Download Lecture 20

Polar Coordinates
There are many ways to mark points in the plane or in 3-dim
space for purposes of navigation. In the familiar rectangular
coordinate system, a point is chosen as the origin and a
perpendicular set of lines is drawn through that point, one
horizontal and one vertical. A unit of Length is chosen, and
every point is given a pair of coordinates (x, y) indicating its
distance, horizontally and vertically from the origin.
The choice of origin, axes, and length is completely arbitrary.
3
(2, 3)
2
1
-3
-2
1
-1
(0,0)
-1
-2
-3
2
3
An alternative method of assigning coordinates is shown below.
(r,θ )
r
θ
Here we draw a straight line from the origin to the point. We
then assign to the point the angle θ that the line makes with
the positive x axis, and the distance r from the origin to the
point.
Some examples are:
(2, π/4)
(1, π/4)
(1, π/2)
(2, 0)
(3, π/6)
(1, 3π/4)
(2, −π/2)
(2, −π/3)
The relationship between rectangular and polar coordinates is
shown in the diagram below.
r
θ
y
x
It is summarized by the equations:
x = rcos(θ)
y = rsin(θ)
r = x 2+ y2
θ = arctan y = tan−1 y
x
x
The angle θ is taken to be between −π and π.
Problem. The polar coordinates of several points are given
below. Find the rectangular coordinates of each of those points.
(a) (3, π/4)
(b) (1, −π/3)
(c) (4, 3π/4)
(d) (1, −π/6)
Solution.
(a) x = 3cos(π /4) = 3 ; y = 3sin(π /4) = 3 ;
2
2
(b) x = cos(−π /3) = 1; y = sin(−π /3) =− 3 ;
2
2
 3
 2,

1
 ,−
2

3 
2 
3

2 

 4 4 
4
4
(c) x = 4cos(3π /4) =− ; y = 4sin(3π /4) = ;  − ,

2
2  2 2
(d) x = cos( −π /6) = 3 ;
2
y = sin(−π /6) =− 1;
2
 3 1
 ,− 
 2 2


Problem. The rectangular coordinates of several points are
given below. Find the polar coordinates of each of those points.
(a) (5, 5)
(b) (1, −√3)
(c) (−3, 3√3) (d) (−2, −2)
Solution.
(a) r =
5 =π ;
2
2
−
1
θ
=
tan
x + y = 50 =5 2;
5 4
(5, 5)
π

5
2,


4


Problem. The rectangular coordinates of several points are
given below. Find the polar coordinates of each of those points.
(a) (5, 5)
(b) (1, −√3)
(c) (−3, 3√3) (d) (−2, −2)
Solution.
(b)
r =
x2 + y 2 = 2;
( )
θ = tan −1 − 3 =−π ;
3
2
(1, −√3)
 π
 2, − 3 


Problem. The rectangular coordinates of several points are
given below. Find the polar coordinates of each of those points.
(a) (5, 5)
(b) (1, −√3)
(c) (−3, 3√3) (d) (−2, −2)
Solution.
(c)
r =
x2 + y 2 = 6;
( )
θ = tan −1 − 3 = 2π ;
3
(− 3, 3√3)
6
 2π 
 6, 3 


Problem. The rectangular coordinates of several points are
given below. Find the polar coordinates of each of those points.
(a) (5, 5)
(b) (1, −√3)
(c) (−3, 3√3) (d) (−2, −2)
Solution.
(d) r =
x2 + y 2 = 8 =2 2; θ = tan−1 −1 =− 3π ;
4
( )
(−2, −2)
3π 

2
2,
−


4


We consider the problem of graphing functions of the form
r = f(θ) in polar coordinates, where the angle is measured in
radians. First we look at two simple cases
Plot the function r = c in polar coordinates, where c is a
positive constant.
This is the case where c = 2.
Plot the function θ = c in polar coordinates, where c is a
constant.
positive r
θ = c radians
negative r
In trigonometric calculations, we usually require r to be positive.
However, when graphing curves r = f(θ) in polar coordinates, we
allow r to be negative (since f(θ) often is) and interpret this to
mean measuring “backward” from the origin along the ray.
We now consider the problem of graphing more complicated
functions.
Plot the function r = sin(θ) in polar coordinates.
There are two ways to see how this plot looks. One is to change
from polar to rectangular coordinates. The equation r = sin(θ)
can also be written as r2 = rsin(θ) , or x2 + y2 = y. By
completing the square, we can write this as
2
2
1
1
1
1




or x2 +  y −  =  
x2 + y 2 − y + =
4 4
 2  2
A second way is to note that as θ increases, the line connecting
the origin to the corresponding point on the curve sweeps
around counterclockwise like the hand of a clock. At each
value of θ the curve is plotted on that line a distance f(θ) from
the origin. Thus as the line revolves around the origin, the
point on the curve slides up and down the line. Proceed as
follows:
Locate the value of r forθ = 0. Then as θ increases to 90
degrees, the point slides in or out until it reaches the correct
location of r corresponding to θ = π/2. This gives an idea of
how the curve looks in the first quadrant.
Continue this process in each quadrant.
For the function r = sin(θ) , the initial position of the point is
r = 0 when θ = 0. At θ = π/2, the point is at a distance
sin(π/2) = 1 along the “clock hand”. Thus as the hand sweeps
out that 90 degree angle, the point moves up the hand from 0 to
1. The result is clearly the picture shown below, at least
approximately.
As the clock hand moves another 90 degrees to π, the point
must move back along the clock hand to sin(π) = 0. Thus the
picture continues as shown below.
At this point we have the entire circle. In the next two
quadrants, r is negative. Thus the point moves to the negative
part of the hand. The circle is painted out again in this way,
and the process then repeats forever.
Now let us look at the similar curve r = 2cos(θ). When θ = 0,
r = 2. At θ = π/2, r = 2cos(π/2) = 0. Therefore, as the angle of
the hand moves from 0 to π/2, the point on the curve moves
down the hand from 2 to 0. This is shown below.
As the angle of the hand moves from π/2 to π, the cosine
becomes negative, so the point is plotted back through the
origin, a negative distance. Thus r continues down the hand
from 0 to −2.
This figure appears to be a circle as well. To verify this we can
translate into rectangular coordinates. If r = 2cos(θ), then
r2 = 2rcos(θ), or x2 + y2 = 2x. As before, we can complete the
square to obtain the equation
x2 − 2 x +1+ y2 =1
or
(x −1)2 + y2 =1
This is clearly the equation of a circle with radius 1 and
center (1, 0).
Graph the curve r = 1 + cos(θ) in polar coordinates.
In this case, the change to rectangular coordinates would not
yield a familiar curve. Thus we proceed in the straightforward
way. At θ = 0, r is 2 and at θ = π/2, r is 1. This is illustrated
below.
Finally, as the angle of the hand moves from 3π/2 to 2π , the
cosine goes from 0 to 1, and so r moves up the hand from 1 to
2.
The result is called a cardioid because of its heart shape.
Here are some other examples.
r = 1 − cos(θ)
r = 1 + sin(θ)
r = 1 − sin(θ)