Differential Geometry/Arc Length

The length of a vector function f on an interval [a,b] is defined as sup{x|t_n∈[a,b], t_n+1>t_n, x=${\sum }_{k=1}^n$ |f(t_k)-f(t_k-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 $\underset{a}{\overset{b}{\int }}|{𝐟}^{\prime }(x)|dx$.

Proof: Consider a partition a=t₀<t₁<t₂<...<t_n=b, and call it P_n. Let P_n+1 be the partition P_n with an additional point, and let the sequence the sequence max{t_n-t_n-1} go into 0 as n goes to infinity, and let l_n 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 t_n' such that

$\sqrt{\sum _{i=1}^3(x_i(t_n)-x_i(t_{n-1}){)}^2}=(t_n-t_{n-1})\sqrt{(\sum _{i=1}^3{x_i}^{\prime }({t_n}^{\prime }))}$.

Hence,

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

which is equal to

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

The amount

$\sqrt{\sum _{i=1}^3{x_i}^{\prime }({t_j}^{\prime })}-\sqrt{\sum _{i=1}^3{x_i}^{\prime }(t_j)}$

shall be denoted d_j. Because of the triangle inequality,

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

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

$|{x_i}^{\prime }(a)-{x_i}^{\prime }(b)|<\frac{ϵ}{3}$ when

|a-b|<δ.

Therefore, if max{t_n-t_n-1}<δ, then d_j<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

$\sum _{j=1}^n(t_j-t_{j-1})\sqrt{\sum _{i=1}^3{x_i}^{\prime }(t_j)}+\sum _{j=1}^n(t_j-t_{j-1})(\sqrt{\sum _{i=1}^3{x_i}^{\prime }({t_j}^{\prime })}-\sqrt{\sum _{i=1}^3{x_i}^{\prime }(t_j)})$

approaches the integral $\underset{a}{\overset{b}{\int }}|{𝐟}^{\prime }(x)|dx$ since the right term approaches 0.

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

$\underset{a^{\prime }}{\overset{b^{\prime }}{\int }}\sqrt{\sum _{i=1}^3(\frac{d{x}_i}{dt}{)}^2}|\frac{dt}{d{t}^{\prime }}|d{t}^{\prime }=\underset{a^{\prime }}{\overset{b^{\prime }}{\int }}\sqrt{\sum _{i=1}^3(\frac{d{x}_i}{d{t}^{\prime }}{)}^2}d{t}^{\prime }$,

indicating that it is the same for any parametric representation.

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