Chapter 10.3, Problem 2E

a)

To determine

To plot: The point with the given polar coordinates and find the corresponding rectangular coordinates.

Answer to Problem 2E

The given polar coordinates are shown in Figure (1) and the corresponding rectangular coordinate is $\left(\frac{3\sqrt{3}}{2},\frac{-3}{2}\right)$ .

Explanation of Solution

Given information: A polar coordinate is given as $\left(-3,\frac{5\pi }{6}\right)$ .

Formula used: Let the point P have polar coordinates $\left(r,\theta \right)$ and rectangular coordinates $\left(x,y\right)$ . Then,

  $\begin{array}{l}x=r\cos \theta \\ y=r\sin \theta \end{array}$

Calculation:

Plot $\left(-3,\frac{5\pi }{6}\right)$ on a coordinate plane as shown below.

  

Figure (1)

Find the $x$ -coordinate of the given point.

  $\begin{array}{c}x=-3\cos \frac{5\pi }{6}\\ =\frac{3\sqrt{3}}{2}\end{array}$

Find the $y$ -coordinate of the given point.

  $\begin{array}{c}y=-3\sin \frac{5\pi }{6}\\ =\frac{-3}{2}\end{array}$

Therefore, the corresponding rectangular coordinate is $\left(\frac{3\sqrt{3}}{2},\frac{-3}{2}\right)$ .

b)

To determine

To plot: The point with the given polar coordinates and find the corresponding rectangular coordinates.

Answer to Problem 2E

The given polar coordinates are shown in Figure (1) and the corresponding rectangular coordinate is $\left(3,4\right)$ .

Explanation of Solution

Given information: A polar coordinate is given as $\left(5,{\tan }^{-1}\frac{4}{3}\right)$ .

Formula used: Let the point P have polar coordinates $\left(r,\theta \right)$ and rectangular coordinates $\left(x,y\right)$ . Then,

  $\begin{array}{l}x=r\cos \theta \\ y=r\sin \theta \end{array}$

Calculation:

Plot $\left(5,{\tan }^{-1}\frac{4}{3}\right)$ on a coordinate plane as shown below.

  

Figure (1)

Find the $x$ -coordinate of the given point.

  $\begin{array}{c}x=5\times \cos \left({\tan }^{-1}\frac{4}{3}\right)\\ =5\cos \left({\cos }^{-1}\frac{3}{\sqrt{3^2+4^2}}\right)\\ =5\cdot \frac{3}{5}\\ =3\end{array}$

Find the $y$ -coordinate of the given point.

  $\begin{array}{c}y=5\times \sin \left({\tan }^{-1}\frac{4}{3}\right)\\ =5\sin \left({\sin }^{-1}\frac{4}{\sqrt{3^2+4^2}}\right)\\ =5\cdot \frac{4}{5}\\ =4\end{array}$

Therefore, the corresponding rectangular coordinate is $\left(3,4\right)$ .

c)

To determine

To plot: The point with the given polar coordinates and find the corresponding rectangular coordinates.

Answer to Problem 2E

The given polar coordinates are shown in Figure (1) and the corresponding rectangular coordinate is $\left(1,0\right)$ .

Explanation of Solution

Given information: A polar coordinate is given as $\left(-1,7\pi \right)$ .

Formula used: Let the point P have polar coordinates $\left(r,\theta \right)$ and rectangular coordinates $\left(x,y\right)$ . Then,

  $\begin{array}{l}x=r\cos \theta \\ y=r\sin \theta \end{array}$

Calculation:

Plot $\left(-1,7\pi \right)$ on polar coordinate system as shown below.

  

Figure (1)

Find the $x$ -coordinate of the given point.

  $\begin{array}{c}x=-1\times \cos 7\pi \\ =1\end{array}$

Find the $y$ -coordinate of the given point.

  $\begin{array}{c}y=-1\times \sin 7\pi \\ =0\end{array}$

Therefore, the corresponding rectangular coordinate is $\left(1,0\right)$ .

d)

To determine

To plot: The point with the given polar coordinates and find the corresponding rectangular coordinates.

Answer to Problem 2E

The given polar coordinates are shown in Figure (1) and the corresponding rectangular coordinate is $\left(-\sqrt{3},3\right)$ .

Explanation of Solution

Given information: A polar coordinate is given as $\left(2\sqrt{3},\theta =\frac{2\pi }{3}\right)$ .

Formula used: Let the point P have polar coordinates $\left(r,\theta \right)$ and rectangular coordinates $\left(x,y\right)$ . Then,

  $\begin{array}{l}x=r\cos \theta \\ y=r\sin \theta \end{array}$

Calculation:

Plot $\left(2\sqrt{3},\frac{2\pi }{3}\right)$ on polar coordinate system as shown below.

  

Figure (1)

Find the $x$ -coordinate of the given point.

  $\begin{array}{c}x=2\sqrt{3}\times \cos \frac{2\pi }{3}\\ =2\sqrt{3}\left(-\frac{1}{2}\right)\\ =-\sqrt{3}\end{array}$

Find the $y$ -coordinate of the given point.

  $\begin{array}{c}y=2\sqrt{3}\times \sin \frac{2\pi }{3}\\ =2\sqrt{3}\left(\frac{\sqrt{3}}{2}\right)\\ =3\end{array}$

Therefore, the corresponding rectangular coordinate is $\left(-\sqrt{3},3\right)$ .