**10.6 Polar Coordinates**:

So far, you have been representing graphs of equations as collections of points (x, y) on the rectangular coordinate system, where x and y represent the directed distances from teh coordinate axes to the point (

*x*,

*y*). In this section, you will study a second system - the Polar Coordinate System.

**I**. To form the polar coordinate system in the plane, fix a point O, called the pole (or origin) and construct from O an initial ray called the polar axis. Then each point P in the plane can be assigned polar coodinates (

*r*,q ) as follows:

1.

*r*= directed distance from O to P.

2. q = directed angle, counterclockwise from polar axis to segment OP.

To find other points in the same place, do the following:

(

*r*, q ) = (

*r*, q ± 2

*n*p )

or

(

*r*, q) = (-

*r*, q ±(2

*n*+ 1)p )

**Example:**Given the polar coordinate (3, p/4)

= (3, p/4 + 2p) = (3, 9p/4)

OR

(-3, p/4 + (2 (0) +1)p) = (-3, 5p/4)

(-3, p/4 - p) = (-3, -3p/4)

**II. Coordinate Conversion**:

Because (

*x*,

*y*) lies on a circle of radius

*r*, it follows that

*r*

^{2}=

*x*

^{2}+

*y*

^{2}. Moreover, for

*r*> 0, the definitions of the trigonometric functions imply:

tan q =

*y*/

*x*

cos q =

*x*/

*r*

sin q =

*y*/

*r*

The Polar Coordinates (

*r*, q) are related to the rectangular coordinates (

*x*,

*y*) as follows:

*x*=

*r*cos q

*y*=

*r*sin q

tan q =

*y*/

*x*

*r*

^{2}=

*x*

^{2}+

*y*

^{2}

**III. Polar to Rectangular Conversion**

**Example 2: Convert each point to rectangular coordinates**

a). (-2, 7p/6) means

*r*= -2 and q = 7p/6

x = (-2)(cos 7p/6) = (-2)(- Ö(3)/2) = Ö3

y = (-2)(sin 7p/6) = (-2)(-1/2) = 1

Therefore the rectangular coordinate is (Ö3 , 1)

b). (8.25, 3.5) means

*r*= 8.25 and q = 3.5 radians

x = (8.25)(cos 3.5) = -7.72576767

y = (8.25)(sin 3.5) = -2.89391628

Therefore the rectangular coordinate is (-7.72576767, -2.89391628)

**IV. Rectangle to Polar Conversion**

**Example 3: Convert each point to polar coordinates**

a) Let 0£ q £ 2p and using the rectangular coordinate (0, -5)

tan q = -5/0 = undefined, therefore q = 270° based on where the point is.

r = Ö (x

^{2}+ y

^{2})

= Ö(0

^{2}+ (-5)

^{2})

= 5

Therefore a polar coordinate is (5, 270°) or (5, 3p/2)

and the negative would be (-5, -90°) or (-5, p/2)

b) (3, -1)

Tan q = -1/3

q = -18.43494882° or -.3217505544 radians

so q = 341.5650512° or 5.961434753 radians

r = Ö(3

^{2}+ (-1)

^{2}) = Ö(9 + 1) = Ö10

Therefore the polar coordinate is (Ö10, 5.961434753 radians)

To find another coordinate for the same location: 5.961434753-p = 2.819842099

so (-Ö10 , 2.819842099 radians)

**V. Equation Convertion**

Recall x = r cos q and y = r sin q and x

^{2}+ y

^{2}= r

^{2}

Given: x

^{2}+ y

^{2}- 6x = 0 substitute in the above equations for x and y

r

^{2}- 6(r cos q) = 0

r ( r - 6 cos q) = 0

r = 0 and r - 6 cos q = 0

r = 6 cos q

so the polar equation is r = 6 cos q

**VI. Convert the polar equation to rectangular form**:

**Example**: r = 4 cos q

r

^{2}= 4r cos q

x

^{2}+ y

^{2}= 4x

x

^{2}-4x + y

^{2}= 0 would be the rectangular equation

**Example:**q = 5p/3

Tan q = tan 5p/3 = -Ö3

y/x = -Ö3/1

y = -Ö3 x

y = xÖ3 = 0

**Example:**r = 6/(2cosq - 3 sin q)

r = 6 / (2(x/r) - 3(y/r))

r = 6r/(2x - 3y)

2x - 3y = 6r/

*r*

2

*x*- 3

*y*= 6