包絡線

設曲線族的每條曲線${\displaystyle C_{s}}$為${\displaystyle t\mapsto (x(s,t),y(s,t))}$。

設存在包絡線。由於包絡線的每點都與曲線族的其中一條曲線的其中一點相切，對於任意的${\displaystyle s}$，設${\displaystyle (x(s,h(s)),y(s,h(s)))}$表示${\displaystyle C_{s}}$和包絡線相切的那點。由此式可見，${\displaystyle s}$是包絡線的變數。要求出包絡線，就即要求出${\displaystyle h(s)}$。

在${\displaystyle C_{s}}$的切向量為${\displaystyle <{\frac {\partial x}{\partial t}},{\frac {\partial y}{\partial t}}>}$，其中${\displaystyle t=h(s)}$。

在E的切向量為${\displaystyle <{\frac {dx}{ds}},{\frac {dy}{ds}}>}$。因為${\displaystyle x}$是${\displaystyle s}$和${\displaystyle t}$的函數，而此處${\displaystyle t=h(s)}$，局部求導有：

${\displaystyle {\frac {dx}{ds}}={\frac {\partial x}{\partial h}}{\frac {dh}{ds}}+{\frac {\partial x}{\partial s}}{\frac {ds}{ds}}={\frac {\partial x}{\partial h}}h'(s)+{\frac {\partial x}{\partial s}}}$

類似地得 ${\displaystyle {\frac {dy}{ds}}={\frac {\partial y}{\partial h}}h'(s)+{\frac {\partial y}{\partial s}}}$。

因為${\displaystyle E}$和${\displaystyle C_{s}}$在該點相切，因此其切向量應平行，故有

${\displaystyle {\frac {\partial x}{\partial t}}=\lambda ({\frac {\partial x}{\partial h}}h'(s)+{\frac {\partial x}{\partial s}})}$

${\displaystyle {\frac {\partial y}{\partial t}}=\lambda ({\frac {\partial y}{\partial h}}h'(s)+{\frac {\partial y}{\partial s}})}$

其中${\displaystyle \lambda \neq 0}$。可用此兩式消去${\displaystyle h'(s)}$。整理後得： ${\displaystyle {\frac {\partial y}{\partial h}}{\frac {\partial x}{\partial s}}={\frac {\partial y}{\partial s}}{\frac {\partial x}{\partial h}}}$

• https://web.archive.org/web/20070621035057/http://www.math.neu.edu/~bridger/Envelope/envelope.htm

• 克萊羅方程

• Special Plane Curves: Envelope
• Envelopes of Lines and Circles