## Tuesday, April 3, 2007

### Precalculus 10.6 Polar Coordinates

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 ± 2np )

or

(r, q) = (-r, q ±(2n + 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 r2 = x2 + y2. 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

r2 = x2 + y2

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 (x2 + y2)
= Ö(02 + (-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(32 + (-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

V. Equation Convertion

Recall x = r cos q and y = r sin q and x2 + y2 = r2

Given: x2 + y2 - 6x = 0 substitute in the above equations for x and y

r2 - 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

r2 = 4r cos q

x2 + y2 = 4x

x2 -4x + y2 = 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

2x - 3y = 6