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DIFFERENTIAL GEOMETRY COURSE

20. *ENVELOPES*

*Explanation 172 :* We saw that a developable surface determines ∞^{1} tangent planes, namely one tangent plane for each rule.

Inversely, let a system of ∞^{1} planes be given: α(t) is the plane with parameter value t. Suppose these planes don't go through one straight line, and ain't all parallel either.

Of course, the normal vector of α(t) is infinitely often differentiable. The same holds for a supporting vector.

We will see these planes α(t) are the tangent planes of a developable surface (they "envelop" this surface).

In special cases, this surface is a cylinder or a cone, but in general it's a surface of tangents. See also 177, 178, 179.

*Definition 173 :* Let α(t) be the plane __x__ **.** __n__(t) = c(t) (__n__ and c sufficiently often differentiable).

The characteristic line with parameter value t is the limit *l*(t) of the intersection line of α(t) and α(t') if t'→ t.

The characteristic point on *l*(t) is the limit of the intersection point of *l*(t) and α(t') if t'→ t.

*Proposition 174 :* The characteristic line is determined by __x__ **.** __n__(t) = c(t), __x__ **.** __n ' __(t) = c ' (t).

The characteristic point is given by __x__ **.** __n__(t) = c(t), __x__ **.** __n ' __(t) = c ' (t),
__x__ **.** __n__ " (t) = c " (t).

*Problem 175 :* Prove proposition 174 by using the Taylor expansion of d(t) = __x__ **.** __n__(t) - c(t).

*Proposition 176 :* The characteristic lines of a system of planes α(t) form a developable surface, of which the planes α(t) are tangent planes. In general, this is a
surface of tangents whose turning curve is formed by the characteristic points.

*Prove:* Let __x__(t) be the characteristic point with parameter value t, and let *l*(t) be the characteristic line.

Differentiation of __x__ **.** __n__ = c gives __x__ ' **.** __n__ + __x__ **.** __n__ ' = c ' , so also
__x__ ' **.** __n__ = 0.

Likewise, differentiation of __x__ **.** __n__ ' = c ' gives __x__ ' **.** __n__ ' = 0. So __x__ ' is a direction vector of the
charakteristic line.

So the characteristic line is tangent in the characteristic point to the curve formed by the characteristic points.

By differentiation of __x__ ' **.** __n__ = 0 we get __x__ " **.** __n__ + __x__ ' **.** __n__ ' = 0, so
__x__ " **.** __n__ = 0.

So the plane α(t) contains the point __x__(t) and has __x__ ' (t) and __x__ " (t) as direction vectors. So it is the osculating plane. According to proposition 142,
this osculating plane is also tangent plane.

*Explanation 177 :* We silently supposed det( __n__, __n__ ' , __n__ " ) ≠ 0 and __x__ ' ≠ __0__. The following
problems 178 and 179 complete the proof.

*Problem 178 :* If __n__ ' and __n__ have the same or opposite directions, then there is fixed vector with the same or opposite directions as __n__ (hint:
show that (__n__/||__n__||) ' = __0__).

If __n__ " = λ__n__ + μ__n__ ' , then __a__ := __n__ ⊗ __n__ ' has the same or opposite directions as __a__ ' ,
so there is fixed vector with the same or opposite directions as __a__. Then the characteristic lines are parallel and form a cylinder.

*Problem 179 :* If __x__ ' = __0__, then the characteristic lines form a cone.

*Problem 180 :* Determine the turning curve of the surface of tangents enveloped by

a) x_{1} sin (t) - x_{2} cos(t) + px_{3} - qt = 0

b) 3t^{2} x_{1} - 3t x_{2} + x_{3} - t^{3} = 0.

In the next problem you have to apply a theory that's analogous to the one we dealt with in this section.

*Problem 181 :* Determine the curve enveloped by the system of straight lines in ℜ^{2} given by x_{1} cos (t) + x_{2} sin(t) = t.

Do you recognize this curve? Make a sketch.

answers

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