Polar Coordinates: Graphs

Learning Objectives

By the end of this section, you will be able to:

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


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)


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)\left(r,\theta \right)
. We interpret
rr
as the distance from the sun and
θ\theta
as the planet’s angular bearing, or its direction from a fixed point on the sun. In this section, we will focus on the polar system and the graphs that are generated directly from polar coordinates.

Testing Polar Equations for Symmetry

Just as a rectangular equation such as
y=x2y={x}^{2}
describes the relationship between
xx
and
yy
on a Cartesian grid, a polar equation describes a relationship between
rr
and
θ\theta
on a polar grid. Recall that the coordinate pair
(r,θ)\left(r,\theta \right)
indicates that we move counterclockwise from the polar axis (positive x-axis) by an angle of
θ\theta
, and extend a ray from the pole (origin)
rr
units in the direction of
θ\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
rr
)

to determine the graph of a polar equation.

In the first test, we consider symmetry with respect to the line
θ=π2\theta =\frac{\pi }{2}
(y-axis). We replace
(r,θ)\left(r,\theta \right)
with
(r,θ)\left(-r,-\theta \right)
to determine if the new equation is equivalent to the original equation. For example, suppose we are given the equation
r=2sinθr=2\sin \theta
;

r=2sinθr=2sin(θ)Replace(r,θ)with (r,θ).r=2sinθIdentity: sin(θ)=sinθ.r=2sinθMultiply both sides by1.\begin{array}{ll}r=2\sin \theta \qquad & \qquad \\ -r=2\sin \left(-\theta \right)\begin{array}{cccc}& & & \end{array}\qquad & \text{Replace}\left(r,\theta \right)\text{with }\left(-r,-\theta \right).\qquad \\ -r=-2\sin \theta \qquad & \text{Identity: }\sin \left(-\theta \right)=-\sin \theta .\qquad \\ r=2\sin \theta \qquad & \text{Multiply both sides by}-1.\qquad \end{array}
This equation exhibits symmetry with respect to the line
θ=π2\theta =\frac{\pi }{2}
.

In the second test, we consider symmetry with respect to the polar axis (
xx
-axis). We replace
(r,θ)\left(r,\theta \right)
with
(r,θ)\left(r,-\theta \right)
or
(r,πθ)\left(-r,\pi -\theta \right)
to determine equivalency between the tested equation and the original. For example, suppose we are given the equation
r=12cosθr=1 - 2\cos \theta
.

r=12cosθr=12cos(θ)Replace (r,θ)with(r,θ).r=12cosθEven/Odd identity\begin{array}{ll}r=1 - 2\cos \theta \qquad & \qquad \\ r=1 - 2\cos \left(-\theta \right)\begin{array}{cccc}& & & \end{array}\qquad & \text{Replace }\left(r,\theta \right)\text{with}\left(r,-\theta \right).\qquad \\ r=1 - 2\cos \theta \qquad & \text{Even/Odd identity}\qquad \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)\left(r,\theta \right)
with
(r,θ)\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=2sin(3θ)r=2\sin \left(3\theta \right)
.

r=2sin(3θ)r=2sin(3θ)\begin{array}{c}r=2\sin \left(3\theta \right)\\ -r=2\sin \left(3\theta \right)\end{array}
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=π2\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.

A General Note: Symmetry Tests

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

3 graphs side by side. (A) shows a ray extending into Q 1 and its symmetric version in Q 2. (B) shows a ray extending into Q 1 and its symmetric version in Q 4. (C) shows a ray extending into Q 1 and its symmetric version in Q 3. See caption for more information. Figure 2. (a) A graph is symmetric with respect to the line
θ=π2\theta =\frac{\pi }{2}
(y-axis) if replacing
(r,θ)\left(r,\theta \right)
with
(r,θ)\left(-r,-\theta \right)
yields an equivalent equation. (b) A graph is symmetric with respect to the polar axis (x-axis) if replacing
(r,θ)\left(r,\theta \right)
with
(r,θ)\left(r,-\theta \right)
or
(r,πθ)\left(-r,\mathrm{\pi -}\theta \right)
yields an equivalent equation. (c) A graph is symmetric with respect to the pole (origin) if replacing
(r,θ)\left(r,\theta \right)
with
(r,θ)\left(-r,\theta \right)
yields an equivalent equation.

How To: Given a polar equation, test for symmetry.

