Tracing of Curves Given in Polar Form
We use following information similar to discussed in the previous lecture for tracing the
curves whose equation is given in polar coordinates r and .
1. Symmetry. Let r = f ( ) be the polar equation of curve.
(a) If on putting either (i) - for or (ii) ( ) for and – r for r, the polar equation
of the curve remains unchanged, the curve is symmetrical about initial line i.e. 0 .
(b) If on putting either (i) for or (ii) – r for r and - for , the equation of
π
curve remains unchanged, the curve is symmetrical about the line θ .
2
(c) If f f , then the curve is symmetrical about the line or y x .
2 4
For example, r = a sin 2 is symmetrical about the line = .
4
(d) If on putting either (i) – r for r or (ii) for , the polar equation of the curve
remains unchanged, the curve is symmetrical about the pole.
2. Pass through origin or pole. The curve passes through origin (pole) if r = 0 for some
real value of , say . In that case is the tangent at pole.
For example, consider the curve r = a (1 – cos ). If we put r = 0, we get = 0. This shows
that the curve passes through the pole and the initial line, i.e. = 0 is the tangent at the pole.
3. Intersection. Find the intersection with the initial line 0 and the line θ π by
2
d
putting these values in the equation of curve. Find tan r . Then φ gives the angle
dr
between the radius vector and the tangent at any point on the curve.
4. Asymptotes. If for some value of , r becomes infinite, we should examine the possibility
of existence of rectilinear asymptote. The equation of such asymptote is obtained by the
following method.
1
If the equation of a curve is f ( ) and be a root of f ( ) = 0, then
r
r as and the equation of an asymptote of the curve is
1 .
r sin θ α (1)
f α
EXAMPLE 1 Find the asymptote of the curve r = a ( sec cos ).
Solution The given equation of curve can be written as
1 1 1 cos
f , say
r a sec cos a cos 2 1
f θ
1 1 cos 2 θ sin θ cos θ 2 cos θ sin θ
a 1 cos θ
2 2
1 sin θ 3 cos 2 θ sin θ
1
sin θ
1 3 cos 2 θ .
a
1 cos 2 θ
2
a
1 cos 2 θ
2
Equating f to zero, we get cos 0 or 2n 1 say , where n is zero or any
2
integer. Then the corresponding asymptote is
1
r sin or r sin 2 n 1 a / sin 2 n 1
f ' 2 2
π
or r sin n π θ a / 1
n
2
or r cos n a 1 as(-1)2n = 1
n
or r 1 cos θ a 1
n n
or r a sec
Therefore the required asymptote is r = a sec .
EXAMPLE 2 Find the asymptote of the curve r = a / (1- cos ).
Solution The given equation of curve can be written as
1 1
1 cos f say
r a
1
f ' sin .
a
Equating f to zero, we get 1- cos = 0 or cos =1 or =2n say , where n is
zero or any integer.
Now f α 1 sin 2 n π 0 or 1
, which is infinite and hence the given curve has
a f α
no asymptote.
EXAMPLE 3 Find the asymptote of the curve r sin = 2 cos2 .
Solution The given equation of curve can be written as
1 sin
f say
r 2 cos 2
1 cos 2 . cos ( 2 sin 2 ). sin 1 cos 2 . cos 2 sin 2 . sin
f ' 2
2 cos 2 2 cos 2 2
Equating f to zero, we get sin = 0 or = n say , where n is any integer.
n n
Now f 1 cos 2 n . cos n 2 sin 2 n . sin n 1 1 .( 1) 0 ( 1) .
2 cos 2 2 n
2
1 2
Hence the equation of the asymptote is
1
r sin
f ( )
2
or r sin n
( 1) n
2
or r ( 1) n sin
( 1) n
or r ( 1) 2 n sin 2 r sin 2
Circular asymptotes to polar curves: If the equation of curve is r = f ( ) and if
lim f ( ) = a, constant
θ
Then r = a is called a circular asymptote or the asymptotic circle of the curve.
EXAMPLE 4 Find the circular asymptote of r ( 1) a
Solution We have r a f
1
Now, lim a lim a
a
θ 1 θ 1
1
The circular asymptote is r = a.
5. Region. Solve the equation for r or . Find the region in which the curve does not exist.
6. Points. Find the values of r for some standard angles and thus prepare a table giving the
value of r corresponding to some convenient angles and plot those points.
