Chapter 14.3, Problem 14WE

To determine

To find: The equivalent polar coordinates for the given rectangular coordinates.

Answer to Problem 14WE

  $\left(2,{225}^0\right)$

Explanation of Solution

Given information:

A rectangular coordinate is given as $\left(-\sqrt{2},-\sqrt{2}\right)$ .

Concepts used: Conversion formula for coordinate systems are given as-

1. From polar to rectangular:

  $x=r\cos \theta$

  $y=r\sin \theta$

2. From rectangular to polar:

  $r=\pm \sqrt{x^2+y^2}$

  $\cos \theta =\frac{x}{r}$ ; $\sin \theta =\frac{y}{r}$ ; $\tan \theta =\frac{x}{y}$ .

Calculation:

Given rectangular coordinate is $\left(-\sqrt{2},-\sqrt{2}\right)$ .

Here $x=-\sqrt{2},y=-\sqrt{2}$ .

Converting into equivalent polar coordinates -

  $\begin{array}{l}r=\pm \sqrt{x^2+y^2}\\ \Rightarrow r=\pm \sqrt{{\left(-\sqrt{2}\right)}^2+{\left(-\sqrt{2}\right)}^2}\\ \Rightarrow r=\pm 2\end{array}$

  $\begin{array}{l}\cos \theta =\frac{x}{r}=\frac{-\sqrt{2}}{2}\left\{\text{Taking positive value of }r\right\}\\ \Rightarrow \theta ={135}^0{\text{,225}}^0\left\{\text{Taking positive value of }r\right\}\\ \sin \theta =\frac{y}{r}=\frac{-\sqrt{2}}{2}\\ \Rightarrow \theta ={225}^0,{315}^0\\ \text{Taking common value of }\theta \text{.}\end{array}$

Hence, desired equivalent polar coordinates would be $\left(2,{225}^0\right)$ .