# Difficult conversion from polar equation to rectangular equation.

How do we convert this into rectangular equation? $r=5\theta$

-

$r=x^2+y^2,\,\tan(\theta)=\frac{y}{x}$,so we have

$\frac{r}{5}=\theta,\tan(\frac{r}{5})=\frac{y}{x}$

the rectangle equation is $$\tan\left(\frac{\sqrt{x^2+y^2}}{5}\right)=\frac{y}{x}$$

-
that should be $r=\sqrt{x^2+y^2}$ in the first line. – Alex R. Mar 30 '12 at 22:00

I get this easily, check that if i am wrong: $r = a\theta$, so in terms of Cartesian coordinates… $$x = a\theta\cos\theta,\qquad y =a\theta\sin\theta.$$

We can also solve for $x$ and $y$ through simple algebraic manipulation

We know: $r^2=x ^2+ y^2$. Let's solve for $x$ first:

\begin{align*} r &= a\theta\\ r^2&=a^2\theta^2\\ x^2+y^2&=a^2\theta^2\\ x^2&=a^2\theta^2-y^2 \end{align*} Square both sides and substitute $r^2=x^2+ y^2$, $y=r\sin\theta$ (hence $y^2=r^2\sin^2\theta$), $x^2=a^2\theta^2-r^2\sin^2\theta$ (hence $x^2=a^2\theta^2- a^2\theta^2\sin^2\theta$) we get
$$x^2=a^2\theta^2(1- \sin^2\theta).$$ $y$ follows easily

-
you should add it to supplement of this question. – noname1014 Mar 30 '12 at 3:30