Chapter 8, Problem 12RE

(a)

To determine

The point $P\left(4,-4\right)$ is to be plotted.

Explanation of Solution

Given:

  $P\left(4,-4\right)$

Graph:

  

Interpretation:

From the given data point:

The polar co-ordinates $\left(x,y\right)$ of the point $P$ are $P\left(4,-4\right)$ . On the graph the point is plotted in the fourth quadrant as first value is positive and second value negative is 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 12RE

The values of polar co-ordinates of the point $P$ are $\left(r=2\sqrt{3},\theta =\frac{5\pi }{6}\right)$

Explanation of Solution

Given:

  $\begin{array}{l}P\left(4,-4\right)\\ r\ge 0\end{array}$

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 $P\left(4,-4\right)$ understands that:

  $\begin{array}{l}x=4\\ y-4\end{array}$

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

  $r=\sqrt{{\left(4\right)}^2+{\left(-4\right)}^2}=\sqrt{16+16}=4\sqrt{2}$

Since, $r\ge 0$ : $r=4\sqrt{2}$

And $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)={\tan }^{-1}\left(\frac{4}{-4}\right)={\tan }^{-1}\left(-1\right)=-\frac{\pi }{4}$

But required to adjust the angle to correspond to the correct quadrant:

  $\theta =-\frac{\pi }{4}+2\pi =\frac{7\pi }{4}$

Conclusion:

Hence, the polar co-ordinates are $\left(r=4\sqrt{2},\theta =\frac{7\pi }{4}\right)$

(c)

To determine

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

Answer to Problem 12RE

The values of polar co-ordinates of the point $P$ are $\left(r=-4\sqrt{2},\theta =\frac{3\pi }{4}\right)$

Explanation of Solution

Given:

  $\begin{array}{l}P\left(4,-4\right)\\ r\le 0\end{array}$

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 $P\left(4,-4\right)$ understands that:

  $\begin{array}{l}x=4\\ y-4\end{array}$

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

  $r=\sqrt{{\left(4\right)}^2+{\left(-4\right)}^2}=\sqrt{16+16}=4\sqrt{2}$

Since, $r\le 0$ : $r=-4\sqrt{2}$

And $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)={\tan }^{-1}\left(\frac{4}{-4}\right)={\tan }^{-1}\left(-1\right)=-\frac{\pi }{4}$

But required to adjust the angle to correspond to the correct quadrant:

  $\theta =-\frac{\pi }{4}+2\pi =\frac{7\pi }{4}$

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

  $\theta =\frac{7\pi }{4}-\pi =\frac{3\pi }{4}$

Conclusion:

Hence, the polar co-ordinates are $\left(r=-4\sqrt{2},\theta =\frac{3\pi }{4}\right)$