Chapter 14.3, Problem 3OE

To determine

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

Answer to Problem 3OE

  $\left(5\sqrt{2},{135}^0\right)$

Explanation of Solution

Given information:

A rectangular coordinate is given as $\left(-5,5\right)$ .

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:

  $r=\pm \sqrt{x^2+y^2}$

  $\cos \theta =\frac{x}{r}$

  $\sin \theta =\frac{y}{r}$

Calculation:

Given rectangular coordinate is $\left(-5,5\right)$ .

Here $x=-5,y=5$ .

Converting into equivalent polar coordinates -

  $\begin{array}{l}r=\pm \sqrt{x^2+y^2}\\ \Rightarrow r=\pm \sqrt{{\left(-5\right)}^2+{\left(5\right)}^2}\\ \Rightarrow r=\pm 5\sqrt{2}\end{array}$

  $\begin{array}{l}\cos \theta =\frac{x}{r}=\frac{-5}{5\sqrt{2}}=-\frac{1}{\sqrt{2}}\left\{\text{Taking positive value of }r\right\}\\ \Rightarrow \theta ={135}^0,{225}^0\\ \sin \theta =\frac{y}{r}=\frac{5}{5\sqrt{2}}=\frac{1}{\sqrt{2}}\left\{\text{Taking positive value of }r\right\}\\ \Rightarrow \theta ={135}^0,{225}^0\\ \Rightarrow \theta ={45}^0,{135}^0\\ \text{Common value of }\theta {\text{ is 135}}^0.\end{array}$

Hence, desired equivalent polar coordinates would be $\left(5\sqrt{2},{135}^0\right)$ .