Chapter 9.5, Problem 40E

To determine

To calculate: To plot the point in rectangular coordinates and find two setsof polar representations of the point, using $0<\theta <2\pi$

Answer to Problem 40E

The rectangular coordinates of point on the graph is plotted and two sets of polar points are $\left(\sqrt{10},-0.321\right),\left(-\sqrt{10},2.819\right)$

Explanation of Solution

Given information: Rectangularis $\left(3,-1\right)$

Formula Used:

Polar coordinate of a point $\left(x,y\right)$ is given as

  $\left(r,\theta \right)$

Where,

  $r=\sqrt{x^2+y^2}$

  $\theta =\arctan \left(\left|\frac{y}{x}\right|\right)$

If both $x,y>0$ , then $\theta =\alpha$

If $x<0,y>0$ , then $\theta =\pi -\alpha$

If $x,y<0$ , then $\theta =\pi +\alpha$

If $x>0,y<0$ , then $\theta =-\alpha$

Point $\left(r,\theta \right)$ can be represented as

  $\left(r,\theta \right)=\left(r,\theta \pm 2n\pi \right)$

Or

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

Calculation:

Rectangular point is given as

  $\left(3,-1\right)$

Plotting the rectangular coordinates on the graph

  

Converting rectangular coordinates into polar coordinates:

Polar coordinate of a point $\left(x,y\right)$ is given as

  $\left(r,\theta \right)$

Calculating the value of $r$ :

  $\begin{array}{l}r=\sqrt{{\left(3\right)}^2+{\left(-1\right)}^2}\\ r=\sqrt{9+1}\\ r=\sqrt{10}\end{array}$

Calculating the value of $\theta$ :

Since $x<0,y>0$ , then $\theta =-\alpha$

  $\begin{array}{l}\theta =\pi -\arctan \left(\left|\frac{-1}{3}\right|\right)\\ \theta \approx -0.321\end{array}$

Thus, polar coordinate of point is $\left(\sqrt{10},-0.321\right)$

Now, let’s find another set polar representations of the point, where $0<\theta <2\pi$

Point $\left(r,\theta \right)$ can be represented as

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

Here,

  $\begin{array}{l}r=\sqrt{10}\\ \theta =-0.321\end{array}$

Substituting the values in above representations, another representation of given point are:

  $\left(-\sqrt{10},-0.321+\pi \right)=\left(-\sqrt{10},2.819\right)$

Conclusion:

Hence, rectangular coordinates of point on the graph is plotted and twosets of polar points are $\left(\sqrt{10},-0.321\right),\left(-\sqrt{10},2.819\right)$