The Polar Coordinate Transformation
If we choose the ray l to be the positive x-axis, then a
point P in the plane has both cartesian coordinates ( x,y) and polar coordinates (r, q).
The definition of the sine and cosine functions imply that (x,y) is given in terms of ( r,q) by
x = r cos(q) , y = r sin(q) |
| (1) |
Solving for r and q then yields the identities
r2 = x2+y2 and tan( q) = |
y
x
|
|
| (2) |
EXAMPLE 3 Convert the point ( 4, p/4) from
polar coordinates into cartesian coordinates, and then show that (2)
converts it back into polar .
Solution: To do so, we let r = 4 and let q = p/4 in (1) to obtain
x = 4cos |
æ è
|
|
p
4
|
ö ø
|
= 2Ö2, y = 4sin |
æ è
|
|
p
4
|
ö ø
|
= 2Ö2 |
|
To map back, we notice that
r2 = x2 + y2 = 8 + 8 = 16, r = 4 |
|
and that y/x = 1 implies that tan(q) = 1, q = p/4.
If we substitute x = r cos(q) and y = r sin(q) into a curve g(x,y) = k, then the result
g( rcos(q), rsin(q) ) = k |
|
is called the pullback of the curve into polar coordinates.
The identity r2 = x2+y2
is often used in pulling a curve back into polar coordinates.For
example, x2+y2 = R2 for a
constant R > 0 has a pullback of
Similarly, lines of the form y = mx become
r sin(q) = m r cos(q) Þ sin(q) = m cos(q) Þ tan(q) = m |
|
This matches example 4 in the last section, in which we saw that the coordinate
curves for the polar coordinate transformation
T( r, q ) =
á r cos(q), r sin(q)
ñ |
|
are circles centered at the origin and lines through the origin,
respectively.
Also, it shows that whenever possible, we should solve for r to
obtain a function of the form r = f(q) .
EXAMPLE 4 Convert the curve x2 +
(y - 1)2 =
1 into polar
coordinates, and then solve for r, if possible.
Solution: Expanding leads to x2 + y2 -
2y + 1 = 1 , so that To do so, we replace y r2 and let x = rcos( q) :
Solving for r then yields
That is, r = 2
sin(q) is a circle of
radius 1 centered at ( 1,0) . In general, a curve of the form
r = 2acos( q) is a circle of radius | a|
centered at ( a,0) and a curve of the form r = 2asin(q) is a circle of radius | a| centered at (0, a) .
Check Your Reading: Where is the center of the circle
r = 2 cos(q)?