# 平面曲线的弧长

1，平面曲线的弧长：我们将曲线用点 $$M_0, M_1,\cdots, M_n$$ 将曲线划分成 $$n$$ 小段，在每一小段上，可以用直线的长度来近似曲线段的长度，

$s=\lim_{{\Delta x\to 0}\atop{\Delta y\to 0}}\sum_{i=1}^n\sqrt{\Delta^2x_i+\Delta^2y_i}$

2，弧微分：若我们记 $ds=\sqrt{dx^2+dy^2}$则 $$s=\int_a^bds$$

3，计算：

（1）曲线 $$L$$ 由参数方程 $$x=\phi(t), y=\psi(t), \alpha\le t\le \beta$$，$$\phi(t),\psi(t)$$ 一阶连续可导，则 $s=\int_{\alpha}^{\beta}\sqrt{x’^2(t)}+y’^2(t)}dt=\int_{\alpha}^{\beta}\sqrt{\phi’^2(t)+\psi’^2(t)}dt$

\begin{align*}s&=\lim_{{\Delta x\to 0}\atop{\Delta y\to 0}}\sum_{i=1}^n\sqrt{\Delta^2x_i+\Delta^2y_i}\\ &=\lim_{\Delta t\to 0}\sum_{i=1}^n\sqrt{\frac{\Delta^2x_i}{\Delta^2t}+\frac{\Delta^2y_i}{\Delta^2 t}}\Delta t\\ &=\int_{\alpha}^{\beta}\sqrt{\phi’^2(t)+\psi’^2(t)}dt\end{align*}

（2）曲线$$L$$ 由函数 $$y=f(x), a\le x\la b$$ 给出，$$f(x)$$ 连续可导，则

$s=\int_a^b\sqrt{1+f’^2(x)}dx$

\begin{align*}s&=\lim_{{\Delta x\to 0}\atop{\Delta y\to 0}}\sum_{i=1}^n\sqrt{\Delta^2x_i+\Delta^2y_i}\\ &=\lim_{\Delta x\to 0}\sum_{i=1}^n\sqrt{1+\frac{\Delta^2y_i}{\Delta^2x_i}}\Delta x_i\\ &=\int_a^b\sqrt{1+f’^2(x)}dx\end{align*}

\begin{align*}s&=\int_{0}^{2\pi}\sqrt{x’^2(\theta)y’^2(\theta)}dt\int_0^{2\pi}\sqrt{(a(1-\cos\theta))^2+(a\sin\theta)^2}d\theta\\ &=a\int_0^{2\pi}\sqrt{1-2\cos\theta+\cos^2\theta+\sin^2\theta}d\theta\\ &=a\int_0^{2\pi}\sqrt{2-2\cos\theta}d\theta=a\int_0^{2\pi}\sqrt{4\sin^\frac{\theta}{2}}d\theta\\ &=2a\int_0^{2\pi}\sin\frac{\theta}{2}d\theta=2a(-a)\cos\frac{\theta}{2}\Big|_0^{2\pi}\\&=8a\end{align*}

$ds=\sqrt{1+\frac{x^2}{dy^2}},\quad dy=\sqrt{1+(2y)^2}dy=\sqrt{1+4y^2}dy$

\begin{align*}s&=\int_0^1\sqrt{1+4y^2}dy\end{align*}

\begin{align*}s&=\int_0^1\sqrt{1+4y^2}dy=\int_0^{\arctan 2}\sqrt{1+\tan^2t}\frac{1}{2}\sec^2tdt\\ &=\int_0^{\arctan 2}\frac{1}{2}\sec^3tdt=\frac{1}{2}\int_0^{\arctan 2}\sec^2t\sec tdt \\ &=\frac{1}{2} \sec t\tan t-\frac{1}{2}\int_0^{\arctan 2}\tan t \cdot \tan t \sec tdt\\ &=\frac{1}{2} \sec t\tan t-\frac{1}{2}\int_0^{\arctan 2}(\sec^2 t-1) \sec tdt\\ &=\frac{1}{2} \sec t\tan t-\frac{1}{2}\int_0^{\arctan 2}(\sec^2 t-1) \sec tdt\\ &=\frac{1}{2} \sec t\tan t-\frac{1}{2}\int_0^{\arctan 2}\sec^3tdt+\frac{1}{2}\int_0^{\arctan 2}\sec tdt\end{align*}

\begin{align*}int_0^{\arctan 2}\sec^3tdt&=\frac{1}{2} \sec t\tan t+\frac{1}{2}\int_0^{\arctan 2}\sec tdt \\ &=\frac{1}{2} \sec t\tan t+\frac{1}{2}\ln|\sec t+\tan t|\Big|_0^{\arctan 2}\\&=\sqrt5+\frac{1}{2}\ln|2+\sqrt5\end{align*}