Chapter 14.3, Problem 6WE

To determine

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

Answer to Problem 6WE

  $\left(\frac{6\sqrt{2}}{2},-\frac{6\sqrt{2}}{2}\right)$

Explanation of Solution

Given information:

A polar coordinate is given as $\left(6,-{45}^0\right)$ .

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 $\left(6,-{45}^0\right)$ .

Here $r=6,\theta =-{45}^0$ .

Plotting the given point in polar coordinate system.

  

Converting into equivalent rectangular coordinates -

  $\begin{array}{l}x=r\cos \theta \\ \Rightarrow x=6\cos \left(-{45}^0\right)\\ \Rightarrow x=6\left(\frac{\sqrt{2}}{2}\right)\\ \Rightarrow x=\frac{6\sqrt{2}}{2}\\ \text{And }y=r\sin \theta \\ \Rightarrow y=6sin\left(-{45}^0\right)\\ \Rightarrow y=6\left(-\frac{\sqrt{2}}{2}\right)\\ \Rightarrow y=-\frac{6\sqrt{2}}{2}\end{array}$

Hence, desired equivalent rectangular coordinates would be $\left(\frac{6\sqrt{2}}{2},-\frac{6\sqrt{2}}{2}\right)$ .