Defining Polar Coordinates

Learning Outcomes

Locate points in a plane by using polar coordinates
Convert points between rectangular and polar coordinates

To find the coordinates of a point in the polar coordinate system, consider Figure 1. The point [latex]P[/latex] has Cartesian coordinates [latex]\left(x,y\right)[/latex]. The line segment connecting the origin to the point [latex]P[/latex] measures the distance from the origin to [latex]P[/latex] and has length [latex]r[/latex]. The angle between the positive [latex]x[/latex] -axis and the line segment has measure [latex]\theta[/latex]. This observation suggests a natural correspondence between the coordinate pair [latex]\left(x,y\right)[/latex] and the values [latex]r[/latex] and [latex]\theta[/latex]. This correspondence is the basis of the polar coordinate system. Note that every point in the Cartesian plane has two values (hence the term ordered pair) associated with it. In the polar coordinate system, each point also has two values associated with it: [latex]r[/latex] and [latex]\theta[/latex].

A point P(x, y) is given in the first quadrant with lines drawn to indicate its x and y values. There is a line from the origin to P(x, y) marked r and this line make an angle θ with the x axis.

As we can observe above, the right triangle pictured above implies a set of relationships between [latex]x, y, r, \:\text{and}\: \theta[/latex]. We first recall those relationships below.

Recall: Right Triangle Trigonometry

Given a right triangle with an acute angle of [latex]\theta[/latex],

[latex]\begin{align}&\sin \left(\theta \right)=\frac{\text{opposite}}{\text{hypotenuse}} \\ &\cos \left(\theta \right)=\frac{\text{adjacent}}{\text{hypotenuse}} \\ &\tan \left(\theta \right)=\frac{\text{opposite}}{\text{adjacent}} \end{align}[/latex]

A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “Sine is opposite over hypotenuse, Cosine is adjacent over hypotenuse, Tangent is opposite over adjacent.”

The side lengths of the right triangle with legs [latex]a[/latex] and [latex]b[/latex] and hypotenuse [latex]c[/latex] are related through the Pythagorean Theorem: [latex]a^2 + b^2 = c^2[/latex]

Using right-triangle trigonometry, the following equations are true for the point [latex]P\text{:}[/latex]

[latex]\cos\theta =\frac{x}{r}\:\:\text{so}\:\:x=r\cos\theta[/latex]

[latex]\sin\theta =\frac{y}{r}\:\:\text{ so }\:\:y=r\sin\theta[/latex].

Furthermore,

[latex]{r}^{2}={x}^{2}+{y}^{2}\text{ and }\tan\theta =\frac{y}{x}[/latex].

Each point [latex]\left(x,y\right)[/latex] in the Cartesian coordinate system can therefore be represented as an ordered pair [latex]\left(r,\theta \right)[/latex] in the polar coordinate system. The first coordinate is called the radial coordinate and the second coordinate is called the angular coordinate. Every point in the plane can be represented in this form.

Note that the equation [latex]\tan\theta =\frac{y}{x}[/latex] has an infinite number of solutions for any ordered pair [latex]\left(x,y\right)[/latex]. However, if we restrict the solutions to values between [latex]0[/latex] and [latex]2\pi[/latex] then we can assign a unique solution to the quadrant in which the original point [latex]\left(x,y\right)[/latex] is located. Then the corresponding value of r is positive, so [latex]{r}^{2}={x}^{2}+{y}^{2}[/latex].

theorem: Converting Points between Coordinate Systems

Given a point [latex]P[/latex] in the plane with Cartesian coordinates [latex]\left(x,y\right)[/latex] and polar coordinates [latex]\left(r,\theta \right)[/latex], the following conversion formulas hold true:

[latex]x=r\cos\theta \:\text{and}\:y=r\sin\theta[/latex],

[latex]{r}^{2}={x}^{2}+{y}^{2}\:\text{and}\:\tan\theta =\frac{y}{x}[/latex].

These formulas can be used to convert from rectangular to polar or from polar to rectangular coordinates.

As we can note carefully above, the equation [latex]\tan\theta =\frac{y}{x}[/latex] is not expressed using the inverse tangent function. The reason for this is because of how the domain restriction of the tangent function leads to a restricted range of the inverse tangent function, reviewed below.

