or equal to θ less than or equal to 2 π

$θ$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$y$

Complete the table and graph the equation

r = 1 - \sin (θ)

in polar coordinates, for

0 less than or equal to θ less than or equal to 2 π

$θ$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$r$

For Problems 25–42, use the catalog of polar graphs to help you identify and sketch the following curves. Check your work by graphing with a graphing utility.

$r equals 3 cos ⁡ open parenthesis θ close parenthesis$

circle

$r equals 2 sin ⁡ open parenthesis θ close parenthesis$

$θ = \frac{π}{4}$

line

$θ = \frac{4 π}{3}$

$r equals 4$

circle

$r equals 2$

$r = 2 + 2 \sin (θ)$

cardioid

$r = 3 + 3 \cos (θ)$

$r = 2 - \cos (θ)$

limaçon

$r = 1 - 3 \sin (θ)$

$r equals 3 sin ⁡ open parenthesis 2 θ close parenthesis$

rose

$r equals 2 cos ⁡ open parenthesis 3 θ close parenthesis$

$r equals 2 cos ⁡ open parenthesis 5 θ close parenthesis$

rose

$r equals 4 sin ⁡ open parenthesis 4 θ close parenthesis$

$r = 2 + 3 \sin (θ)$

limaçon

$r = 3 + 2 \sin (θ)$

$r squared equals cos ⁡ open parenthesis 2 θ close parenthesis$

lemniscate

$r squared equals 4 sin ⁡ open parenthesis 2 θ close parenthesis$

For Problems 43–52, identify each curve, and graph it.

$r csc ⁡ open parenthesis θ close parenthesis equals 2$

circle

$r equals 2 sec ⁡ open parenthesis θ close parenthesis$

$r^{2} = 4, 0 \leq θ \leq \frac{3 π}{4}$

arcs of a circle

$θ = \frac{π}{4}, | r | < 2$

$r = \sin (θ), \frac{3 π}{4} \leq θ \leq \frac{5 π}{4}$

semicircle

$r = \cos (θ), 0 \leq θ \leq \frac{π}{2}$

$r equals 2 sin ⁡ open parenthesis 2 θ close parenthesis cos ⁡ open parenthesis 2 θ close parenthesis$

rose

$r = \cos^{2} (θ) - \sin^{2} (θ)$

$r (1 - \cos (θ)) = \sin^{2} (θ)$

cardioid

$r \sec (θ) = \sec (θ) - \tan (θ)$

For Problems 53–58, graph the following polar curves. Do you recognize them?

$r = \frac{2}{1 - \cos (θ)}$

parabola

$r = \frac{6}{2 + \sin (θ)}$

$r = \frac{2}{2 - \cos (θ)}$

ellipse

$r = \frac{1}{1 + \sin (θ)}$

$r = \frac{1}{1 + 2 \sin (θ)}$

hyperbola

$r = \frac{3}{2 - 3 \cos (θ)}$

For Problems 59–66, write a polar equation for the graph.

$r = 2 + 2 \cos (θ)$

$r equals 3 sin ⁡ open parenthesis 5 θ close parenthesis$

$r equals 5 sin ⁡ open parenthesis θ close parenthesis$

$r = 1 + 2 \cos (θ)$

For Problems 67–74, find the coordinates of the intersection points of the two curves analytically. Then graph the curves to verify your answers.

$r = \cos (θ), r = 1 - \cos (θ)$

$open parenthesis 0, 0 close parenthesis$ , $(\frac{1}{2}, \frac{π}{3})$ , $(\frac{1}{2}, \frac{5 π}{3})$

$r = \sin (θ), r = \cos (θ)$

$r = 3 \sin (θ), r = 3 \cos (θ)$

$open parenthesis 0, 0 close parenthesis$ , $(\frac{3}{\sqrt{2}}, \frac{π}{4})$ , $(\frac{- 3}{\sqrt{2}}, \frac{5 π}{4})$

$r = \sin (2 θ), r = \cos (2 θ)$

$r = 1, r = 1 - \cos (θ)$

$(1, \frac{π}{2})$ , $(1, \frac{3 π}{2})$

$r = 3 \cos (θ), r = 1 + \cos (θ)$

$r = 2 + \sin (θ), r = 2 - \cos (θ)$

$(\frac{4 + \sqrt{2}}{2}, \frac{3 π}{4})$ , $(\frac{4 - \sqrt{2}}{2}, \frac{7 π}{4})$

$r = \sin (θ), r = \sin (2 θ)$

For Problems 75–82, graph the polar curve.

$r^{2} = \tan (θ)$

$r^{2} = \cot (θ)$

$r = \csc (θ) - 2$ (conchoid)

$r = \tan (θ)$ (kappa curve)

$r equals cos ⁡ open parenthesis 2 θ close parenthesis sec ⁡ open parenthesis θ close parenthesis$ (strophoid)

$r = \sin (θ) \tan (θ)$ (cissoid)

$r = \frac{1}{\sqrt{θ}}$

$r = \cos (\frac{θ}{2}), 0 \leq θ \leq 4 π$

Graph the polar curves $r = 1 - 2 \sin (n θ)$ for $n equals 2, 3, 4, 5, 6$ . Explain how the value of the parameter $n$ affects the curve.

The curve has $n$ large loops and $n$ small loops.

Graph the polar curves $r = 1 - 3 \cos (n θ)$ for $n equals 2, 3, 4, 5, 6$ . Explain how the value of the parameter $n$ affects the curve.

Trigonometry by Katherine Yoshiwara (yoshiwarabooks.org), GNU Free Documentation License 1.2 or later. Adapted for the XYZ HTML edition with the authors' permission (recorded 2026-07-04). License: GFDL-1.2-or-later.