# integral representation of length of smooth curve

Suppose $\gamma :[0,1]\to {\mathbb{R}}^{m}$ is a continuously differentiable curve. Then the definition of its length as a rectifiable curve

$$ |

is equal to its length as computed in differential geometry:

$${\int}_{0}^{1}\parallel {\gamma}^{\prime}(t)\parallel \mathit{d}t.$$ |

###### Proof.

Let the partition $\{{t}_{i}\}$ of $[0,1]$ be arbitrary. Then

$\sum _{i=1}^{n}}\parallel \gamma ({t}_{i})-\gamma ({t}_{i-1})\parallel $ | $={\displaystyle \sum _{i=1}^{n}}\parallel {\displaystyle {\int}_{{t}_{i-1}}^{{t}_{i}}}{\gamma}^{\prime}(t)\mathit{d}t\parallel $ | (fundamental theorem of calculus^{}) |
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$\le {\displaystyle \sum _{i=1}^{n}}{\displaystyle {\int}_{{t}_{i-1}}^{{t}_{i}}}\parallel {\gamma}^{\prime}(t)\parallel \mathit{d}t$ | (triangle inequality^{} for integrals) |
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$={\displaystyle {\int}_{0}^{1}}\parallel {\gamma}^{\prime}(t)\parallel \mathit{d}t.$ |

Hence $L\le {\int}_{0}^{1}\parallel {\gamma}^{\prime}(t)\parallel \mathit{d}t$. (By the way, this also shows that $\gamma $ is rectifiable in the first place.)

The inequality^{} in the other direction is more tricky.
Given $\u03f5>0$, we know that
${\int}_{0}^{1}\parallel {\gamma}^{\prime}(t)\parallel \mathit{d}t$ can be approximated up to $\u03f5$
by a Riemann sum^{} of the form

$$\sum _{i=1}^{n}\parallel {\gamma}^{\prime}({t}_{i-1})\parallel ({t}_{i}-{t}_{i-1})$$ |

provided the partition $\{{t}_{i}\}$ is fine enough, i.e. has mesh width $\le \mathrm{\Delta}$ for some small $\mathrm{\Delta}>0$. We want to approximate ${\gamma}^{\prime}({t}_{i-1})$ with $[\gamma ({t}_{i})-\gamma ({t}_{i-1})]/({t}_{i}-{t}_{i-1})$, but this only works if ${t}_{i}-{t}_{i-1}$ is small.

To get the precise estimates, use uniform continuity of ${\gamma}^{\prime}$ on $[0,1]$ to obtain a $\delta >0$ such that $\parallel {\gamma}^{\prime}(\tau )-{\gamma}^{\prime}(t)\parallel \le \u03f5$ whenever $|\tau -t|\le \delta $. Then for all $$ and $t\in [0,1]$,

$$\parallel \frac{\gamma (t+h)-\gamma (t)}{h}-{\gamma}^{\prime}(t)\parallel \le \frac{1}{h}{\int}_{t}^{t+h}\parallel {\gamma}^{\prime}(\tau )-{\gamma}^{\prime}(t)\parallel \mathit{d}\tau \le \frac{h}{h}\u03f5=\u03f5.$$ |

Let the partition $\{{t}_{i}\}$ have a mesh width less than both $\delta $ and $\mathrm{\Delta}$. Then setting $h={t}_{i}-{t}_{i-1}$ successively in each summand, we have

${\int}_{0}^{1}}\parallel {\gamma}^{\prime}(t)\parallel \mathit{d}t$ | $\le {\displaystyle \sum _{i=1}^{n}}\parallel {\gamma}^{\prime}({t}_{i-1})\parallel ({t}_{i}-{t}_{i-1})+\u03f5$ | ||

$\le {\displaystyle \sum _{i=1}^{n}}{\displaystyle \frac{\parallel \gamma ({t}_{i})-\gamma ({t}_{i-1})\parallel}{{t}_{i}-{t}_{i-1}}}({t}_{i}-{t}_{i-1})+{\displaystyle \sum _{i=1}^{n}}\u03f5({t}_{i}-{t}_{i-1})+\u03f5$ | |||

$={\displaystyle \sum _{i=1}^{n}}\parallel \gamma ({t}_{i})-\gamma ({t}_{i-1})\parallel +2\u03f5$ | |||

$\le L+2\u03f5.$ |

Taking $\u03f5\to 0$ yields ${\int}_{0}^{1}\parallel {\gamma}^{\prime}(t)\parallel \mathit{d}t\le L$. ∎

We remark that $L={\int}_{0}^{1}\parallel {\gamma}^{\prime}(t)\parallel \mathit{d}t$ is true for piecewise smooth curves $\gamma $ also, simply by adding together the results for each smooth segment of $\gamma $.

Title | integral representation of length of smooth curve |
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Canonical name | IntegralRepresentationOfLengthOfSmoothCurve |

Date of creation | 2013-03-22 15:39:39 |

Last modified on | 2013-03-22 15:39:39 |

Owner | stevecheng (10074) |

Last modified by | stevecheng (10074) |

Numerical id | 11 |

Author | stevecheng (10074) |

Entry type | Derivation |

Classification | msc 51N05 |

Related topic | ArcLength |

Related topic | Rectifiable |

Related topic | TotalVariation |