Converting a rectangular equation to polar form

[Lizzy]

The book isn't really clear on how to do this. It explains how to convert rectangular points to polar points but not the equations. Is it similar? I need help with the process. I would appreciate it if I got help for this problem as an example. I think I can figure out the rest.

Change this equation to polar form: 3x-y+2=0

Thanks!

 

[stapel]

Didn't the book give you some formulas and some relationships...?

Eliz.

 

[Lizzy]

Clarification on received reply

Didn't the book give you some formulas and some relationships...?

Eliz.

Well the book gives these relationships:

tangent of the angle is equal to y/x
cosine of the angle is = to x/r
sine of the angle is = to y/r

x= r (cosine of the angle)
y= r (sine of the angle)
r^2 = (x^2) + (y^2)

I tried getting the rectangular points of the equation and then tried to change them to polar points but I'm not sure if I'm in the right track or how to apply these relationships. Help!!!

 

[stapel]

You know that "y = 3x + 2" is a straight line, which naturally isn't a good match to polar coordinates. You may find this document (it's a PDF file) to be helpful. Follow the worked examples.

Eliz.

 

[soroban]

Hello, Lizzy!

You have the equations: \(\displaystyle \:\begin{Bmatrix}x \:=\:r\cos\theta \\ y \:=\:r\sin\theta\end{Bmatrix}\)

Use them . . .

Write in polar form: \(\displaystyle \:3x\,-\,y\,+\,2\;=\;0\)

Just make the conversion:

. . . . . \(\displaystyle 3x\;\;-\;\;y\;\;+\;\;2\;\;=\;\;0\)
. . . . - .\(\displaystyle \downarrow\;\;\;\;\;\downarrow\)
. . \(\displaystyle 3(r\cos\theta) \,-\,(r\sin\theta) \,+\,2\;=\;0\)


Solve for \(\displaystyle r\): \(\displaystyle \:3r\cos\theta\,-\,r\sin\theta\:=\:-2\)

Factor: . . . .\(\displaystyle r(3\cos\theta\,-\,\sin\theta)\;=\;-2\)

Therefore: \(\displaystyle \L\:r\;=\;\frac{-2}{3\cos\theta\,-\,\sin\theta}\;\;\Rightarrow\;\;\fbox{r\;=\;\frac{2}{\sin\theta\,-\,3\cos\theta}}\)