# 6.4: More with Polar Curves

**At Grade**Created by: CK-12

## Learning Objectives

- Graph polar curves to see the points of intersection of the curves.
- Graph equivalent polar curves.
- Recognize equivalent polar curves from their equations.

## Intersections of Polar Curves

When you worked with a system of linear equations with two unknowns, finding the point of intersection of the equations meant finding the coordinates of the point that satisfied both equations. If the equations are rectangular equations for curves, determining the point(s) of intersection of the curves involves solving the equations algebraically since each point will have one ordered pair of coordinates associated with it.

**Example 1:** Solve the following system of equations algebraically:

\begin{align*}x^2 + 4y^2 - 36 & = 0 \\
x^2 + y & = 3\end{align*}

**Solution:** Before solving the system, graph the equations to determine the number of points of intersection.

The graph of \begin{align*}x^2 + 4y^2 - 36 = 0\end{align*}

\begin{align*}& x^2 + 4y^2 - 36 = 0 \to x^2 + 4y^2 = 36 && x^2 + 4y^2 + 0y = 36 && \quad x^2 + 4y^2 + 0y = 36 \\
& x^2 + y = 3 \to x^2 + 0y^2 + y = 3 && -1(x^2 + 0y^2 + y = 3) && -x^2 - 0y^2 - y = -3 \\
&&& && \quad \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \\
&&& && \quad 4y^2 - y = 33\end{align*}

Using the quadratic formula, \begin{align*}a = 4 \ b = -1 \ c = -33\end{align*}

\begin{align*}y & = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(4)(-33)}}{2(4)} \\
y & = \frac{1 + 23}{8} = 3 \qquad y = \frac{1 - 23}{8} = -2.75\end{align*}

These values must be substituted into one of the original equations.

\begin{align*}x^2 + y &= 3 && \qquad \ \ x^2 + y = 3 \\
x^2 +3 & = 3 && x^2 + (-2.75) = 3 \\
x^2 & = 0 && \qquad \qquad \ \ x^2 = 5.75 \\
x & = 0 && x = \pm \sqrt{5.75} \ \approx 2.4 \end{align*}

The three points of intersection as determined algebraically in Cartesian representation are \begin{align*}\text{A} (0, 3), \text{B} (2.4, -2.75)\end{align*}

If we are working with polar equations to determine the polar coordinates of a point of intersection, we must remember that there are many polar coordinates that represent the same point. Remember that switching to polar form changes a great deal more than the notation. Unlike the Cartesian system which has one name for each point, the polar system has an infinite number of names for each point. One option would be to convert the polar coordinates to rectangular form and then to convert the coordinates for the intersection points back to polar form. Perhaps the best option would be to explore some examples. As these examples are presented, be sure to use your graphing calculator to create your own visual representations of the equations presented.

**Example 2:** Determine the polar coordinates for the intersection point(s) of the following polar equations: \begin{align*}r = 1\end{align*}

**Solution:** Begin with the graph. Using the process described in the technology section in this chapter; create the graph of these polar equations on your graphing calculator. Once the graphs are on the screen, use the **trace** function and the arrow keys to move the cursor around each graph. As the cursor is moved, you will notice that the equation of the curve is shown in the upper left corner and the values of \begin{align*}\theta, x, y\end{align*}

\begin{align*}& r = 1 && 2 \cos \theta = 1 \\
& r = 2 \cos \theta && \cos \theta = \frac{1}{2} \\
& && \cos^{-1}(\cos \theta) = \cos^{-1} \frac{1}{2} \end{align*}

\begin{align*}\theta = \frac{\pi}{3}\end{align*}

The points of intersection are \begin{align*}\left(1, \frac{\pi}{3} \right)\end{align*} and \begin{align*}\left(1, \frac{5 \pi}{3} \right)\end{align*}. However, these two solutions only cover the possible values \begin{align*}0 \le \theta \le 2 \pi\end{align*}. If you consider that \begin{align*}\cos \theta = \frac{1}{2}\end{align*} is true for an infinite number of theta these solutions must be extended to include \begin{align*}\left (1 , \frac{\pi}{3} \right )\end{align*} and \begin{align*}\left (1, \frac{5 \pi}{3} \right ) + 2 \pi k, k \varepsilon Z.\end{align*} Now the solutions include all possible rotations.

This example was solved as any system of rectangular equations would be solved. Does this approach work all the time?

