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4.3: Systems of Polar Equations

Difficulty Level: At Grade Created by: CK-12
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Learning Objectives

  • Solve systems of polar equations graphically, numerically, and analytically
  • Find all intersections of two polar equations graphed on the same set of axis
  • Plot polar functions using a table of values
  • Recognize equivalent polar equations


Polar equations can be graphed using polar coordinates. Graphing two polar equations on the same set of axis may result in having point(s) of intersection.

All points on a polar graph are coordinates that make the equation valid. The coordinates of point(s) of intersection when substituted into each equation will make both of the equations valid.

Graph and Calculate Intersections of Polar Curves

One method to find point(s) of intersection for two polar graphs is by setting the equations equal to each other. Call the first equation r1 and the second equation r2 .Points of intersection are when r1 = r2, so set the equations equal and then solve the resulting trigonometric equation.

Example: Find the intersection of \begin{align*}r_1 = 3 \ \mbox{sin} \theta \qquad r_2=\sqrt{3} \mbox{ cos}\ \theta \end{align*}r1=3 sinθr2=3 cos θ

First set them equal

\begin{align*}3\ \mbox{sin}\ \theta =\sqrt{3}\ \mbox{cos}\ \theta \end{align*}3 sin θ=3 cos θ

divide both sides of the equation by cos θ and 3:

\begin{align*}\frac{3\ \mbox{sin}\ \theta}{3\ \mbox{cos}\ \theta} = \frac{\sqrt{3}\ \mbox{cos}\ \theta}{3\ \mbox{cos}\ \theta}\end{align*}3 sin θ3 cos θ=3 cos θ3 cos θ

and this simplifies to

\begin{align*}\frac{\mbox{sin}\ \theta}{\mbox{cos}\ \theta} = \frac{\sqrt{3}}{3}\end{align*}sin θcos θ=33

and then use the identity \begin{align*}\frac{\mbox{sin}\ \theta}{\mbox{cos}\ \theta} = \mbox{tan}\ \theta\end{align*}sin θcos θ=tan θ

\begin{align*}\mbox{tan}\ \theta = \frac{\sqrt{3}}{3}\end{align*}tan θ=33

\begin{align*}\theta = \frac{\pi}{6}\end{align*}θ=π6 or \begin{align*}\frac{7 \pi}{6}\end{align*}7π6

substitute \begin{align*}\frac{\pi}{6}\end{align*}π6 in either equation to obtain r = 1.5

substitute \begin{align*}\frac{7\pi}{6}\end{align*}7π6 in either equation to obtain -1.5

Note: the coordinates\begin{align*}\left (1.5,\frac{\pi}{6}\right )\end{align*}(1.5,π6) and \begin{align*}\left (-1.5,\frac{7\pi}{6}\right)\end{align*} represent the SAME polar point so there is only one solution to this equation.

Are we done? If we look at the graphs of r1 and r2, there is another point of intersection

when θ = 0,

r1 = 3 sin θ = 3 sin(0) = 0

that is r1 = 3 sin θ goes through the pole (0, 0).

For r2 when \begin{align*} \theta = \frac{\pi}{2}\end{align*} , r2 = 0 that is \begin{align*}r_2 = \sqrt{3}\ \mbox{cos}\ \theta\end{align*} goes through the point \begin{align*}\left (0,\frac{\pi}{2}\right )\end{align*}. Therefore, both graphs go through the pole and the pole is a point of intersection.

The pole was NOT revealed as a point of intersection using the first step! (Why? Hint: How many ways are there to represent the pole in polar coordinates?) This shows us that after you use algebraic methods to find intersections at points other than the pole, you should also check for intersections at the pole.

Example: Find the point(s) of intersection for the two graphs:

r1 = 1

r2 = 2 sin 2θ

1. First set r1 = r2 and solve:

\begin{align*}1 = 2\ \mbox{sin}\ 2\theta\end{align*}

\begin{align*}\frac{1}{2} = \mbox{sin}\ 2\theta\end{align*}

Now, use a substitution α = 2θ to solve

\begin{align*}\frac{1}{2} = \mbox{sin}\ \alpha \end{align*}

\begin{align*}\alpha = \frac{\pi}{6}, \frac{5\pi}{6}\end{align*}

Since α = 2θ , solving for θ gives us

\begin{align*}\theta = \frac{\pi}{12}, \frac{5\pi}{12}\end{align*}

But, recall that θ has the range 0 ≤ θ ≤ 2π. Since we solved with 0 ≤ α ≤ 2π we actually need to consider values of θ with 0 ≤ θ ≤ 4π. Why? Recall that sin(2θ) has two cycles between 0 and 2π, and so we add two more solutions,

