Polar coordinates and curves described using polar coordinates

Polar coordinates.
Polar coordinates provide an alternative system of coordinates for describing the location of points in a plane. In this system there is an
origin or pole , O, and a polar axis, which is a half-line or ray extending horizontally to the right of the pole.
The position of a point P in the polar coordinate plane is determined by its polar coordinates ( ) , r , where:
r is the "distance" of the point P from the pole (origin), and
is the angle between the ray OP and the polar axis, following the convention that angles measured in the counterclockwise direction are

considered to be positive and angles measured in the clockwise direction are considered to be negative.


Unlike rectangular coordinates, the polar coordinates are not unique. The point ( ) , r + 2 k , where k is an integer, is the same as the
point ( ) , r . For example,

|
\

|
| , 2

6
=
|
\

|
| , 2
13
6

and

|
\

|
|
, 1
5
4
=
|
\

|
|
, 1
3
4
.
It is legitimate to allow negative values of r whereby the point ( ) , r is at a distance of r from the pole on the opposite side of the
pole as ( ) , r along the same infinite line through the pole and as ( ) , r . Thus
( ) , r = ( ) , r + .

Because of the familiarity of the standard rectangular cartesian coordinate system, the pole of the polar coordinate system is often identified
with the origin in the standard x-y coordinate plane and then the polar axis can be identified with the section of the x axis associated with
positive x values.

When converting the coordinates of a point from polar form to rectangular form, the following equations may be used.
{
= x r cos
= y r sin

When converting the coordinates of a point from rectangular form to polar form, the following equations may be used.

= + x
2
y
2
r
2

=
|
\

|
|
arctan
y
x
x 0

If = x 0, then =

2
or =

2
as appropriate.
For example, the point with rectangular coordinates ( ) , 1 3 has (possible) polar coordinates
|
\

|
|
, 2

6
, while the point with rectangular
coordinates ( ) , 0 2 has (possible) polar coordinates
|
\

|
|
, 2

2
or
|
\

|
|
, 2

2
.
Polar curves.
Typically, an equation relating the variables r and describes a curve in polar coordinates or simply a polar curve.
Example 1:
Sketch the polar curve given by the equation = r 4 cos .
Solution:
We can find the corresponding cartesian equation as follows.
Multiplying both sides of the equation
= r 4 cos
by r gives the equation
= r
2
4 r cos ,
so that, if a point on the curve has rectangular coordinates ( ) , x y we can use the equations
{
= + x
2
y
2
r
2

= x r cos

to obtain the cartesian equation:
= + x
2
y
2
4 x.
This equation is equivalent to
= + x
2
4 x y
2
0,
which, in turn, by completing the square for the x terms, is equivalent to
= + + x
2
4 x 4 y
2
4,
that is,
= + ( ) x 2
2
y
2
4.
This is the cartesian equation of a circle with centre at the point ( ) , 2 0 and radius 2.


Notes:
Alternatively, one can see directly that a point P with polar coordinates ( ) , r is on the circle with its centre at the point ( ) , 0 2 and radius
2 exactly when
= r 4 cos
from the fact that the triangle formed from the diameter of the circle that lies along the x axis, the line segment from the origin to the point
P and the line segment from P to the point ( ) , 4 0 always has a right angle at P.

One can always resort to the construction of a table of values.

(
(
(
(
|

6
0

2
2
3
3
4
5
6

= r 4 cos | 0 2 2 2 2 3 4 2 3 2 2 2 0 2 2 2 2 3 4


The point A has polar coordinates
|
\

|
|
,

6
2 3 and rectangular coordinates ( ) , 3 3 .
The point B has polar coordinates
|
\

|
|
,

4
2 2 and rectangular coordinates ( ) , 2 2 .
The point C has polar coordinates
|
\

|
|
,

3
2 and rectangular coordinates ( ) , 1 3 .
The points D has polar coordinates
|
\

|
|
,
2
3
2 or
|
\

|
|
,

3
2 and rectangular coordinates ( ) , 1 3 .
The points E has polar coordinates
|
\

|
|
,
3
4
2 2 or
|
\

|
|
,

4
2 2 and rectangular coordinates ( ) , 2 2 .
The points F has polar coordinates
|
\

|
|
,
3
4
2 2 or
|
\

|
|
,

6
2 3 and rectangular coordinates ( ) , 3 3 .
A useful aid to to sketching the polar graph of an equation of the form = r ( ) f is to first plot of the graph of = r ( ) f in cartesian
coordinates with the axis horizontal and the r axis vertical.
Plotting = r 4 cos in this way indicates that as increases from