**Example 3:** Find the intersection of the graphs of \begin{align*}r = \sin \theta\end{align*} and \begin{align*}r = 1 - \sin \theta\end{align*}

**Solution:** Begin with the graph. You can create these graphs using your graphing calculator.

\begin{align*}& r = \sin \theta && \quad \ \sin \theta = 1 - \sin \theta \\ & r = 1 - \sin \theta && \ \ 2 \sin \theta = 1 \\ & && \quad \sin \theta \ = \frac{1}{2}\\ & r = \sin \theta && \theta = \frac{\pi}{6} \ \text{in the first quadrant and}\ \theta = \frac{5 \pi}{6} \ \text{in the second quadrant.}\\ & r = \sin \frac{\pi}{6} && \text{The intersection points are}\ \left ( \frac{1}{2}, \frac{\pi}{6} \right ) \ \text{and} \ \left ( \frac{1}{2}, \frac{5 \pi}{6} \right )\\ & r = \frac{1}{2} && \text{Another intersection point seems to be the origin} \ (0, 0).\end{align*}

If you consider that \begin{align*}\sin \theta = \frac{1}{2}\end{align*} is true for an infinite number of theta as was \begin{align*}\cos \theta = \frac{1}{2}\end{align*} in the previous example, the same consideration must be applied to include all possible solutions. To prove if the origin is indeed an intersection point, we must determine whether or not both curves pass through (0, 0).

\begin{align*}& r = \sin \theta && r = 1 - \sin \theta \\ & 0 = \sin \theta && 0 = 1 - \sin \theta \\ & r = 0 && 1 = \sin \theta \\ &&& \frac{\pi}{2} = 0\end{align*}

From this investigation, the point (0, 0) was on the curve \begin{align*}r = \sin \theta\end{align*} and the point \begin{align*} \left ( 0, \frac{ \pi}{2} \right )\end{align*} was on the curve \begin{align*}r = 1 - \sin \theta\end{align*}. Because the second coordinates are different, it seems that they are two different points. However, the coordinates represent the same point (0,0). The intersection points are \begin{align*} \left( \frac{1}{2}, \frac{\pi}{6} \right), \left( \frac{1}{2}, \frac{5 \pi}{6} \right )\end{align*} and (0,0).

Sometimes it is helpful to convert the equations to rectangular form, solve the system and then convert the polar coordinates back to polar form.

**Example 4:** Find the intersection of the graphs of \begin{align*}r = 2 \cos \theta\end{align*} and \begin{align*}r = 1 + \cos \theta\end{align*}

**Solution:** Begin with the graph:

\begin{align*}r = 2 \cos \theta\end{align*} expressed in rectangular form

\begin{align*}& r = 2 \cos \theta \\ & r^2 = 2r \cos \theta && Multiply \ by \ r \\ & x^2 + y^2 = 2x && Substitution & \end{align*}

\begin{align*}r = 1 + cos \theta\end{align*} expressed in rectangular form

\begin{align*}& r = 1 + \cos \theta \\ & r^2 = r + r \cos \theta && Multiply \ by \ r \\ & x^2 + y^2 = \sqrt{x^2 + y^2} + x && Substitution\end{align*}

The equations are now in rectangular form. Solve the system of equations.

\begin{align*}x^2 + y^2 & = 2x \\ x^2 + y^2 & = \sqrt{x^2 + y^2} + x \\ 2x & = \sqrt{2x} + x \\ x & = \sqrt{2x} \\ x^2 & = 2x \\ x^2 - 2x & = 0 \\ x(x - 2) & = 0 \\ x & = 0 && x - 2 = 0 \\ &&& x = 2\end{align*}

Substituting these values into the first equation:

\begin{align*}x^2 + y^2 & = 2x && \quad x^2 + y^2 = 2x \\ (0)^2 + y^2 & =2(0) && (2)^2 + y^2 =2(2) \\ y^2 & = 0 && \quad 4 + y^2 = 4 \\ y & = 0 && \qquad \quad y^2 = 0 \\ &&& \qquad \quad y = 0\end{align*}

**The points of intersection are** \begin{align*}(0, 0)\end{align*} and \begin{align*}(2, 0)\end{align*}.

The rectangular coordinates are \begin{align*}(0, 0)\end{align*} and \begin{align*}(2, 0)\end{align*}. Converting these coordinates to polar coordinates give the same coordinates in polar form. The points can be converted by using the angle menu of the TI calculator.