  1. Substitute the appropriate combination of components for
    (r,θ):\left(r,\theta \right):
    (r,θ)\left(-r,-\theta \right)
    for
    θ=π2\theta =\frac{\pi }{2}
    symmetry;
    (r,θ)\left(r,-\theta \right)
    for polar axis symmetry; and
    (r,θ)\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.


Example 1: Testing a Polar Equation for Symmetry

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

Solution

Test for each of the three types of symmetry.

1) Replacing
(r,θ)\left(r,\theta \right)
with
(r,θ)\left(-r,-\theta \right)
yields the same result. Thus, the graph is symmetric with respect to the line
θ=π2\theta =\frac{\pi }{2}
.
r=2sin(θ)r=2sinθEven-odd identityr=2sinθMultiplyby1Passed\begin{array}{ll}-r=2\sin \left(-\theta \right)\qquad & \qquad \\ -r=-2\sin \theta \qquad & \text{Even-odd identity}\qquad \\ r=2\sin \theta \qquad & \text{Multiply}\text{by}-1\qquad \\ \text{Passed}\qquad & \qquad \end{array}
2) Replacing
θ\theta
with
θ-\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.
r=2sin(θ)r=2sinθEven-odd identityr=2sinθ2sinθFailed\begin{array}{ll}r=2\sin \left(-\theta \right)\qquad & \qquad \\ r=-2\sin \theta \qquad & \text{Even-odd identity}\qquad \\ r=-2\sin \theta \ne 2\sin \theta \qquad & \qquad \\ \text{Failed}\qquad & \qquad \end{array}
3) Replacing
rr
with
r-r
changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole.
r=2sinθ r=2sinθ2sinθFailed\begin{array}{l}-r=2\sin \theta \qquad \\ \text{ }r=-2\sin \theta \ne 2\sin \theta \qquad \\ \text{Failed}\qquad \end{array}

Analysis of the Solution

Using a graphing calculator, we can see that the equation
r=2sinθr=2\sin \theta
is a circle centered at
(0,1)\left(0,1\right)
with radius
r=1r=1
and is indeed symmetric to the line
θ=π2\theta =\frac{\pi }{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


Try It 1

Test the equation for symmetry:
r=2cosθr=-2\cos \theta
.

Solution

 Graphing Polar Equations by Plotting Points

To graph in the rectangular coordinate system we construct a table of
xx
and
yy
values. To graph in the polar coordinate system we construct a table of
θ\theta
and
rr
values. We enter values of
θ\theta
 into a polar equation and calculate
rr
. However, using the properties of symmetry and finding key values of
θ\theta
and
rr
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\theta
that result in
r=0r=0
. Recall that, to find the zeros of polynomial functions, we set the equation equal to zero and then solve for
xx
. We use the same process for polar equations. Set
r=0r=0
, and solve for
θ\theta
.

For many of the forms we will encounter, the maximum value of a polar equation is found by substituting those values of
θ\theta
into the equation that result in the maximum value of the trigonometric functions. Consider
r=5cosθr=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
θ=0\theta =0
, so our polar equation is
5cosθ5\cos \theta
, and the value
θ=0\theta =0
will yield the maximum
r|r|
.

Similarly, the maximum value of the sine function is 1 when
θ=π2\theta =\frac{\pi }{2}
, and if our polar equation is
r=5sinθr=5\sin \theta
, the value
θ=π2\theta =\frac{\pi }{2}
will yield the maximum
r|r|
. We may find additional information by calculating values of
rr
when
θ=0\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.

Example 2: Finding Zeros and Maximum Values for a Polar Equation

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

Solution

To find the zeros, set
rr
equal to zero and solve for
θ\theta
.

2sinθ=0sinθ=0θ=sin10θ=nπwhere n is an integer\begin{array}{ll}2\sin \theta =0\qquad & \qquad \\ \sin \theta =0\qquad & \qquad \\ \theta ={\sin }^{-1}0\qquad & \qquad \\ \theta =n\pi \qquad & \text{where }n\text{ is an integer}\qquad \end{array}
Substitute any one of the
θ\theta
values into the equation. We will use
00
.

r=2sin(0)r=0\begin{array}{l}\begin{array}{l}\\ r=2\sin \left(0\right)\end{array}\qquad \\ r=0\qquad \end{array}
The points
(0,0)\left(0,0\right)
and
(0,±nπ)\left(0,\pm n\pi \right)
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
sinθ\sin \theta
, which occurs when
θ=π2±2kπ\theta =\frac{\pi }{2}\pm 2k\pi
resulting in
sin(π2)=1\sin \left(\frac{\pi }{2}\right)=1
. Substitute
π2\frac{\pi }{2}
for
θ.\mathrm{\theta .}


r=2sin(π2)r=2(1)r=2\begin{array}{l}r=2\sin \left(\frac{\pi }{2}\right)\qquad \\ r=2\left(1\right)\qquad \\ r=2\qquad \end{array}

Analysis of the Solution

The point
(2,π2)\left(2,\frac{\pi }{2}\right)
will be the maximum value on the graph. Let’s plot a few more points to verify the graph of a circle.

