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# Polar Curves – Definition, Types of Polar Curves, and Examples

**Polar curves**give us a better understanding of how we graph equations with a different coordinate system than what we’re used to. In the past, we’ve learned about polar coordinates, so it’s time for us to expand our graphing skills on the polar coordinate system. By the end of our discussion, you’ll see how complex rectangular equations can be graphed easily with the help of common polar curves.

**As we have mentioned, we can’t graph polar curves without understanding how the polar coordinate system works. Head over to this link in case you need a quick refresher. In this article, we’ll cover all the fundamental concepts we need to understand polar curves:**

*Polar curves are graphs of equations that are defined by polar coordinates. As with regular equations and curves, the polar curve consists of all polar coordinates that satisfy the given equation.*- We’ll do a quick refresher on how we plot polar coordinates on polar grids and extend this knowledge to graph curves.
- You’ll also learn how to test a given polar equation’s symmetry and know how to use the result to graph polar curves.
- We’ll provide you a brief discussion on each of the common polar graphs that you’ll be encountering.

**What is a polar curve?**

**A polar curve is**simply the resulting

**graph of a polar equation defined by**$\boldsymbol{r}$ and $\boldsymbol{\theta}$. We use polar grids or polar planes to plot the polar curve and

**this graph is defined by all sets of**$\boldsymbol{(r, \theta)}$,

**that satisfy the given polar equation,**$\boldsymbol{r = f(\theta)}$.

**the pole**and the horizontal axis extending to the right is called the

**polar axis**. The angles outside are the angles that return exact values and this is why they are great references for polar grids as well. Similar to rectangular coordinate systems, the polar curve of $r = f(\theta)$ will be defined by the polar coordinates satisfying the equation. However, this time, we’re graphing the curve on the polar grid.

**How to graph polar curves?**

Polar curves can be graphed by **plotting key polar coordinates on the polar grid**. This method is similar to graphing rectangular equations –

**we find input and output values that satisfy the given equation**and

**set up a table of values**.

**We’ll break down the steps of graphing polar curves through this method and remember these pointers when graphing polar curves:**

__Graphing polar curves by plotting points__- Construct a partial table of values by assigning different values of $\theta$ from $0$ to $\pi$.
- Since trigonometric functions are periodic, see if you the values for $r$ has repeated.
- Stop when it does and
**start plotting each polar coordinate**(approximate the values of $r$ whenever needed). **Connect the polar coordinates with a curve**and that becomes the polar equation’s graph.

\begin{aligned}\boldsymbol{\theta}\end{aligned} | \begin{aligned}\boldsymbol{r = 2\cos \theta }\end{aligned} | \begin{aligned}\boldsymbol{(r, \theta)}\end{aligned} |

\begin{aligned} 0\end{aligned} | \begin{aligned} 2 \cos 0 = 2(1) = 2\end{aligned} | \begin{aligned}(4, 0) \end{aligned} |

\begin{aligned} \dfrac{\pi}{6}\end{aligned} | \begin{aligned} 2 \cos \dfrac{\pi}{6} = 2\left(\dfrac{\sqrt{3}}{2}\right) = \sqrt{3}\end{aligned} | \begin{aligned} \left(\sqrt{3}, \dfrac{\pi}{6} \right) \approx \left(1.73, \dfrac{\pi}{6}\right)\end{aligned} |

\begin{aligned} \dfrac{\pi}{3}\end{aligned} | \begin{aligned} 2 \cos \dfrac{\pi}{3} = 2\left(\dfrac{1}{2}\right) = 1\end{aligned} | \begin{aligned} \left(1, \dfrac{\pi}{3} \right) \end{aligned} |

\begin{aligned} \dfrac{\pi}{2}\end{aligned} | \begin{aligned} 2 \cos \dfrac{\pi}{2} = 2\left(0\right) = 0\end{aligned} | \begin{aligned} \left(0, \dfrac{\pi}{2} \right) \end{aligned} |

\begin{aligned} \dfrac{2\pi}{3}\end{aligned} | \begin{aligned} 2 \cos \dfrac{2\pi}{3} = 2\left(-\dfrac{1}{2}\right) = -1\end{aligned} | \begin{aligned} \left(-1, \dfrac{2\pi}{3} \right) \end{aligned} |