**Example 5:** Josie is drawing a mural with polar equations. One mural is represented by the equation \begin{align*}r = 3 \cos \theta\end{align*} and the other by \begin{align*}r = 2 - \cos \theta\end{align*}. She wants to see where they will intersect before she transfers her image onto the wall where she is painting.

**Solution:** To determine where they will intersect, we will begin with a graph.

\begin{align*}r & = 3 \cos \theta \\ r & = 2 - \cos \theta \\ 3 \cos \theta & = 2 - \cos \theta \\ 3 \cos \theta + \cos \theta & = 2 \\ 4 \cos \theta & = 2 \\ \cos \theta & = \frac{2}{4} = \frac{1}{2} \\ \theta & = \frac{\pi}{3} \qquad \text{and} \qquad \theta = \frac{5 \pi}{3}\end{align*}

\begin{align*}& r = 3 \cos \theta && r = 3 \cos \theta \\ & r = 3 \cos \frac{\pi}{3} && r = 3 \cos \frac{5 \pi}{3} \\ & r = 3 \cdot \frac{1}{2} = \frac{3}{2} && r = 3 \cdot \frac{1}{2} = \frac{3}{2}\end{align*}

Josie’s murals would intersect and two points \begin{align*}\left(\frac{3}{2}, \frac{\pi}{3} \right)\end{align*} and \begin{align*}\left(\frac{3}{2}, \frac{5 \pi}{3} \right)\end{align*}.

## Equivalent Polar Curves

The expression “same only different” comes into play in this lesson. We will graph two distinct polar equations that will produce two equivalent graphs. Use your graphing calculator and create these curves as the equations are presented.

Previously, graphs were generated of a limaçon, a dimpled limaçon, a looped limaçon and a cardioid. All of these were of the form \begin{align*}r = a \pm b \sin \theta\end{align*} or \begin{align*}r = a \pm b \cos \theta\end{align*}. The easiest way to see what polar equations produce equivalent curves is to use either a graphing calculator or a software program to generate the graphs of various polar equations.

**Example 6:** Plot the following polar equations and compare the graphs.

a) \begin{align*}r & = 1 + 2 \sin \theta \\ r & = -1 + 2 \sin \theta\end{align*}

b) \begin{align*}r & = 4 \cos \theta \\ r & = 4 \cos (-\theta)\end{align*}

**Solution:** By looking at the graphs, the result is the same. So, even though \begin{align*}a\end{align*} is different in both, they have the same graph. We can assume that the sign of \begin{align*}a\end{align*} does not matter.

b) These functions also result in the same graph. Here, \begin{align*}\theta\end{align*} differed by a negative. So we can assume that the sign of \begin{align*}\theta\end{align*} does not change the appearance of the graph.

**Example 7:**

Graph the equations \begin{align*}x^2 + y^2 = 16\end{align*} and \begin{align*}r = 4\end{align*}. Describe the graphs.

**Solution:**

Both equations, one in rectangular form and one in polar form, are circles with a radius of 4 and center at the origin.

**Example 8:** Graph the equations \begin{align*}(x - 2)^2 + (y + 2)^2 = 8\end{align*} and \begin{align*}r = 4 \cos \theta - 4 \sin \theta\end{align*}. Describe the graphs.

**Solution:** There is not a visual representation shown here, but on your calculator you should see that the graphs are circles centered at (2, -2) with a radius \begin{align*}2 \sqrt{2} \approx 2.8\end{align*}.

## Points to Consider

- When looking for intersections, which representation is easier to work with? Look over the examples and find some in which doing the algebra in polar coordinates is more direct than finding intersections in Cartesian form.
- Will polar curves always intersect?
- If not, when will intersection not occur?
- If two polar curves have different equations, can they be the same curve?

