L
=
∫
t
f
t
i
√
(
d
x
d
t
)
2
+
(
d
y
d
t
)
2
d
t
. Since
x
and
y
are perpendicular, it's not difficult to see why this computes the arclength.
It isn't very different from the arclength of a regular function:
L
=
∫
b
a
√
1
+
(
d
y
d
x
)
2
d
x
. If you need the derivation of the parametric formula, please ask it as a separate question.
We find the 2 derivatives:
d
x
d
t
=
3
−
3
t
2
d
y
d
t
=
6
t
And we substitute these into the integral:
L
=
∫
√
3
0
√
(
3
−
3
t
2
)
2
+
(
6
t
)
2
d
t
And solve:
=
∫
√
3
0
√
9
−
18
t
2
+
9
t
4
+
36
t
2
d
t
=
∫
√
3
0
√
9
+
18
t
2
+
9
t
4
d
t
=
∫
√
3
0
√
(
3
+
3
t
2
)
2
d
t
=
∫
√
3
0
(
3
+
3
t
2
)
d
t
=
3
t
+
t
3
∣
∣
√
3
0
=
3
√
3
+
3
√
3
=6The arclength of a parametric curve can be found using the formula:
L
=
∫
t
f
t
i
√
(
d
x
d
t
)
2
+
(
d
y
d
t
)
2
d
t
. Since
x
and
y
are perpendicular, it's not difficult to see why this computes the arclength.
It isn't very different from the arclength of a regular function:
L
=
∫
b
a
√
1
+
(
d
y
d
x
)
2
d
x
. If you need the derivation of the parametric formula, please ask it as a separate question.
We find the 2 derivatives:
d
x
d
t
=
3
−
3
t
2
d
y
d
t
=
6
t
And we substitute these into the integral:
L
=
∫
√
3
0
√
(
3
−
3
t
2
)
2
+
(
6
t
)
2
d
t
And solve:
=
∫
√
3
0
√
9
−
18
t
2
+
9
t
4
+
36
t
2
d
t
=
∫
√
3
0
√
9
+
18
t
2
+
9
t
4
d
t
=
∫
√
3
0
√
(
3
+
3
t
2
)
2
d
t
=
∫
√
3
0
(
3
+
3
t
2
)
d
t
=
3
t
+
t
3
∣
∣
√
3
0
=
3
√
3
+
3
√
3
=
6
√
3
Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.
Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.