\begin{aligned} \dfrac{5\pi}{6}\end{aligned} | \begin{aligned} 2 \cos \dfrac{5\pi}{6} = 2\left(-\dfrac{\sqrt{3}}{2}\right) = -\sqrt{3}\end{aligned} | \begin{aligned} \left(-\sqrt{3}, \dfrac{5\pi}{6} \right) \approx \left(-1.73, \dfrac{5\pi}{6}\right) \end{aligned} |

\begin{aligned}\pi\end{aligned} | \begin{aligned} 2 \cos \pi = 2(1) = 2\end{aligned} | \begin{aligned}(2, \pi) \end{aligned} |

**will be a circle with a radius of**$\boldsymbol{\dfrac{a}{2}}$. Don’t worry, we’ve allotted a separate section for these general forms. For now, let’s see how we can use the symmetry of polar curves to graph them faster.

**will be a circle with a radius of**$\boldsymbol{\dfrac{a}{2}}$. Don’t worry, we’ve allotted a separate section for these general forms. For now, let’s see how we can use the symmetry of polar curves to graph them faster.

**Understanding the symmetry of polar curves**

Polar curves may exhibit symmetry over the**polar axis**, the

**line**$\boldsymbol{\theta = \dfrac{\pi}{2}}$, or the

**pole**. Once we establish the rules of symmetry for a polar equation, we can evaluate fewer polar coordinates and simply reflect half the curve over whichever it exhibits symmetry on. These are three symmetry tests you can apply to determine which type of symmetry the polar equation you’re working on exhibits.

Symmetric with Respect
Ty}|to the Polar Axis |
Symmetric with Respect
to the Line $\boldsymbol{\theta = \dfrac{\pi}{2}}$ |
Symmetric with Respect
to the Pole |

\begin{aligned}(r, \theta) \rightarrow (r, -\theta) \end{aligned} · Simply substitute $\theta$ with its negative equivalent, $-\theta$. · If the resulting expression is identical with our original one, the polar curve is symmetric with respect to the polar axis. | \begin{aligned}(-r, \theta) \rightarrow (-r, -\theta) \end{aligned} · Simply substitute $r$ and $\theta$ with their negative equivalents,$-r$ and $-\theta$. · If the resulting expression is identical with our original one, the polar curve is symmetric with respect to the line $\theta = \dfrac{\pi}{2}$. | \begin{aligned}(r, \theta) \rightarrow (-r, \theta) \end{aligned} · Simply substitute $r$ with its negative equivalent, $-r$. · If the resulting expression is the negative equivalent of our original one, the polar curve is symmetric with respect to the pole. |

**1) symmetry with respect to the polar axis**,

**2) symmetry with respect to the line**$\boldsymbol{\theta = \dfrac{\theta}{2}}$, and

**3) symmetry with respect to the pole**.

**Let’s apply what we’ve just learned about symmetries to graph the polar equation, $r = 2\cos 3\theta$. Moving forward, let’s make it a habit to check the polar equation’s symmetry first.**

__Graphing polar curves by its symmetry first__- For the
**polar axis symmetry**:**replace**$\boldsymbol{\theta}$**with**$\boldsymbol{-\theta}$ and see if the resulting expression is still the same. - For the
**line**$\boldsymbol{\theta = \dfrac{\pi}{2}}$:**replace**$\boldsymbol{r}$**and**$\boldsymbol{\theta}$**with**$\boldsymbol{-r}$ and $\boldsymbol{-\theta}$. - For the
**pole**:**replace**$\boldsymbol{r}$**with**$\boldsymbol{-r}$ and see if the resulting expression is still the same.

