# Why do we need the partial derivative $\frac {\partial F}{\partial t}$ in the definition of an envelope?

Suppose we have a family $F(x,y,z,t)=0$ (where $t$ is our parameter).

My reasoning: If we would now like to create the envelope $E$ of the family $F_t$, we would need to find out what part of $F(x,y,z,t)$ contributes to $E$ for eacht value of $t$.

My algebra: The way I would do this is to consider the intersection $$F(x,y,z,t)=F(x,y,z,u).$$ Now I would just let $u\to t$ and I would expect to find exactly the part of $F(x,y,z,t)=0$, which contributes to $E$.

This does not appear to work for several examples I’ve tried. Looking at Wikipedia, it would appear that I need to find a term $\dfrac{F(x,y,z,u)-F(x,y,z,t)} {u-t}$ at some point.

I don’t see where this term could come from (in particular its denominator). I believe that my reasoning is correct, but I can’t figure out why my algebra doesn’t seem to be.

#### Solutions Collecting From Web of "Why do we need the partial derivative $\frac {\partial F}{\partial t}$ in the definition of an envelope?"

I’ll write $E_t$ for the part of $F(x,y,z,t)$ which contributes to $E$.

The assumption that the solution set of $F(x,y,z,t)=F(x,y,z,u)$ describes the part $E_t$ when $u\to t$ is wrong. We do have that $E_t$ is contained in this solution set, but there is no equality. So one could say that $F(x,y,z,t)=F(x,y,z,u)$ is a minimal condition that we want to satisfy. To examin its behaviour when $u\to t$, we exploit that
$$F(x,y,z,t)=F(x,y,z,u)\iff \frac{F(x,y,z,u)-F(x,y,z,t)}{u-t}=0,$$
as long as $t\neq u$. Fortunately, we explicitely avoid $t=u$ when we take $\lim\limits_{u\to t}$.

This means that we may say $(x,y,z)\in E_t\implies \dfrac{\partial F(x,y,z,t)}{\partial t}=0$. Furthermore we must still have that $F(x,y,z,t)=0$ and thus come to find that $E_t$ is described by the solution set of $$\begin{cases}F(x,y,z,t)=0\\ \\\dfrac{\partial F(x,y,z,t)}{\partial t}=0\end{cases}$$