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
![{\displaystyle {\sqrt {\sum _{i=1}^{3}{\Big (}x_{i}(t_{n})-x_{i}(t_{n-1}){\Big )}^{2}}}=(t_{n}-t_{n-1}){\sqrt {\sum _{i=1}^{3}x_{i}'(t_{n}')}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b4ee9861aa3d9273bf5dd072fa6d623ae84b298)
Hence,
![{\displaystyle l_{n}=\sum _{j=1}^{n}{\sqrt {\sum _{i=1}^{3}{\Big (}x_{i}(t_{j})-x_{i}(t_{j-1}){\Big )}^{2}}}=\sum _{j=1}^{n}(t_{j}-t_{j-1}){\sqrt {\sum _{i=1}^{3}x_{i}'(t_{j}')}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1647845b43c959f08cffa1af4d8bdaf08783785e)
which is equal to
![{\displaystyle \sum _{j=1}^{n}(t_{j}-t_{j-1}){\sqrt {\sum _{i=1}^{3}x_{i}'(t_{j})}}+\sum _{j=1}^{n}(t_{j}-t_{j-1})\left({\sqrt {\sum _{i=1}^{3}x_{i}'(t_{j}')}}-{\sqrt {\sum _{i=1}^{3}x_{i}'(t_{j})}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/46451bf49b29b23efb099c1966145ee63853a334)
The amount
![{\displaystyle {\sqrt {\sum _{i=1}^{3}x_{i}'(t_{j}')}}-{\sqrt {\sum _{i=1}^{3}x_{i}'(t_{j})}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae5d2da207a61b3675bb9197bb5da48aaf2db54a)
shall be denoted
. Because of the triangle inequality,
![{\displaystyle d_{j}\leq {\sqrt {\sum _{i=1}^{3}(x_{i}'(t_{j}')-x_{i}'(t_{j}))^{2}}}\leq \sum _{i=1}^{3}{\Big |}x_{i}'(t_{j}')-x_{i}'(t_{j}){\Big |}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9edb213c7be6e9a964e22ff19729225022a2027)
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
![{\displaystyle \sum _{j=1}^{n}(t_{j}-t_{j-1}){\sqrt {\sum _{i=1}^{3}x_{i}'(t_{j})}}+\sum _{j=1}^{n}(t_{j}-t_{j-1})\left({\sqrt {\sum _{i=1}^{3}x_{i}'(t_{j}')}}-{\sqrt {\sum _{i=1}^{3}x_{i}'(t_{j})}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/46451bf49b29b23efb099c1966145ee63853a334)
approaches the integral
since the right term approaches 0.
If there is another parametric representation from
, and one obtains another arc length, then
![{\displaystyle \int \limits _{a'}^{b'}{\sqrt {\sum _{i=1}^{3}\left({\frac {dx_{i}}{dt}}\right)^{2}}}\left|{\frac {dt}{dt'}}\right|dt'=\int \limits _{a'}^{b'}{\sqrt {\sum _{i=1}^{3}\left({\frac {dx_{i}}{dt'}}\right)^{2}}}dt'}](https://wikimedia.org/api/rest_v1/media/math/render/svg/612a8f5290d9e9bfd5becfa8d170d1aa47efba49)
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
.