Understanding Polar Coordinates: Learn It 1

Giselle Miller

Understanding Polar Coordinates: Learn It 1

Plot points using polar coordinates (r,θ)
Switch back and forth between polar and rectangular (x,y) coordinates
Draw polar curves from their equations
Identify when polar curves have symmetry

Polar Coordinates

The rectangular coordinate system maps points to ordered pairs using perpendicular axes. The polar coordinate system offers an alternative approach that can be more useful in certain situations, particularly when dealing with circular or rotational patterns.

Instead of describing a point’s location using horizontal and vertical distances, polar coordinates use:

Distance from the origin (called [latex]r[/latex])
Angle from the positive [latex]x[/latex]-axis (called [latex]\theta[/latex])

The rectangular coordinate system (or Cartesian plane) provides a means of mapping points to ordered pairs and ordered pairs to points. This is called a one-to-one mapping from points in the plane to ordered pairs. The polar coordinate system provides an alternative method of mapping points to ordered pairs. In this section we see that in some circumstances, polar coordinates can be more useful than rectangular coordinates.

Defining Polar Coordinates

Consider any point [latex]P[/latex] in the coordinate plane with rectangular coordinates [latex]\left(x,y\right)[/latex].

To find its polar coordinates, we measure the radius [latex]r[/latex] as the distance from the origin to point [latex]P[/latex], and the angle [latex]\theta[/latex] as the angle between the positive [latex]x[/latex]-axis and the line segment from the origin to [latex]P[/latex]. This creates a one-to-one mapping between points and the ordered pair [latex](r,\theta )[/latex], just as rectangular coordinates use [latex]\left(x,y\right)[/latex].

polar coordinate system

Each point in the plane corresponds to an ordered pair [latex](r,\theta )[/latex], where [latex]r[/latex] is the distance from the origin and [latex]\theta[/latex] is the angle from the positive [latex]x[/latex]-axis.

Converting Between Coordinate Systems

The connection between rectangular and polar coordinates relies on right triangle relationships. When you draw a line from the origin to point [latex]P[/latex], you create a right triangle where the hypotenuse equals [latex]r[/latex], the horizontal leg equals [latex]x[/latex], the vertical leg equals [latex]y[/latex], and one angle equals [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.

Recall: Right Triangle Trigonometry
[latex]\\[/latex]
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: 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 the right triangle formed by point [latex]P[/latex], the origin, and the coordinate axes, we can derive relationships between rectangular and polar coordinates.

From the basic trigonometric ratios, we get:

[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].

Additionally, the Pythagorean theorem and tangent ratio give us:

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

These relationships allow us to convert any point [latex]\left(x,y\right)[/latex] in rectangular coordinates to polar coordinates [latex]\left(r,\theta \right)[/latex]. The first coordinate [latex]r[/latex] is called the radial coordinate, and the second coordinate [latex]\theta[/latex] is called the angular coordinate.

converting points between coordinate systems

For a point [latex]P[/latex] with rectangular coordinates [latex]\left(x,y\right)[/latex] and polar coordinates [latex]\left(r,\theta \right)[/latex]:

Polar to Rectangular: [latex]x=r\cos\theta \:\text{and}\:y=r\sin\theta[/latex]
Rectangular to Polar: [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.

The equation [latex]\tan\theta =\frac{y}{x}[/latex] has infinitely many solutions for any point [latex]\left(x,y\right)[/latex]. However, when we restrict solutions to values between [latex]0[/latex] and [latex]2\pi[/latex], we can find a unique angle that places the point in the correct quadrant. This ensures that the corresponding value of [latex]r[/latex] is positive.

Notice that we don’t simply write [latex]\theta = \tan^{-1}\left(\frac{y}{x}\right)[/latex]. This is because the inverse tangent function has a restricted range that can lead to incorrect quadrant placement.

Recall: Using The Inverse Tangent Function in the coordinate plane

The inverse tangent function has range [latex]\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)[/latex], so it only produces angles in Quadrants I and IV.

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

When the point [latex]\left(x, y \right)[/latex] is in Quadrant II or III, you must add [latex]\pi[/latex] to the inverse tangent result to get the correct angle.

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]

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 y-axis. The angle between the positive x-axis and the positive y-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 the example above.

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).