Polar functions are described in the form r = f\left( \theta \right), where r is the output and \theta is the input. The output r corresponds to the radius or the distance of the point from the origin, while the input \theta corresponds to the angle from the positive x-axis.

Now, when we are graphing circles and roses using polar functions, there are specific forms these functions take.

A circle on a polar grid centered at the origin with radius labeled r=3

For example, a simple circle with radius a centered at the origin can be represented as r=a. This means for all angles \theta, the distance from the origin remains constant, which is the radius a. This is why it forms a circle.

See the example shown on the graph of the circle r=3.

More generally, circles can be written in polar form as:

r=a \cos \theta \\ \text{or} \\ r=a \sin \theta

Where |a| is the diameter of the circle.

A circle on a polar grid with one side crossing through the origin and the other side crossing through 2 on the horizontal axis. Diameter labeled d=2

r\left( \theta \right)=2 \cos \theta has a diameter of 2 on the horizontal axis

A circle on a polar grid with one side crossing through the origin and the other side crossing through 2 on the vertical axis. Diameter labeled d=2

r\left( \theta \right)=2 \sin \theta has a diameter of 2 on the vertical axis

	r=a \cos \theta	r=a \sin \theta
a>0	\text{right of the origin}	\text{above the origin}
a<0	\text{left of the origin}	\text{below the origin}
\text{symmetry}	\text{across the horizontal axis}	\text{across the vertical axis}

Examples

Example 1

Graph each of the following without using a calculator.

r=4 \sin \theta

Worked Solution

Create a strategy

a=4 which means the circle has a diameter of 4 units. Because 4>0 and this circle uses the sine function this circle will be located above the origin, symmetric about the vertical axis.

Apply the idea

A circle on a polar grid with one side crossing through the origin and the other side crossing through 4 on the vertical axis.

r=-8 \cos \theta

Worked Solution

Create a strategy

a=-8 which means the circle has a diameter of 8 units. Because -8<0 and this circle uses the cosine function this circle will be located to the left of the origin, symmetric about the horizontal axis.

Apply the idea

A circle on a polar grid with one side crossing through the origin and the other side crossing through negative 8 on the horizontal axis.

Idea summary

Equations of circles in polar form are written as: r=a \cos \theta or r= a \sin \theta

	r=a \cos \theta	r=a \sin \theta
a>0	\text{right of the origin}	\text{above the origin}
a<0	\text{left of the origin}	\text{below the origin}
\text{symmetry}	\text{across the horizontal axis}	\text{across the vertical axis}

Roses

Roses are a bit more complex than circles. They are represented by polar functions in the form r = a \cos \left(k\theta \right) or r = a \sin \left(k\theta \right). Here, k is an integer and it determines the number of petals on the rose. If k is odd, the rose will have k petals, but if k is even, the rose will have 2k petals.

A rose with 3 petals graphed on a polar grid with a radius of 2. Centered at the origin with one petal in the first quadrant, one in the second quadrant and one partially in the third and fourth quadrants symmetric about the vertical axis.

r\left( \theta \right)=2 \sin \left(3 \theta \right) has 3 petals and a max radius of 2 units

A rose with 3 petals graphed on a polar grid with a radius of 2. Centered at the origin with one petal in the second quadrant, one in the third quadrant and one partially in the first and fourth quadrants symmetric about the horizontal axis.

r\left( \theta \right)=2 \cos \left(3 \theta \right) has 3 petals and a max radius of 2 units

So when you're given a polar function, remember that the input (the \theta value) affects the angle at which we're looking from the positive x-axis and the output (the r value) affects the distance from the origin. For circles, this distance is constant, but for roses, this distance changes depending on the angle, creating beautiful petal-like shapes. Since the petals are evenly spaced, the angle between the petals can be determined by dividing the full turn around the circle \left(360 \degree \text{ or } 2 \pi \right) by the number of petals.

