# 最速降線問題

## 歷史

1638年，伽利略在《論兩種新科學》中以為此線是圓弧。約翰·伯努利參考之前分析過的等時降落軌跡，證明了此線是擺線，並在1696年6月的《博學通報》發表。艾薩克·牛頓雅各布·伯努利萊布尼茲洛必達都得出同一結論，即正确的答案应该是摆线的一段。除了洛必達的解外，其他人的解都在1697年5月的《博學通報》出現。

## 证明

### 约翰·伯努利的证明

${\displaystyle v={\sqrt {2gy}}}$,

${\displaystyle {\frac {\sin {\theta }}{v}}={\frac {1}{v}}{\frac {dx}{ds}}={\frac {1}{v_{m}}}}$,

1. 在刚开始，当质点的速度为零时，夹角也必然是零。因此，最速降线在起始处与竖直方向相切
1. 当轨迹变为水平即夹角变为90°时，速度达到最大。

${\displaystyle v_{m}={\sqrt {2gD}}}$.

${\displaystyle v_{m}^{2}(dx)^{2}=v^{2}(ds)^{2}=v^{2}((dx)^{2}+(dy)^{2})}$

${\displaystyle dx={\frac {vdy}{\sqrt {v_{m}^{2}-v^{2}}}}}$.

${\displaystyle dx={\sqrt {\frac {y}{D-y}}}dy}$

### 雅各布·伯努利的证明

${\displaystyle ds^{2}=dx^{2}+dy^{2}}$.

dy不变求微分，得到

${\displaystyle 2ds\ d^{2}s=2dx\ d^{2}x}$

${\displaystyle {\frac {dx}{ds}}d^{2}x=d^{2}s=v\ d^{2}t}$

${\displaystyle d^{2}t_{1}={\frac {1}{v_{1}}}{\frac {dx_{1}}{ds_{1}}}d^{2}x}$
${\displaystyle d^{2}t_{2}={\frac {1}{v_{2}}}{\frac {dx_{2}}{ds_{2}}}d^{2}x}$

${\displaystyle d^{2}t_{2}-d^{2}t_{1}=0={\bigg (}{\frac {1}{v_{2}}}{\frac {dx_{2}}{ds_{2}}}-{\frac {1}{v_{1}}}{\frac {dx_{1}}{ds_{1}}}{\bigg )}d^{2}x}$

${\displaystyle {\frac {1}{v_{2}}}{\frac {dx_{2}}{ds_{2}}}={\frac {1}{v_{1}}}{\frac {dx_{1}}{ds_{1}}}}$

## 最速降線的數學形式與最短時間

${\displaystyle x={\frac {1}{2}}k^{2}\left(\theta -\sin \theta \right),\ y={\frac {1}{2}}k^{2}\left(1-\cos \theta \right).}$[1]

${\displaystyle t=\int _{\theta =0}^{\theta =\theta _{1}}\mathrm {d} t=\int _{\theta =0}^{\theta =\theta _{1}}{\frac {\mathrm {d} s}{v}}}$

${\displaystyle {\frac {ds}{v}}={\frac {k}{\sqrt {2g}}}\mathrm {d} \theta }$
${\displaystyle t={\frac {k}{\sqrt {2g}}}\theta _{1}}$

${\displaystyle k={\sqrt {2r}}}$
${\displaystyle t=\theta _{1}{\sqrt {\frac {r}{g}}}}$

${\displaystyle l={\sqrt {x_{1}^{2}+y_{1}^{2}}}=r{\sqrt {\left(\theta -\sin \theta \right)^{2}+\left(1-\cos \theta \right)^{2}}}}$
${\displaystyle \tan \phi ={\frac {y_{1}}{x_{1}}}={\frac {1-\cos \theta }{\theta -\sin \theta }}}$

${\displaystyle t\,\left(\,l,\,\theta \right)={\sqrt {\frac {l}{g}}}{\frac {\theta }{\sqrt[{4}]{\left(\theta -\sin \theta \right)^{2}+\left(1-\cos \theta \right)^{2}}}}}$

${\displaystyle \left(\,\phi ,\,t\right)=\left(\,\arctan {\frac {1-\cos \theta }{\theta -\sin \theta }},\,{\sqrt {\frac {l}{g}}}{\frac {\theta }{\sqrt[{4}]{\left(\theta -\sin \theta \right)^{2}+\left(1-\cos \theta \right)^{2}}}}\right)}$

## 參考資料

1. ^ Brachistochrone Problem -- from Wolfram MathWorld. [2014-08-10]. （原始内容存档于2020-11-12）.