Envelope

Have you ever seen this, a little bright curve formed when light rays reflect off the circular walls of your cup and overlap?

That curve so happens to be the curve of a Cardioid, and is the caustic envelope of a circle. But how do they know this? In this wiki, we aim to show a method of deriving such shapes, or for finding the envelope of a function.

Compactly it can be said that an envelope of a family of curves in the plane is a curve that is tangent to each member of the family at some point. The envelope can also be seen as a set of points which a family of curves, $F(x,y,k)=0$, intersects $F(x,y,k+n)=0$, as n approaches 0.

The black point is a point on the envelope. The red line is $F(x,y,k)=0$ for a particular value of $k,$ and the blue line is $F(x,y,k+n)=0$ as $n$ approaches 0.

The envelope is defined as the set of points for which

$$\frac { \partial }{ \partial k } F(x,y,k)=F(x,y,k)=0.$$

The intersection between $F(x,y,k)$ and $F(x,y,k+n)$ is given by

$$\begin{aligned} F(x,y,k)&=F(x,y,k+n)=0\\ \Rightarrow F(x,y,k)&=\frac{F(x,y,k+n)-F(x,y,k)}{n}. \end{aligned}$$

Limiting $n$ to 0 gives the definition above.

Find the area under which a ladder of length 1 occupies while sliding down a wall.

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Our first step is to define $F(x,y,k)$, which in this case models the behavior of the ladder.

The red line between the $y$-axis and the $x$-axis represents the ladder:

$$\begin{aligned} F(x,y,k) &=-\frac{\sqrt{1-k^2}}{k}x+\sqrt{1-k^2}-y\\ &=0 \\ \frac { \partial }{ \partial k } F(x,y,k) &=\frac{x-k^3}{k^2\sqrt{1-k^2}}\\ &=0. \end{aligned}$$

Solving the above equations, we get the envelope in parametric form as

$$\left(k^3,\left(1-k^2\right)\sqrt{1-k^2}\right).$$