Converting Polar Equation to Cartesian Equation problem

Question

So I have

1. $$\frac{r}{3\tan \theta} = \sin \theta$$

2. $$r=3\cos \theta$$

What would be the Cartesian equation???

Answer 1

You could substitute for $r$ and $\theta$ their expressions in terms of $x$ and $y$. But remember that trigonometric functions of $\theta$ have easier expressions than $\theta$ itself.

Answer 2

the first $$\frac{r}{3\tan \theta} = \sin \theta$$ $$\frac{r\cos \theta}{3\sin \theta} = \sin \theta$$ $$\frac{x}{3\sin \theta} =\sin \theta$$ $$x =3\sin^2 \theta$$ $$r^2x =3r^2\sin^2 \theta$$ $$x(x^2+y^2)=3y^2$$

The second $$r=3\cos \theta$$ $$r^2=3r\cos \theta$$ $$r^2=3x$$ $$x^2+y^2=3x$$

Answer 3

First note that

$$\left\{ \matrix{ x = r\cos \theta \hfill \cr y = r\sin \theta \hfill \cr} \right.$$

the general approach will be to solve for $r$ and $\theta$ and replace in your polar equation. However, in most times there are some shortcuts. See the following for the second one

$$\eqalign{ & r = 3\cos \theta \cr & r = 3{x \over r} \cr & x = {1 \over 3}{r^2} \cr & 3x = {x^2} + {y^2} \cr} $$

and hence your final equation will be

$${x^2} + {y^2} - 3x = 0$$

which is a conic section. Specifically, it is a circle. I leave the first one for you. :)