Editing Envelope (mathematics) (section)

=== Example 4 ===
[[File:Envelope_astroid.svg|thumb|An [[astroid]] as the envelope of the family of lines connecting points (''s'',0), (0,''t'') with ''s''<sup>2</sup>&nbsp;+&nbsp;''t''<sup>2</sup>&nbsp;=&nbsp;1]]
The following example shows that in some cases the envelope of a family of curves may be seen as the topologic boundary of a union of sets, whose boundaries are the curves of the envelope. For <math>s>0</math> and <math>t>0</math> consider the (open) right triangle in a Cartesian plane with vertices <math>(0,0)</math>, <math>(s,0)</math> and <math>(0,t)</math>
:<math>T_{s,t}:=\left\{(x,y)\in\R_+^2:\ \frac{x}{s}+\frac{y}{t}<1\right\}.
</math>
Fix an exponent <math>\alpha>0</math>, and consider the union of all the triangles <math>T_{s,t} </math> subjected to the constraint <math>\textstyle s^\alpha+t^\alpha=1 </math>, that is the open set
:<math>\Delta_\alpha:=\bigcup_ {s^\alpha+t^\alpha=1}  T_{s,t}.</math>
To write a Cartesian representation for <math>\textstyle\Delta_\alpha</math>, start with any <math>\textstyle s>0</math>, <math>\textstyle t>0</math> satisfying <math>\textstyle s^\alpha+t^\alpha=1 </math> and any <math>\textstyle(x,y)\in\R_+^2</math>. The [[Hölder inequality#Notable special cases|Hölder inequality]] in <math>\textstyle\R^2</math> with respect to the conjugated exponents <math>p:=1+\frac{1}{\alpha}</math> and <math>\textstyle q:={1+\alpha}</math> gives:
:<math>x^\frac{\alpha}{\alpha+1}+y^\frac{\alpha}{\alpha+1}\leq \left(\frac{x}{s}+\frac{y}{t}\right)^\frac{\alpha}{\alpha+1}\Big(s^\alpha+t^\alpha\Big)^\frac{1}{\alpha+1}=\left(\frac{x}{s}+\frac{y}{t}\right)^\frac{\alpha}{\alpha+1}</math>,
with equality if and only if <math>\textstyle s:\,t=x^\frac{1}{1+\alpha}:\,y^\frac{1}{1+\alpha}</math>.
In terms of a union of sets the latter inequality reads: the point <math>(x,y)\in\R_+^2</math> belongs to the set <math>\textstyle\Delta_\alpha</math>, that is, it belongs to some <math>\textstyle T_{s,t}</math> with <math>\textstyle s^\alpha+t^\alpha=1</math>, if and only if it satisfies
:<math>x^\frac{\alpha}{\alpha+1}+y^\frac{\alpha}{\alpha+1}<1.</math>
Moreover, the boundary in <math>\R_+^2</math> of the set <math>\textstyle \Delta_\alpha</math> is the envelope of the corresponding family of line segments
:<math>\left\{(x,y)\in\R_+^2:\ \frac{x}{s}+\frac{y}{t}=1\right\}\ ,\qquad s^\alpha+t^\alpha=1</math>
(that is, the hypotenuses of the triangles), and has Cartesian equation
:<math>x^\frac{\alpha}{\alpha+1}+y^\frac{\alpha}{\alpha+1}=1.</math>
Notice that, in particular, the value <math>\alpha=1</math> gives the arc of parabola of the [[#Example_2|Example 2]], and the value <math>\alpha=2</math> (meaning that all hypotenuses are unit length segments) gives the [[astroid]].