Differential Geometry/Arc Length

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


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

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 $$\int\limits_a^b\left|\vec{f}'(x)\right|dx$$.

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 $$\lim_{n\to\infty}\max\{t_n-t_{n-1}\}=0$$, 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
 * Proof


 * $$\sqrt{\sum_{i=1}^3\Big(x_i(t_n)-x_i(t_{n-1})\Big)^2}=(t_n-t_{n-1})\sqrt{\sum_{i=1}^3 x_i'(t_n')}$$

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,


 * $$d_j\le\sqrt{\sum_{i=1}^3(x_i'(t_j')-x_i'(t_j))^2}\le\sum_{i=1}^3\Big|x_i'(t_j')-x_i'(t_j)\Big|$$

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 $$\Big|f'(x)\Big|$$.