θ\theta
r=2sinθr=2\sin \theta
rr
0
r=2sin(0)=0r=2\sin \left(0\right)=0
00
π6\frac{\pi }{6}
r=2sin(π6)=1r=2\sin \left(\frac{\pi }{6}\right)=1
11
π3\frac{\pi }{3}
r=2sin(π3)1.73r=2\sin \left(\frac{\pi }{3}\right)\approx 1.73
1.731.73
π2\frac{\pi }{2}
r=2sin(π2)=2r=2\sin \left(\frac{\pi }{2}\right)=2
22
2π3\frac{2\pi }{3}
r=2sin(2π3)1.73r=2\sin \left(\frac{2\pi }{3}\right)\approx 1.73
1.731.73
5π6\frac{5\pi }{6}
r=2sin(5π6)=1r=2\sin \left(\frac{5\pi }{6}\right)=1
11
π\pi
r=2sin(π)=0r=2\sin \left(\pi \right)=0
00
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


Try It 2

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

Solution

 Graphing Circles and the 5 Classic Polar Curves

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.

A General Note: Formulas for the Equation of a Circle

Some of the formulas that produce the graph of a circle in polar coordinates are given by
r=acosθr=a\cos \theta
and
r=asinθr=a\sin \theta
, where
aa
is the diameter of the circle or the distance from the pole to the farthest point on the circumference. The radius is
a2\frac{|a|}{2}
, or one-half the diameter. For
r=acosθ,r=a\cos \theta ,
the center is
(a2,0)\left(\frac{a}{2},0\right)
. For
r=asinθr=a\sin \theta
, the center is
(a2,π)\left(\frac{a}{2},\pi \right)
. Figure 5 shows the graphs of these four circles.

Example 3: Sketching the Graph of a Polar Equation for a Circle

Sketch the graph of
r=4cosθr=4\cos \theta
.

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|r|
for
r=4cosθr=4\cos \theta
. First, set
r=0r=0
, and solve for
θ\theta
. Thus, a zero occurs at
θ=π2±kπ\theta =\frac{\pi }{2}\pm k\pi
. A key point to plot is
(0,  π2)\left(0,\text{ }\text{ }\frac{\pi }{2}\right)
.

To find the maximum value of
rr
, note that the maximum value of the cosine function is 1 when
θ=0±2kπ\theta =0\pm 2k\pi
. Substitute
θ=0\theta =0
into the equation:

r=4cosθr=4cos(0)r=4(1)=4\begin{array}{c}r=4\cos \theta \\ r=4\cos \left(0\right)\\ r=4\left(1\right)=4\end{array}
The maximum value of the equation is 4. A key point to plot is
(4,0)\left(4,0\right)
.

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

θ\theta
0
π6\frac{\pi }{6}
π4\frac{\pi }{4}
π3\frac{\pi }{3}
π2\frac{\pi }{2}
2π3\frac{2\pi }{3}
3π4\frac{3\pi }{4}
5π6\frac{5\pi }{6}
π\pi
rr
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.

A General Note: Formulas for a Cardioid

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

Graph of four cardioids. (A) is r = a + bcos(theta). Cardioid extending to the right. (B) is r=a-bcos(theta). Cardioid extending to the left. (C) is r=a+bsin(theta). Cardioid extending up. (D) is r=a-bsin(theta). Cardioid extending down. Figure 7


How 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=0r=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
    rr
    and
    θ\theta
    .
  5. Plot the points and sketch the graph.


Example 4: Sketching the Graph of a Cardioid

Sketch the graph of
r=2+2cosθr=2+2\cos \theta
.

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=0r=0
, we have
θ=π+2kπ\theta =\pi +2k\pi
. The zero of the equation is located at
(0,π)\left(0,\pi \right)
. The graph passes through this point.

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

r=2+2cos(0)r=2+2(1)=4\begin{array}{l}\begin{array}{l}\\ r=2+2\cos \left(0\right)\end{array}\qquad \\ r=2+2\left(1\right)=4\qquad \end{array}
The point
(4,0)\left(4,0\right)
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]\left[0,\pi \right]
. The upper portion of the graph is then reflected over the polar axis. Next, we make a table of values, as in the table below, and then we plot the points and draw the graph. See Figure 8.

