text: ### 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 " = λn + μn ' , then a := nn ' 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) x1 sin (t) - x2 cos(t) + px3 - qt = 0
  b) 3t2 x1 - 3t x2 + x3 - t3 = 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 ℜ2 given by x1 cos (t) + x2 sin(t) = t.
  Do you recognize this curve? Make a sketch.
explanation: The document is a section from a differential geometry course, specifically discussing envelopes and developable surfaces. It includes several propositions, definitions, and problems. The text contains multiple questions (problems) that require proofs or solutions, but no answers are provided.
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distilabel_metadata.content: ### 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 " = λn + μn ' , then a := nn ' 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) x1 sin (t) - x2 cos(t) + px3 - qt = 0
  b) 3t2 x1 - 3t x2 + x3 - t3 = 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 ℜ2 given by x1 cos (t) + x2 sin(t) = t.
  Do you recognize this curve? Make a sketch.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a section from a differential geometry course, specifically discussing envelopes and developable surfaces. It includes several propositions, definitions, and problems. The text contains multiple questions (problems) that require proofs or solutions, but no answers are provided.
  
  Label: multiple-question