NOTE 1 Sometimes it is convenient to change the problem into Cartesian coordinates by
taking x = r cos and y = r sin .
NOTE 2 In polar coordinates, r is taken positive if measured away from O along the line
bounding the vectorial angle and is taken negative if measured in the direction opposite to
above. Also is taken positive when measured in the anti-clockwise direction and negative
when measured in the clockwise direction. The sign of r and will be clear from the
following figure 2.31
Figure-2.31
EXAMPLE 1 Trace the curve r = a (1- cos ) ( Cardiod ).
Solution We have following information about the curve:
1. Putting = - , the equation of curve remains unchanged, therefore the curve is
symmetrical about initial line.
2. Putting r = 0 in the equation of curve, we get
cos =1 or = 0.
Hence the curve passes through pole and the equation of tangent at the pole is = 0.
3. We have r = 0 for = 0; r = a for = / 2 and r = 2a for .
Figure-2.32
dr
Also a sin θ
dθ
dθ 1 θ
tan r a 1 cos θ tan or .
dr a sin θ 2 2
At , , and r = 2a.
2
At (2a, π) the tangent is perpendicular to radius vector.
4. Clearly the curve has no asymptote.
5. As increases from 0 to , r increases for 0 to 2a.
6. Form the following the table of values of r and .
0 2
3 2 3
r = a (1- cos ) 0 a/2 a (3/2)a 2a
With the above data the shape of the curve is shown in the Figure 2.32.
EXAMPLE 2 Trace the curve r 2 cos 2 a 2 (Hyperbola).
Solution We have following information about the curve:
1. The curve is symmetrical about the pole, about the initial line = 0 and about the line
π
θ .
2
Changing to Cartesian coordinates by x = r cos , y = r sin , the given equation curve
becomes x 2 y 2 a 2 .
2. The curve does not pass through pole since for r = 0, we are not getting the finite value
of (in terms of Cartesian coordinates, (0,0) does not satisfy the equation of curve).
3. When =0, r = ± a i.e. points (a, 0) and (-a, 0) lie on the curve.
Differentiating the given equation of curve w. r. t. r, we get
dθ 2a 2
2 sin 2 θ 3
dr r
d a2
or r 2
dr r sin 2
a2
or tan cos 2 θ cot 2θ
a 2 sin 2 θ
or 2.
2
At 0, and r a .
2
At points (a, 0) and (-a, 0), the tangent is perpendicular to radius vector.
4. Equation of the asymptotes are y = ± x or = ± in polar coordinates.
4
5. Solving for r, we get r 2 a2 / cos2 . This shows that as increases from 0 to ,r
4
increases from a to .
3 2
For the values of lying between and , r is negative i.e. r is imaginary. So the curve
4 4
3
does not exist for . With these data the shape of the curve is shown in the
4 4
adjoining figure 2.33.
EXAMPLE 3 Trace the curve r = a + b cos , when a < b (Limacon).
Solution We have the following information about the curve:
1. The curve is symmetrical about the initial line.
2. The curve passes through the pole (origin) when = cos 1 a / b . The line
is tangent to the curve at pole.
3. The curve has no asymptote.
4. When =0, r = a + b; when = , r a and when = ; r a b which is negative
2
as a < b.
5.Consider the following table of values of r and .
0 π 2 /3 3 /4
4 3 2
r a +b b b a b b a-b
a+ a+ a- a-
2 2 2 2
b
When a > then r is positive for all
2
2
values of lying between 0 to but r
3
is negative when 3 or i.e. r
4
is positive in the above data upto 2 and
3
negative from 3 onward. Hence r must Figure 2.34
4
vanish some where between 2 and 3 . Let (lying between 2 and 3 ) for
3 4 3 4
which r = 0.
For values of lying between and , r is negative and points corresponding to such values
of will be marked in the opposite direction on these lines as r is negative for them. With
the above information the shape of curve is shown in the adjoining Figure 2.34.
EXAMPLE 4 Trace the curve r2 = a2cos 2 (Leminscate of Bernoulli).
Solution We have following information about the curve:
1. The curve is symmetrical about the initial line 0 and the . Also the curve is
2
symmetrical about pole.
2. Putting r = 0 in the equation of curve, we have cos 2 =0
or 2 = ± or .
2 4
Hence the straight line = ± are the tangents to the curve at the pole.
4
3.At = 0, r = ± a i.e. the curve cuts the initial line at (a, 0) and (- a, 0).
