# Polar to rectangular

• Apr 24th 2013, 06:41 PM
dgable
Polar to rectangular
I am trying to help grand daughter express this is rectangular form:

thetha = pi/3
• Apr 24th 2013, 06:46 PM
Prove It
Re: Polar to rectangular
If \displaystyle \displaystyle \begin{align*} \theta = \frac{\pi}{3} \end{align*} then you have a ray starting at the origin extending out infinitely long making this angle with the positive x-axis.

The Cartesian (rectangular) form for the equation of a line is $\displaystyle \displaystyle y = mx + c$ where m is the gradient and c is the y intercept. c is clearly 0. As for m, it can be calculated using \displaystyle \displaystyle \begin{align*} m = \tan{(\theta)} = \tan{\left( \frac{\pi}{3} \right)} = \sqrt{3} \end{align*}.

So the rectangular form of this equation is \displaystyle \displaystyle \begin{align*} y = \sqrt{3}\,x \end{align*} where \displaystyle \displaystyle \begin{align*} x \geq 0 \end{align*}.
• Apr 24th 2013, 06:58 PM
dgable
Re: Polar to rectangular
• Apr 24th 2013, 08:20 PM
Soroban
Re: Polar to rectangular
Hello, dgable!

Quote:

$\displaystyle \text{Express in rectangular form: }\:\theta \,=\,\tfrac{\pi}{3}$

There are a few identities for converting polar to rectangular:

. . $\displaystyle \begin{array}{ccccccccc}r^2 &=&x^2+y^2 &\quad& r\cos\theta &=& x \\ \tan\theta &=&\dfrac{y}{x} && r\sin\theta &=& y \end{array}$

$\displaystyle \text{We are given: }\qquad\;\;\theta \:=\:\tfrac{\pi}{3}$

$\displaystyle \text{Take tangents: }\,\tan\theta \:=\:\tan\left(\tfrac{\pi}{3}\right)$

$\displaystyle \text{Substitute: }\qquad\quad\,\frac{y}{x} \:=\:\sqrt{3}$

Therefore: . . . . . $\displaystyle y \:=\:\sqrt{3}\,x$