# 4.3 Polar coordinates  (Page 2/8)

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$\begin{array}{l}\begin{array}{l}\\ \mathrm{cos}\text{\hspace{0.17em}}\theta =\frac{x}{r}\to x=r\mathrm{cos}\text{\hspace{0.17em}}\theta \end{array}\hfill \\ \mathrm{sin}\text{\hspace{0.17em}}\theta =\frac{y}{r}\to y=r\mathrm{sin}\text{\hspace{0.17em}}\theta \hfill \end{array}$

Dropping a perpendicular from the point in the plane to the x- axis forms a right triangle, as illustrated in [link] . An easy way to remember the equations above is to think of $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ as the adjacent side over the hypotenuse and $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ as the opposite side over the hypotenuse.

## Converting from polar coordinates to rectangular coordinates

To convert polar coordinates $\text{\hspace{0.17em}}\left(r,\text{\hspace{0.17em}}\theta \right)\text{\hspace{0.17em}}$ to rectangular coordinates $\text{\hspace{0.17em}}\left(x,\text{\hspace{0.17em}}y\right),$ let

$\mathrm{cos}\text{\hspace{0.17em}}\theta =\frac{x}{r}\to x=r\mathrm{cos}\text{\hspace{0.17em}}\theta$
$\mathrm{sin}\text{\hspace{0.17em}}\theta =\frac{y}{r}\to y=r\mathrm{sin}\text{\hspace{0.17em}}\theta$

Given polar coordinates, convert to rectangular coordinates.

1. Given the polar coordinate $\text{\hspace{0.17em}}\left(r,\theta \right),$ write $\text{\hspace{0.17em}}x=r\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=r\mathrm{sin}\text{\hspace{0.17em}}\theta .$
2. Evaluate $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta .$
3. Multiply $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ to find the x- coordinate of the rectangular form.
4. Multiply $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ to find the y- coordinate of the rectangular form.

## Writing polar coordinates as rectangular coordinates

Write the polar coordinates $\text{\hspace{0.17em}}\left(3,\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ as rectangular coordinates.

Use the equivalent relationships.

$\begin{array}{l}\begin{array}{l}\\ x=r\mathrm{cos}\text{\hspace{0.17em}}\theta \end{array}\hfill \\ x=3\mathrm{cos}\text{\hspace{0.17em}}\frac{\pi }{2}=0\hfill \\ y=r\mathrm{sin}\text{\hspace{0.17em}}\theta \hfill \\ y=3\mathrm{sin}\text{\hspace{0.17em}}\frac{\pi }{2}=3\hfill \end{array}$

The rectangular coordinates are $\text{\hspace{0.17em}}\left(0,3\right).\text{\hspace{0.17em}}$ See [link] .

## Writing polar coordinates as rectangular coordinates

Write the polar coordinates $\text{\hspace{0.17em}}\left(-2,0\right)\text{\hspace{0.17em}}$ as rectangular coordinates.

See [link] . Writing the polar coordinates as rectangular, we have

$\begin{array}{l}x=r\mathrm{cos}\text{\hspace{0.17em}}\theta \hfill \\ x=-2\mathrm{cos}\left(0\right)=-2\hfill \\ \hfill \\ y=r\mathrm{sin}\text{\hspace{0.17em}}\theta \hfill \\ y=-2\mathrm{sin}\left(0\right)=0\hfill \end{array}$

The rectangular coordinates are also $\text{\hspace{0.17em}}\left(-2,0\right).$

Write the polar coordinates $\text{\hspace{0.17em}}\left(-1,\frac{2\pi }{3}\right)\text{\hspace{0.17em}}$ as rectangular coordinates.

$\left(x,y\right)=\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$

## Converting from rectangular coordinates to polar coordinates

To convert rectangular coordinates to polar coordinates    , we will use two other familiar relationships. With this conversion, however, we need to be aware that a set of rectangular coordinates will yield more than one polar point.

## Converting from rectangular coordinates to polar coordinates

Converting from rectangular coordinates to polar coordinates requires the use of one or more of the relationships illustrated in [link] .

## Writing rectangular coordinates as polar coordinates

Convert the rectangular coordinates $\text{\hspace{0.17em}}\left(3,3\right)\text{\hspace{0.17em}}$ to polar coordinates.

We see that the original point $\text{\hspace{0.17em}}\left(3,3\right)\text{\hspace{0.17em}}$ is in the first quadrant. To find $\text{\hspace{0.17em}}\theta ,\text{\hspace{0.17em}}$ use the formula $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{y}{x}.\text{\hspace{0.17em}}$ This gives

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{3}{3}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =1\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\mathrm{tan}}^{-1}\left(1\right)=\frac{\pi }{4}\hfill \end{array}$

To find $\text{\hspace{0.17em}}r,\text{\hspace{0.17em}}$ we substitute the values for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ into the formula $\text{\hspace{0.17em}}r=\sqrt{{x}^{2}+{y}^{2}}.\text{\hspace{0.17em}}$ We know that $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ must be positive, as $\text{\hspace{0.17em}}\frac{\pi }{4}\text{\hspace{0.17em}}$ is in the first quadrant. Thus

$\begin{array}{l}\begin{array}{l}\\ r=\sqrt{{3}^{2}+{3}^{2}}\end{array}\hfill \\ r=\sqrt{9+9}\hfill \\ r=\sqrt{18}=3\sqrt{2}\hfill \end{array}$

So, $\text{\hspace{0.17em}}r=3\sqrt{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\theta \text{=}\frac{\pi }{4},\text{\hspace{0.17em}}$ giving us the polar point $\text{\hspace{0.17em}}\left(3\sqrt{2},\frac{\pi }{4}\right).\text{\hspace{0.17em}}$ See [link] .

## Transforming equations between polar and rectangular forms

We can now convert coordinates between polar and rectangular form. Converting equations can be more difficult, but it can be beneficial to be able to convert between the two forms. Since there are a number of polar equations that cannot be expressed clearly in Cartesian form, and vice versa, we can use the same procedures we used to convert points between the coordinate systems. We can then use a graphing calculator to graph either the rectangular form or the polar form of the equation.

Given an equation in polar form, graph it using a graphing calculator.

1. Change the MODE to POL , representing polar form.
2. Press the Y= button to bring up a screen allowing the input of six equations: $\text{\hspace{0.17em}}{r}_{1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{r}_{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\text{\hspace{0.17em}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{r}_{6}.$
3. Enter the polar equation, set equal to $\text{\hspace{0.17em}}r.$
4. Press GRAPH .

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