At = , r 2 a 2 i.e. r is imaginary and hence the curve does not cuts the y-axis.
2
dr dr
Also 2r 2 a 2 sin 2θ or r a 2 sin 2 θ
dθ dθ
dθ r2
tan r 2 cot 2 θ
dr a sin 2 θ
tan tan 2 or 2
2 2
At 0 i.e. at a, 0 , the tangents are perpendicular to the initial line.
4. Clearly, the curve has no asymptote.
π 3π
5. As increases from 0 to , r 2 is positive and decreases from a to 0. When θ ,
4 4 4
r 2 is negative i.e. r is imaginary, therefore the curve does not exist between
π 3π
= and .With the help of these points, the curve is shown in the figure 2.35.
4 4
Figure-2.35
EXAMPLE 5 Trace the curve r = a ( sec cos ).
Solution Let us change the equation to Cartesian coordinates as
r2
r2 a r cos r sec a r cos
r cos
x2 y2
or x 2 y 2 a x
x
or
x x2 y 2 a 2x2 y 2
or y2 x2
2a x
xa
Now for tracing this curve, see Example.3 of same Section 2.4.13 with asymptote as x = a in
place of x = 0. The shape of the curve is shown in the figure 2.36.
Figure-2.36
Tracing of curves when the equation of curve is r = a sin n or r = a cos n , where n is
positive integer and a > 0.
In general, the curve r = a sin n or r = a cos n consists of n loops if n is odd and 2n loops
if n is even. It should be noted that all loops lie within a circle of radius a. In tracing the
curves of the above types we divide each quadrant into n equal parts and give these values of
and find the corresponding values of r.
EXAMPLE 6 Trace the curve r = a cos2 .
Solution Let O be the origin, Ox and Oy being the x-axis and y-axis respectively (see,
Figure 2.37). Here we shall divide the each quadrant into two equal parts i.e. the first
quadrant is divided into two part 00to 450 and 450 to 900. Since n is 2 i.e. even, therefore the
number of loops in the curve is 4.
1. The curve is symmetrical about the initial line = 0 and the line = .
2
2. The curve passes through pole where cos 2 = 0 or 2 = ± or = ± .
2 4
The straight lines = ± are tangents at the pole.
4
3.At θ 0 ,r a and at θ π/ 2 , r a i.e. r will be marked in opposite direction for
/ 2 .
Also dr
2 a sin 2 θ
dθ
dθ a cos 2 θ 1
tan r cot 2 θ
dr 2 a sin 2 θ 2
At 0 , / 2 ; / 2 , / 2 ,
i.e at (a, 0) and (0, -a), the tangents are perpendicular to radius vector.
Figure-2.37
4. Clearly, the curve has no asymptote.
5.The following is the of values of θ and r :
0 /6 /4 /2 2 /3 3 π/4 5 π/4
r = a cos2 a a/2 0 -a - a/2 0 a/2 a
Plot these points and due to symmetry about the initial line the other portion can be traced.
EXAMPLE 7 Trace the curve r a sin 4 θ.
Solution Here we shall divide the each quadrant into 4 equal parts i.e first quadrant is divided
into 4 parts 0 to / 8 , / 8 to / 4 , / 4 to 3 / 8 and 3 / 8 to / 2 . Since n is 4 i.e even
therefore the number of loops in the curve is 8.
1. The given curve is symmetrical about the pole, initial line and the line / 2.
Figure-2.38
2. Putting r 0 , sin 4 0 or 4 θ 0 , π , 2 π , 3 π , ... ,8 π
or 0 , / 4 , , 3 , , 5 / 4 , 3 / 2 , 7 / 4, 2 which are tangent lines at the pole.
2 4
In fact 0 , / 4 , , 3 are the four tangents at the pole to the curve and others are just
2 4
extension of these due to symmetry.
3. Clearly the curve has no asymptote.
dr
4. Equating to zero, we have cos 4 0
d
or 4 θ π/ 2 , 3π/ 2 , 5π/ 2 ,...or θ π/ 8, 3π/ 8, 5π/ 8,...
These values of give the maximum value of r which is a.
5.The corresponding values of and r are given below:
0 /8 /4 3 / 8 /2 5 / 8 6/8 7 / 8
r 0 a 0 -a 0 a 0 -a 0
With the above data, the shape of the curve is shown in the figure 2.38