θ\theta
00
π4\frac{\pi }{4}
π2\frac{\pi }{2}
2π3\frac{2\pi }{3}
π\pi
rr
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<ab<21<\frac{a}{b}<2
and convex limaçons when
ab2\frac{a}{b}\ge 2
.

A General Note: Formulas for One-Loop Limaçons

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

Four dimpled limaçons side by side. (A) is r=a+bcos(theta). Extending to the right. (B) is r=a-bcos(theta). Extending to the left. (C) is r=a+bsin(theta). Extending up. (D) is r=a-bsin(theta). Extending down. Figure 9. Dimpled limaçons


How 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.


Example 5: Sketching the Graph of a One-Loop Limaçon

Graph the equation
r=43sinθr=4 - 3\sin \theta
.

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=π2\theta =\frac{\pi }{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=0r=0
results in
θ\theta
being undefined. What does this mean? How could
θ\theta
be undefined? The angle
θ\theta
is undefined for any value of
sinθ>1\sin \theta >1
. Therefore,
θ\theta
is undefined because there is no value of
θ\theta
for which
sinθ>1\sin \theta >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
rr
when
θ=0\theta =0
.

r(0)=43sin(0)r=430=4\begin{array}{l}r\left(0\right)=4 - 3\sin \left(0\right)\qquad \\ r=4 - 3\cdot 0=4\qquad \end{array}
So, there is at least one polar axis intercept at
(4,0)\left(4,0\right)
.

Next, as the maximum value of the sine function is 1 when
θ=π2\theta =\frac{\pi }{2}
, we will substitute
θ=π2\theta =\frac{\pi }{2}
 into the equation and solve for
rr
. Thus,
r=1r=1
.

Make a table of the coordinates similar to the table below.

θ\theta
00
π6\frac{\pi }{6}
π3\frac{\pi }{3}
π2\frac{\pi }{2}
2π3\frac{2\pi }{3}
5π6\frac{5\pi }{6}
π\pi
7π6\frac{7\pi }{6}
4π3\frac{4\pi }{3}
3π2\frac{3\pi }{2}
5π3\frac{5\pi }{3}
11π6\frac{11\pi }{6}
2π2\pi
rr
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 of the Solution

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
sinθ\sin \theta
is likely symmetric with respect to the line
θ=π2\theta =\frac{\pi }{2}
, evaluating more points helps to verify that the graph is correct.

Try It 3

Sketch the graph of
r=32cosθr=3 - 2\cos \theta
.

Solution
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.

A General Note: Formulas for Inner-Loop Limaçons

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

Graph of four inner loop limaçons side by side. (A) is r=a+bcos(theta),a<b. Extended to the right. (B) is a-bcos(theta), a<b. Extends to the left. (C) is r=a+bsin(theta), a<b. Extends up. (D) is r=a-bsin(theta), a<b. Extends down. Figure 11


Example 6: Sketching the Graph of an Inner-Loop Limaçon

Sketch the graph of
r=2+5cosθr=2+5\text{cos}\theta
.

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

θ\theta
00
π6\frac{\pi }{6}
π3\frac{\pi }{3}
π2\frac{\pi }{2}
2π3\frac{2\pi }{3}
5π6\frac{5\pi }{6}
π\pi
7π6\frac{7\pi }{6}
4π3\frac{4\pi }{3}
3π2\frac{3\pi }{2}
5π3\frac{5\pi }{3}
11π6\frac{11\pi }{6}
2π2\pi
rr
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\theta =\pi
. 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.

A General Note: Formulas for Lemniscates

The formulas that generate the graph of a lemniscate are given by
r2=a2cos2θ{r}^{2}={a}^{2}\cos 2\theta
and
r2=a2sin2θ{r}^{2}={a}^{2}\sin 2\theta
where
a0a\ne 0
. The formula
r2=a2sin2θ{r}^{2}={a}^{2}\sin 2\theta
is symmetric with respect to the pole. The formula
r2=a2cos2θ{r}^{2}={a}^{2}\cos 2\theta
is symmetric with respect to the pole, the line
θ=π2\theta =\frac{\pi }{2}
, and the polar axis. See Figure 13 for the graphs.

