Chapter 8.1, Problem 42E

To determine

To convert the given rectangular coordinates to polar coordinates with $r>0$ and $0\le \theta <2\pi$ -

  $\left(0,-\sqrt{3}\right)$

Answer to Problem 42E

The polar coordinates are $\left(\sqrt{3},\frac{\pi }{2}\right)$ .

Explanation of Solution

Given:

Rectangular coordinates - $\left(0,-\sqrt{3}\right)$

  $r>0$ and $0\le \theta <2\pi$

Formula Used:

A polar equation is any equation that describes a relation between r and θ , where r represents the distance from the pole (origin) to a point on a curve, and θ represents the counter-clockwise angle made by a point on a curve, the pole, and the positive x-axis.

Also, $x=r\cos \theta$ and $y=r\sin \theta$

  $r=\sqrt{x^2+y^2}$ and $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)$

Calculation:

Given rectangular coordinates are $\left(0,-\sqrt{3}\right)$

Using the above definition, we have:

  $x=0$ and $y=-\sqrt{3}$

  $\begin{array}{l}r=\sqrt{{\left(0\right)}^2+{\left(-\sqrt{3}\right)}^2}\\ r=\sqrt{0+3}\\ r=\sqrt{3}\end{array}$

Now,

  $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)$

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

Thus, the polar coordinates are $\left(\sqrt{3},\frac{\pi }{2}\right)$ .

Conclusion:

The polar coordinates are $\left(\sqrt{3},\frac{\pi }{2}\right)$ .