Everyone has dreamed of flying at one time or another. Not only would there be much less traffic to worry about, but directions would be so much simpler!

Walking or driving: "Go East 2 blocks, turn left, then North 6 blocks. Wait for the train. Turn right, East 3 more blocks, careful of the cow! Turn left, go North 4 more blocks and park."

Flying: "Fly 30 deg East of North for a little less than 11 and 1/4 blocks. Land."

Nice daydream, but what does it have to do with polar coordinates?

### Polar Coordinates

The **polar coordinate system** is an alternative to the Cartesian coordinate system you have used in the past to graph functions. The polar coordinate system is specialized for visualizing and manipulating angles.

Angles are identified by travelling counter-clockwise around the circular graph from the 0 deg line, or *r-axis* (where the + *x* axis would be) to a specified angle.

To plot a specific point, first go along the *r*-axis by *r* units. Then, rotate counterclockwise by the given angle, commonly represented "θ". Be careful to use the correct units for the angle measure (either radians or degrees).

#### Radians

Usually polar plots are done with radians (especially if they include trigonometric functions), but sometimes degrees are used.

A **radian** is the angle formed between the *r* axis and a polar axis drawn to meet a section of the circumference that is the same length as the radius of a circle.

Given that the circumference of a circle is \begin{align*}2 \pi \cdot r\end{align*}*r* is the radius, that means there are \begin{align*}2 \pi\end{align*} radians in a complete circle, and \begin{align*}1 \pi\end{align*} radians in 1/2 of a circle.

If 1/2 of a circle is \begin{align*}\pi\end{align*} radians, and is 180 deg, that means that there are \begin{align*}\frac{180}{\pi}\end{align*} degrees in each radian.

That translates to approximately 57.3 degrees = 1 radian.

#### Graphing Using Technology

Polar equations can be graphed using a graphing calculator: With the graphing calculator- go to **MODE**. There select **RADIAN** for the angle measure and **POL** (for Polar) on the **FUNC** (function)line. When Y = is pressed, note that the equation has changed from y = to r = . There input the polar equation. After pressing graph, if you can’t see the full graph, adjust *x*- and *y*- max/min, etc in **WINDOW.**

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### Examples

#### Example 1

Plot the points on a polar coordinate graph: Point A \begin{align*}\left (2,\frac{\pi}{3}\right )\end{align*}, Point B \begin{align*}(4, 135^{o})\end{align*}, and Point C \begin{align*}\left (-2,\frac{\pi}{6}\right )\end{align*}.

Below is the pole, polar axis and the points A, B and C.

#### Example 2

Plot the following points.

- \begin{align*}(4, 30^o)\end{align*}
- \begin{align*}(2.5, \pi)\end{align*}
- \begin{align*}\left(-1,\frac{\pi}{3}\right )\end{align*}
- \begin{align*}\left(3,\frac{5\pi}{6}\right )\end{align*}
- \begin{align*}(-2, 300^o)\end{align*}

#### Example 3

Use a graphing calculator or plotting program to plot the following equations.

- \begin{align*}r = 1 + 3 sin \theta\end{align*}

- \begin{align*}r = 1 + 2 cos \theta\end{align*}

Review the steps above under **graphing using technology** if you are having trouble.

#### Example 4

Convert from radians to degrees.

Recall that \begin{align*}\pi rad = 180^o\end{align*} and \begin{align*}1rad = \frac{180}{\pi} \approx 57.3^o\end{align*}.

- \begin{align*}\frac{\pi}{2}\end{align*}

If \begin{align*}\pi rad = 180^o\end{align*} then \begin{align*}\frac{\pi}{2}rad = 90^o\end{align*}

- \begin{align*}5.17\end{align*}

If \begin{align*}1rad \approx 57.3^o\end{align*} then \begin{align*}5.17rad \approx 296^o\end{align*}

- \begin{align*}\frac{3\pi}{2}\end{align*}

If \begin{align*}\pi rad = 180^o\end{align*} then \begin{align*}\frac{3\pi}{2}rad = 270^o\end{align*}

#### Example 5

Convert from degrees to radians.

Recall that \begin{align*}\frac{180^o}{\pi} = 57.3^o \approx 1rad\end{align*}.

- \begin{align*}251^o\end{align*}

If \begin{align*}57.3^o \approx 1rad\end{align*} then \begin{align*}251^o \approx 4.38rad \approx 1.4\pi rad\end{align*}

- \begin{align*}360^o\end{align*}

If \begin{align*}57.3^o \approx 1rad\end{align*} then \begin{align*}360^o \approx 6.28rad\end{align*}

- \begin{align*}327^o\end{align*}

If \begin{align*}57.3^o \approx 1rad\end{align*} then \begin{align*}\frac{327^o}{57.3^o} \approx 5.71 rad\end{align*}

#### Example 6

Convert from degrees to radians, answer in terms of \begin{align*}\pi\end{align*}.

Recall that \begin{align*}2\pi rad = 360^o\end{align*} and therefore \begin{align*}\pi rad = 180^o\end{align*}.

- \begin{align*}90^o\end{align*}

If \begin{align*}\pi rad = 180^o\end{align*} then \begin{align*}\frac{\pi}{2} rad = 90^o\end{align*}

- \begin{align*}270^o\end{align*}

If \begin{align*} \pi rad = 180^o\end{align*} and \begin{align*}\frac{\pi}{2} rad = 90^o\end{align*} then \begin{align*}1\frac{1}{2}\pi rad \to \frac{3}{2}\pi \to \frac{3\pi}{2} rad = 270^o\end{align*}

- \begin{align*}45^o\end{align*}

If \begin{align*}\frac{\pi}{2}rad = 90^o\end{align*} then \begin{align*}\frac{\pi}{4} rad = 45^o\end{align*}

### Review

- Why can a point on the plane not be labeled using a unique ordered pair \begin{align*}(r, \theta)\end{align*}.
- Explain how to graph \begin{align*}(r, \theta)\end{align*} if \begin{align*}r < 0\end{align*} and/or \begin{align*}\theta > 360\end{align*}.

Graph each point on the polar plane.

- A \begin{align*}(6, 145^o)\end{align*}
- B \begin{align*}\left(-2, \frac{13\pi}{6} \right)\end{align*}
- C \begin{align*}\left(\frac{7}{4}, -210^o\right)\end{align*}
- D \begin{align*}\left(5, \frac{\pi}{2}\right)\end{align*}
- E \begin{align*}\left(3.5, \frac{-\pi}{8}\right)\end{align*}

Name two other pairs of polar coordinates for each point.

- \begin{align*}(1.5, 170^o)\end{align*}
- \begin{align*}\left(-5, \frac{\pi}{-3}\right)\end{align*}
- \begin{align*}(3, 305^o)\end{align*}

Graph each polar equation.

- \begin{align*}r = 3\end{align*}
- \begin{align*}\theta = \frac{\pi}{5}\end{align*}
- \begin{align*}r = 15.5\end{align*}
- \begin{align*}r = 1.5\end{align*}
- \begin{align*}\theta = -175^o\end{align*}

Find the distance between the given points.

- \begin{align*}P_1 \left(5, \frac{\pi}{2}\right)\end{align*} and \begin{align*}P_2 \left(7, \frac{3\pi}{9}\right)\end{align*}
- \begin{align*}P_1 (1.3, -52^o) \end{align*} and \begin{align*}P_2 (-13.6, -162^o)\end{align*}
- \begin{align*}P_1 (3, 250^o)\end{align*} and \begin{align*} P_2 (7, 90^o)\end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 4.1.