\begin{align*}\alpha = \frac{13\pi}{6}, \frac{17\pi}{6}\end{align*}

and since α = 2θ,

\begin{align*}\theta = \frac{13\pi}{12}, \frac{17\pi}{12}\end{align*}

Finally, we need to consider solutions when r = -1 because r = 1 and r = -1 are the same polar equation. So, solving

\begin{align*}-\frac{1}{2} = \mbox{sin}\ \alpha\end{align*}

\begin{align*}\alpha = \frac{7\pi}{6}, \frac{11\pi}{6}\end{align*}

Again, using α = 2θ and adding solutions for the repetition gives us four more solutions,

\begin{align*}\theta = \frac{7\pi}{12}, \frac{11\pi}{12}, \frac{19\pi}{12}, \frac{21\pi}{12}\end{align*}

So in total, there are eight solutions to this set of equations.

Note: Recall-when solving trigonometric equations where the angle is θ requires looking at all potential values between 0 and 2π. When the angle is 2θ as it is in this case, be sure to look for all potential values between 0 and 4π. When the angle is 3θ as it is in this case, be sure to look for all potential values between 0 and 6π., and so on.

2. Since r1 cannot equal 0, the pole is not on its graph and not a point of intersection.

3. The graph reveals eight points of intersection which were found in step (1).

Equivalent Polar Curves

Recall from trigonometry that the same trigonometric function can often be written in multiple ways. For example, above you worked with r = 2 sin (2θ). Recall the identity

sin (2θ) = 2 sin θ cos θ.

Substituting that identity into the equation yields

r = 2 sin (2θ)

r = 2 (2 sin θ cos θ)

r = 4 sin θ cos θ

Thus, the two equations

r = 2 sin (2θ) and r = 4 sin θ cos θ are equivalent. To verify this, try graphing both equations on the same axes.

We also discovered another interesting issue with polar graphing above: the equation r = 1 makes a circle of radius 1 centered on the origin, and so does the equation r = -1. Thus r = 1 is the same curve as r = -1. In general r = k is equivalent to r = -k.

Both of these examples highlight that polar equations can be tricky. When you solve problems involving polar equations, you need to consider ALL solutions to the equation, not just the immediate solutions from your first attempt at solving. Graphing polar equations is a good way to check for all solutions.

Given Two Polar Curves, Find All Intersection Points

Example: Find point(s) of intersection, if any exist, for the following pair of equations:

a. r1 = 2 and r2 = secθ

Here we will use a table of values for each function, solving by quadrant. Recall that the period of sec θ is 2π.

For the first quadrant

θ (angle) 0 π/6 π/4 π/3 π/2
r1 (distance) 2 2 2 2 2
r2 1 1.15 1.4 2 und

For the second quadrant

θ (angle) 2π/3 3π/4 5π/6 π
r1 (distance) 2 2 2 2
r2 -2 -1.4 -1.15 -1

For the third quadrant:

θ (angle) 7π/6 5π/4 4π/3 3π/2
r1 (distance) 2 2 2 2
r2 -1.15 -1.4 -2 und

For the fourth quadrant

θ (angle) 5π/3 7π/4 11π/6
r1 (distance) 2 2 2 2
r2 2 1.4 1.15 1

Note that the 3rd and 4th quadrant repeat 1st and 2nd quadrant values.

Observe in the table of values that (2, π/3) and (2, 5π/3) are the points of intersection. Look at the curves- the first equation yields a circle while the second yields a line. The maximum number of intersecting points for a line and a circle is 2. The two points have been found.

Another way to solve polar equations analytically is to set both of the equations equal to each other, that is let r1 = r2 , or equivalently

2 = sec θ.

Using rules from trigonometry this equation can be written as

½ = cos θ,

by taking the reciprocal of both sides.

Solving for cos θ = ½ , θ = π/3 or 5π/3. These are the same values for θ the table of values yield. Point(s) of intersection for Polar curves drawn on the same axis can be found by equating their equations and then solving using the methods for solving trigonometric equations.

Example: Find the point(s) of intersection for these pair of Polar equations:

b. r = 2 + 4 sinθ and θ = 60°

The equation θ = 60o is a line making a 60° angle with the r axis. Now we make a table of values for r = 2 + 4 sin θ

For the first quadrant:

θ (angle) 0 30 45 60 90
R (distance) 2 4 4.83 5.46 6

For the second quadrant:

θ (angle) 120 135 150 180
R (distance) 5.46 4.83 4 2

For the third quadrant:

θ (angle) 210 225 240 270
R (distance) 0 -.83 -1.46 -2

For the fourth quadrant

θ (angle) 300 315 330 360
R (distance) -1.46 -.83 0 2

Notice there are two solutions in the table., (60, 5.46) and (240, -1.46) = (60, 1.46). Recall that when r < 0, you plot a point (r, θ), by rotating 180o (or π).