Four graphs of lemniscates side by side. (A) is r^2 = a^2 * cos(2theta). Horizonatal figure eight, on x-axis. (B) is r^2 = - a^2 * cos(2theta). Vertical figure eight, on y axis. (C) is r^2 = a^2 * sin(2theta). Diagonal figure eight on line y=x. (D) is r^2 = -a^2 *sin(2theta). Diagonal figure eight on line y=-x. Figure 13


Example 7: Sketching the Graph of a Lemniscate

Sketch the graph of
r2=4cos2θ{r}^{2}=4\cos 2\theta
.

Solution

The equation exhibits symmetry with respect to the line
θ=π2\theta =\frac{\pi }{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=2θu=2\theta
.

0=4cos2θ0=4cosu0=cosucos10=π2u=π2Substitute 2θ back in for u.2θ=π2θ=π4\begin{array}{ll}0=4\cos 2\theta \qquad & \qquad \\ 0=4\cos u\qquad & \qquad \\ 0=\cos u\qquad & \qquad \\ {\cos }^{-1}0=\frac{\pi }{2}\qquad & \qquad \\ u=\frac{\pi }{2}\qquad & \text{Substitute }2\theta \text{ back in for }u.\qquad \\ 2\theta =\frac{\pi }{2}\qquad & \qquad \\ \theta =\frac{\pi }{4}\qquad & \qquad \end{array}
So, the point
(0,π4)\left(0,\frac{\pi }{4}\right)
is a zero of the equation.

Now let’s find the maximum value. Since the maximum of
cosu=1\cos u=1
when
u=0u=0
, the maximum
cos2θ=1\cos 2\theta =1
when
2θ=02\theta =0
. Thus,

r2=4cos(0)r2=4(1)=4r=±4=2\begin{array}{c}{r}^{2}=4\cos \left(0\right)\\ {r}^{2}=4\left(1\right)=4\\ r=\pm \sqrt{4}=2\end{array}
We have a maximum at (2, 0). Since this graph is symmetric with respect to the pole, the line
θ=π2\theta =\frac{\pi }{2}
, and the polar axis, we only need to plot points in the first quadrant.

Make a table similar to the table below.

θ\theta
0
π6\frac{\pi }{6}
π4\frac{\pi }{4}
π3\frac{\pi }{3}
π2\frac{\pi }{2}
rr
2
2\sqrt{2}
0
2\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 of the Solution

Making a substitution such as
u=2θu=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
rr
. This is because there are no real square roots for these values of
θ\theta
. In other words, the corresponding r-values of
4cos(2θ)\sqrt{4\cos \left(2\theta \right)}


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.

A General Note: Rose Curves

The formulas that generate the graph of a rose curve are given by
r=acosnθr=a\cos n\theta
and
r=asinnθr=a\sin n\theta
where
a0a\ne 0
. If
nn
is even, the curve has
2n2n
petals. If
nn
is odd, the curve has
nn
petals.

Graph of two rose curves side by side. (A) is r=acos(ntheta), where n is even. Eight petals extending from origin, equally spaced. (B) is r=asin(ntheta) where n is odd. Three petals extending from the origin, equally spaced. Figure 15


Example 8: Sketching the Graph of a Rose Curve (n Even)

Sketch the graph of
r=2cos4θr=2\cos 4\theta
.

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=π2\theta =\frac{\pi }{2}
and the pole.

Now we will find the zeros. First make the substitution
u=4θu=4\theta
.

0=2cos4θ0=cos4θ0=cosucos10=uu=π24θ=π2θ=π8\begin{array}{c}0=2\cos 4\theta \\ 0=\cos 4\theta \\ 0=\cos u\\ {\cos }^{-1}0=u\\ u=\frac{\pi }{2}\\ 4\theta =\frac{\pi }{2}\\ \theta =\frac{\pi }{8}\end{array}
The zero is
θ=π8\theta =\frac{\pi }{8}
. The point
(0,π8)\left(0,\frac{\pi }{8}\right)
is on the curve.

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

r=2cos(40)r=2cos(0)r=2(1)=2\begin{array}{l}\\ \begin{array}{l}r=2\cos \left(4\cdot 0\right)\qquad \\ r=2\cos \left(0\right)\qquad \\ r=2\left(1\right)=2\qquad \end{array}\end{array}
The point
(2,0)\left(2,0\right)
is on the curve.

The graph of the rose curve has unique properties, which are revealed in the table below.

