Question

How do I convert from rectangular to polar coordinates?

Answer 1

\[\LARGE r(x,y)=\sqrt{x^2+y^2} \\ \\ \LARGE \theta(x,y)=\arctan( \frac{y}{x})\]

This is clearly wrong because what if we look at the point (x,y) = (1,1) and the point (x,y) = (-1,-1) we will get: \[\LARGE \arctan \frac{1}{1}=\arctan \frac{-1}{-1}= \frac{\pi}{4}\] But clearly both of these points have different angles.

Answer 2

Next thing you know, everything you thought was true in math will also be wrong just wait lol.

Answer 3

hheeh what are the rectangular coordinates anyhow?

Answer 4

just keep in mind that a polar will be \(\large {(r,\theta)\qquad
\begin{cases}
x=rcos(\theta)\\
y=rsin(\theta)
\end{cases}\qquad (x,y)}\)

Answer 5

Sure, I mean it's not like I really have an issue with the concept, I'm just sort of saying that the math doesn't actually represent what people are doing when the convert from one to the other because it's different. It's just nobody really notices or cares because they know what they mean when they write it.

Answer 6

It's probably better to utilize \(\tan\theta=\dfrac{y}{x}\) to avoid the problem with arctangent's domain

Answer 7

will about theta = arctan y/x
we always consider where is theta suppose to be according to @jdoe0001
so its like consederind sin theta and cos theta sighns at the same time

to me there is no harm as long as u aware to this

Answer 8

for example for 1,1 and -1,-1

theta =pi /4
but were should theta be ? ( two options )