1. I have a question about envelopes of surfaces. In a book I am reading the following:

    Suppose $S_a$ is a one parameter family of surfaces in $R^3$ given by $z=w(x,y;a)$ where $w$ depends smoothly on $x,y$ and the real parameter $a$. Consider also the equation $\partial_a w(x,y;a)=0$. For a fixed values of $a$, these two equations determine a curve $\gamma_a$. The envelope $E$ of the family of surfaces $S_a$ is just the union of these curves $\gamma_a$. The equation for $E$ is found simply by solving $\partial_a w(x,y,a)=0$ for $a$ as a function of $x$ and $y$, $a=f(x,y)$, and then substituting into $z=w(x,y,f(x,y))$. Moreover, along $\gamma_a$, $a$ is constant and we have $$dz = w_xdx + w_ydy \\0 = w_{ax}dx + w_{ay}dy$$ For instance, if $S_a$ is a one-parameter family of 2-spheres: $(x-a)^2+y^2+z^2 = 1$, then the envelope is a cylinder of radius 1.

Can anyone provide an "intuitive" geometric reason for why taking the derivative with respect to the parameter, setting it equal to zero, and plugging it back into $F$ gives the envelope? I see that it works in the example of the sphere, I obtain a cylinder $y^2 + z^2 = 1$.

  1. In the procedure descirbed above they use the notation $$dz = w_xdx + w_ydy \\0 = w_{ax}dx + w_{ay}dy$$ Is this a formal expression? When I read it as $$\frac{dz}{dt} = w_x\frac{dx}{dt} + w_y\frac{dy}{dt} \\0 = w_{ax}\frac{dx}{dt} + w_{ay}\frac{dy}{dt}$$ It makes sense to me. I looked this up and saw some junk on cotangent spaces, but couldn't understand how it was related to the discussion above.

  2. They introduce a notion of Monge Cone in the following way:

    Consider the 1st order PDE: $F(x,y,z,p,q)=0$. At any point $(x_0,y_0,z_0)$, $F$ establishes a functional relation between $p$ and $q$. Assuming $F_q(x_0,y_0,z_0,p,q)\neq 0$, implicit function theorem gives us: $F(x_0,y_0,z_0,p,q(p))=0$ for all $p$. The possible tangent planes to the graph $z=u(x,y)$ are given by: $$(z-z_0) = p(x-x_0)+q(p)(y-y_0)$$ which, as $p$ varies, describe a one-parameter family of planes through the point $(x_0,y_0,z_0)$.

Using the equations in #2, they solve for the envelope of planes at $(x_0,y_0,z_0)$ (parameter is $p$) and find: $$dz = pdx + qdy \\ 0 = dx + \frac{dq}{dp}dy$$

How do I see this is a cone at $(x_0,y_0,z_0)$?

  • $\begingroup$ Which book? Which page? $\endgroup$
    – Qmechanic
    Commented Oct 25, 2022 at 13:21


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