	\text{both}	r=a \cos \left(k\theta\right)	r=a \sin \left(k\theta\right)
\text{petal length}	\|a\|	\text{ }	\text{ }
\text{odd } k	k \text{ petals}	\text{ }	\text{ }
\text{even } k	2k \text{ petals}	\text{ }	\text{ }
\text{angle between petals}	\dfrac{2 \pi}{k}	\text{ }	\text{ }
\text{1st petal location}	\text{ }	\theta = 0	\theta=\dfrac{k \pi}{2}
\text{symmetry}	\text{ }	\text{across the horizontal axis}	\text{across the vertical axis}

Make sure to keep any reflections in mind when determining characteristics of a graph. A reflection occurs when a is negative.

Examples

Example 2

Without graphing, identify the following information about r=3 \cos \left(5 \theta \right).

Number of petals

Worked Solution

Create a strategy

Recall that k affects the number of petals. k is 5 which is an odd number, so there are k petals.

Apply the idea

5 petals

Angle between petals

Worked Solution

Create a strategy

Since there are 5 petals we will divide a full turn around the circle \left(2 \pi \right) by 5 to find the angle.

Apply the idea

\dfrac{2 \pi}{5} or 72 \degree

Location of 1st petal

Worked Solution

Apply the idea

Because this equation uses cosine the first petal is lotacted at \theta=0.

Symmetry

Worked Solution

Apply the idea

Because this equation uses cosine, it has symmetry across the horizontal axis.

Length of each petal

Worked Solution

Create a strategy

The length of each petal for a rose is a.

Apply the idea

5 units.

Example 3

Consider the equation: r\left(\theta \right)=-4 \sin \left(3 \theta\right)

Identify the characteristics of the r\left(\theta \right) without graphing.

Worked Solution

Apply the idea

We can identify many characteristics by analyzing the equation.

Number of petals: k so there are 3 petals
Petal length: |a|=4 units
Angle between petals: \dfrac{2\pi}{k}=\dfrac{2\pi}{3}
Location of 1st petal: \theta=-\dfrac{k\pi}{2}=-\dfrac{3\pi}{2}=\dfrac{\pi}{2}

Reflect and check

Notice when identifying the location of the 1st petal, the negative comes from the reflection that occurs as a result of a being negative.

Graph r\left(\theta \right).

Worked Solution

Create a strategy

We need to find the location of each petal so we will use the location of the first petal and the angle between petals.

From part (a) the first petal is located at \dfrac{\pi}{2} so we will use that as our starting point.

The angle between petals is \dfrac{2\pi}{3}.

Apply the idea

1st petal:

Located at an angle of \dfrac{\pi}{2} which is on the positive vertical axis. The tip of the first petal lies at a value of 4 because that is the petal length identified in part (a). So the tip of the first petal has coordinates \left(4,\dfrac{\pi}{2}\right).

2nd petal:

Starting at the angle of the first petal, \dfrac{\pi}{2} we add \dfrac{2\pi}{3} to find the angle of the second petal.

\dfrac{\pi}{2}+\dfrac{2\pi}{3}=\dfrac{3\pi}{6}+\dfrac{4\pi}{6}=\dfrac{7\pi}{6} This angle falls in the 3rd quadrant.

this gives the coordinates \left(4,\dfrac{7\pi}{6}\right).

3rd petal:

Starting at the angle of the second petal, \dfrac{7\pi}{6} we add \dfrac{2\pi}{3} to find the angle of the second petal.

\dfrac{7\pi}{6}+\dfrac{2\pi}{3}=\dfrac{7\pi}{6}+\dfrac{4\pi}{6}=\dfrac{11\pi}{6} This angle falls in the 4th quadrant.

this gives the coordinates \left(4,\dfrac{11\pi}{6}\right).

Now that we know where all of the petals are, we can sketch the graph.