## Review Questions

- Find the intersection of the graphs of \begin{align*}r = \sin 3 \theta\end{align*} and \begin{align*}r = 3 \sin \theta\end{align*}.
- Find the intersection of the graphs of \begin{align*}r = 2 + 2 \sin \theta\end{align*} and \begin{align*}r = 2 - 2 \cos \theta\end{align*}
- Write the rectangular equation \begin{align*}x^2 + y^2 = 6x\end{align*} in polar form and graph both equations. Should they be equivalent?
- Determine if \begin{align*}r = -2 + \sin \theta\end{align*} and \begin{align*}r = 2 - \sin \theta\end{align*} are equivalent
*without*graphing. - Determine if \begin{align*}r = -3 + 4 \cos (- \pi)\end{align*} and \begin{align*}r = 3 + 4 \cos \pi\end{align*} are equivalent
*without*graphing. - Graph the equations \begin{align*}r = 7 - 3 \cos \frac{\pi}{3} \end{align*} and \begin{align*}r = 7 - 3 \cos \left(- \frac{\pi}{3} \right)\end{align*}. Are they equivalent?
- Formulate a theorem about equivalent polar curves resulting from the equations \begin{align*}r = a \pm b \cos \theta\end{align*} or \begin{align*}r = a \pm b \sin \theta\end{align*}. What can be different to yield the same graph? What must be the same? Explain your answer and show graphs to support your conclusions.
- Determine two polar curves that will never intersect.

## Review Answers

There appears to be one point of intersection.

\begin{align*}r & = \sin 3 \ \theta && r = 3 \sin \theta \\ 0 & = \sin 3 \ \theta && 0 = 3 \sin \theta \\ 0 & = \theta && 0 = \sin \theta \\ &&& 0 = \theta\end{align*}

**The point of intersection is** (0, 0)

- \begin{align*}r & = 2 + 2 \sin \theta && r = 2 + 2 \sin \theta \\ r & = 2 + 2 \sin \left( \frac{3 \pi}{4} \right) && r = 2 + 2 \sin \frac{7 \pi}{4} \\ r & \approx 3.4 && r \approx 0.59 \\ r & = 2 + 2 \sin \theta && r = 2 - 2 \cos \theta \\ 0 & = 2 + 2 \sin \theta && 0 = 2 - 2 \cos \theta \\ -1 & = \sin \theta && 1 = \cos \theta \\ \theta & = \frac{3\pi}{2} && \theta = 0\end{align*} Since both equations have a solution at \begin{align*}r = 0\end{align*} , that is (0, \begin{align*}\frac{3\pi}{2}\end{align*}) and (0, 0), respectively, and these two points are equivalent, the two equations will intersect at (0, 0). \begin{align*}r & = 2 + 2 \sin \theta \\ r & = 2 - 2 \cos \theta \\ 2 + 2 \sin \theta & = 2 - 2 \cos \theta \\ 2 \sin \theta & = -2 \cos \theta \\ \frac{2 \sin \theta}{2 \cos \theta} & = -\frac{2 \cos \theta}{2 \cos \theta} \\ \frac{\sin \theta}{\cos \theta} & = -1 \\ \tan \theta & = -1 \\ \theta & = \frac{3 \pi}{4} \ \text{and} \ \theta = \frac{7 \pi}{4}\end{align*} The points of intersection are \begin{align*} \left(3.4, \frac{3 \pi}{4} \right), \left(0.59, \frac{7 \pi}{4} \right)\end{align*} and \begin{align*}(0, 0)\end{align*}.
- \begin{align*}x^2 + y^2 & = 6x \\ r^2 & = 6(r \cos \theta) && r^2 = x^2 + y^2 \qquad \ \text{and} \qquad \ x = y \cos \theta \\ r & = 6 \cos \theta && \text{divide by}\ r \end{align*} Both equations produced a circle with center \begin{align*}(3, 0)\end{align*} and a radius of \begin{align*}3\end{align*}.
- \begin{align*}r = - 2 + \sin \theta\end{align*} and \begin{align*}r = 2 - \sin \theta\end{align*} are not equivalent because the sine has the opposite sign. \begin{align*}r = - 2 + \sin \theta\end{align*} will be primarily above the horizontal axis and \begin{align*}r = 2 - \sin \theta\end{align*} will be mostly below. However, the two do have the same pole axis intercepts.
- \begin{align*}r = -3 + 4 \cos ( - \pi)\end{align*} and \begin{align*}r = 3 + 4 \cos \pi\end{align*} are equivalent because the sign of a does not matter, nor does the sign of \begin{align*}\theta\end{align*}.
- Yes, the equations produced the same graph so they are equivalent.
- Students answers will vary, but they need to include that \begin{align*}b\end{align*} must be the same sign. They should also mention that the sign of \begin{align*}a\end{align*} does not matter, nor does the sign of \begin{align*}\theta\end{align*}.
- There are several answers here. The most obvious are any two pairs of circles, for example \begin{align*}r = 3\end{align*} and \begin{align*}r = 9\end{align*}.

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