How do you convert rectangular coordinates to polar coordinates?

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How do you convert rectangular coordinates to polar coordinates?

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Answer 1 · 2014-12-01T11:21:09

To convert from polar to rectangular:

$x = r cos θ$ $x=rcos theta$
$y = r sin θ$ $y=rsin theta$

To convert from rectangular to polar:

$r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$
$tan θ = \frac{y}{x}$ $tan theta= y/x$

This is where these equations come from:
$tutorial.math.lamar.edu$

Basically, if you are given an $(r, θ)$ $(r,theta)$ -a polar coordinate- , you can plug your $r$ $r$ and $θ$ $theta$ into your equation for $x = r cos θ$ $x=rcos theta$ and $y = r sin θ$ $y=rsin theta$ to get your $(x, y)$ $(x,y)$ .

The same holds true for if you are given an $(x, y)$ $(x,y)$ -a rectangular coordinate- instead. You can solve for $r$ $r$ in $r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$ to get $r = \sqrt{x^{2} + y^{2}}$ $r=sqrt(x^2+y^2)$ and solve for $θ$ $theta$ in $tan θ = \frac{y}{x}$ $tan theta= y/x$ to get $θ = arctan (\frac{y}{x})$ $theta=arctan (y/x)$ (arctan is just tan inverse, or ${tan}^{- 1}$ $tan^-1$ ). Note that there can be infinitely many polar coordinates that mean the same thing. For example, $(5, \frac{π}{3}) = (5, - 5 \frac{π}{3}) = (- 5, 4 \frac{π}{3}) = (- 5, - 2 \frac{π}{3})$ $(5, pi/3)=(5,-5pi/3)=(-5,4pi/3)=(-5,-2pi/3)$ ...However, by convention, we are always measuring positive $θ$ $theta$ COUNTERCLOCKWISE from the x-axis, even if our $r$ $r$ is negative.

Let's look at a couple examples.

( 1)Convert $(4, 2 \frac{π}{3})$ $(4,2pi/3)$ into Cartesian coordinates.

So we just plug in our $r = 4$ $r=4$ and $θ = 2 \frac{π}{3}$ $theta= 2pi/3$ into

$x = 4 cos 2 \frac{π}{3} = - 2$ $x=4cos 2pi/3=-2$
$y = 4 sin 2 \frac{π}{3} = 2 \sqrt{3}$ $y=4sin 2pi/3=2sqrt3$

The cartersian coordinate is $(- 2, 2 \sqrt{3})$ $(-2,2sqrt3)$

(2) Convert $(1, 1)$ $(1,1)$ into polar coordinates. ( since there are many posibilites of this, the restriction here is that $r$ $r$ must be positive and $θ$ $theta$ must be between 0 and $π$ $pi$ )

So, $x = 1$ $x=1$ and $y = 1$ $y=1$ . We can find $r$ $r$ and $θ$ $theta$ from:
$r = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$ $r=sqrt(1^2+1^2)=sqrt2$
$θ = arctan (\frac{y}{x}) = arctan (1) = \frac{π}{4}$ $theta=arctan (y/x)=arctan(1)=pi/4$

The polar coordinate is $(\sqrt{2}, \frac{π}{4})$ $(sqrt2,pi/4)$

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How do you convert rectangular coordinates to polar coordinates?

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