Question

Rectangular Coordinates to Polar Coordinates

Convert the rectangular coordinates to polar coordinates with$r>0$and

$0\le \theta <2\pi$$(0,-\sqrt{3})$

Short answer

The polar coordinates are$(r,\theta )=(\sqrt{3},\frac{3\pi }{2})$

Step-by-step solution

Step 1. Given information

A point is given with rectangular coordinates.$(0,-\sqrt{3})$

Step 2. Concept used

Substitute the values in the equations and run the numbers. It's possible that more than one polar point will emerge as an answer. However, the right response is that which falls inside the question's range.

Step 3. Calculation

The point given is$(x,y)=(0,-\sqrt{3})$, we use formulae as-

$\begin{array}{l}{r}^2=x^2+y^2\\ \theta ={\tan }^{-1}\left(\frac{y}{x}\right)\end{array}$

Now substitute the values in formulae,

$\begin{array}{l}{r}^2=x^2+y^2\\ \Rightarrow r^2=\text{Invalid <m:msup> element}{0}^2+\\ \Rightarrow r^2=3\\ \Rightarrow r=\pm \sqrt{3}\end{array}$

$\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{y}{x}\right)\\ \theta ={\tan }^{-1}\left(\frac{-\sqrt{3}}{0}\right)\\ \theta ={\tan }^{-1}\left(\infty \right)\\ \Rightarrow \theta =\frac{\pi }{2},\frac{3\pi }{2}\end{array}$

Since,$r>0$ we consider$r=6$ only and this occurs at.$\theta =\pi$