θ\theta
0
π8\frac{\pi }{8}
π4\frac{\pi }{4}
3π8\frac{3\pi }{8}
π2\frac{\pi }{2}
5π8\frac{5\pi }{8}
3π4\frac{3\pi }{4}
rr
2 0 −2 0 2 0 −2
As
r=0r=0
when
θ=π8\theta =\frac{\pi }{8}
, it makes sense to divide values in the table by
π8\frac{\pi }{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=0r=0
, each petal extends out a distance of
r=2r=2
, and then turns back to zero
2n2n
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,
nn
even

Analysis of the Solution

When these curves are drawn, it is best to plot the points in order, as in the table of Example 8's solution. 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.

Try It 4

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

Solution

Example 9: Sketching the Graph of a Rose Curve (n Odd)

Sketch the graph of
r=2sin(5θ)r=2\sin \left(5\theta \right)
.

Solution

The graph of the equation shows symmetry with respect to the line
θ=π2\theta =\frac{\pi }{2}
. Next, find the zeros and maximum. We will want to make the substitution
u=5θu=5\theta
.

0=2sin(5θ)0=sinusin10=0u=05θ=0θ=0\begin{array}{c}0=2\sin \left(5\theta \right)\\ 0=\sin u\\ {\sin }^{-1}0=0\\ u=0\\ 5\theta =0\\ \theta =0\end{array}
The maximum value is calculated at the angle where
sinθ\sin \theta
is a maximum. Therefore,

r=2sin(5π2)r=2(1)=2\begin{array}{l}\begin{array}{l}\\ r=2\sin \left(5\cdot \frac{\pi }{2}\right)\end{array}\qquad \\ r=2\left(1\right)=2\qquad \end{array}
Thus, the maximum value of the polar equation is 2. This is the length of each petal. As the curve for
nn
odd yields the same number of petals as
nn
, there will be five petals on the graph.

Create a table of values similar to the table below.

θ\theta
0
π6\frac{\pi }{6}
π3\frac{\pi }{3}
π2\frac{\pi }{2}
2π3\frac{2\pi }{3}
5π6\frac{5\pi }{6}
π\pi
rr
0 1 −1.73 2 −1.73 1 0
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,
nn
odd

Try It 5

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

Solution

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.

A General Note: Archimedes’ Spiral

The formula that generates the graph of the Archimedes’ spiral is given by
r=θr=\theta
 for
θ0\theta \ge 0
. As
θ\theta
increases,
rr
 increases at a constant rate in an ever-widening, never-ending, spiraling path.

Two graphs side by side of Archimedes' spiral. (A) is r= theta, [0, 2pi]. (B) is r=theta, [0, 4pi]. Both start at origin and spiral out counterclockwise. The second has two spirals out while the first has one. Figure 18

How To: Given an Archimedes’ spiral over
[0,2π]\left[0,2\pi \right]
, sketch the graph.

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


Example 10: Sketching the Graph of an Archimedes’ Spiral

Sketch the graph of
r=θr=\theta
over
[0,2π]\left[0,2\pi \right]
.

Solution

As
rr
is equal to
θ\theta
, the plot of the Archimedes’ spiral begins at the pole at the point (0, 0). While the graph hints of symmetry, there is no formal symmetry with regard to passing the symmetry tests. Further, there is no maximum value, unless the domain is restricted.

Create a table such as the one below.

θ\theta
π4\frac{\pi }{4}
π2\frac{\pi }{2}
π\pi
3π2\frac{3\pi }{2}
7π4\frac{7\pi }{4}
2π2\pi
rr
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 of the Solution

The domain of this polar curve is
[0,2π]\left[0,2\pi \right]
. 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.

Try It 6

Sketch the graph of
r=θr=-\theta
over the interval
[0,4π]\left[0,4\pi \right]
.

