# integral representation of length of smooth curve

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

 $L=\sup\Bigl{\{}\sum_{i=1}^{n}\lVert\gamma(t_{i})-\gamma(t_{i-1})\rVert\colon 0% =t_{0}

is equal to its length as computed in differential geometry:

 $\int_{0}^{1}\lVert\gamma^{\prime}(t)\rVert\,dt\,.$
###### Proof.

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

 $\displaystyle\sum_{i=1}^{n}\lVert\gamma(t_{i})-\gamma(t_{i-1})\rVert$ $\displaystyle=\sum_{i=1}^{n}\Bigl{\lVert}\int_{t_{i-1}}^{t_{i}}\gamma^{\prime}% (t)\,dt\Bigr{\rVert}$ $\displaystyle\leq\sum_{i=1}^{n}\int_{t_{i-1}}^{t_{i}}\lVert\gamma^{\prime}(t)% \rVert\,dt$ $\displaystyle=\int_{0}^{1}\lVert\gamma^{\prime}(t)\rVert\,dt\,.$

Hence $L\leq\int_{0}^{1}\lVert\gamma^{\prime}(t)\rVert\,dt$. (By the way, this also shows that $\gamma$ is rectifiable in the first place.)

The inequality in the other direction is more tricky. Given $\epsilon>0$, we know that $\int_{0}^{1}\lVert\gamma^{\prime}(t)\rVert\,dt$ can be approximated up to $\epsilon$ by a Riemann sum of the form

 $\sum_{i=1}^{n}\lVert\gamma^{\prime}(t_{i-1})\rVert(t_{i}-t_{i-1})$

provided the partition $\{t_{i}\}$ is fine enough, i.e. has mesh width $\leq\Delta$ for some small $\Delta>0$. 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 $\lVert\gamma^{\prime}(\tau)-\gamma^{\prime}(t)\rVert\leq\epsilon$ whenever $\lvert\tau-t\rvert\leq\delta$. Then for all $0 and $t\in[0,1]$,

 $\left\lVert\frac{\gamma(t+h)-\gamma(t)}{h}-\gamma^{\prime}(t)\right\rVert\leq% \frac{1}{h}\int_{t}^{t+h}\lVert\gamma^{\prime}(\tau)-\gamma^{\prime}(t)\rVert% \,d\tau\leq\frac{h}{h}\,\epsilon=\epsilon\,.$

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

 $\displaystyle\int_{0}^{1}\lVert\gamma^{\prime}(t)\rVert\,dt$ $\displaystyle\leq\sum_{i=1}^{n}\lVert\gamma^{\prime}(t_{i-1})\rVert(t_{i}-t_{i% -1})+\epsilon$ $\displaystyle\leq\sum_{i=1}^{n}\frac{\lVert\gamma(t_{i})-\gamma(t_{i-1})\rVert% }{t_{i}-t_{i-1}}\,(t_{i}-t_{i-1})+\sum_{i=1}^{n}\epsilon(t_{i}-t_{i-1})+\epsilon$ $\displaystyle=\sum_{i=1}^{n}\lVert\gamma(t_{i})-\gamma(t_{i-1})\rVert+2\epsilon$ $\displaystyle\leq L+2\epsilon\,.$

Taking $\epsilon\to 0$ yields $\int_{0}^{1}\lVert\gamma^{\prime}(t)\rVert\,dt\leq L$. ∎

We remark that $L=\int_{0}^{1}\lVert\gamma^{\prime}(t)\rVert\,dt$ 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 IntegralRepresentationOfLengthOfSmoothCurve 2013-03-22 15:39:39 2013-03-22 15:39:39 stevecheng (10074) stevecheng (10074) 11 stevecheng (10074) Derivation msc 51N05 ArcLength Rectifiable TotalVariation