6.1: Polar Coordinates
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
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
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 circlescircles that share a common center. The common center
These coordinates are the result of assuming that the angle is rotated counterclockwise. If the angle were rotated clockwise then the coordinates of
Example 1: Plot the point
Solution, A: To plot
Solution, B: To plot
These points that you have plotted have
The point is reflected across the pole to point
There are multiple representations for the coordinates of a polar point
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 (315250)^\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
 Graph each point:
 \begin{align*}M (2.5, 210^\circ)\end{align*}
 \begin{align*}S \left (3.5, \frac{5 \pi}{6} \right )\end{align*}
 \begin{align*}A \left (1, \frac{3 \pi}{4} \right )\end{align*}
 \begin{align*}Y \left (5.25,  \frac{\pi}{3} \right )\end{align*}
 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*}.
 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*}.
 Given \begin{align*}P_1\end{align*} and \begin{align*}P_2\end{align*}, calculate the distance between the points.
 \begin{align*}P_1 (1,30^\circ)\end{align*} and \begin{align*}P_2 (6,135^\circ)\end{align*}
 \begin{align*}P_1 (2,65^\circ)\end{align*} and \begin{align*}P_2 (9,85^\circ)\end{align*}
 \begin{align*}P_1 (3,142^\circ)\end{align*} and \begin{align*}P_2 (10,88^\circ)\end{align*}
 \begin{align*}P_1 (5,160^\circ)\end{align*} and \begin{align*}P_2 (16, 335^\circ)\end{align*}
Review Answers

 \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*}
 \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*}
 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*}.
 \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*}
 \begin{align*}P_1 P_2 & = \sqrt{2^2 + 9^2  2(2)(9) \cos 150^\circ} \\ & = 10.78\end{align*}
 \begin{align*}P_1 P_2 & = \sqrt{3^2 + 10^2  2(3)(10) \cos (322  272)^\circ} \\ & = 8.39\end{align*}
 \begin{align*}P_1 P_2 & = \sqrt{5^2 + 16^2  2(5)(16) \cos (200  25)^\circ} \\ & = 20.99\end{align*}