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Given a family of curves $F(x, y, c) = 0$ for $c$ in a range of real numbers, the envelope $E$ is the curve tangent to every member of the family.

I see that it is defined by the solution of $F(x, y, c) = 0 = \dfrac{\partial F}{\partial c} (x, y, c)$.

How do I prove this ? I know that if $\dfrac{\partial F}{\partial c} (x, y, c) \neq 0$, then we can solve $F(x, y, c) = 0$ at least locally for $c = c(x, y)$ but then I cannot see how $\dfrac{\partial F}{\partial c} (x, y, c) = 0$.

I saw a similar question with an answer that involved some differential geometry and "straightening of fields" that I do not understand. Can I have an answer based on simple analysis that explains

a) How does the solution of the two equations define $E$

b) What is the problem if both $\dfrac{\partial F}{\partial x} = \dfrac{\partial F}{\partial y} = 0$

c) What happens if $F(x, y, c) = f(x, y) + \phi(c)$ ?

me10240
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1 Answers1

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First of all, you want the individual curves $F_c(x,y)=F(x,y,c)=0$ for fixed $c$ to be smooth, so you'd better assume that $\dfrac{\partial F}{\partial x}$ and $\dfrac{\partial F}{\partial y}$ are never simultaneously zero. Then you want the implicit function theorem to guarantee you that your two equations will define a smooth curve (locally) parametrized by $c$, so you want the matrix $$\begin{bmatrix} \frac{\partial F}{\partial x} & \frac{\partial F}{\partial y} \\ \frac{\partial^2 F}{\partial x\partial c} & \frac{\partial^2 F}{\partial y\partial c}\end{bmatrix}$$ to be nonsingular at some point $(x_0,y_0,c_0)$.

Then you have to check that the curve your equations define locally parametrically as $(x,y)=\phi(c)$ has the property that it is tangent to the curve $F_c(x,y)=0$ at $\phi(c)$. I'll let you check this yourself by differentiating $F(\phi(c),c) = 0$.

In answer to your third question, you'll get parallel level curves and there will be no envelope. You might want to try doing some examples. Here's one for you: $F(x,y,c)=y+c^2x-c$.

Ted Shifrin
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  • If it were a family of surfaces instead, say $F_c(x, y, z)$ then the envelope would also be a surface. How would the implicit function theorem work and what would be the condition required ? – me10240 Apr 28 '16 at 14:43
  • You similarly will need to know that at each point the matrix $$\begin{bmatrix} \frac{\partial F}{\partial x} & \frac{\partial F}{\partial y} & \frac{\partial F}{\partial z} \ \frac{\partial^2 F}{\partial x\partial c} & \frac{\partial^2 F}{\partial y\partial c}& \frac{\partial^2 F}{\partial z\partial c}\end{bmatrix}$$ has rank $2$. – Ted Shifrin Apr 28 '16 at 16:24
  • I see. Then would it imply that locally, WLOG, $x = x(z,c)$, $y=y(z,c), z= z$. But from there, I am unable to see how the tangency would be proved in this case? – me10240 Apr 28 '16 at 16:37
  • Do the same calculation as in the curve case. Show that the gradient of $F_c$ is orthogonal to the tangent plane of the parametric surface $\Phi(z,c) = \big(x(z,c),y(z,c),z\big)$. – Ted Shifrin Apr 28 '16 at 16:38
  • The normal to $\Phi$ is $y_{,c} i - x_{,c} j +(x_{,c} y_{,z} - y_{,c} x_{,z})k$ where $y_{,c} := \frac{\partial y}{\partial c}$. But the normal to $F_c$ isnt parallel based on this calculation. Where am I going wrong ? – me10240 Apr 28 '16 at 19:44
  • Do the dot product of $\nabla F_c$ with the partial derivatives of $\Phi$. – Ted Shifrin Apr 28 '16 at 20:32
  • Am still unable to get it for the 3d case. Could you post an explicit answer for this part ? – me10240 Apr 28 '16 at 22:17
  • Do you mean c) in your original question? There is no envelope in that case. – Ted Shifrin Apr 28 '16 at 23:03
  • No, I meant the calculation of $\nabla F_c$ with the tangent. – me10240 Apr 28 '16 at 23:07
  • I got this far: Every tangent in the tangent plane of $\Phi$ is a linear combination of $(x_{,c}, y_{,c}, 0)$ and $(x_{,z}, y_{,z}, 1)$. Thus I need to show dot product with both is zero ? – me10240 Apr 28 '16 at 23:10
  • Yes, perfect. Now remember that, as above, $F_c(x(z,c),y(z,c),z) = 0$, so you get two equations by differentiating with respect to $z$ and to $c$. – Ted Shifrin Apr 29 '16 at 01:13