Chapter 9, Problem 16CT

To determine

To find:Polar coordinates of point $\left(2,-2\right)$

Answer to Problem 16CT

  $\left(2\sqrt{2},-\frac{\pi }{4}\right)$

And two additional polar representations: $\left(2\sqrt{2},\frac{7\pi }{4}\right)$ and $\left(-2\sqrt{2},\frac{3\pi }{4}\right)$

Explanation of Solution

Given information:

  $\left(2,-2\right)$

  $\left(2,-2\right)$

ThePolar coordinates of rectangular form of point $\left(x,y\right)$ is represented as $\left(r,\theta \right)$ .

  $\begin{array}{l}x=r\cos \theta \\ y=r\sin \theta \end{array}$

Here

  $\left(x,y\right)=\left(2,-2\right)$

Therefore

  $2=r\cos \theta$ --------(1)

  $-2=r\sin \theta$ ----------(2)

Squaring and adding on both sides,

We get

  $\begin{array}{l}{r}^2{\sin }^2\left(\theta \right)+r^2{\cos }^2\left(\theta \right)=2^2+{\left(-2\right)}^2\\ \Rightarrow r^2\left({\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)\right)=4+4\\ \Rightarrow r^2\left(1\right)=8\\ \Rightarrow r^2=8\\ \Rightarrow r=\pm \sqrt{8}\\ \Rightarrow r=\pm 2\sqrt{2}\end{array}$

Substituting $r=2\sqrt{2}$ in equation (1)

  $\begin{array}{l}2=r\cos \theta \\ \Rightarrow 2=2\sqrt{2}\cos \theta \\ \Rightarrow \frac{2}{2\sqrt{2}}=\cos \theta \\ \Rightarrow \cos \theta =\frac{1}{\sqrt{2}}\\ \Rightarrow \cos \theta =\left(-\frac{\pi }{4}\right)\\ \Rightarrow \theta =-\frac{\pi }{4}\end{array}$

Since the point $\left(2,-2\right)$ lies in 4^th quadrant, $\theta$ also represents angle in 4^th quadrant.

Therefore, the polar coordinates of point $\left(2,-2\right)$ is $\left(2\sqrt{2},-\frac{\pi }{4}\right)$

The other possible value of $\theta$ for value of $r=2\sqrt{2}$ is $\frac{7\pi }{4}$ .

Therefore, the other polar coordinates of point $\left(2,-2\right)$ is $\left(2\sqrt{2},\frac{7\pi }{4}\right)$

Substitute $r=-2\sqrt{2}$ in equation (1),

We get

  $\begin{array}{l}2=r\cos \theta \\ \Rightarrow 2=-2\sqrt{2}\cos \theta \\ \Rightarrow \frac{2}{-2\sqrt{2}}=\cos \theta \\ \Rightarrow \cos \theta =-\frac{1}{\sqrt{2}}\\ \Rightarrow \cos \theta =\cos \left(\frac{3\pi }{4}\right)\\ \Rightarrow \theta =\frac{3\pi }{4}\end{array}$

The other polar coordinate of point $\left(2,-2\right)$ is $\left(-2\sqrt{2},\frac{3\pi }{4}\right)$

Therefore, the two additional polar representations of point $\left(2,-2\right)$ are

  $\left(2\sqrt{2},\frac{7\pi }{4}\right)$ and $\left(-2\sqrt{2},\frac{3\pi }{4}\right)$