Recall: Using The Inverse Tangent Function in the coordinate plane

If [latex]-\frac{\pi}{2} < \theta < \frac{\pi}{2}[/latex], and [latex]\tan \theta = \frac{y}{x}[/latex], then [latex]\theta = \tan^{-1} \left( \frac{y}{x} \right)[/latex]

That is, the inverse tangent function has a range of [latex]\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)[/latex], meaning that it always produces a positive angle in Quadrant I or a negative angle in Quadrant IV.

If [latex]\frac{\pi}{2} < \theta < \frac{3\pi}{2}[/latex] and [latex]\tan \theta = \frac{y}{x}[/latex], then [latex]\theta = \tan^{-1} \left( \frac{y}{x} \right) + \pi[/latex]

In other words, if the point [latex]\left(x, y \right)[/latex] is in Quadrant II or III, the preceding rule means that you must add [latex]\pi[/latex] to the output of the inverse tangent function to produce an angle in the correct quadrant.

Example: Converting between Rectangular and Polar Coordinates

Convert each of the following points into polar coordinates.

[latex]\left(1,1\right)[/latex]
[latex]\left(-3,4\right)[/latex]
[latex]\left(0,3\right)[/latex]
[latex]\left(5\sqrt{3},-5\right)[/latex]

Convert each of the following points into rectangular coordinates.

[latex]\left(3,\frac{\pi}{3}\right)[/latex]
[latex]\left(2,\frac{3\pi}{2}\right)[/latex]
[latex]\left(6,\frac{-5\pi}{6}\right)[/latex]

Show Solution

Use [latex]x=1[/latex] and [latex]y=1[/latex] in the theorem:

[latex]\begin{array}{ccccccc}\begin{array}{ccc}\hfill {r}^{2}& =\hfill & {x}^{2}+{y}^{2}\hfill \\ & =\hfill & {1}^{2}+{1}^{2}\hfill \\ \hfill r& =\hfill & \sqrt{2}\hfill \end{array}\hfill & & & \text{and}\hfill & & & \begin{array}{ccc}\hfill \tan\theta & =\hfill & \frac{y}{x}\hfill \\ & =\hfill & \frac{1}{1}=1\hfill \\ \hfill \theta & =\hfill & \frac{\pi }{4}.\hfill \end{array}\hfill \end{array}[/latex]

Therefore this point can be represented as [latex]\left(\sqrt{2},\frac{\pi }{4}\right)[/latex] in polar coordinates.
Use [latex]x=-3[/latex] and [latex]y=4[/latex] in the theorem:

[latex]\begin{array}{ccccccc}\begin{array}{ccc}\hfill {r}^{2}& =\hfill & {x}^{2}+{y}^{2}\hfill \\ & =\hfill & {\left(-3\right)}^{2}+{\left(4\right)}^{2}\hfill \\ \hfill r& =\hfill & 5\hfill \end{array}\hfill & & & \text{and}\hfill & & & \begin{array}{ccc}\hfill \tan\theta & =\hfill & \frac{y}{x}\hfill \\ & =\hfill & -\frac{4}{3}\hfill \\ \hfill \theta & =\hfill & -\text{arctan}\left(\frac{4}{3}\right)\hfill \\ & \approx \hfill & 2.21.\hfill \end{array}\hfill \end{array}[/latex]

Therefore this point can be represented as [latex]\left(5,2.21\right)[/latex] in polar coordinates.
Use [latex]x=0[/latex] and [latex]y=3[/latex] in the theorem:

[latex]\begin{array}{ccccccc}\begin{array}{ccc}\hfill {r}^{2}& =\hfill & {x}^{2}+{y}^{2}\hfill \\ & =\hfill & {\left(3\right)}^{2}+{\left(0\right)}^{2}\hfill \\ & =\hfill & 9+0\hfill \\ \hfill r& =\hfill & 3\hfill \end{array}\hfill & & & \text{and}\hfill & & & \begin{array}{ccc}\hfill \tan\theta & =\hfill & \frac{y}{x}\hfill \\ & =\hfill & \frac{3}{0}.\hfill \end{array}\hfill \end{array}[/latex]

