# 弧长

（重定向自可求长曲线

## 定义

${\displaystyle C}$ 欧几里德空间${\displaystyle S=\mathbb {R} ^{n}}$ （或某个有限维度量空间）中的一条曲线。它是某个从实数区间映射S连续函数${\displaystyle f:[a,b]\rightarrow S}$ 图像。考虑区间${\displaystyle [a,b]}$ 的一个分割：${\displaystyle a=t_{0} ${\displaystyle f(t_{0}),f(t_{1}),\ldots ,f(t_{n-1}),f(t_{n})}$ 是曲线${\displaystyle C}$ 上的${\displaystyle n+1}$ 个点。将${\displaystyle f(t_{i})}$ ${\displaystyle f(t_{i+1})}$ 两点之间的距离记为${\displaystyle d\left(f(t_{i}),f(t_{i+1})\right)}$ ，这也是从${\displaystyle f(t_{i})}$ 连到${\displaystyle f(t_{i+1})}$ 的线段的长度。而曲线${\displaystyle C}$ 的弧长${\displaystyle L(C)}$ 定义为：

${\displaystyle L(C)=\sup _{a=t_{0}

## 用积分计算弧长

${\displaystyle L_{T}=\sum _{i=0}^{n-1}\|\Delta r_{i}\|=\sum _{i=0}^{n-1}\|r'(t_{i})\|\Delta t_{i}.}$

${\displaystyle s=\int _{a}^{b}\|r'(t)\|dt.}$

${\displaystyle ds^{2}=dX^{2}+dY^{2}.}$

${\displaystyle ds}$ 足够接近0的时候，${\displaystyle dx}$ ${\displaystyle dy}$ 也足够接近0. 所以在给定的时刻${\displaystyle t}$ ，在${\displaystyle \left(X(t),Y(t)\right)}$ 附近有：

${\displaystyle ds={\sqrt {\left({dX(t) \over dt}\right)^{2}+\left({dY(t) \over dt}\right)^{2}}}dt={\sqrt {X'(t)^{2}+Y'(t)^{2}}}\,dt.}$

${\displaystyle s=\int _{0}^{1}{\sqrt {X'(t)^{2}+Y'(t)^{2}}}\,dt.}$

${\displaystyle s=\int _{0}^{1}{\sqrt {X'(t)^{2}+Y'(t)^{2}+Z'(t)^{2}}}\,dt.}$

${\displaystyle s=\int _{0}^{1}{\sqrt {X'(t)^{2}+Y'(t)^{2}}}\,dt.=\int _{0}^{1}{\sqrt {1+[f'(x)]^{2}}}\,dx}$

${\displaystyle s=\int _{a}^{b}{\sqrt {r^{2}+({dr \over d\theta })^{2}}}\,d\theta .}$

## 例子

### 半立方抛物线的弧长

${\displaystyle 2yy\prime =9(x-1)^{2}}$

${\displaystyle (y\prime )^{2}=\left({\frac {9(x-1)^{2}}{2y}}\right)^{2}=\left({\frac {9(x-1)^{2}}{2{\sqrt {3}}(x-1)^{\frac {3}{2}}}}\right)^{2}=\left({\frac {3{\sqrt {3}}(x-1)^{\frac {1}{2}}}{2}}\right)^{2}={\frac {27(x-1)}{4}}.}$

${\displaystyle s=\int _{1}^{4}{\sqrt {1+(y\prime )^{2}}}\,dx=\int _{1}^{4}{\sqrt {1+{\frac {27(x-1)}{4}}}}\,dx={\frac {85{\sqrt {85}}-8}{81}}}$