Skills Review for Cylindrical and Spherical Coordinates

Learning Outcomes

Identify reference angles for angles measured in both radians and degrees
Convert points between rectangular and polar coordinates

In the Cylindrical and Spherical Coordinates section, we will learn about two other coordinates systems, both extensions of the polar coordinate system. Here we will review how to evaluate the sine and cosine functions and convert points between rectangular and polar form.

Find Reference Angles

(See Module 2, Skills Review for Vectors in the Plane)

Converting between Rectangular and Polar Coordinates

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.

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 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 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

Vectors in Space: Prerequisite Material