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# 6.1: Polar Coordinates

Difficulty Level: At Grade Created by: CK-12

## Introduction

This chapter introduces and explores the polar coordinate system, which is based on a radius and theta. Students will learn how to plot points and basic graphs in this system as well as convert x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} coordinates into polar coordinates and vise versa. We will explore the different graphs that can be generated in the polar system and also use polar coordinates to better understand different aspects of complex numbers.

## Learning Objectives

• Distinguish between and understand the difference between a rectangular coordinate system and a polar coordinate system.
• Plot points with polar coordinates on a polar plane.

## Plotting Polar Coordinates

The graph paper that you have used for plotting points and sketching graphs has been rectangular grid paper. All points were plotted in a rectangular form (x,y)\begin{align*}(x, y)\end{align*} by referring to a perpendicular x\begin{align*}x-\end{align*} and y\begin{align*}y-\end{align*}axis. In this section you will discover an alternative to graphing on rectangular grid paper – graphing on circular grid paper.

Look at the two options below:

You are all familiar with the rectangular grid paper shown above. However, the circular paper lends itself to new discoveries. The paper consists of a series of concentric circles-circles that share a common center. The common center O\begin{align*}O\end{align*}, is known as the pole or origin and the polar axis is the horizontal line r\begin{align*}r\end{align*} that is drawn from the pole in a positive direction. The point P\begin{align*}P\end{align*} that is plotted is described as a directed distance r\begin{align*}r\end{align*} from the pole and by the angle that OP¯¯¯¯¯¯¯¯\begin{align*}\overline{OP}\end{align*} makes with the polar axis. The coordinates of P\begin{align*}P\end{align*} are (r,θ)\begin{align*}(r, \theta)\end{align*}.

These coordinates are the result of assuming that the angle is rotated counterclockwise. If the angle were rotated clockwise then the coordinates of P\begin{align*}P\end{align*} would be (r,θ)\begin{align*}(r, -\theta)\end{align*}. These values for P\begin{align*}P\end{align*} are called polar coordinates and are of the form P(r,θ)\begin{align*}P(r, \theta)\end{align*} where r\begin{align*}r\end{align*} is the absolute value of the distance from the pole to P\begin{align*}P\end{align*} and θ\begin{align*}\theta\end{align*} is the angle formed by the polar axis and the terminal arm OP¯¯¯¯¯¯¯¯\begin{align*}\overline{OP}\end{align*}.

Example 1: Plot the point A(5,255)\begin{align*}A (5, -255^\circ)\end{align*} and the point B(3,60)\begin{align*}B (3, 60^\circ)\end{align*}

Solution, A: To plot A\begin{align*}A\end{align*}, move from the pole to the circle that has r=5\begin{align*}r = 5\end{align*} and then rotate 255\begin{align*}255^\circ\end{align*} clockwise from the polar axis and plot the point on the circle. Label it A\begin{align*}A\end{align*}.

Solution, B: To plot B\begin{align*}B\end{align*}, move from the pole to the circle that has r=3\begin{align*}r = 3\end{align*} and then rotate 60\begin{align*}60^\circ\end{align*} counter clockwise from the polar axis and plot the point on the circle. Label it B\begin{align*}B\end{align*}.

These points that you have plotted have r\begin{align*}r\end{align*} values that are greater than zero. How would you plot a polar point in which the value of r\begin{align*}r\end{align*} is less than zero? How could you plot these points if you did not have polar paper? If you were asked to plot the point (1,135)\begin{align*}(-1, 135^\circ)\end{align*} or (1,3π4)\begin{align*}\left (-1, \frac{3\pi}{4} \right )\end{align*} you would rotate the terminal arm OP¯¯¯¯¯¯¯¯\begin{align*}\overline{OP}\end{align*} counterclockwise 135\begin{align*}135^\circ\end{align*} or 3π4\begin{align*}\frac{3\pi}{4}\end{align*}. (Remember that the angle can be expressed in either degrees or radians). To accommodate r=1\begin{align*}r = -1\end{align*}, extend the terminal arm OP¯¯¯¯¯¯¯¯\begin{align*}\overline{OP}\end{align*} in the opposite direction the number of units equal to |r|\begin{align*}|r|\end{align*}. Label this point M\begin{align*}M\end{align*} or whatever letter you choose. The point can be plotted, without polar paper, as a rotation about the pole as shown below.