Direct application of the second equation leads to division by zero. Graphing the point [latex]\left(0,3\right)[/latex] on the rectangular coordinate system reveals that the point is located on the positive [latex]y[/latex]-axis. The angle between the positive [latex]x[/latex]-axis and the positive [latex]y[/latex]-axis is [latex]\frac{\pi }{2}[/latex]. Therefore this point can be represented as [latex]\left(3,\frac{\pi }{2}\right)[/latex] in polar coordinates.
Use [latex]x=5\sqrt{3}[/latex] and [latex]y=-5[/latex] in the theorem:

[latex]\begin{array}{ccccccc}\begin{array}{ccc}\hfill {r}^{2}& =\hfill & {x}^{2}+{y}^{2}\hfill \\ & =\hfill & {\left(5\sqrt{3}\right)}^{2}+{\left(-5\right)}^{2}\hfill \\ & =\hfill & 75+25\hfill \\ \hfill r& =\hfill & 10\hfill \end{array}\hfill & & & \text{and}\hfill & & & \begin{array}{ccc}\hfill \tan\theta & =\hfill & \frac{y}{x}\hfill \\ & =\hfill & \frac{-5}{5\sqrt{3}}=-\frac{\sqrt{3}}{3}\hfill \\ \hfill \theta & =\hfill & -\frac{\pi }{6}.\hfill \end{array}\hfill \end{array}[/latex]

Therefore this point can be represented as [latex]\left(10,-\frac{\pi }{6}\right)[/latex] in polar coordinates.
Use [latex]r=3[/latex] and [latex]\theta =\frac{\pi }{3}[/latex] in the theorem:

[latex]\begin{array}{ccccccc}\begin{array}{ccc}\hfill x& =\hfill & r\cos\theta \hfill \\ & =\hfill & 3\cos\left(\frac{\pi }{3}\right)\hfill \\ & =\hfill & 3\left(\frac{1}{2}\right)=\frac{3}{2}\hfill \end{array}\hfill & & & \text{and}\hfill & & & \begin{array}{ccc}\hfill y& =\hfill & r\sin\theta \hfill \\ & =\hfill & 3\sin\left(\frac{\pi }{3}\right)\hfill \\ & =\hfill & 3\left(\frac{\sqrt{3}}{2}\right)=\frac{3\sqrt{3}}{2}.\hfill \end{array}\hfill \end{array}[/latex]

Therefore this point can be represented as [latex]\left(\frac{3}{2},\frac{3\sqrt{3}}{2}\right)[/latex] in rectangular coordinates.
Use [latex]r=2[/latex] and [latex]\theta =\frac{3\pi }{2}[/latex] in the theorem:

[latex]\begin{array}{ccccccc}\begin{array}{ccc}\hfill x& =\hfill & r\cos\theta \hfill \\ & =\hfill & 2\cos\left(\frac{3\pi }{2}\right)\hfill \\ & =\hfill & 2\left(0\right)=0\hfill \end{array}\hfill & & & \text{and}\hfill & & & \begin{array}{ccc}\hfill y& =\hfill & r\sin\theta \hfill \\ & =\hfill & 2\sin\left(\frac{3\pi }{2}\right)\hfill \\ & =\hfill & 2\left(-1\right)=-2.\hfill \end{array}\hfill \end{array}[/latex]

Therefore this point can be represented as [latex]\left(0,-2\right)[/latex] in rectangular coordinates.
Use [latex]r=6[/latex] and [latex]\theta =-\frac{5\pi }{6}[/latex] in the theorem:

[latex]\begin{array}{ccccccc}\begin{array}{ccc}\hfill x& =\hfill & r\cos\theta \hfill \\ & =\hfill & 6\cos\left(-\frac{5\pi }{6}\right)\hfill \\ & =\hfill & 6\left(-\frac{\sqrt{3}}{2}\right)\hfill \\ & =\hfill & -3\sqrt{3}\hfill \end{array}\hfill & & & \text{and}\hfill & & & \begin{array}{ccc}\hfill y& =\hfill & r\sin\theta \hfill \\ & =\hfill & 6\sin\left(-\frac{5\pi }{6}\right)\hfill \\ & =\hfill & 6\left(-\frac{1}{2}\right)\hfill \\ & =\hfill & -3.\hfill \end{array}\hfill \end{array}[/latex]

Therefore this point can be represented as [latex]\left(-3\sqrt{3},-3\right)[/latex] in rectangular coordinates.