2
to +

2
, the distance r of the point ( ) , r from the origin first
increases from 0 to a maximum value of 4 as increases from

2
to 0, and the decreases to 0 as theta increases from 0 to

2
.

One can imagine a ray sweeping around the origin with a point that traces out the polar graph first moving away from the origin and then
back towards the origin.

Example 2:
Sketch the polar curve given by the equation = r cos 2 .
Solution:

The cartesian graph of indicates that r reaches the maximum value of 1 when = 0, and 2 for 0 2 and r has the
minimum value 1 when =

2
and
3
2
for 0 2
Also = r 0 when when =

4
,
3
4
,
5
4
,
3
2
and
7
4
for 0 2 .

The following pictures indicate how the curve is traced out as increases.

Example 3:
Sketch the polar curve given by the equation = r 1 cos .
Solution:
The following graph and table of values can be used to help sketch the graph.

(
(
(
(
(
(
(
| 0

2
3
4

5
4
3
2
7
4
2
= r 1 cos | 0
2 2
2
1
+ 2 2
2
2
+ 2 2
2
1
2 2
2
0


2 2
2
~ 0.29289 and
+ 2 2
2
~ 1.7071.

The following pictures indicate how the curve is traced out as increases.

Note: This curve is called a cardioid.
Example 4:
Sketch the polar curve given by the equation = r 1 sin 2 .
Solution:
The following graph and table of values can be used to help sketch the graph.

(
(
(
(
| 0

2
3
4

5
4
3
2
7
4
2
= r 1 sin 2 | 1 0 1 2 1 0 1 2 1


The following pictures indicate how the curve is traced out as increases.

Tangents to polar curves.
Given a polar curve = r ( ) f , we can make the substitution = r ( ) f in the equations
{
= x r cos
= y r sin

to obtain the parametric equations:
{
= x ( ) f cos
= y ( ) f sin

for the curve.
Then

d x
d
= ( ) f ' cos ( ) f sin =
d r
d
cos r sin
and

d y
d
= + ( ) f ' sin ( ) f cos =
d r
d
+ sin r cos ,
so that

dy
dx
=
|
\

|
|
d y
d
|
\

|
|
d x
d
=
+
|
\

|
|
d r
d
sin r cos

|
\

|
|
d r
d
cos r sin
------- (i).
If the curve passes though the pole then we can take = r 0 in (i) to see that the slope of the tangent at such a point is given by
=
dy
dx
tan .
provided that
d r
d
0.
Example 5:
Find the polar coordinates of the points on the cardioid given by the equation = r 1 cos where the tangent line to the curve is (a)
horizontal (b) vertical.
Solution:
= r 1 cos
implies that
=
d r
d
sin .
Hence

d x
d
=
d r
d
cos r sin = sin cos ( ) 1 cos sin
= sin ( ) cos ( ) 1 cos
= sin ( ) 2 cos 1 .
Also

d y
d
=
d r
d
+ sin r cos = + sin
2
( ) 1 cos cos
= + 1 cos
2
cos cos
2

= + 1 cos 2 cos
2

= ( ) + 2 cos 1 ( ) cos 1 .
so that

dy
dx
=
|
\

|
|
d y
d
|
\

|
|
d x
d
=
( ) + 2 cos 1 ( ) cos 1
sin ( ) 2 cos 1
.
Tangent lines to the curve are horizontal where =
d y
d
0 and
d x
d
0, and tangent lines are vertical where =
d x
d
0 and
d y
d
0.
=
d y
d
0
<=> = ( ) + 2 cos 1 ( ) cos 1 0
<=> = cos
1
2
or = cos 1.
Now = cos
1
2
if and only if =
2
3
or =
4
3
for 0 < 2 and = cos 1 if and only if = 0 for 0 < 2 .
Hence =
dy
d
0 when = 0,
2
3
or
4
3
for 0 < 2 .
=
d x
d
0
<=> = sin ( ) 2 cos 1 0
<=> = sin 0 or = cos
1
2
.
Now = sin 0 if and only if = 0 or = for 0 < 2 and = cos
1
2
if and only if =

