# Thread: Converting polar equations to parametric

1. ## Converting polar equations to parametric

Hello-- I have four equations I simply can't figure out:

r=cot(theta)csc(theta)
r=8sin(theta)
r=2cos(theta)sin(theta)
rsin(theta+pi/6)=2

I tried squaring each of these in hopes of finding something to work with, but I'm lost...please help!!

2. Hello, riverjib!

Do you mean "polar to rectangular"?

We need the conversion formulas:

. $\begin{array}{ccc}x & = & r\cos\theta \\ y & = & r\sin\theta \\ r & = & \sqrt{x^2+y^2}\end{array}$

$(1)\;\;r\:=\:\cot\theta\csc\theta$
We have: . $r \;=\;\frac{\cos\theta}{\sin\theta} + \frac{1}{\sin\theta}\quad\Rightarrow\quad r\sin\theta \:=\:\cos\theta + 1
$

Multiply by $r\!:\;\;r(r\sin\theta) \:=\:r\cos\theta + r$

Substitute: . $\sqrt{x^2+y^2}\cdot y \:=\:x + \sqrt{x^2+y^2}$

Then: . $\sqrt{x^2+y^2}\cdot y - \sqrt{x^2+y^2} \:=\:x\quad\Rightarrow\quad(y-1)\sqrt{x^2+y^2} \:=\:x$

Square both sides: . $\boxed{(y-1)^2(x^2+y^2) \:=\:x^2}$

$(2)\;\;r\:=\:8\sin\theta)$
Multiply by $r\!:\;\;r^2\:=\:8\cdot r\sin\theta$

Substitute: . $\boxed{x^2+y^2\:=\:8y}$

$(3)\;\;r\:=\:2\cos\theta\sin\theta$

Multiply by $r^2\!:\;\;r^3 \:=\:2(r\cos\theta)(r\sin\theta)$

Substitute: . $(x^2+y^2)^{\frac{3}{2}} \:=\:2xy
$

Square both sides: . $\boxed{(x^2+y^2)^3\:=\:4x^2y^2}$

$(4)\;\;r\sin\left(\theta + \frac{\pi}{6}\right)\:=\:2$
We have: . $r\left(\sin\theta\cos\frac{\pi}{6} + \sin\frac{\pi}{6}\cos\theta\right) \:=\:2 \quad\Rightarrow\quad r\left(\frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta\right) \:=\:2$

Multiply by 2: . $r\left(\sqrt{3}\sin\theta + \cos\theta\right) \:=\:4\quad\Rightarrow\quad \sqrt{3}(r\sin\theta) + r\cos\theta \:=\:4$

Substitute: . $\boxed{\sqrt{3}\,y + x \:=\:4}$

3. ## Thank you so much!

Thank you! I did 40 problems like this successfully, and got stumped on these four almost two hours ago. I actually figured out just before you posted that I could multiply one side by r as long as I did the same to the other side (big duh! on my part), and that solved the problem of the missing r's. But I was still stuck on the fourth equation, and you helped me just as I was beginning to lose my mind.