Finally, we need to check the pole: r = 2 + 4 sin θ passes through the pole for θ = 330o, and θ = 60o also passes through the pole. Thus, the third point of intersection is (0, 0).

Example: r1 = 2 cosθ and r2 = 1.

Here we make a table:

For the first quadrant:

θ (angle) 0 π/6 π/4 π/3 π/2
r1 (distance) 2 \begin{align*}\sqrt{3}\end{align*} \begin{align*}\sqrt{2}\end{align*} 1 0
r2 1 1 1 1 1

For the second quadrant:

θ (angle) 2π/3 3π/4 5π/6 π
r1 (distance) -1 \begin{align*}-\sqrt{2}\end{align*} \begin{align*}-\sqrt{3}\end{align*} -2
r2 1 1 1 1

For the third quadrant:

θ (angle) 7π/6 5π/4 4π/3 3π/2
r1 (distance) \begin{align*}-\sqrt{3}\end{align*} \begin{align*}-\sqrt{2}\end{align*} -1 0
r2 1 1 1 1

For the fourth quadrant:

θ (angle) 5π/3 7π/4 11π/6
r1 (distance) 1 \begin{align*}\sqrt2\end{align*} \begin{align*}\sqrt3\end{align*} 2
r2 1 1 1 1

So the unique solutions are at \begin{align*}\theta = \frac{\pi}{3} , \frac{4\pi}{3}\end{align*}. There are also two repeated solutions in this set (can you find them?).

Here is a graph showing the two solutions:

Applications, Technological Tools

To graph a Polar equation using a TI calculator:

Press: “[MODE]” On the 4th line be sure to Press “[POL]” (for Polar)

Then Press “[Y=]” Note: the “y” has now been replaced with “r”. Input the equation. Note: When in polar mode, pressing “X, T, θ, n” key, θ is the default

Note: There may be a need to adjust, as always, the window values, in order to see the graph within the constraints of the calculator window.

Lesson Summary

1. One method for finding point(s) if intersection for two graphs is to plot points for both graphs on the same axis. A second method, is to let the value for r1 equation equal value for r2 .

The procedure for finding the point or points of intersection for two polar graphs is almost the same as it is for rectangular graphs.

2. The pole can be represented by many different polar coordinates (example: (0, 0) or (0, 90º) or \begin{align*}\left (0 \ \frac{-\pi}{4} ,\right )\end{align*}, it is important to check if both graphs go through the pole.

3. Analytically graph the two equations on the same polar axis to check if all of the point(s) of intersection have been found in steps (1) and (2).

Points to Consider

The rectangular equation x = 2 and the polar equation r = 2 cscθ, reveal the same curve. Yet in this example, the rectangular form has only one variable in the equation, whereas the polar form has two variables. Although the polar form may be useful at times, the rectangular form is easier to work with.

x2 + y2 = 1 and r = 1 also reveal the same curve. Here, the polar form is more simplified since it only contains one variable. What this reveals is that sometimes one form of an equation may be more useful in terms of simplicity to work with, than the other form.

Being able to recognize that two apparently different equations may reveal the same curve can be difficult at first, but having this knowledge provides two potentially useful ways to work with a problem. Either form can be used, depending on which provides the simplest or fastest method. Furthermore, having two equivalent methods offers a unified understanding of the mathematics involved which can provide insights that may not be achieved when thinking within the context of any one individual system.

Review Questions

  1. Find the Point(s) of intersection for each system of equations. Graph to verify your solution.
  2. a. r1 = csc θ r2 = 2 sin θ
    b. r1 = cos θ r2 = 1 + sin θ
    c. r1 = sin θ r2 = sin 2θ
    d. r1 = -4 sin θ r2 = -4 cos θ
    e. r1 = 1 - 2 sin θ \begin{align*}r_2 =\sqrt {9\ \mbox{cos}(\theta)}\end{align*}
    f. r1 = 1 - cos θ r = 4 cos(3θ)

Review Answers

  1. a. \begin{align*}\left ({\sqrt2}, \frac{\pi}{4}\right ), \ \left ({\sqrt2}, \frac{3\pi}{4}\right )\end{align*} b. (0, 0), (1, 0) c. (0, 0), \begin{align*}\left (\frac{\sqrt3}{2}, \frac{2\pi}{3}\right ), \ \left (\frac{\sqrt3}{2}, \frac{\pi}{3}\right )\end{align*} d. (0, 0), \begin{align*}\left ({2\sqrt2}, \ \frac{5\pi}{4}\right )\end{align*} e. (1, 276o), (2.44, 313o) f. (0, 0), (1.08, 95o), (1.77, 142o), (1.77, 218o), (1.08, 265o)

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