A rose with 3 petals graphed on a polar grid with a radius of 4. Centered at the origin with one petal in the third quadrant, one in the fourth quadrant and one partially in the first and second quadrants symmetric about the vertical axis

Idea summary

Equations of roses take the form: r=a \cos \left(k \theta \right) or r=a \sin \left(k \theta \right)

	\text{both}	r=a \cos \left(k\theta\right)	r=a \sin \left(k\theta\right)
\text{petal length}	\|a\|	\text{ }	\text{ }
\text{odd } k	k \text{ petals}	\text{ }	\text{ }
\text{even } k	2k \text{ petals}	\text{ }	\text{ }
\text{angle between petals}	\dfrac{2 \pi}{k}	\text{ }	\text{ }
\text{1st petal location}	\text{ }	\theta = 0	\theta=\dfrac{k \pi}{2}
\text{symmetry}	\text{ }	\text{horizontal axis}	\text{vertical axis}

Limacons

Another type of polar graph is called a limacon. A limacon is a special type of polar graph that is defined by a function of the form r = a \pm b \cos \theta or r = a \pm b \sin \theta, where a and b are constants. Depending on the values of a and b, the limacon can either have a loop ("inner loop") or a dimple ("cardioid").

The input value \theta corresponds to the angle made with the positive x-axis. As we increase or decrease \theta, we effectively rotate around the origin. On the other hand, the output value r corresponds to the distance from the origin. This distance will change based on the value of our function for a given \theta.

A table with columns theta, cosine theta, and r of theta equals 1 plus 2 cosine theta. The first row has values 0, 1, and 3. Second row values are pi over 2, 0 and 1. Third row has values pi, negative 1 and negative 1. Fourth row has values 3 pi over 2, zero, and 3. And the fifth row has values 2 pi, 1, and 3.

Let's make it a bit clearer with an example. Suppose we have the function r\left(\theta\right) = 1 + 2 \cos \theta. Here, as \theta varies from 0 to 2 \pi, the value of \cos \theta will cycle between -1 and 1. This variation in the \cos \theta value affects r, causing it to fluctuate between -1 and 3.

The graph of a limacon on a polar grid. It has a loop that crosses through the origin and the point pi, negative 1. The outer part of the limacon crosses through the points pi over 2, 1. 0, 3. And 3pi over 2, 1. It is symmetric about the horizontal axis.

This fluctuation in r is what gives the limacon its unique shape.

The parameters a and b determine the shape of the limacon.

The graph of a limacon on a polar grid. It has a dent at 3 pi over 2, 1. It crosses the horizontal axis at 6 on both sides and crosses the vertical axis at pi over 2, 11. It is symmetric about the vertical axis.

If |a|>|b| the limacon has a "dent" but no loop.

For example: r=6+5 \sin \theta

The graph of a heart shaped limacon on a polar grid. It crosses the horizontal axis at the origin where the dent of the heart is and at 0,10. It crosses the vertical axis at pi over 2, 5 and at 3pi over 2, 5. It is symmetric about the horizontal axis.

If |a|=|b| the limacon has a heart shape called a cartioid.

For example: r=5+5 \cos \theta

The graph of a limacon on a polar grid. It has a loop that crosses through the origin and the point pi, negative 2. The outer part of the limacon crosses through the points pi over 2, 3. 0, 8. And 3pi over 2, 3. It is symmetric about the horizontal axis.

If |a|<|b| the limacon has a loop.

For example: r=3+5 \cos \theta

The values of a and b also tell us about the radius.

Maximum radius =|a|+|b|

Minimum radius =|a|-|b|

We can also see that \sin and \cos create the same shape of graph, they are just rotations of each other by an angle of \dfrac{\pi}{2}.

r=3+5 \cos \theta

The graph of a limacon on a polar grid. It has a loop that crosses through the origin and the point 3 pi over 2 , negative 2. The outer part of the limacon crosses through the points pi over 2, 8. 0, 3. And pi, 3. It is symmetric about the vertical axis.

r=3+5 \sin \theta

Examples

Example 4

Sketch a graph of the equation: r=4 - 3 \cos \theta

Worked Solution

Create a strategy

We can start by making a table of values to find some key points on the graph. Then we can compare |a| to |b| determine whether the limacon has a dent, a loop, or a heart shape.