Symmetry Test:
Polar Axis |
Symmetry Test:
Line $\boldsymbol{\theta = \dfrac{\pi}{2}}$ |
Symmetry Test:
Pole |

\begin{aligned}r &= 2 \cos(-3\theta)\\&= 3 \cos 3\theta \\&\Rightarrow \textbf{Passed}\end{aligned} | \begin{aligned}(-r) &= 3 \cos(-3\theta)\\-r &= 3 \cos 3\theta \\r &= -3\cos 3\theta \\&\Rightarrow \textbf{Failed}\end{aligned} | \begin{aligned}(-r) &= 3 \cos(3\theta)\\-r &= 3 \cos 3\theta \\r &= -3\cos 3\theta \\&\Rightarrow \textbf{Failed}\end{aligned} |

$\boldsymbol{\theta}$ | $0$ | $\dfrac{\pi}{6}$ | $\dfrac{\pi}{4}$ | $\dfrac{\pi}{3}$ | $\dfrac{\pi}{2}$ |

$\boldsymbol{r}$ | $3$ | $0$ | $-2.1$ | $-3$ | $0$ |

**rose**. This shows that the polar equation, $r = 3\cos 3\theta$, is a

**rose with three petals**and is symmetric with respect to the polar axis. In fact, all polar equations of the form, $\boldsymbol{r = a \cos n\theta}$, will

**return a rose with**$\boldsymbol{n}$

**petals**.

**Types of polar graphs**

We’ve already shown you examples of polar curves: a circle, a limacon, and a rose, to be exact. In this section, we’ll summarize all common polar graphs you’ll encounter along with their general forms.
**: If the polar equation has a general form of $r = \dfrac{a}{\sin \theta}$ or $r = \dfrac{a}{\cos \theta}$, where $a \neq 0$, its curve will be a horizontal line or a vertical line, respectively. The line passes through either $(0, a)$ or $(a, 0)$.**

__Lines__**: If the polar equation has a general form of $r = a\cos \theta$ or $r = a\sin \theta$, where $a > 0$, its curve will be a circle with a radius of $\dfrac{a}{2}$.**

__Circles__**: If the polar equation has a general form of $r = a \pm b\cos \theta$ or $r = a \pm b\sin \theta$, where $a,b > 0$, its curve will be a curve called the limacon (it’s French for snail). You’ll encounter four variations of limacons and their shape depends on the value of $\dfrac{a}{b}$.**

__Limacons__**If the polar equation has a general form of $r = a \cos n\theta$ or $r = a\sin n\theta$, where $a \neq 0$, its graph will be a polar curve called the rose. We call each segment of the rose petals and the number of petals will depend on whether $n$ is even or odd.**

__Rose:__- If $n$ is even, the polar curve (rose) will show $2n$ petals.
- If $n$ is odd, the polar curve (rose) will show $n$ petals.

**Lemniscate**

**If the polar equation has a general form of $r^2 = a^2 \cos 2\theta$ or $r = a^2 \sin 2\theta$, where $a \neq 0$, its curve will be called a lemniscate.**

__:__**their general forms and properties will definitely help you save time when graphing different polar equations**. We’ve prepared more sample problems for you to work on and test your understanding of polar graphs. When you’re ready, head over to the next section!

**\begin{aligned}r =2- 4\cos\theta &, r= 1+ \sin \theta \\r = 4\cos 6\theta&, r =3\cos\theta \end{aligned} The polar curves of these four polar equations are as shown below. Match the polar equations with their corresponding polar curve.**

*Example 1*__Solution__There is no need for us to graph each polar equation individually. That’s the benefit of knowing the common polar graphs’ general forms. We’ll identify the polar equation’s general form then use that to determine its polar curve’s shape. Starting with $\boldsymbol{r =2- 4\cos\theta}$, we can see that its general form is $r = a – b \cos \theta$.

- This general form would return a limacon. Now, let’s take a look at $\dfrac{a}{b} = \dfrac{1}{2}$.
- Since this is less than $1$, the limacon will have an inner loop with a length of $1$ unit.
- This means that $r = 2 – 4\cos \theta$ will have a limacon with an inner loop making
**C the correct polar curve**representing it.

Polar Equation |
General Form |
Polar Curve |

\begin{aligned} r= 2- 4\cos \theta\end{aligned} | \begin{aligned}a + b \cos \theta &\Rightarrow \text{Limacon}\\ \dfrac{a}{b} = \dfrac{1}{2} &\Rightarrow \text{with Inner Loop} \end{aligned} | \begin{aligned}\textbf{C}\end{aligned} |

\begin{aligned} r= 1+ \sin \theta\end{aligned} | \begin{aligned}a + b \sin \theta &\Rightarrow \text{Limacon}\\ \dfrac{a}{b} = 1 &\Rightarrow \text{Dimpled Limacon} \end{aligned} | \begin{aligned}\textbf{A}\end{aligned} |

\begin{aligned} r =4\cos 6\theta\end{aligned} | \begin{aligned} a\cos n\theta &\Rightarrow \text{Rose}\\ n = 6 &\Rightarrow \text{12 Petals} \end{aligned} | \begin{aligned}\textbf{D}\end{aligned} |

\begin{aligned} r = 3\cos \theta\end{aligned} | \begin{aligned} a\cos \theta &\Rightarrow \text{Circle}\\ a = 3 &\Rightarrow \text{Radius of 1.5} \end{aligned} | \begin{aligned}\textbf{B}\end{aligned} |

