find the length of a loop of the curve r = a(pie^2-1)
Answers
Step-by-step explanation:
Increasing or decreasing the value of
a
will only change the radius of the curve.
To find when the curve begins and ends, set
r
=
0
, since this is where the curve is at the origin.
If
a
sin
3
θ
=
0
, then
sin
3
θ
=
0
. Since
sin
θ
=
0
at
θ
=
0
,
π
,
2
π
...
we see that for
sin
3
θ
, it will be
0
at
0
,
π
/
3
,
2
π
/
3
...
So, the curve in the first quadrant varies from
θ
=
0
to
θ
=
π
/
3
.
The expression for the area of any polar equation
r
from
θ
=
α
to
θ
=
β
is given by
1
2
∫
β
α
r
2
d
θ
.
For one loop of the given equation, the corresponding integral is then
1
2
∫
π
/
3
0
(
a
sin
3
θ
)
2
d
θ
.
Working this integral:
1
2
∫
π
/
3
0
(
a
sin
3
θ
)
2
d
θ
=
1
2
∫
π
/
3
0
a
2
(
sin
2
3
θ
)
d
θ
Use the identity
cos
2
α
=
1
−
2
sin
2
α
to rewrite for
sin
2
α
, showing that
sin
2
α
=
1
2
(
1
−
cos
2
α
)
.
We can use this to say that
sin
2
3
θ
=
1
2
(
1
−
cos
6
θ
)
. Then the integral reduces:
=
1
2
∫
π
/
3
0
a
2
(
1
2
(
1
−
cos
6
θ
)
)
d
θ
=
a
2
4
∫
π
/
3
0
(
1
−
cos
6
θ
)
d
θ
Integrating term by term:
=
a
2
4
(
θ
−
1
6
sin
6
θ
)
∣
∣
∣
π
/
3
0
=
a
2
4
[
π
3
−
1
6
sin
2
π
−
(
0
−
1
6
sin
0
)
]
=
a
2
4
(
π
3
)
=
π
a
2
12