Chapter 8, Problem 10RE

(a)

To determine

The point $P\left(3\sqrt{3},3\right)$ is to be plotted.

Explanation of Solution

Given:

  $P\left(3\sqrt{3},3\right)$

Graph:

  

Interpretation:

From the given data point:

The polar co-ordinates $\left(x,y\right)$ of the point $P$ are $P\left(3\sqrt{3},3\right)=P\left(5.196,1.732\right)$ . On the graph the point is plotted in the second quadrant as both values are positive in nature.

(b)

To determine

The values of polar co-ordinates of the point $P$ are to be estimated at $r\ge 0$ .

Answer to Problem 10RE

The values of polar co-ordinates of the point $P$ are $\left(r=30,\theta ={\tan }^{-1}\left(3\right)\right)$

Explanation of Solution

Given:

  $\begin{array}{l}P\left(3\sqrt{3},3\right)\\ r\ge 0\end{array}$

Concept Used:

When a point is given with its rectangular co-ordinates $\left(x,y\right)$ , its polar co-ordinates $\left(r,\theta \right)$ are defined as:

  $\begin{array}{l}r=\sqrt{x^2+y^2}\\ \theta ={\tan }^{-1}\left(\frac{y}{x}\right)\end{array}$

Calculation:

From the given point $P\left(3\sqrt{3},3\right)$ understands that:

  $\begin{array}{l}x=3\sqrt{3}\\ y=\sqrt{3}\end{array}$

Now, polar co-ordinates of $P\left(r,\theta \right)$ which are expressed as:

  $r=\sqrt{{\left(3\sqrt{3}\right)}^2+{\left(\sqrt{3}\right)}^2}=\sqrt{27+3}=\sqrt{30}$

Since, $r\ge 0$ : $r=\sqrt{30}$

And $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)={\tan }^{-1}\left(\frac{3\sqrt{3}}{\sqrt{3}}\right)={\tan }^{-1}\left(3\right)$

Conclusion:

Hence, the polar co-ordinates are $\left(r=30,\theta ={\tan }^{-1}\left(3\right)\right)$

(c)

To determine

The values of polar co-ordinates of the point $P$ are to be estimated at $r\le 0$ .

Answer to Problem 10RE

The values of polar co-ordinates of the point $P$ are $\left(r=-\sqrt{30},\theta =\pi +{\tan }^{-1}\left(3\right)\right)$

Explanation of Solution

Given:

  $\begin{array}{l}P\left(8,8\right)\\ r\le 0\end{array}$

Concept Used:

When a point is given with its rectangular co-ordinates $\left(x,y\right)$ , its polar co-ordinates $\left(r,\theta \right)$ are defined as:

  $\begin{array}{l}r=\sqrt{x^2+y^2}\\ \theta ={\tan }^{-1}\left(\frac{y}{x}\right)\end{array}$

Calculation:

From the given point $P\left(3\sqrt{3},3\right)$ understands that:

  $\begin{array}{l}x=3\sqrt{3}\\ y=\sqrt{3}\end{array}$

Now, polar co-ordinates of $P\left(r,\theta \right)$ which are expressed as:

  $r=\sqrt{{\left(3\sqrt{3}\right)}^2+{\left(\sqrt{3}\right)}^2}=\sqrt{27+3}=\sqrt{30}$

Since, $r\le 0$ : $r=-\sqrt{30}$

And $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)={\tan }^{-1}\left(\frac{3\sqrt{3}}{\sqrt{3}}\right)={\tan }^{-1}\left(3\right)$

Add or subtract the $\pi$ from the value of $\theta$ as $r\le 0$

  $\theta =\pi +{\tan }^{-1}\left(3\right)$

Conclusion:

Hence, the polar co-ordinates are $\left(r=-\sqrt{30},\theta =\pi +{\tan }^{-1}\left(3\right)\right)$