Apply the idea

Complete a table of values for \theta=0, \dfrac{\pi}{2}, \pi, \text{ and } \dfrac{3 \pi}{2}

\theta	\cos \theta	r=4-3 \cos \theta
0	1	4-3\cdot1=1
\dfrac{\pi}{2}	0	4-3\cdot0=4
\pi	-1	4-3\cdot -1=7
\dfrac{3 \pi}{2}	0	4-3\cdot0=4
2 \pi	1	4-3\cdot1=1

Now we can graph each of the points \left(0,1 \right),\text{ }\left(\dfrac{\pi}{2},4 \right),\text{ }\left(\pi,7 \right),\text{ }\left(\dfrac{3\pi}{2},4 \right),\text{ }\left(2\pi,1 \right)

A polar grid with the following points graphed: 0,1. pi over 2, 4. pi, 7. 3 pi over 2, 4. And 2pi, 1.

Now, we can compare |a| and |b| to see what kind of shape the graph has before we connect the points.

|a|=|4|=4 and |b|=|3|=3 and 4>3 so this limacon has a dent but no loop. Now we can connect the points on the graph with that in mind.

The graph of a limacon on a polar grid. It has a dent at 0, 1. ANd crosses through the points 0,1. pi over 2, 4. pi, 7. 3 pi over 2, 4. And 2pi, 1. It is symmetric about the horizontal axis.

Example 5

Sketch a graph of the equation: r=-3 + 4\sin \theta

Worked Solution

Create a strategy

We can start by making a table of values to find some key points on the graph. Then we can compare |a| to |b| determine whether the limacon has a dent, a loop, or a heart shape.

Apply the idea

Complete a table of values for \theta=0, \dfrac{\pi}{2}, \pi, \text{ and } \dfrac{3 \pi}{2}

\theta	\sin \theta	r=-3+4 \sin \theta
0	0	-3+4\cdot0=-3
\dfrac{\pi}{2}	1	-3+4\cdot1=1
\pi	0	-3+4\cdot 0=-3
\dfrac{3 \pi}{2}	-1	-3+4\cdot -1=-7
2 \pi	0	-3+4\cdot0=-3

Now we can graph each of the points \left(0,-3 \right),\text{ }\left(\dfrac{\pi}{2},1 \right),\text{ }\left(\pi,-3 \right),\text{ }\left(\dfrac{3\pi}{2},-7 \right),\text{ }\left(2\pi,-3 \right)

A polar grid with the following points graphed: 0,negative 3. pi over 2, 1. pi, negative 3. And 3 pi over 2, negative 7.

Now, we can compare |a| and |b| to see what kind of shape the graph has before we connect the points.

|a|=|-3|=3 and |b|=|4|=4 and 3<4 so this limacon has a loop. Now we can connect the points on the graph with that in mind.

The graph of a limacon on a polar grid. It has a loop that crosses through the origin and the point pi over 2 ,1. The outer part of the limacon crosses through the points 0, negative 3. Pi, negative 3. And 3 pi over 2, negative 7. It is symmetric about the vertical axis.

Idea summary

Equations of limacons take the form: r=a \pm b \cos \theta or r=a \pm b \sin \theta

The values a and b determine the shape of the graph.

If |a|>|b| there is a "dent" but no loop.
If |a|=|b| it is a heart shaped cardioid.
If |a|<|b| there is a loop.
Maximum radius =|a|+|b|
Minimum radius =|a|-|b|

3.14 Polar function graphs

Introduction

Ideas

Circles

Examples

Example 1

Roses

Examples

Example 2

Example 3

Limacons

Examples

Example 4

Example 5

Outcomes

3.14.A

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