# Differential Geometry/Arc Length

The length of a vector function f on an interval [a,b] is defined as sup{x|tn∈[a,b], tn+1>tn, x=$\textstyle \sum_{k=1}^n$ |f(tk)-f(tk-1)|}. 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 [a,b] would be equal to $\int_{a}^{b} |\mathbf{f}'(x)|dx$.

Proof: Consider a partition a=t0<t1<t2<...<tn=b, and call it Pn. Let Pn+1 be the partition Pn with an additional point, and let the sequence max{tn-tn-1} go into 0 as n goes to infinity, and let ln be the arc length of the segments by joining the f(x) of the vector function. By the mean value theorem, there exists in the nth partition a number tn' such that

$\sqrt{\sum_{i=1}^3 (x_i(t_n)-x_i(t_{n-1}))^2}=(t_n-t_{n-1})\sqrt{(\sum_{i=1}^3 x_i'(t_n'))}$.

Hence,

$l_n=\sum_{j=1}^n \sqrt{\sum_{i=1}^3 (x_i(t_j)-x_i(t_{j-1}))^2}=\sum_{j=1}^n(t_j-t_{j-1})\sqrt{\sum_{i=1}^3 x_i'(t_j')}$,

which is equal to

$\sum_{j=1}^n(t_j-t_{j-1})\sqrt{\sum_{i=1}^3 x_i'(t_j)}+\sum_{j=1}^n(t_j-t_{j-1})(\sqrt{\sum_{i=1}^3 x_i'(t_j')}-\sqrt{\sum_{i=1}^3 x_i'(t_j)})$.

The amount

$\sqrt{\sum_{i=1}^3 x_i'(t_j')}-\sqrt{\sum_{i=1}^3 x_i'(t_j)}$

shall be denoted dj. Because of the triangle inequality,

$d_j\le \sqrt{\sum_{i=1}^3(x_i'(t_j')-x_i'(t_j))^2}\le \sum_{i=1}^3 |x_i'(t_j')-x_i'(t_j)|$.

Each component is at least once continuously differentiable. There exists thus for any ε>0, there is a δ>0 such that

$|x_i'(a)-x_i'(b)|<\frac{\epsilon}{3}$ when

|a-b|<δ.

Therefore, if max{tn-tn-1}<δ, then dj<ε, so that

$|\sum_{j=1}^n (t_n-t_{n-1})|d_j<$ε(b-a) which approaches 0 when n approaches infinity.

Thus, the amount

$\sum_{j=1}^n(t_j-t_{j-1})\sqrt{\sum_{i=1}^3 x_i'(t_j)}+\sum_{j=1}^n(t_j-t_{j-1})(\sqrt{\sum_{i=1}^3 x_i'(t_j')}-\sqrt{\sum_{i=1}^3 x_i'(t_j)})$

approaches the integral $\int_{a}^{b} |\mathbf{f}'(x)|dx$ since the right term approaches 0.

If there is another parametric representation from [a',b'], and one obtains another arc length, then

$\int_{a'}^{b'} \sqrt{\sum_{i=1}^3 (\frac {dx_i}{dt})^2} |\frac{dt}{dt'}| dt'= \int_{a'}^{b'} \sqrt{\sum_{i=1}^3 (\frac {dx_i}{dt'})^2} dt'$,

indicating that it is the same for any parametric representation.

The function s(t)=$\int_{t_0}^{t} |\mathbf{f}'(x)|dx$ where t0 is a constant is called the arc length parameter of the curve. Its derivative turns out to be |f'(x)|.