Solution
 

Key Concepts

  • It is easier to graph polar equations if we can test the equations for symmetry with respect to the line
    θ=π2\theta =\frac{\pi }{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.
  • Polar equations may be graphed by making a table of values for
    θ\theta
    and
    rr
    .
  • The maximum value of a polar equation is found by substituting the value
    θ\theta
    that leads to the maximum value of the trigonometric expression.
  • The zeros of a polar equation are found by setting
    r=0r=0
    and solving for
    θ\theta
    .
  • Some formulas that produce the graph of a circle in polar coordinates are given by
    r=acosθr=a\cos \theta
    and
    r=asinθr=a\sin \theta
    .
  • The formulas that produce the graphs of a cardioid are given by
    r=a±bcosθr=a\pm b\cos \theta
    and
    r=a±bsinθr=a\pm b\sin \theta
    , for
    a>0,b>0a>0,b>0
    , and
    ab=1\frac{a}{b}=1
    .
  • The formulas that produce the graphs of a one-loop limaçon are given by
    r=a±bcosθr=a\pm b\cos \theta
    and
    r=a±bsinθr=a\pm b\sin \theta
    for
    1<ab<21<\frac{a}{b}<2
    .
  • The formulas that produce the graphs of an inner-loop limaçon are given by
    r=a±bcosθr=a\pm b\cos \theta
    and
    r=a±bsinθr=a\pm b\sin \theta
    for
    a>0,b>0a>0,b>0
    , and
    a<ba<b
    .
  • The formulas that produce the graphs of a lemniscates are given by
    r2=a2cos2θ{r}^{2}={a}^{2}\cos 2\theta
    and
    r2=a2sin2θ{r}^{2}={a}^{2}\sin 2\theta
    , where
    a0a\ne 0
    .
  • The formulas that produce the graphs of rose curves are given by
    r=acosnθr=a\cos n\theta
    and
    r=asinnθr=a\sin n\theta
    , where
    a0a\ne 0
    ; if
    nn
    is even, there are
    2n2n
    petals, and if
    nn
    is odd, there are
    nn
    petals.
  • The formula that produces the graph of an Archimedes’ spiral is given by
    r=θ,θ0r=\theta ,\theta \ge 0
    .


Summary of Curves

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

Glossary

Archimedes’ spiral
a polar curve given by
r=θr=\theta
. When multiplied by a constant, the equation appears as
r=aθr=a\theta
. As
r=θr=\theta
, 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±bcosθr=a\pm b\cos \theta
and
r=a±bsinθr=a\pm b\sin \theta
, where
ab=1\frac{a}{b}=1


convex limaҫon
a type of one-loop limaçon represented by
r=a±bcosθr=a\pm b\cos \theta
and
r=a±bsinθr=a\pm b\sin \theta
such that
ab2\frac{a}{b}\ge 2


dimpled limaҫon
a type of one-loop limaçon represented by
r=a±bcosθr=a\pm b\cos \theta
and
r=a±bsinθr=a\pm b\sin \theta
such that
1<ab<21<\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±bcosθr=a\pm b\cos \theta
and
 r=a±bsinθ \text{ }r=a\pm b\sin \theta \text{ }
where
a<ba<b


lemniscate
a polar curve resembling a figure 8 and given by the equation
r2=a2cos2θ{r}^{2}={a}^{2}\cos 2\theta
and
r2=a2sin2θ,a0{r}^{2}={a}^{2}\sin 2\theta ,a\ne 0


one-loop limaҫon
a polar curve represented by
r=a±bcosθr=a\pm b\cos \theta
and
r=a±bsinθr=a\pm b\sin \theta
such that
a>0,b>0a>0,b>0
, and
ab>1\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=acosnθr=a\cos n\theta
and
r=asinnθr=a\sin n\theta
; when
nn
is even there are
2n2n
petals, and the curve is highly symmetrical; when
nn
is odd there are
nn
petals.


Section Exercises

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

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?

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?

For the following exercises, test the equation for symmetry.

6.
r=5cos3θr=5\cos 3\theta


7.
r=33cosθr=3 - 3\cos \theta


8. 
r=3+2sinθr=3+2\sin \theta


9.
r=3sin2θr=3\sin 2\theta


10. 
r=4r=4


11.
r=2θr=2\theta


12. 
r=4cosθ2r=4\cos \frac{\theta }{2}


13.
r=2θr=\frac{2}{\theta }


14. 
r=31cos2θr=3\sqrt{1-{\cos }^{2}\theta }


15.
r=5sin2θr=\sqrt{5\sin 2\theta }


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

16.
r=3cosθr=3\cos \theta


17.
r=4sinθr=4\sin \theta


18.
r=2+2cosθr=2+2\cos \theta


19.
r=22cosθr=2 - 2\cos \theta


20.
r=55sinθr=5 - 5\sin \theta


21.
r=3+3sinθr=3+3\sin \theta


22.
r=3+2sinθr=3+2\sin \theta


23.
r=7+4sinθr=7+4\sin \theta


24.
r=4+3cosθr=4+3\cos \theta


25.
r=5+4cosθr=5+4\cos \theta


26.
r=10+9cosθr=10+9\cos \theta


27.
r=1+3sinθr=1+3\sin \theta


28.
r=2+5sinθr=2+5\sin \theta


29.
r=5+7sinθr=5+7\sin \theta


30.
r=2+4cosθr=2+4\cos \theta


31.
r=5+6cosθr=5+6\cos \theta


32.
r2=36cos(2θ){r}^{2}=36\cos \left(2\theta \right)


