Chapter 6, Problem 23RE

Solution

a)

To find: the polar coordinates of a point $P=\left(5,0\right)$ that satisfy $0\le \theta \le 2\pi$ .

The polar coordinates for the point $\left(5,0\right)$ that satisfy $0\le \theta \le 2\pi ,$ are $\left(5,0°\right)$ , $\left(-5,\pi \right)$ and $\left(5,2\pi \right)$ .

Given information:

The rectangular coordinates of the point $P$ are $\left(5,0\right)$ and $0\le \theta \le 2\pi$ .

Formula used:

• Conversion of rectangular coordinates to polar coordinates:

Let the point $P$ have rectangular coordinates $\left(x,y\right)$ . The polar coordinates $\left(r,\theta \right)$ of $P$ are computed by the following equations:

  $r=\pm \sqrt{x^2+y^2},$ and $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)+n\pi ,$ where $n$ is an integer

• The coordinates $\left(r,\theta \right),$ $\left(r,\theta +2n\pi \right)$ , and $\left(-r,\theta +\pi \right)$ all the same point.

Calculation:

Substitute $5$ for $x$ and $0$ for $y$ in the equations: $r=\pm \sqrt{x^2+y^2},$ and $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)+n\pi ,$ where $n$ is an integer.

  $\begin{array}{c}r=\pm \sqrt{{\left(5\right)}^2+{\left(0\right)}^2}\\ =\pm \sqrt{25}\\ =\pm 5\end{array}$

Thus, $r=\pm 5$

And

  $\begin{array}{c}\theta ={\tan }^{-1}\left(0/\left(5\right)\right)+n\pi \\ ={\tan }^{-1}\left(0\right)+n\pi \\ =n\pi \end{array}$

Given $0\le \theta \le 2\pi ,$ then possible values for $\theta$ are $0,$

  $\pi$ and $2\pi$ .

Thus, the polar coordinates for the point $\left(5,0\right)$ that satisfy $0\le \theta \le 2\pi ,$ are $\left(5,0°\right)$ or $\left(5,2\pi \right)$ , and $\left(-5,\pi \right)$ .

b)

To find: the polar coordinates of point $P=\left(5,0\right)$ that satisfy $-\pi \le \theta \le \pi$ .

The polar coordinates for the point $\left(5,0\right)$ that satisfy $-\pi \le \theta \le \pi ,$ are $\left(5,0°\right)$ , $\left(-5,-\pi \right)$ , and $\left(-5,\pi \right)$ .

Given information:

The rectangular coordinates of the point $P$ are $\left(5,0\right)$ and $-\pi \le \theta \le \pi$ .

Formula used:

• Conversion of rectangular coordinates to polar coordinates:

Let the point $P$ have rectangular coordinates $\left(x,y\right)$ . The polar coordinates $\left(r,\theta \right)$ of $P$ are computed by following equations:

  $r=\pm \sqrt{x^2+y^2},$ and $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)+n\pi ,$ where $n$ is an integer.

• The coordinates $\left(r,\theta \right),$ $\left(r,\theta +2n\pi \right)$ , and $\left(-r,\theta +\pi \right)$ all the same point.

Calculation:

As calculated in part (a), $r=\pm 5$ and $\theta =n\pi$ , where $n$ is the integer.

Given $-\pi \le \theta \le \pi ,$ then possible values for $\theta$ are $-\pi ,$

  $0,$ and $\pi$ .

Thus, the polar coordinates for the point $\left(5,0\right)$ that satisfy $-\pi \le \theta \le \pi ,$ are $\left(5,0°\right)$ , $\left(-5,-\pi \right),$ and $\left(-5,\pi \right)$ .

c)

To find: the polar coordinates of point $P=\left(5,0\right)$ that satisfy $0\le \theta \le 4\pi$ .

The polar coordinates for the point $\left(5,0\right)$ that satisfy $0\le \theta \le 4\pi ,$ are $\left(5,0°\right),$

  $\left(5,2\pi \right),$

  $\left(5,4\pi \right),$

  $\left(-5,\pi \right),$ and $\left(-5,3\pi \right)$ .

Given information:

The rectangular coordinates of the point $P$ are $\left(5,0\right)$ and $0\le \theta \le 4\pi$ .

Formula used:

• Conversion of rectangular coordinates to polar coordinates:

Let the point $P$ have rectangular coordinates $\left(x,y\right)$ . The polar coordinates $\left(r,\theta \right)$ of $P$ are computed by following equations:

  $r=\pm \sqrt{x^2+y^2},$ and $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)+n\pi ,$ where $n$ is an integer.

• The coordinates $\left(r,\theta \right),$ $\left(r,\theta +2n\pi \right)$ , and $\left(-r,\theta +\pi \right)$ all the same point.

Calculation:

As calculated in part (a), $r=\pm 5$ and $\theta =n\pi$ , where $n$ is the integer.

Given $0\le \theta \le 4\pi ,$ the possible values for $\theta$ are $0,$ $\pi ,$ $2\pi ,$ $3\pi$ and $4\pi$ .

Thus, the polar coordinates for the point $\left(5,0\right)$ that satisfy $0\le \theta \le 4\pi ,$ are $\left(5,0°\right),$ $\left(5,2\pi \right),$ $\left(5,4\pi \right),$ $\left(-5,\pi \right),$ and $\left(-5,3\pi \right)$ .