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Boundless Algebra
Complex Numbers and Polar Coordinates
The Polar Coordinate System
Algebra Textbooks Boundless Algebra Complex Numbers and Polar Coordinates The Polar Coordinate System
Algebra Textbooks Boundless Algebra Complex Numbers and Polar Coordinates
Algebra Textbooks Boundless Algebra
Algebra Textbooks
Algebra
Concept Version 5
Created by Boundless

Converting Between Polar and Cartesian Coordinates

Polar and Cartesian coordinates can be interconverted using the Pythagorean Theorem and trigonometry.

Learning Objective

  • Derive and use the formulae for converting between Polar and Cartesian coordinates


Key Points

    • To convert from polar to rectangular (Cartesian) coordinates use the following formulas (derived from their trigonometric function definitions): 
    • $\cos \theta =\frac{x}{r}\rightarrow x=r\cos \theta\\\sin \theta =\frac{y}{r}\rightarrow y=r\sin \theta \\r^2=x^2+y^2\\\tan\theta=\frac{y}{x}$

Full Text

Polar Coordinates to Rectangular (Cartesian) Coordinates

When given a set of polar coordinates, we may need to convert them to rectangular coordinates. To do so, we can recall the relationships that exist among the variables $x$, $y$, $r$, and $θ$, from the definitions of $\cos \theta$ and $\sin \theta$.  Solving for the variables $x$ and $y$ yields the following formulas:

$\displaystyle \cos \theta =\frac{x}{r}\quad\Rightarrow\quad x=r\cos \theta $                      

 $\displaystyle \sin \theta =\frac{y}{r}\quad\Rightarrow\quad y=r\sin \theta $

An easy way to remember the equations above is to think of $\cos\theta$ as the adjacent side over the hypotenuse and $\sin\theta$ as the opposite side over the hypotenuse.  Dropping a perpendicular from the point in the plane to the $x$-axis forms a right triangle, as illustrated in Figure below.

Trigonometry Right Triangle 

A right triangle with rectangular (Cartesian) coordinates and equivalent polar coordinates. 

To convert polar coordinates $(r,θ)$ to rectangular coordinates $(x,y)$ follow these steps:

1) Write $\cos \theta =\frac{x}{r}\Rightarrow x=r\cos \theta $ and $\sin \theta =\frac{y}{r}\Rightarrow y=r\sin \theta $.

2) Evaluate $\cos\theta$ and $\sin\theta$.

3) Multiply $\cos\theta$ by $r$ to find the $x$-coordinate of the rectangular form.

4) Multiply $\sin\theta$ by $r$ to find the $y$-coordinate of the rectangular form.

Example:   Write the polar coordinates $(3,\frac {\pi}{2})$ as rectangular coordinates.

$\displaystyle \begin{aligned} x &= r\cos \theta \\ &= 3cos \frac{\pi}{2}\\ &= 0 \end{aligned} $

$\displaystyle \begin{aligned} y&=r\sin \theta\\&=3\sin\frac{\pi}{2}\\&=3 \end{aligned}$

The rectangular coordinates are $(0,3)$.

Polar and Coordinate Grid of Equivalent Points

The rectangular coordinate $(0,3)$ is the same as the polar coordinate $(3,\frac {\pi}{2})$ as plotted on the two grids above.

Rectangular (Cartesian) Coordinates to Polar Coordinates

To convert rectangular coordinates to polar coordinates, we will use two other familiar relationships. With this conversion, however, we need to be aware that a set of rectangular coordinates will yield more than one polar point.

Converting from rectangular coordinates to polar coordinates requires the use of one or more of the relationships illustrated below. Recall:

$\displaystyle \begin{aligned} \cos \theta &=\frac{x}{r}\quad\Rightarrow\quad x=r\cos \theta \\\sin \theta &=\frac{y}{r}\quad\Rightarrow\quad y=r\sin \theta \\ r^2&=x^2+y^2\\\tan\theta&=\frac{y}{x} \end{aligned}$

Trigonometry Right Triangle

A right triangle with rectangular (Cartesian) coordinates and equivalent polar coordinates.

Example: Convert the rectangular coordinates $(3,3)$ to polar coordinates.

We are given the values of $x$ and $y$ and need to solve for $\theta$ and $r$.  Start by solving for $\theta$ using the $\tan$ function:

$\displaystyle \begin{aligned} \tan \theta&=\frac{y}{x}\\&=\frac{3}{3}\\&=1\\ \end{aligned}$

So:

$\displaystyle \begin{aligned} \theta &= \tan^{-1}\left( 1 \right)\\ &=\frac{\pi}{4} \end{aligned} $

Next substitute the values of $x$ and $y$ into the formula $r^2=x^2+y^2$ and solve for $r$.

$\displaystyle \begin{aligned} r^2&=x^2+y^2\\ &=3^2+3^2\\ &=18\\ \end{aligned}$

So:

$\displaystyle \begin{aligned} r&=\sqrt{18}\\ &=3\sqrt{2} \end{aligned} $

The polar coordinates are $(3\sqrt2,\frac{\pi}{4})$. 

Note that $r^2 = 18$ implies $r=\pm\sqrt{18}$. We chose to ignore the negative $r$ value. Also note that $\tan^{-1}\left( 1 \right)$ has many answers. This corresponds to the non-uniqueness of polar coordinates. Multiple sets of polar coordinates can have the same location as our first solution. For example, the points  $(-3\sqrt2,\frac{5\pi}{4})$ and $(3\sqrt2,-\frac{7\pi}{2})$ will coincide with the original solution of $(3\sqrt2,\frac{\pi}{4})$.

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