33.
r2=10cos(2θ){r}^{2}=10\cos \left(2\theta \right)


34.
r2=4sin(2θ){r}^{2}=4\sin \left(2\theta \right)


35.
r2=10sin(2θ){r}^{2}=10\sin \left(2\theta \right)


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


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


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


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


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


41.
r=θr=-\theta


42.
r=2θr=2\theta


43.
r=3θr=-3\theta


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

44.
r=1θr=\frac{1}{\theta }


45. 
r=1θr=\frac{1}{\sqrt{\theta }}


46.
r=2sinθtanθr=2\sin \theta \tan \theta
, a cissoid

47.
r=21sin2θr=2\sqrt{1-{\sin }^{2}\theta }
, a hippopede

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


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


50.
r=θ2r={\theta }^{2}


51.
r=θ+1r=\theta +1


52.
r=θsinθr=\theta \sin \theta


53.
r=θcosθr=\theta \cos \theta


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

54.
r=θ,r=θr=\theta ,r=-\theta


55.
r=θ,r=θ+sinθr=\theta ,r=\theta +\sin \theta


56. 
r=sinθ+θ,r=sinθθr=\sin \theta +\theta ,r=\sin \theta -\theta


57.
r=2sin(θ2),r=θsin(θ2)r=2\sin \left(\frac{\theta }{2}\right),r=\theta \sin \left(\frac{\theta }{2}\right)


58. 
r=sin(cos(3θ))r=sin(3θ)r=\sin \left(\cos \left(3\theta \right)\right)r=\sin \left(3\theta \right)


59. On a graphing utility, graph
r=sin(165θ)r=\sin \left(\frac{16}{5}\theta \right)
on
[0,4π],[0,8π],[0,12π]\left[0,4\pi \right],\left[0,8\pi \right],\left[0,12\pi \right]
, and
[0,16π]\left[0,16\pi \right]
. Describe the effect of increasing the width of the domain.

60. On a graphing utility, graph and sketch
r=sinθ+(sin(52θ))3r=\sin \theta +{\left(\sin \left(\frac{5}{2}\theta \right)\right)}^{3}
on
[0,4π]\left[0,4\pi \right]
.

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

r1=3sin(3θ)r2=2sin(3θ)r3=sin(3θ)\begin{array}{l}\begin{array}{l}\\ {r}_{1}=3\sin \left(3\theta \right)\end{array}\qquad \\ {r}_{2}=2\sin \left(3\theta \right)\qquad \\ {r}_{3}=\sin \left(3\theta \right)\qquad \end{array}


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

r1=3+3cosθr2=2+2cosθr3=1+cosθ\begin{array}{l}\begin{array}{l}\\ {r}_{1}=3+3\cos \theta \end{array}\qquad \\ {r}_{2}=2+2\cos \theta \qquad \\ {r}_{3}=1+\cos \theta \qquad \end{array}


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

r1=3θr2=2θr3=θ\begin{array}{l}\begin{array}{l}\\ {r}_{1}=3\theta \end{array}\qquad \\ {r}_{2}=2\theta \qquad \\ {r}_{3}=\theta \qquad \end{array}


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

64.
r1=3+2sinθ,r2=2{r}_{1}=3+2\sin \theta ,{r}_{2}=2


65.
r1=64cosθ,r2=4{r}_{1}=6 - 4\cos \theta ,{r}_{2}=4


66. 
r1=1+sinθ,r2=3sinθ{r}_{1}=1+\sin \theta ,{r}_{2}=3\sin \theta


67.
r1=1+cosθ,r2=3cosθ{r}_{1}=1+\cos \theta ,{r}_{2}=3\cos \theta


68. 
r1=cos(2θ),r2=sin(2θ){r}_{1}=\cos \left(2\theta \right),{r}_{2}=\sin \left(2\theta \right)


69.
r1=sin2(2θ),r2=1cos(4θ){r}_{1}={\sin }^{2}\left(2\theta \right),{r}_{2}=1-\cos \left(4\theta \right)


70. 
r1=3,r2=2sin(θ){r}_{1}=\sqrt{3},{r}_{2}=2\sin \left(\theta \right)


71.
r12=sinθ,r22=cosθ{r}_{1}{}^{2}=\sin \theta ,{r}_{2}{}^{2}=\cos \theta


72. 
r1=1+cosθ,r2=1sinθ{r}_{1}=1+\cos \theta ,{r}_{2}=1-\sin \theta


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