Differential Geometry/Arc Length

The length of a vector function $f$ on an interval $[a,b]$ is defined as

$\sup \left\{x\right|t_n\in [a,b],t_n<t_{n+1},x=\sum _{k=1}^n|f(t_k)-f(t_{k-1})|\}$

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 $\underset{a}{\overset{b}{\int }}|{\overrightarrow{f}}^{\prime }(x)|dx$ .

Proof

Consider a partition $a=t_0<t_1<\cdots <t_n=b$ , and call it $P_n$ . Let $P_{n+1}$ be the partition $P_n$ with an additional point, and let $\underset{n\to \mathrm{\infty }}{\lim }\max \{t_n-t_{n-1}\}=0$ , 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 ${t_n}^{\prime }$ 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 $\epsilon >0$ , there is a $\delta >0$ such that

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

Therefore, if $\max \{t_n-t_{n-1}\}<\delta$ then $d_j<\epsilon$ , so that

$|\sum _{j=1}^n(t_n-t_{n-1})|d_j<\epsilon (b-a)$ 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 $\underset{a}{\overset{b}{\int }}|{\overrightarrow{f}}^{\prime }(x)|dx$ since the right term approaches 0.

If there is another parametric representation from $[a^{\prime },b^{\prime }]$ , and one obtains another arc length, then

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

indicating that it is the same for any parametric representation.

The function $s(t)=\underset{t_0}{\overset{t}{\int }}|{\overrightarrow{f}}^{\prime }(x)|dx$ where $t_0$ is a constant is called the arc length parameter of the curve. Its derivative turns out to be $|f^{\prime }(x)|$ .