The point is reflected across the pole to point M\begin{align*}M\end{align*}.

There are multiple representations for the coordinates of a polar point P(r,θ)\begin{align*}P(r, \theta)\end{align*}. If the point P\begin{align*}P\end{align*} has polar coordinates (r,θ)\begin{align*}(r, \theta)\end{align*}, then P\begin{align*}P\end{align*} can also be represented by polar coordinates (r,θ+360k)\begin{align*}(r, \theta + 360k^\circ)\end{align*} or (r,θ+[2k+1]180)\begin{align*}(-r, \theta + [2k + 1] 180^\circ)\end{align*} if θ\begin{align*}\theta\end{align*} is measured in degrees or by (r,θ+2πk)\begin{align*}(r, \theta + 2 \pi k)\end{align*} or \begin{align*}(-r, \theta + [2k + 1] \pi)\end{align*} if \begin{align*}\theta\end{align*} is measured in radians. Remember that \begin{align*}k\end{align*} is any integer and represents the number of rotations around the pole. Unless there is a restriction placed upon \begin{align*}\theta\end{align*}, there will be an infinite number of polar coordinates for \begin{align*}P(r, \theta)\end{align*}.

Example 2: Determine four pairs of polar coordinates that represent the following point \begin{align*}P(r, \theta)\end{align*} such that \begin{align*}-360^\circ \le \theta \le 360^\circ\end{align*}.

Solution: Pair 1 \begin{align*}\rightarrow (4, 120^\circ)\end{align*}. Pair 2 \begin{align*}\rightarrow (4, -240^\circ)\end{align*} comes from using \begin{align*}k = -1\end{align*} and \begin{align*}(r, \theta + 360^\circ k), (4, 120^\circ + 360(-1))\end{align*}. Pair 3 \begin{align*}\rightarrow (-4, 300^\circ)\end{align*} comes from using \begin{align*}k = 0\end{align*} and \begin{align*}(-r, \theta + [2k + 1] 180^\circ), (-4, 120^\circ + [2(0) + 1] 180^\circ)\end{align*}. Pair 4 \begin{align*}\rightarrow (-4, -60^\circ)\end{align*} comes from using \begin{align*}k = -1\end{align*} and \begin{align*}(-r, \theta + [2k + 1] 180^\circ), (-4, 120^\circ + [2(-1) + 1] 180^\circ)\end{align*}.

These four pairs of polar coordinates all represent the same point \begin{align*}P\end{align*}. You can apply the same procedure to determine polar coordinates of points that have \begin{align*}\theta\end{align*} measured in radians. This will be an exercise for you to do at the end of the lesson.

## The Distance between Two Polar Coordinates

Just like the Distance Formula for \begin{align*}x\end{align*} and \begin{align*}y\end{align*} coordinates, there is a way to find the distance between two polar coordinates. One way that we know how to find distance, or length, is the Law of Cosines, \begin{align*}a^2 = b^2 + c^2 - 2bc \cos A\end{align*} or \begin{align*}a = \sqrt{b^2 + c^2 - 2bc \cos A}\end{align*}. If we have two points \begin{align*}(r_1, \theta_1)\end{align*} and \begin{align*}(r_2, \theta_2)\end{align*}, we can easily substitute \begin{align*}r_1\end{align*} for \begin{align*}b\end{align*} and \begin{align*}r_2\end{align*} for \begin{align*}c\end{align*}. As for \begin{align*}A\end{align*}, it needs to be the angle between the two radii, or \begin{align*}(\theta_2 - \theta_1)\end{align*}. Finally, \begin{align*}a\end{align*} is now distance and you have \begin{align*}d = \sqrt{r^2_1 + r^2_2 - 2 r_1 r_2 \cos (\theta_2 - \theta_1)}\end{align*}.

Example 3: Find the distance between \begin{align*}(3, 60^\circ)\end{align*} and \begin{align*}(5, 145^\circ)\end{align*}.

Solution: After graphing these two points, we have a triangle. Using the new Polar Distance Formula, we have \begin{align*}d = \sqrt{3^2 + 5^2 - 2(3)(5) \cos 85^\circ} \approx 5.6\end{align*}.

