Chapter 9, Problem 8RE

To determine

To find:the rectangular coordinates of the given point.And also find two pairs of polar coordinates.

Answer to Problem 8RE

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

Explanation of Solution

Given:

The givenpoint is $\left(1,-1\right)$ .

Calculation:

According to the question, it is to find two pairs of polar coordinates for ach point, one with $r>0$ and other with $r<0$ and express $\theta$ in radians.

And the given point is $\left(1,-1\right)$ .

Therefore,

  $\begin{array}{l}P\left(1,-1\right)=\left(x,y\right)\\ x=1,y=-1\\ \therefore r=\sqrt{x^2+y^2}\\ =\sqrt{{\left(1\right)}^2+{\left(-1\right)}^2}=\sqrt{1+1}=\sqrt{2}=\sqrt{2}\end{array}$

Now compute the value of $\tan \theta$ so that to find $\theta$ .

  $\begin{array}{l}As,\tan \theta =\frac{y}{x}=\frac{1}{-1}=-1\\ \therefore \theta ={\tan }^{-1}\left(-1\right)\\ \Rightarrow \theta =-\frac{\pi }{4}\end{array}$

Now, to find pair of polar coordinates with $r>0$ and other with $r<0$ .

  $\begin{array}{l}P\left(\sqrt{2},-\frac{\pi }{4}\right)\left\{\text{as},r>0\right\}\\ P\left(-3\sqrt{2},\frac{3\pi }{4}\right)\left\{\text{as},r<0\right\}\end{array}$

Hence, the rectangular coordinates of the given point is $P\left(\sqrt{2},-\frac{\pi }{4}\right),P\left(-3\sqrt{2},\frac{3\pi }{4}\right)$ .