- 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 (read: has a picture in their head) 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 $y^2 + z^2 = 1$.

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 (Namely they are ODEs valid on a characteristic curve parameterized by $t$). I looked this up and saw some stuff on cotangent spaces, but couldn't understand how it was related to the discussion above.

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 $(x_0,y_0,z_0)$?

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