Currently, I am reading Differential Geometry of Curves and Surfaces by Manfredo P. Do Carmo. While solving exercises problems, I came across this question and I have tried for a while and came to conclusion that the only problem for this set to be regular surface is it boundary, but could not prove rigorously. Have a look at the question below.
Let two points $p(t)$ and $q(t)$ move with the same speed, $p$ starting from $(0,0,0)$ and moving along the $z$-axis whereas $q$ starting at $(a,0,0)$, $a$ nonzero real number and moving parallel to $y$-axis. The line through $p(t)$ and $q(t)$ describes a set in $\mathbb R^3$. And I am not clear whether the set is a regular surface or not. I think that the boundary of the set is the only problem but cannot prove rigorously. It is clear that for $t>0$ and $0<s<1$, the parametrization $X(t,s)=(sa,st,t(1-s))$ satisfies all the conditions for it to be a chart of a regular surface. So, I have concluded that $s=0,t>=0$ and $s=1,t>=0$ are the only problem. Does anybody have any suggestion. Any hint would be appreciated.
EDIT 1:
Not sure if this would be useful for working on it further.
Some velocity ( here 4) or any other link function $f(v,t)$ may need to be given extra as an $x$ coordinate to the given speeds say 2, in order to span/connect two skew lines along $(y,z) $ directions to define a flat surface which in the first case makes up triangular patches through origin.