Chapter H.1, Problem 2E

(a)

To determine

To graph:

The following polar coordinates and others pairs of polar coordinates for $r>0$ and $r<0$ conditions:

  $\left(1,\frac{7\pi }{4}\right)$

Explanation of Solution

Given information:

  $\left(r,\theta \right)=\left(1,\frac{7\pi }{4}\right)$

Graph:

  

Interpretation:

The given polar co-ordinate $\left(r,\theta \right)=\left(1,\frac{7\pi }{4}\right)$ is plotted as above. Now, need to find the two others pairs of polar coordinates:

The point represented by $\left(r,\theta \right)$ is the same point which is represented by $\left(r,\theta +2\pi n\right)$ . So, the first new coordinate is:

  $\begin{array}{l}\left(r,\theta +2\pi n\right)=\left(1,\frac{7\pi }{4}+2\pi \right)\\ \left(r,\theta +2\pi n\right)=\left(1,\frac{15\pi }{4}\right)\end{array}$

The same point is represented by $\left(r,\theta \right)$ and $\left(-r,\theta +\left(2n+1\right)n\right)$ . So, the second coordinate pair is:

  $\begin{array}{l}\left(-r,\theta +\left(2n+1\right)\pi \right)=\left(-1,\frac{7\pi }{4}-2\pi \right)\\ \left(-r,\theta +\left(2n+1\right)\pi \right)=\left(-1,-\frac{\pi }{4}\right)\end{array}$

Hence,

The coordinate for the condition $r>0$ is $\left(1,\frac{15\pi }{4}\right)$ and for $r<0$ is $\left(-1,-\frac{\pi }{4}\right)$ .

(b)

To determine

To graph:

The following polar coordinates and others pairs of polar coordinates for $r>0$ and $r<0$ conditions:

  $\left(-3,\frac{\pi }{6}\right)$

Explanation of Solution

Given information:

  $\left(r,\theta \right)=\left(-3,\frac{\pi }{6}\right)$

Graph:

  

Interpretation:

The given polar co-ordinate $\left(r,\theta \right)=\left(-3,\frac{\pi }{6}\right)$ is plotted as above. Now, need to find the two others pairs of polar coordinates:

The point represented by $\left(r,\theta \right)$ is the same point which is represented by $\left(r,\theta +2\pi n\right)$ . So, the first new coordinate is:

  $\begin{array}{l}\left(r,\theta +2\pi n\right)=\left(-3,\frac{\pi }{6}+2\pi \right)\\ \left(r,\theta +2\pi n\right)=\left(-3,\frac{13\pi }{6}\right)\end{array}$

The same point is represented by $\left(r,\theta \right)$ and $\left(-r,\theta +\left(2n+1\right)n\right)$ . So, the second coordinate pair is:

  $\begin{array}{l}\left(-r,\theta +\left(2n+1\right)\pi \right)=\left(-3,\frac{\pi }{6}-2\pi \right)\\ \left(-r,\theta +\left(2n+1\right)\pi \right)=\left(-3,\frac{7\pi }{6}\right)\end{array}$

Hence,

The coordinate for the condition $r>0$ is $\left(-3,\frac{13\pi }{6}\right)$ and for $r<0$ is $\left(-3,\frac{7\pi }{6}\right)$ .

(c)

To determine

To graph:

The following polar coordinates and others pairs of polar coordinates for $r>0$ and $r<0$ conditions:

  $\left(1,-1\right)$

Explanation of Solution

Given information:

  $\left(r,\theta \right)=\left(1,-1\right)$

Graph:

  

Interpretation:

The given polar co-ordinate $\left(r,\theta \right)=\left(1,-1\right)$ is plotted as above. Now, need to find the two others pairs of polar coordinates:

The point represented by $\left(r,\theta \right)$ is the same point which is represented by $\left(r,\theta +2\pi n\right)$ . So, the first new coordinate is:

  $\left(r,\theta +2\pi n\right)=\left(1,-1+2\pi \right)$

The same point is represented by $\left(r,\theta \right)$ and $\left(-r,\theta +\left(2n+1\right)n\right)$ . So, the second coordinate pair is:

  $\left(-r,\theta +\left(2n+1\right)\pi \right)=\left(-1,-1-2\pi \right)$

Hence,

The coordinate for the condition $r>0$ is $\left(1,-1+2\pi \right)$ and for $r<0$ is $\left(-1,-1-2\pi \right)$ .