3
or =
5
3
for 0 < 2 .
Hence =
dx
d
0 when = 0,

3
, , or
5
3
for 0 < 2 .
Thus we see that the tangent line to the curve are horizontal at the points
|
\

|
|
,
3
2
2
3
and
|
\

|
|
,
3
2
4
3
and the tangent line to the curve are
vertical at the points
|
\

|
| ,
1
2

3
, ( ) , 2 and
|
\

|
| ,
1
2
5
3
.
Notes:
The following picture shows the location of the horizontal and vertical tangent lines.

The slope of tangent line to the curve tends to 0 as tends to 0, since
lim
0

( ) + 2 cos 1 ( ) cos 1
sin ( ) 2 cos 1
= lim
0
+ 2 cos 1
2 cos 1
. lim
0
1 cos
sin

= 3 lim
0
+

2
! 2

4
! 4
. . .
+

3
! 3
. . .
= 0.
This might suggest that there is a tangent line to the curve at the pole with zero slope. However, the picture indicates that this is not
possible and this can be confirmed from the fact that the section of the curve in the first quadrant given by 0

3
is the graph of the
function f given by = ( ) f x
2 4 x 4 x
2
2 1 4 x
2
and the section of the curve for
5
3
2 , or equivalently

3
0,
is the graph of = y ( ) f x .
The formula for the function f can be confirmed by first obtaining a cartesian equation for the curve.
If r 0, the equation = r 1 cos is equivalent to
= r
2
r r cos .
Substituting = r cos x in this equation gives
= r
2
r x
which is equivalent to
= + r
2
x r.
This equation implies that
= ( ) + r
2
x
2
r
2
,
which then gives the following cartesian equation for the cardioid.
= ( ) + + x
2
y
2
x
2
+ x
2
y
2
.
This equation is equivalent to
= + + + + + x
4
y
4
x
2
2 x
2
y
2
2 x
3
2 x y
2
+ x
2
y
2

and to
= + + + + x
4
y
4
2 x
2
y
2
2 x
3
2 x y
2
y
2
0 ------- (ii).
This equation can be arranged as a quadratic equation for y
2
, that is, in the form
= + + + y
4
( ) + 2 x
2
2 x 1 y
2
x
4
2 x
3
0.
The quadratic formula gives
y
2
=
1 2 x 2 x
2
+/ ( ) + 2 x
2
2 x 1
2
4 x
4
8 x
2
2
------- (iii).
Since = ( ) + 2 x
2
2 x 1
2
+ + + 4 x
4
4 x
2
1 8 x
3
4 x
2
4 x = + + 4 x
4
1 8 x
3
4 x, equation (iii) gives
= y
2
1 2 x 2 x
2
+/ 1 4 x
2
=
2 4 x 4 x
2
+/ 2 1 4 x
4

which, in turn, gives
y = +
2 4 x 4 x
2
+/ 2 1 4 x
2
.
The cardioid can be broken up into four sections given by the four equations:
= y
+ 2 4 x 4 x
2
2 1 4 x
2
, = y
2 4 x 4 x
2
2 1 4 x
2
,
= y
+ 2 4 x 4 x
2
2 1 4 x
2
, = y
2 4 x 4 x
2
2 1 4 x
2
.
The first of these four equations decribes the function f mentioned previously.
The following picture shows the graphs of these four equations in red, green, maroon and blue respectively.

