Chapter H.1, Problem 5E

(a)

To determine

To find:The polar coordinates $\left(r,\theta \right)$

Answer to Problem 5E

The polar coordinates of $\left(2,-2\right)$ are,

  $\left(2\sqrt{2,}\frac{7\pi }{4}\right),$ Where $r>0$

  $\left(-2\sqrt{2,}\frac{3\pi }{4}\right),$ Where $r<0$

Explanation of Solution

Given:

The point $\left(2,-2\right)$ where $r>0$ and $0\le \theta <2\pi$

Calculation:

Consider the Cartesian coordinates be $\left(x,y\right)$ .

The polar coordinates $\left(r,\theta \right)$ is defines as,

  $r^2=x^2+y^2$

  $x=r\cos \theta$

  $y=r\sin \theta$

This implies that, $\tan \theta =\frac{y}{x}$ .

The Cartesian coordinates be $\left(2,-2\right)$ .

Calculate $r$ and $\theta$ as follows:

  $r^2=2^2+{\left(-2\right)}^2$

  $=8$

  $r=\pm 2\sqrt{2}$

  $\tan \theta =\frac{y}{x}$

  $=\frac{-2}{2}$

  $=-1$

Thus, the angle $\theta$ is $\frac{3\pi }{4},\frac{7\pi }{4}$ as the angle $0\le \theta \le 2\pi$ .

The polar coordinates of $\left(2,-2\right)$ are,

  $\left(2\sqrt{2,}\frac{7\pi }{4}\right),$ Where $r>0$

  $\left(-2\sqrt{2,}\frac{3\pi }{4}\right),$ Where $r<0$

Conclusion:

Therefore, thepolar coordinates of $\left(2,-2\right)$ are,

  $\left(2\sqrt{2,}\frac{7\pi }{4}\right),$ Where $r>0$

  $\left(-2\sqrt{2,}\frac{3\pi }{4}\right),$ Where $r<0$

(b)

To determine

To find: The polar coordinates $\left(r,\theta \right)$

Answer to Problem 5E

The polar coordinates of $\left(-1,\sqrt{3}\right)$ are,

  $\left(2,\frac{2\pi }{3}\right),$ where $r>0$

  $\left(-2,\frac{5\pi }{3}\right),$ where $r<0$

Explanation of Solution

Given:

The point $\left(-1,\sqrt{3}\right)$ where $r<0$ and $0\le \theta <2\pi$

Calculation:

The Cartesian coordinates be $\left(-1,\sqrt{3}\right)$ .

Calculate $r$ and $\theta$ as follows:

  $r^2={\left(-1\right)}^2+{\left(\sqrt{3}\right)}^2$

  $=4$

  $r=\pm 2\sqrt{2}$

  $\tan \theta =\frac{y}{x}$

  $=\frac{\sqrt{3}}{-1}$

  $=-\sqrt{3}$

Thus, the angle $\theta$ is $\frac{2\pi }{3},\frac{5\pi }{3}$ as the angle $0\le \theta \le 2\pi$ .

The polar coordinates of $\left(-1,\sqrt{3}\right)$ are,

  $\left(2,\frac{2\pi }{3}\right),$ Where $r>0$

  $\left(-2,\frac{5\pi }{3}\right),$ Where $r<0$

Conclusion:

Therefore, the polar coordinates of $\left(-1,\sqrt{3}\right)$ are,

  $\left(2,\frac{2\pi }{3}\right),$ Where $r>0$

  $\left(-2,\frac{5\pi }{3}\right),$ Where $r<0$