# Differential Geometry/Arc Length

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

${\displaystyle \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 ${\displaystyle [a,b]}$ would be equal to ${\displaystyle \int \limits _{a}^{b}\left|{\vec {f}}'(x)\right|dx}$ .

Proof

Consider a partition ${\displaystyle a=t_{0} , and call it ${\displaystyle P_{n}}$ . Let ${\displaystyle P_{n+1}}$ be the partition ${\displaystyle P_{n}}$ with an additional point, and let ${\displaystyle \lim _{n\to \infty }\max\{t_{n}-t_{n-1}\}=0}$ , and let ${\displaystyle l_{n}}$ be the arc length of the segments by joining the ${\displaystyle f(x)}$ of the vector function. By the mean value theorem, there exists in the nth partition a number ${\displaystyle t_{n}'}$ such that

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

Hence,

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

which is equal to

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

The amount

${\displaystyle {\sqrt {\sum _{i=1}^{3}x_{i}'(t_{j}')}}-{\sqrt {\sum _{i=1}^{3}x_{i}'(t_{j})}}}$

shall be denoted ${\displaystyle d_{j}}$ . Because of the triangle inequality,

${\displaystyle d_{j}\leq {\sqrt {\sum _{i=1}^{3}(x_{i}'(t_{j}')-x_{i}'(t_{j}))^{2}}}\leq \sum _{i=1}^{3}{\Big |}x_{i}'(t_{j}')-x_{i}'(t_{j}){\Big |}}$

Each component is at least once continuously differentiable. There exists thus for any ${\displaystyle \varepsilon >0}$ , there is a ${\displaystyle \delta >0}$ such that

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

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

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

Thus, the amount

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

approaches the integral ${\displaystyle \int \limits _{a}^{b}\left|{\vec {f}}'(x)\right|dx}$ since the right term approaches 0.

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

${\displaystyle \int \limits _{a'}^{b'}{\sqrt {\sum _{i=1}^{3}\left({\frac {dx_{i}}{dt}}\right)^{2}}}\left|{\frac {dt}{dt'}}\right|dt'=\int \limits _{a'}^{b'}{\sqrt {\sum _{i=1}^{3}\left({\frac {dx_{i}}{dt'}}\right)^{2}}}dt'}$

indicating that it is the same for any parametric representation.

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