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Consider a family of curves parameterised by $t$, where each member of the family is described by $$F(t,x,y) = 0.$$ Define an envelope of the family to be a curve $E$ such that every point on $E$ is tangent to some member of the family. When do we have that $$\frac{\partial F}{\partial t}(t,x,y) = 0$$ for all $(x,y) \in E$, and in those cases, why?

Mr. Chip
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1 Answers1

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Yes, the envelope $E$ is precisely the solution set of the system \begin{align*} F(t,x,y) &= 0 \\ \frac{\partial F}{\partial t}(t,x,y) &= 0\,, \end{align*} if we have a sufficient condition given by the Implicit Function Theorem. That is, in order for these equations to be guaranteed to define $(x,y)$ locally as a $C^1$ function of $t$, we need the $2\times 2$ matrix $$\begin{bmatrix} \frac{\partial F}{\partial x} & \frac{\partial F}{\partial y} \\ \frac{\partial^2 F}{\partial x\partial t} & \frac{\partial^2 F}{\partial y\partial t} \end{bmatrix}$$ to be nonsingular.

Here's a hint to explain where the condition $\frac{\partial F}{\partial t}=0$ comes from. At the point $(x(t),y(t))\in E$ the tangent vector to $E$ should be tangent to the curve $$F(x,y,t)=0\tag{$\star$}$$ (with $t$ fixed), hence orthogonal to the gradient vector $\big(\frac{\partial F}{\partial x}(x(t),y(t)),\frac{\partial F}{\partial y}(x(t),y(t))\big)$. Now differentiate equation $(\star)$ with $t$ varying and $x=x(t)$, $y=y(t)$.

Ted Shifrin
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  • I'm not sure I follow. Do you mean the total derivative $\frac{dF}{dt}$? Then we'd get something of the form $F_x dx/dt + F_y dy/dt + F_t = 0$. I suppose... hmm. Do the first terms add to zero because of the dot product? – Mr. Chip May 25 '13 at 23:26
  • Yep, that makes sense, because the tangent to E at (x(t),y(t)) has gradient (x'(t),y'(t)). This is nice. I tried a different approach; what we've done here involves thinking of the mapping as $t \mapsto (x(t),y(t))$, whereas I approached it as $(x,y) \mapsto t(x,y)$, and then used the smoothness assumptions to write $t(x)$. I wonder if there's a reason to prefer either method? – Mr. Chip May 25 '13 at 23:32
  • No, you want a point on each curve, and there is one curve for each $t$. – Ted Shifrin May 25 '13 at 23:51
  • I meant a mapping $E \to T, (x,y) \mapsto t(x,y)$ where $T$ is an indexing set for the family. If there's more than one choice for a given $(x,y)$, choose it arbitrarily. Is that still flawed? – Mr. Chip May 26 '13 at 00:07
  • Actually I can see why it would be... you'd need an extra assumption that $t$ itself is smooth, on top of everything else. – Mr. Chip May 26 '13 at 00:08
  • I guess you're thinking of the inverse function of mine, then. But dimensions here dictate that the envelope should be a curve, and so you want a single parameter for the curve. This is the natural setup of the Implicit Function Theorem. – Ted Shifrin May 26 '13 at 00:17
  • @TedShifrin Hi! Can you please tell the reason for why you said "the tangent vector to should be tangent to the curve"? Is it the definition of envelope? - I am new to this topic. – Vicrobot Mar 26 '19 at 09:53
  • @TedShifrin And if you don't mind; can you please help me for this post related to envelopes? – Vicrobot Mar 26 '19 at 09:58
  • @Vicrobot Please read the question and answer very carefully. – Ted Shifrin Mar 26 '19 at 20:33
  • @TedShifrin At this time I don't know about that Implicit function theorem. May be you're approach in this answer is going from the idea of tangency, to limit of intersections of infinitesimally near curves(As asked by OP). I read about that in wikipedia. But I couldn't relate that how locus of those intersection points together form a curve which is tangent to all members at some point. – Vicrobot Mar 26 '19 at 20:45
  • @TedShifrin I am out of resources right now(In matter of offline books). If not a big elaboration; you can just provide me any link to book or website where I could go; which would be really helpful for me. – Vicrobot Mar 26 '19 at 20:49
  • @TedShifrin: What are the necessary and sufficient conditions for one parametric (implicit) family of surfaces embedded in $\mathbb{R}^3$ to have an envelope? – Bumblebee Sep 21 '22 at 09:14