Polar coordinates problem: put 3r = sinA into rect. coords

[chiurox]

in 3r = sin A, r and A represent polar coordinates, write each polar equation as an equation in rectangular coordinates (x,y).

I normally see these kinds of question in the form of r = 3sinA etc... but I'm stuck on how to solve this. Could anyone help me?
So far I got 3x^2 + 3y^2 - x = 0 is this correct?

Thanks

 

[stapel]

If you're okay with "r = (whatever)" formatting, then would it help to divide through by 3 to get r = (1/3)sin(A)...? :?:

Eliz.

 

[soroban]

Re: Polar coordinates problem

Hello, chiurox

In: \(\displaystyle \, 3r \:= \:\sin\theta\), \(\displaystyle r\) and \(\displaystyle \theta\) represent polar coordinates.
Write the polar equation as an equation in rectangular coordinates (x,y).

I normally see these kinds of question in the form of: \(\displaystyle \,r \:= \:3\sin\theta\), etc.
. . . . . and you don't know to divide by 3 ?

We have the convertions: \(\displaystyle \:\begin{array}{ccc}x & = & r\cdot\cos\theta \\ y & = & r\cdot\sin\theta \\ r^2 & = & x^2\,+\,y^2\end{array}\)

\(\displaystyle \text{Multiply the equation by }r:\;\;3\underbrace{r^2}_{\downarrow} \;=\;\underbrace{r\cdot\sin\theta}_{\downarrow}\)
. . - . . . . . . . . . . . . . . . . . \(\displaystyle 3(x^2\,+\,y^2) \;=\:y\)

The rectangular equation is: \(\displaystyle \:3x^2\,+\,3y^2 \,-\,y \;=\;0\)

 

[chiurox]

Thanks a lot stapel and soroban!
Yeah, I always have the problem of starting the problem or doing other unnecessary stuff. So yeah, rr = (xx + yy) forgot about using this identity.