**Test whether $r = 4 \cos \theta$ is symmetric with respect to the polar axis, the line $\theta = \dfrac{\theta}{2}$, or the pole.**

*Example 2*__Solution__When testing for symmetry with respect to the pole, we replace $\theta$ with $-\theta$ and see if it still returns the original polar equation. \begin{aligned}r &= 4 \cos (-\theta)\\&= 4\cos \theta\\ &\Rightarrow \textbf{Passed} \end{aligned} This means that $r = 4\cos\theta $ is symmetric with respect to the polar axis. Now, let’s test $r = 4\cos \theta$ for symmetry with respect to the line $\theta = \dfrac{\pi}{2}$ by replacing both $r$ and $\theta$ with $-r$ and $-\theta$. \begin{aligned}(-r) &= 4 \cos (-\theta)\\ -r &= 4\cos \theta\\r &= -4\cos \theta &\Rightarrow \textbf{Failed} \end{aligned} Using the symmetry tests, we can’t guarantee that the polar equation is symmetric with respect to the pole. Keep in mind that failing the particular symmetry test is inconclusive. We’ll now work on the third symmetry test for the pole – replace $r$ with $-r$. \begin{aligned}(-r) &= 4 \cos \theta\\ r &= -4\cos \theta &\Rightarrow \textbf{Failed} \end{aligned} The polar equation, $r =4\cos \theta$ didn’t pass the third symmetry test. This means that we can only guarantee that the polar equation is symmetric with

**respect to the polar axis**.

**Test $r^2 = 9\cos 2\theta$ for symmetry then graph its polar curve.**

*Example 3*__Solution__Use the three symmetry tests to see if $r^2 =9\cos2\theta$ is symmetric with respect to either the polar axis, the line $\theta=\dfrac{\pi}{2}$, or the pole.

Symmetry Test:
Polar Axis |
Symmetry Test:
Line $\boldsymbol{\theta = \dfrac{\pi}{2}}$ |
Symmetry Test:
Pole |

\begin{aligned}r^2 &= 9 \cos 2(-\theta)\\&= 9 \cos 2\theta \\&\Rightarrow \textbf{Passed}\end{aligned} | \begin{aligned}(-r)^2&= 9 \cos 2(-\theta)\\r^2 &= 9\cos 2\theta \\&\Rightarrow \textbf{Passed}\end{aligned} | \begin{aligned}(-r)^2 &= 9 \cos 2(-\theta)\\r^2 &= 9\cos 2\theta \\&\Rightarrow \textbf{Passed}\end{aligned} |

$\boldsymbol{\theta}$ | $0$ | $\dfrac{\pi}{6}$ | $\dfrac{\pi}{4}$ |

$\boldsymbol{r}$ | $\pm3$ | $\pm2.1$ | $0$ |

### Practice Questions

1. \begin{aligned}r =3\sin 5\theta &, r= 3\sin 2\theta \\r = 1 – 3\sin \theta&, r^2 = 25\sin 2\theta \end{aligned} The polar curves of these four polar equations are as shown below. Match the polar equations with their corresponding polar curve.### Answer Key

1. $\begin{aligned}r &=3\sin 5\theta \Rightarrow\textbf{A} \\r &= 3\sin 2\theta\Rightarrow\textbf{D} \\r &= 1 – 3\sin \theta \Rightarrow\textbf{C}\\r^2 &= 25\sin 2\theta\Rightarrow\textbf{B} \end{aligned}$ 2. $r^2 = 16\sin 2\theta$ is symmetric with respect to the pole. 3. a. Symmetric with respect to the polar axis.*Images/mathematical drawings are created with GeoGebra.*

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