# Differential Geometry/Arc Length

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

$\sup \left\{x{\Bigg |}t_{n}\in [a,b],t_{n} 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 $\int \limits _{a}^{b}\left|{\vec {f}}'(x)\right|dx$ .

Proof

Consider a partition $a=t_{0} , and call it $P_{n}$ . Let $P_{n+1}$ be the partition $P_{n}$ with an additional point, and let $\lim _{n\to \infty }\max\{t_{n}-t_{n-1}\}=0$ , 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 $t_{n}'$ such that

${\sqrt {\sum _{i=1}^{3}{\Big (}x_{i}(t_{n})-x_{i}(t_{n-1}){\Big )}^{2}}}=(t_{n}-t_{n-1}){\sqrt {\sum _{i=1}^{3}x_{i}'(t_{n}')}}$ Hence,

$l_{n}=\sum _{j=1}^{n}{\sqrt {\sum _{i=1}^{3}{\Big (}x_{i}(t_{j})-x_{i}(t_{j-1}){\Big )}^{2}}}=\sum _{j=1}^{n}(t_{j}-t_{j-1}){\sqrt {\sum _{i=1}^{3}x_{i}'(t_{j}')}}$ which is equal to

$\sum _{j=1}^{n}(t_{j}-t_{j-1}){\sqrt {\sum _{i=1}^{3}x_{i}'(t_{j})}}+\sum _{j=1}^{n}(t_{j}-t_{j-1})\left({\sqrt {\sum _{i=1}^{3}x_{i}'(t_{j}')}}-{\sqrt {\sum _{i=1}^{3}x_{i}'(t_{j})}}\right)$ The amount

${\sqrt {\sum _{i=1}^{3}x_{i}'(t_{j}')}}-{\sqrt {\sum _{i=1}^{3}x_{i}'(t_{j})}}$ shall be denoted $d_{j}$ . Because of the triangle inequality,

$d_{j}\leq {\sqrt {\sum _{i=1}^{3}(x_{i}'(t_{j}')-x_{i}'(t_{j}))^{2}}}\leq \sum _{i=1}^{3}{\Big |}x_{i}'(t_{j}')-x_{i}'(t_{j}){\Big |}$ Each component is at least once continuously differentiable. There exists thus for any $\varepsilon >0$ , there is a $\delta >0$ such that

${\Big |}x_{i}'(a)-x_{i}'(b){\Big |}<{\frac {\varepsilon }{3}}$ when $|a-b|<\delta$ .

Therefore, if $\max\{t_{n}-t_{n-1}\}<\delta$ then $d_{j}<\varepsilon$ , so that

$\left|\sum _{j=1}^{n}(t_{n}-t_{n-1})\right|d_{j}<\varepsilon (b-a)$ which approaches 0 when n approaches infinity.

Thus, the amount

$\sum _{j=1}^{n}(t_{j}-t_{j-1}){\sqrt {\sum _{i=1}^{3}x_{i}'(t_{j})}}+\sum _{j=1}^{n}(t_{j}-t_{j-1})\left({\sqrt {\sum _{i=1}^{3}x_{i}'(t_{j}')}}-{\sqrt {\sum _{i=1}^{3}x_{i}'(t_{j})}}\right)$ approaches the integral $\int \limits _{a}^{b}\left|{\vec {f}}'(x)\right|dx$ since the right term approaches 0.

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

$\int \limits _{a'}^{b'}{\sqrt {\sum _{i=1}^{3}\left({\frac {dx_{i}}{dt}}\right)^{2}}}\left|{\frac {dt}{dt'}}\right|dt'=\int \limits _{a'}^{b'}{\sqrt {\sum _{i=1}^{3}\left({\frac {dx_{i}}{dt'}}\right)^{2}}}dt'$ indicating that it is the same for any parametric representation.

The function $s(t)=\int \limits _{t_{0}}^{t}\left|{\vec {f}}'(x)\right|dx$ where $t_{0}$ is a constant is called the arc length parameter of the curve. Its derivative turns out to be ${\Big |}f'(x){\Big |}$ .