# Converting from polar to Cartesian coordinates.

I'm looking at some notes that I was given for my Calculus II class on converting from Cartesian to polar coordinates. Now I understand how to solve for r and $\theta$ but I'm looking at how she solved these two and maybe some can clear the air on these two problems.

Qaere: Cartesian coordinates are given. Write in polar coordinates with $0 \le \theta \le 2\pi$ and (i) $r>0$ and (ii) $r<0$.

(a) $\bigl(-1, \sqrt{3}\bigr)$

So, $(-1)^2+\bigl(-\sqrt{3}\bigr)^2=r^2$ gets us $r=\pm 2$.

(i) $\bigl(2,\frac{4\pi}3\bigr)$

(ii) $\bigl(-2, \frac \pi 3\bigr)$

So actually as I'm writing this (a) makes sense.

Now for (b) $(-2,3)$

$r=\sqrt{13}$

(i) $\left(\sqrt{13},\arctan\left(-\frac 3 2\right)+\pi\right)$

(ii) $\left(-\sqrt{13}, \arctan\left(-\frac 3 2\right)+2\pi\right)$

Now that I look at the values for each that end of in the corresponding quadrants. But maybe someone can just give me a better way to do these problem so I'm not wasting time graphing them out. Or maybe that's the best way to go about them to reduce any error. Thanks!

• I have actually answered my own question. – EhBabay Jun 25 '13 at 7:33

## 1 Answer

Some points for the first one (a). As you have achieved correctly $\theta=\frac{-\pi}3$ and $r=\pm 2$. Now consider the following graphs: • One can tell you teach well, by merely looking at your posts! + – Namaste Jun 26 '13 at 0:07
• @amWhy: Sometimes, it is hard to me to tell what is happening inside the proof. Indeed, it needs a high level writing Amy. These times, handy drawings help me well. – mrs Jun 26 '13 at 0:24
• Yes, there's no replacing "blackboards/whiteboards" (or illustrations, etc)...It's too bad we can't work "live, in action, with pointers to emphasize, etc. – Namaste Jun 26 '13 at 0:27