# 5.2.5: Equivalent Polar Curves

- Page ID
- 4181

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)While working on a problem in math class, you get a solution with a certain equation. In this case, your solution is \(3+2\cos (\theta )\). Your friend comes over and tells you that he thinks he has solved the problem. However, when he shows you his paper, his equation looks different from yours. His solution is \(−3+2\cos (−\theta )\). Is there a way you can determine if the two equations are equivalent?

The expression “same only different” comes into play in this section. 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.

In some other section, graphs were generated of a limaçon, a dimpled limaçon, a looped limaçon and a cardioid. All of these were of the form \(r=a\pm b\sin \theta \) or \(r=a\pm b\cos \theta \). 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.

#### Comparing Graphs of Polar Equations

Plot the following polar equations and compare the graphs.

\(\begin{aligned} r &=1+2\sin \theta \\ r&=−1+2\sin \theta \end{aligned}\)

By looking at the graphs, the result is the same. So, even though a is different in both, they have the same graph. We can assume that the sign of a does not matter.

\(\begin{aligned} r&=4\cos \theta \\ r&=4\cos (−\theta )\end{aligned} \)

These functions also result in the same graph. Here, \(\theta \) differed by a negative. So we can assume that the sign of \(\theta \) does not change the appearance of the graph.

#### Describing Graphs

1. Graph the equations \(x^2+y^2=16\) and \(r=4\). Describe the graphs.

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

2. Graph the equations \((x−2)^2+(y+2)^2=8\) and \(r=4\cos \theta −4\sin \theta \). Describe the graphs.

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 \(2\sqrt{2}\approx 2.8\).

Earlier, you were asked if there is a way you can determine if the two equations are equivalent.

**Solution**

As you learned in this section, we can compare graphs of equations to see if the equations are the same or not.

A graph of \(3+2\cos (\theta )\) looks like this:

And a graph of \(−3+2\cos (−\theta )\) looks like this:

As you can see from the plots, your friend is correct. Your graph and his are the same, therefore the equations are equivalent.

Write the rectangular equation \(x^2+y^2=6x\) in polar form and graph both equations. Should they be equivalent?

**Solution**

\(\begin{aligned}

x^{2}+y^{2} &=6 x & & \\

r^{2} &=6(r \cos \theta) & & r^{2}=x^{2}+y^{2} \quad \text { and } \quad x=y \cos \theta \\

r &=6 \cos \theta & & \text { divide by } r

\end{aligned}\)

Both equations produced a circle with center \((3,0)\) and a radius of 3.

Determine if \(r=−2+\sin \theta \) and \(r=2−\sin \theta\) are equivalent * without* graphing.

**Solution**

\(r=−2+\sin \theta \) and \(r=2−\sin \theta \) are not equivalent because the sine has the opposite sign. \(r=−2+\sin \theta \) will be primarily above the horizontal axis and \(r=2−\sin \theta \) will be mostly below. However, the two do have the same pole axis intercepts.

Determine if \(r=−3+4\cos (−\pi )\) and \(r=3+4\cos \pi\) are equivalent * without* graphing.

**Solution**

\(r=−3+4\cos (−\pi )\) and \(r=3+4\cos \pi \) are equivalent because the sign of "a" does not matter, nor does the sign of \(\theta \).

### Review

For each equation in rectangular form given below, write the equivalent equation in polar form.

- \(x^2+y^2=4\)
- \(x^2+y^2=6y\)
- \((x−1)^2+y^2=1\)
- \((x−4)^2+(y−1)^2=17\)
- \(x^2+y^2=9\)

For each equation below in polar form, write another equation in polar form that will produce the same graph.

- \(r=4+3\sin \theta\)
- \(r=2−\sin \theta\)
- \(r=2+2\cos \theta\)
- \(r=3−\cos \theta\)
- \(r=2+\sin \theta\)

Determine whether each of the following sets of equations produce equivalent graphs * without* graphing.

- \(r=3−\sin \theta \) and \(r=3+\sin \theta\)
- \(r=1+2\sin \theta \) and \(r=−1+2\sin \theta\)
- \(r=3\sin \theta \) and \(r=3\sin (−\theta )\)
- \(r=2\cos \theta \) and \(r=2\cos (−\theta )\)
- \(r=1+3\cos \theta \) and \(r=1−3\cos \theta\)

### Review (Answers)

To see the Review answers, open this PDF file and look for section 6.8.

## Additional Resources

Practice: Equivalent Polar Curves