飞蛾扑火问题

$\begin{aligned} & dr=-rd\theta\sin(\frac{\pi}{2}-\alpha) \\ & r=e^{-\theta\cos\alpha} \\ \end{aligned}$

最速降线

$\begin{aligned} & T=\int_{t_1}^{t_2}\frac{\text{d}s}{v}=\int_{x_1}^{x_2}\frac{\sqrt{1+y'^2}}{\sqrt{2gy}}\text{d}x \\ & f(y,y')=\frac{\sqrt{1+y'^2}}{\sqrt{2gy}} \\ & y(1+y'^2)=C \\ & \text{d}x=\sqrt{\frac{y}{2a-y}}\text{d}y \xlongequal{y=a-az}-a\sqrt{\frac{1-z}{1+z}}\text{d}z \xlongequal{z=\cos\theta}2a\sin^2\frac{\theta}{2}\text{d}\theta \\ x&=a(\theta-\sin\theta) \\ y&=a(1-\cos\theta) \end{aligned}$

悬链线

微分法

$\begin{aligned} & T\sin\theta=\rho gs,T\cos\theta=H \\ & y'=\frac{1}{a}\int_0^x\sqrt{1+y'^2}\text{d}x \\ & \frac{\text{d}p}{\text{d}x}=\frac{1}{a}\sqrt{1+p^2} \\ & y=\frac{a}{2}(\text{e}^{\frac{x}{a}}+\text{e}^{-\frac{x}{a}}) \\ \end{aligned}$

变分法

续浅谈变分原理：运动常数、拉格朗日乘子法

相同周长时圆的面积最大

设从$(-r,0)$点到$(r,0)$点的曲线函数为$y(x)$，曲线长度为$L$，即在

$\int_{-r}^r\sqrt{1+y'^2}\text{d}x=L$

的约束条件下，求

$J=\int_{-r}^ry\text{d}x$

的最大值。
设

$z(x)=\int_{-r}^x\sqrt{1+y'^2}\text{d}x$

则

$z'(x)=\sqrt{1+y'^2}$

拉格朗日乘子法取

$H=y+\lambda(z'^2-y'^2-1)$

则

$\begin{aligned} & \frac{\partial H}{\partial y}-\frac{\text{d}}{\text{d}x}\frac{\partial H}{\partial y'}=0 \\ & 1=-2(\lambda y')' \\ & \lambda y'=ax \\ & \frac{\partial H}{\partial z}-\frac{\text{d}}{\text{d}x}\frac{\partial H}{\partial z'}=0 \\ & \lambda z'=b \\ & \frac{b^2}{\lambda^2}=\frac{a^2x^2}{\lambda^2}+1 \\ & y'=\pm\frac{ax}{\sqrt{b^2-a^2x^2}} \\ & y=\sqrt{\frac{b^2}{a^2}-x^2} \\ \end{aligned}$

代入$y(-r)=y(r)=0$得$y=\sqrt{r^2-x^2}$。