Chapter 8, Problem 1T

(a)

To determine

To find: Rectangular coordinates of a given point whose polar coordinates are known.

Answer to Problem 1T

The rectangular coordinates of the given point are $\left(-4\sqrt{2},-4\sqrt{2}\right)$.

Explanation of Solution

Given:

The given polar coordinates of the point are $\left(r,\theta \right)=\left(8,\frac{5\pi }{4}\right)$.

The value of $r$ is 8 and value of $\theta$ is $\frac{5\pi }{4}$.

Calculation:

The formula to calculate $x-$ coordinate from polar coordinates $\left(r,\theta \right)$ is, $x=r\cos \theta$.

Substitute 8 for $r$ and $\frac{5\pi }{4}$ for $\theta$ in the above equation,

$\begin{array}{c}x=8\cos \left(\frac{5\pi }{4}\right)\\ =8\left(-\frac{1}{\sqrt{2}}\right)\\ =-4\sqrt{2}\end{array}$

The $x-$ coordinate is $-4\sqrt{2}$.

The formula to calculate $y-$ coordinate from polar coordinates $\left(r,\theta \right)$ is, $y=r\sin \theta$.

Substitute 8 for $r$ and $\frac{5\pi }{4}$ for $\theta$ in the above equation,

$\begin{array}{c}y=8\sin \left(\frac{5\pi }{4}\right)\\ =8\left(-\frac{1}{\sqrt{2}}\right)\\ =-4\sqrt{2}\end{array}$

The $y$-coordinate is $-4\sqrt{2}$.

Thus, the rectangular coordinates of the given point are $\left(-4\sqrt{2},-4\sqrt{2}\right)$.

(b)

To determine

To evaluate: The polar coordinates of a given point with $r>0$ and $r<0$ whose rectangular coordinates are known.

Answer to Problem 1T

The polar coordinates of a given point with $r>0$ is $\left(4\sqrt{3},\frac{5\pi }{6}\right)$ and polar coordinates of a given point with $r<0$ is $\left(-4\sqrt{3},\frac{11\pi }{6}\right)$.

Explanation of Solution

Given:

The rectangular coordinates of the point are $\left(-6,2\sqrt{3}\right)$.

Calculation:

Section 1:

The given rectangular coordinates of the point are $\left(-6,2\sqrt{3}\right)$.

The value of x is $-6$ and value of y is $2\sqrt{3}$.

Use the equations $r^2=x^2+y^2$ and $\tan \theta =\frac{y}{x}\left(x\ne 0\right)$ to convert rectangular coordinate system of a point to its corresponding polar coordinate system $\left(r,\theta \right)$.

The value of r is,

$\begin{array}{c}{r}^2=x^2+y^2\\ r=\sqrt{x^2+y^2}\end{array}$

Substitute $-6$ for x and $2\sqrt{3}$ for y in the above formula,

$\begin{array}{c}r=\sqrt{{\left(-6\right)}^2+{\left(2\sqrt{3}\right)}^2}\\ =\sqrt{36+12}\\ =\sqrt{48}\\ =\pm 4\sqrt{3}\end{array}$

The value of r is $4\sqrt{3}$ $\left(r>0\right)$.

The value of $\theta$ is,

$\begin{array}{c}\tan \theta =\frac{y}{x}\left(x\ne 0\right)\\ \theta ={\tan }^{-1}\frac{y}{x}\end{array}$

Substitute $-6$ for x and $2\sqrt{3}$ for y in the above formula,

$\begin{array}{c}\theta ={\tan }^{-1}\frac{y}{x}\\ ={\tan }^{-1}\left(\frac{2\sqrt{3}}{-6}\right)\\ ={\tan }^{-1}\left(-\frac{1}{\sqrt{3}}\right)\\ =\frac{5\pi }{6},\frac{11\pi }{6}\end{array}$

The value of $\theta$ is $\frac{5\pi }{6}$.

Thus, the polar coordinates for point $\left(-6,2\sqrt{3}\right)$ is $\left(4\sqrt{3},\frac{5\pi }{6}\right)$.

Section 2:

From part (a) the value of r is $-4\sqrt{3}$ for $\left(r<0\right)$.

From part (a) the value of $\theta$ is $\frac{11\pi }{6}$.

Thus, the polar coordinates for point $\left(-6,2\sqrt{3}\right)$ with $r<0$ is $\left(-4\sqrt{3},\frac{11\pi }{6}\right)$.