Chapter 6.4, Problem 28E

Solution

a)

To find: The polar coordinates from the rectangular coordinates that satisfy $0\le \theta \le 2\pi$ .

The polar coordinates are $\left(\sqrt{10},1.25\right)$ or $\left(-\sqrt{10},4.39\right)$ .

Given information:

The point is $P=\left(1,3\right)$ .

Formula used:

Use the following conversion relationships between a polar coordinate $\left(r,\theta \right)$ and its corresponding rectangular coordinate $\left(x,y\right)$ .

  $\begin{array}{c}{r}^2=x^2+y^2\\ \tan \theta =\frac{y}{x}\end{array}$

Calculation:

Substitute $1$ for $x$ and $3$ for $y$ in equation $r^2=x^2+y^2$ to find $r$ .

  $\begin{array}{c}{r}^2={\left(1\right)}^2+{\left(3\right)}^2\\ r^2=10\\ r=\pm \sqrt{10}\end{array}$

Substitute $1$ for $x$ and $3$ for $y$ in equation $\tan \theta =\frac{y}{x}$ to find $\theta$ .

  $\begin{array}{c}\tan \theta =\frac{3}{1}\\ \tan \theta =3\\ \theta ={\tan }^{-1}\left(3\right)\\ \theta =1.25+\pi n\end{array}$

On interval $0\le \theta \le 2\pi$, the possible angles are,

  $\theta =1.25,4.39$

For $\theta =1.25$, use $r=\sqrt{10}$ since $P$ is on Quadrant I. So the polar coordinate is,

  $\left(\sqrt{10},1.25\right)$

For $\theta =4.39$, use $r=-\sqrt{10}$ since $P$ is on Quadrant I (so that the point is opposite the terminal side of the angle). So the polar coordinate is,

  $\left(-\sqrt{10},4.39\right)$

Therefore, the polar coordinates are $\left(\sqrt{10},1.25\right)$ or $\left(-\sqrt{10},4.39\right)$ .

b)

To find: The polar coordinates from the rectangular coordinates that satisfy $-\pi \le \theta \le \pi$ .

The polar coordinates are $\left(-\sqrt{10},-1.89\right)$ or $\left(\sqrt{10},1.25\right)$ .

Calculation:

On interval $-\pi \le \theta \le \pi$, the possible angles are,

  $\theta =-1.89,1.25$

For $\theta =-1.89$, use $r=-\sqrt{10}$ since $P$ is on Quadrant I (so that the point is opposite the terminal side of the angle):. So the polar coordinate is,

  $\left(-\sqrt{10},-1.89\right)$

For $\theta =1.25$, use $r=\sqrt{10}$ since $P$ is on Quadrant I. So the polar coordinate is,

  $\left(\sqrt{10},1.25\right)$

Therefore, the polar coordinates are $\left(-\sqrt{10},-1.89\right)$ or $\left(\sqrt{10},1.25\right)$ .

c)

To find: The polar coordinates from the rectangular coordinates that satisfy $0\le \theta \le 4\pi$ .

The polar coordinates are $\left(\sqrt{10},1.25\right)$,

  $\left(-\sqrt{10},4.39\right)$, $\left(\sqrt{10},7.53\right)$ or $\left(-\sqrt{10},10.67\right)$ .

Calculation:

On interval $0\le \theta \le 4\pi$, the possible angles are,

  $\theta =1.25,4.39,7.53,10.67$

For $\theta =1.25$, use $r=\sqrt{10}$ since $P$ is on Quadrant I. So the polar coordinate is,

  $\left(\sqrt{10},1.25\right)$

For $\theta =4.39$, use $r=-\sqrt{10}$ since $P$ is on Quadrant I (so that the point is opposite the terminal side of the angle). So the polar coordinate is,

  $\left(-\sqrt{10},4.39\right)$

For $\theta =7.53$, use $r=\sqrt{10}$ . So the polar coordinate is,

  $\left(\sqrt{10},7.53\right)$

For $\theta =10.67$, use $r=-\sqrt{10}$ . So the polar coordinate is,

  $\left(-\sqrt{10},10.67\right)$

Therefore, the polar coordinates are $\left(\sqrt{10},1.25\right)$, $\left(-\sqrt{10},4.39\right)$, $\left(\sqrt{10},7.53\right)$ or $\left(-\sqrt{10},10.67\right)$ .