Example 4: Find the distance between \begin{align*}(9, -45^\circ)\end{align*} and \begin{align*}(-4, 70^\circ)\end{align*}.

Solution: This one is a little trickier than the last example because we have negatives. The first point would be plotted in the fourth quadrant and is equivalent to \begin{align*}(9, 315^\circ)\end{align*}. The second point would be \begin{align*}(4, 70^\circ)\end{align*} reflected across the pole, or \begin{align*}(4, 250^\circ)\end{align*}. Use these two values of \begin{align*}\theta\end{align*} for the formula. Also, the radii should always be positive when put into the formula. That being said, the distance is \begin{align*}d = \sqrt{9^2 + 4^2 - 2(9)(4) \cos (315-250)^\circ} \approx 8.16\end{align*}.

## Points to Consider

• How is the polar coordinate system similar/different from the rectangular coordinate system?
• How do you plot a point on a polar coordinate grid?
• How do you determine the coordinates of a point on a polar grid?
• How do you calculate the distance between two points that have polar coordinates?

## Review Questions

1. Graph each point:
1. \begin{align*}M (2.5, 210^\circ)\end{align*}
2. \begin{align*}S \left (-3.5, \frac{5 \pi}{6} \right )\end{align*}
3. \begin{align*}A \left (1, \frac{3 \pi}{4} \right )\end{align*}
4. \begin{align*}Y \left (5.25, - \frac{\pi}{3} \right )\end{align*}
2. For the given point \begin{align*}A \left (-4, \frac{\pi}{4} \right )\end{align*}, list three different pairs of polar coordinates that represent this point such that \begin{align*}-2\pi \le \theta \le 2\pi\end{align*}.
3. For the given point \begin{align*}B (2, 120^\circ)\end{align*}, list three different pairs of polar coordinates that represent this point such that \begin{align*}-2\pi < \theta < 2 \pi\end{align*}.
4. Given \begin{align*}P_1\end{align*} and \begin{align*}P_2\end{align*}, calculate the distance between the points.
1. \begin{align*}P_1 (1,30^\circ)\end{align*} and \begin{align*}P_2 (6,135^\circ)\end{align*}
2. \begin{align*}P_1 (2,-65^\circ)\end{align*} and \begin{align*}P_2 (9,85^\circ)\end{align*}
3. \begin{align*}P_1 (-3,142^\circ)\end{align*} and \begin{align*}P_2 (10,-88^\circ)\end{align*}
4. \begin{align*}P_1 (5,-160^\circ)\end{align*} and \begin{align*}P_2 (16, -335^\circ)\end{align*}

1. \begin{align*}& \left (-4, \frac{\pi}{4} \right ) && \text{all positive} \rightarrow && \left (4, \frac{5\pi}{4} \right ) \\ & && \text{both negative} \rightarrow && \left (-4, \frac{-7\pi}{4} \right ) \\ & && \text{negative angle} \rightarrow && \left (4, \frac{-3\pi}{4} \right )\end{align*}
2. \begin{align*}& (2, 120^\circ) && \text{negative angle only} \rightarrow && (2, -240^\circ) \\ & && \text{negative radius only} \rightarrow && (-2, 300^\circ) \\ & && \text{both negative} \rightarrow && (-2, -60^\circ)\end{align*}
3. Use \begin{align*}P_1 P_2 = \sqrt{r^2_1 + r^2_2 - 2r_1 r_2 \cos (\theta_2 - \theta_1)}\end{align*}.
1. \begin{align*}P_1 P_2 & = \sqrt{1^2 + 6^2 - 2(1)(6) \cos (135^\circ - 30^\circ)} \\ P_1 P_2 & \approx 6.33\ units\end{align*}
2. \begin{align*}P_1 P_2 & = \sqrt{2^2 + 9^2 - 2(2)(9) \cos 150^\circ} \\ & = 10.78\end{align*}
3. \begin{align*}P_1 P_2 & = \sqrt{3^2 + 10^2 - 2(3)(10) \cos (322 - 272)^\circ} \\ & = 8.39\end{align*}
4. \begin{align*}P_1 P_2 & = \sqrt{5^2 + 16^2 - 2(5)(16) \cos (200 - 25)^\circ} \\ & = 20.99\end{align*}

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