The function f given by
= ( ) f x
2 4 x 4 x
2
2 1 4 x
2

does not have a derivative at 0, although for x > 0 we have
= ( ) f ' x

4
1 4 x
8 x 4
4 2 4 x 4 x
2
2 1 4 x
.
It is possible to check that ( ) f ' x tends to 0 as x 0
+
.
As a further investigation of the nature of the curve for x close to 0 note that the binomial expansion
= 1 4 x 1 2 x 2 x
2
4 x
3
10 x
4
28 x
5
. . .
gives
2 4 x 4 x
2
2 1 4 x = + + + 8 x
3
20 x
4
56 x
5
. . . ,
so that when x is close to 0 we have
( ) f x ~ 2 x
|
\

|
|
3
2
.
.
The following picture shows the graphs of = y ( ) f x and = y ( ) f x in red and the graphs of = y 2 x
|
\

|
|
3
2
and = y 2 x
|
\

|
|
3
2
in blue for x
close to 0.

As a final remark, we can find a formula for
dy
dx
from equation (iii) by implicit differentiation.
Differentiating both sides of equation (ii) with respect to x gives:
+ 4 x
3
4 y
3
+ +
dy
dx
4 x y
2
4 x
2
y + + +
dy
dx
6 x
2
2 y
2
4 x y
dy
dx
2 y =
dy
dx
0.
This implies that

dy
dx
( ) + + 4 y
3
4 x
2
y 4 x y 2 y = ( ) + + + 4 x
3
4 x y
2
6 x
2
2 y
2

so that

dy
dx
=
+ + + 4 x
3
4 x y
2
6 x
2
2 y
2
+ + 4 y
3
4 x
2
y 4 x y 2 y
=
+ + + 2 x
3
2 x y
2
3 x
2
y
2
y ( ) + + 2 y
2
2 x
2
2 x 1
.
Note that the derivative
dy
dx
does not exist at the point ( ) , 0 0 .
It is possible to use this formula to find the coordinates of the points where =
dy
dx
0 and where =
dx
dy
0, but a lot more work is needed than
when working with polar coordinates.
For example, =
dy
dx
0 when = + + + 2 x
3
2 x y
2
3 x
2
y
2
0, that is, when = y
2

+ 2 x
3
3 x
2
+ 2 x 1
. Using this equation to substitute for y
2
in
equation (ii) gives (after some work)
=
x
2
( ) + 4 x 3
( ) + 2 x 1
2
0.
Hence =
dy
dx
0 when = x
3
4
and = y +
3 3
4
. (After some more work.)
Questions:
Q1. Find a cartesian equation for each of the following polar curves and also describe the curve.
(a) = r 3 sin (b) = r + 2 sin 2 cos (c) = r csc (d) = r tan sec

Q2. In each of the following cases find a polar equation for the curve with the given cartesian equation.
(a) = y
2
x (b) = + x y 4 (c) = + x
2
y
2
2 x (d) = x
2
y
2
1
Q3. Sketch the curve with the given polar equation.
(a) = r sin (b) = r 1 sin (c) = r sin 2 (d) = r cos 3
(e) = r 1 2 cos (f) = r
|
\

|
|
sin

2
(g) = r + 1 2 cos 2
Q4. Find the slope of the tangent line to the curve = r 2 sin at the point where =

6
.
Q5. Find polar coordinates of the points on the curve = r 3 cos where the tangent line is horizontal or vertical.
Answers:
Q1. (a) = + x
2
y
2
3 y <=> = + x
2
|
\

|
| x
3
2
2
9
4
... a circle with centre at
|
\

|
| , 0
3
2
and radius
3
2
.
(b) = + x
2
y
2
+ 2 y 2 x <=> = + ( ) x 1
2
( ) y 1
2
2 ... a circle with centre at ( ) , 1 1 and radius 2.
(c) = y 1 ... a horizontal line through ( ) , 0 1 .
(d) = y x
2
... a parabola with its axis vertical and vertex at ( ) , 0 0 .
Q2. (a) = r csc cot (b) = r
4
+ cos sin
(c) = r 2 cos (d) = r
1
cos 2

Q3. (a)


(b)

(c)

(d)

(e)


(f)

(g)


Q4. =
dy
dx
tan 2 so =
dy
dx =

6
|
\

|
|
tan

3
= 3.
Q5. The tangent line is horizontal at the points
|
\

|
|
,
3
2

4
and
|
\

|
|
,
3
2

4
, and vertical at the points ( ) , 3 0 and
|
\

|
|
, 0

2
.
Version: 16/4/2008 Peter Stone