Watch the following video to see the worked solution to Example: Converting between Rectangular and Polar Coordinates.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “7.3 Polar Coordinates” here (opens in new window).

try it

Convert [latex]\left(-8,-8\right)[/latex] into polar coordinates and [latex]\left(4,\frac{2\pi }{3}\right)[/latex] into rectangular coordinates.

Hint

Show Solution

Try It

The polar representation of a point is not unique. For example, the polar coordinates [latex]\left(2,\frac{\pi }{3}\right)[/latex] and [latex]\left(2,\frac{7\pi }{3}\right)[/latex] both represent the point [latex]\left(1,\sqrt{3}\right)[/latex] in the rectangular system. Also, the value of [latex]r[/latex] can be negative. Therefore, the point with polar coordinates [latex]\left(-2,\frac{4\pi }{3}\right)[/latex] also represents the point [latex]\left(1,\sqrt{3}\right)[/latex] in the rectangular system, as we can see by using the theorem:

[latex]\begin{array}{ccccccc}\begin{array}{ccc}\hfill x& =\hfill & r\cos\theta \hfill \\ & =\hfill & -2\cos\left(\frac{4\pi }{3}\right)\hfill \\ & =\hfill & -2\left(-\frac{1}{2}\right)=1\hfill \end{array}\hfill & & & \text{and}\hfill & & & \begin{array}{ccc}\hfill y& =\hfill & r\sin\theta \hfill \\ & =\hfill & -2\sin\left(\frac{4\pi }{3}\right)\hfill \\ & =\hfill & -2\left(-\frac{\sqrt{3}}{2}\right)=\sqrt{3}.\hfill \end{array}\hfill \end{array}[/latex]

Every point in the plane has an infinite number of representations in polar coordinates. However, each point in the plane has only one representation in the rectangular coordinate system.

Note that the polar representation of a point in the plane also has a visual interpretation. In particular, [latex]r[/latex] is the directed distance that the point lies from the origin, and [latex]\theta[/latex] measures the angle that the line segment from the origin to the point makes with the positive [latex]x[/latex] -axis. Positive angles are measured in a counterclockwise direction and negative angles are measured in a clockwise direction. The polar coordinate system appears in the following figure.

A series of concentric circles is drawn with spokes indicating different values between 0 and 2π in increments of π/12. The first quadrant starts with 0 where the [latex]x[/latex]-axis would be, then the next spoke is marked π/12, then π/6, π/4, π/3, 5π/12, π/2, and so on into the second, third, and fourth quadrants. The polar axis is noted near the former [latex]x[/latex]-axis line.

The line segment starting from the center of the graph going to the right (called the positive [latex]x[/latex]-axis in the Cartesian system) is the polar axis. The center point is the pole, or origin, of the coordinate system, and corresponds to [latex]r=0[/latex]. The innermost circle shown in Figure 2 contains all points a distance of 1 unit from the pole, and is represented by the equation [latex]r=1[/latex]. Then [latex]r=2[/latex] is the set of points 2 units from the pole, and so on. The line segments emanating from the pole correspond to fixed angles. To plot a point in the polar coordinate system, start with the angle. If the angle is positive, then measure the angle from the polar axis in a counterclockwise direction. If it is negative, then measure it clockwise. If the value of [latex]r[/latex] is positive, move that distance along the terminal ray of the angle. If it is negative, move along the ray that is opposite the terminal ray of the given angle.

Example: Plotting Points in the Polar Plane

Plot each of the following points on the polar plane.

[latex]\left(2,\frac{\pi }{4}\right)[/latex]
[latex]\left(-3,\frac{2\pi }{3}\right)[/latex]
[latex]\left(4,\frac{5\pi }{4}\right)[/latex]

Show Solution

Watch the following video to see the worked solution to Example: Plotting Points in the Polar Plane.

You can view the transcript for this segmented clip of “7.3 Polar Coordinates” here (opens in new window).

try it

Plot [latex]\left(4,\frac{5\pi }{3}\right)[/latex] and [latex]\left(-3,-\frac{7\pi }{2}\right)[/latex] on the polar plane.

Hint

Show Solution

Module 1: Parametric Equations and Polar Coordinates