Converting polar equation to rectangular equation

Question

I'm doing exercises of converting polar equations to rectangular. I would like to check if what I'm doing is right.

1. $$\frac{\sin^2(\theta)}{\cos(\theta)}=\frac{1}{r}$$ $$r\sin^2(\theta)=\cos(\theta)$$ $$r\cdot r\sin^2(\theta)=\cos(\theta)\cdot r$$ $$r^2\sin^2(\theta)=r\cos(\theta)$$ $$(r\sin(\theta))^2=r\cos(\theta)$$ $$y^2=x$$

Is this right? Do I have to take the square root on both sides?

2. $$r=\frac{\theta}{\sin(\theta)}$$ $$ \sin(\theta)\cdot r=\frac{\theta}{\sin(\theta)}\cdot \sin(\theta)$$ $$r\sin(\theta)=\theta$$ $$y=\theta$$

I have my doubts here because I think $\theta$ must disappear.

Answer 1

As $y=r\cos\theta,x=?$

$\dfrac yx=\tan\theta$

$\theta=\arctan\dfrac yx$ where we need to consider general values of $\arctan$

as $\theta$ can assume any finite and real value

Answer 2

For the first one you are correct:

I am assuming you are using the following identities:

$$x=r\text{cos}(\theta)\hspace{4mm}y=r\text{sin}(\theta)\hspace{4mm}r=\sqrt{x^2+y^2}$$

As for if you have to take the square root to the equation $x=y^2$ not that the ordered pairs $(x,y)$ that satisfy this equation are the same if you consider $y^2=x$ or $y=\pm \sqrt{x}$ with the later being more customary. What you do have to be mindful of is since there is no restriction on $\theta$ are we accounting for all the quadrants of our graph. Note that $y^2=x$ falls in quadrant I and IV. The polar equation $r=\frac{\cos(\theta))}{\sin^2(\theta)}$ is in quadrant I and IV so all is good.