The length of a vector function
on an interval
is defined as
![{\displaystyle \sup \left\{x{\Bigg |}t_{n}\in [a,b],t_{n}<t_{n+1},x=\sum _{k=1}^{n}{\Big |}f(t_{k})-f(t_{k-1}){\Big |}\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/668fdc24465c1c0687a4a5873a1c13cbbdbd22c0)
If this number is finite, then this function is rectifiable.
For continuously differentiable vector functions, the arc length of that vector function on the interval
would be equal to
.
- Proof
Consider a partition
, and call it
. Let
be the partition
with an additional point, and let
, and let
be the arc length of the segments by joining the
of the vector function. By the mean value theorem, there exists in the nth partition a number
such that

Hence,

which is equal to

The amount

shall be denoted
. Because of the triangle inequality,

Each component is at least once continuously differentiable. There exists thus for any
, there is a
such that
when
.
Therefore, if
then
, so that
which approaches 0 when n approaches infinity.
Thus, the amount

approaches the integral
since the right term approaches 0.
If there is another parametric representation from
, and one obtains another arc length, then

indicating that it is the same for any parametric representation.
The function
where
is a constant is called the arc length parameter of the curve. Its derivative turns out to be
.