How do you find the length of the cardioid r=1+sin(θ)?
Answers
This is the equation of a cardioid. The polar equation is expressed in terms of a periodic function, so we have to know its period to get the exact length (and not just a portion or some excess of it).
1
+
sin
θ
has period
T
=
2
π
(the function can be obtained as a vertical translation of the sine function in the plane of coordinates
(
θ
,
r
)
).
The length of a periodic polar curve can be computed by integrating the arc length on a complete period of the function, i.e. on an interval
I
of length
T
=
2
π
:
l
=
∫
I
d
s
where
d
s
=
√
r
2
+
(
d
r
d
θ
)
2
d
θ
So we have to compute the derivative:
d
r
d
θ
=
d
d
θ
(
1
+
sin
θ
)
=
cos
θ
and this implies
d
s
=
√
(
1
+
sin
θ
)
2
+
(
cos
θ
)
2
d
θ
=
√
1
+
2
sin
θ
+
sin
2
θ
+
cos
2
θ
d
θ
By the Pythagorean trigonometric identity
cos
2
θ
+
sin
2
θ
=
1
:
d
s
=
√
2
+
2
sin
θ
d
θ
=
√
2
(
1
+
sin
θ
)
d
θ
Integrating that square root is not easy... we'd better take advantage of trigonometric identities. We have
√
1
+
sin
θ
(multiplied by a numerical coefficient) and we would like to get rid of that square root. The half angle formula for cosine states that
cos
2
(
α
2
)
=
1
+
cos
α
2
. We can use this with sine too, because if
α
=
θ
−
π
2
then
cos
α
=
cos
(
θ
−
π
2
)
=
sin
θ
. So we substitute in the formula and get
cos
2
(
θ
−
π
2
2
)
=
1
+
cos
(
θ
−
π
2
)
2
and then
2
cos
2
(
θ
2
−
π
4
)
=
1
+
sin
θ
By substitution in the last
d
s
expression that we wrote
d
s
=
√
4
cos
2
(
θ
2
−
π
4
)
d
θ
=
2
∣
∣
∣
cos
(
θ
2
−
π
4
)
∣
∣
∣
d
θ
Now we can integrate this on an interval of length
2
π
. To do that, we would like to get rid of the absolute value. The graph of the integrand function is the following
graph{2*abs(cos(x/2-pi/4)) [-10, 10, -5, 5]}
We know that
cos
(
x
2
−
π
4
)
≥
0
⇔
−
π
2
+
2
k
π
≤
x
2
−
π
4
≤
π
2
+
2
k
π
for
k
any integer value, so we get that
−
π
2
π
+
4
k
π
≤
x
≤
3
2
π
+
4
k
π
, and in particular
−
π
2
π
≤
x
≤
3
2
π
.
In the end, we can integrate the arc length on
I
=
[
−
π
2
,
3
2
π
]
, since this is an interval of length
2
π
. We can remove the absolute value, since
cos
(
x
2
−
π
4
)
≥
0
on
I
.
∫
3
2
π
−
π
2
2
cos
(
θ
2
−
π
4
)
d
θ
=
4
∫
3
2
π
−
π
2
1
2
cos
(
θ
2
−
π
4
)
d
θ
=
4
sin
(
θ
2
−
π
4
)
∣
3
2
π
−
π
2
=
4
sin
(
π
2
)
−
4
sin
(
−
π
2
)
=
4
+
4
=
8