Editing Envelope (mathematics) (section)

==Examples==

===Example 1===
These definitions ''E''<sub>1</sub>, ''E''<sub>2</sub>, and ''E''<sub>3</sub> of the envelope may be different sets. Consider for instance the curve {{nowrap|1=''y'' = ''x''<sup>3</sup>}} parametrised by {{nowrap|γ : '''R''' → '''R'''<sup>2</sup>}} where {{nowrap|1=γ(''t'') = (''t'',''t''<sup>3</sup>)}}. The one-parameter family of curves will be given by the tangent lines to γ.

First we calculate the discriminant <math>\mathcal D</math>. The generating function is
:<math> F(t,(x,y)) = 3t^2x - y - 2t^3.</math>
Calculating the partial derivative {{nowrap|1=''F''<sub>''t''</sub> = 6''t''(''x'' – ''t'')}}. It follows that either {{nowrap|1=''x'' = ''t''}} or {{nowrap|1=''t'' = 0}}. First assume that {{nowrap|1=''x'' = ''t'' and ''t'' ≠ 0}}. Substituting into F: <math>F(t,(t,y)) = t^3 - y \, </math>
and so, assuming that ''t'' ≠ 0, it follows that {{nowrap|1=''F'' = ''F''<sub>''t''</sub> = 0}} if and only if {{nowrap|1=(''x'',''y'') = (''t'',''t''<sup>3</sup>)}}. Next, assuming that {{nowrap|1=''t'' = 0}} and substituting into ''F'' gives {{nowrap|1=''F''(0,(''x'',''y'')) = &minus;''y''}}.  So, assuming {{nowrap|1=''t'' = 0}}, it follows that {{nowrap|1=''F'' = ''F''<sub>''t''</sub> = 0}} if and only if {{nowrap|1=''y'' = 0}}. Thus the discriminant is the original curve and its tangent line at γ(0):
:<math> \mathcal{D} = \{(x,y) \in \R^2 : y = x^3\} \cup \{(x,y) \in \R^2 : y = 0 \} \ . </math>

Next we calculate ''E''<sub>1</sub>. One curve is given by {{nowrap|1=''F''(''t'',(''x'',''y'')) = 0}} and a nearby curve is given by {{nowrap|''F''(''t'' + &epsilon;,(''x'',''y''))}} where ε is some very small number. The intersection point comes from looking at the limit of {{nowrap|1=''F''(''t'',(''x'',''y''))&nbsp;=&nbsp;''F''(''t'' + &epsilon;,(''x'',''y''))}} as ε tends to zero. Notice that {{nowrap|1=''F''(''t'',(''x'',''y''))&nbsp;=&nbsp;''F''(''t'' + &epsilon;,(''x'',''y''))}} if and only if
:<math> L := F(t,(x,y)) - F(t+\varepsilon,(x,y)) = 2\varepsilon^3+6\varepsilon t^2+6\varepsilon^2t-(3\varepsilon^2+6\varepsilon t)x = 0. </math>
If {{nowrap|1=''t'' ≠ 0}} then ''L'' has only a single factor of ε. Assuming that {{nowrap|1=''t'' ≠ 0}} then the intersection is given by
:<math>\lim_{\varepsilon \to 0} \frac{1}{\varepsilon} L = 6t(t-x) \ . </math>
Since {{nowrap|''t'' ≠ 0}} it follows that {{nowrap|1=''x'' = ''t''}}. The ''y'' value is calculated by knowing that this point must lie on a tangent line to the original curve γ: that {{nowrap|1=''F''(''t'',(''x'',''y'')) = 0}}. Substituting and solving gives ''y'' = ''t''<sup>3</sup>. When {{nowrap|1=''t'' = 0}}, ''L'' is divisible by ε<sup>2</sup>. Assuming that {{nowrap|1=''t'' = 0}} then the intersection is given by
:<math>\lim_{\varepsilon \to 0} \frac{1}{\varepsilon^2} L = 3x \ . </math>
It follows that {{nowrap|1=''x'' = 0}}, and knowing that {{nowrap|1=''F''(''t'',(''x'',''y'')) = 0}} gives {{nowrap|1=''y'' = 0}}. It follows that
:<math> E_1 = \{(x,y) \in \R^2 : y = x^3 \} \ . </math>

Next we calculate ''E''<sub>2</sub>. The curve itself is the curve that is tangent to all of its own tangent lines. It follows that
:<math> E_2 = \{(x,y) \in \R^2 : y = x^3 \} \ . </math>

Finally we calculate ''E''<sub>3</sub>.  Every point in the plane has at least one tangent line to γ passing through it, and so region filled by the tangent lines is the whole plane. The boundary ''E''<sub>3</sub> is therefore the empty set. Indeed, consider a point in the plane, say (''x''<sub>0</sub>,''y''<sub>0</sub>). This point lies on a tangent line if and only if there exists a ''t'' such that
:<math>F(t,(x_0,y_0)) = 3t^2x_0 - y_0 - 2t^3 = 0 \ . </math>
This is a cubic in ''t'' and as such has at least one real solution. It follows that at least one tangent line to γ must pass through any given point in the plane. If {{nowrap|''y'' > ''x''<sup>3</sup>}} and {{nowrap|''y'' > 0}} then each point (''x'',''y'') has exactly one tangent line to γ passing through it. The same is true if {{nowrap|''y'' < ''x''<sup>3</sup>}} {{nowrap|''y'' < 0}}. If {{nowrap|''y'' < ''x''<sup>3</sup>}} and {{nowrap|''y'' > 0}} then each point (''x'',''y'') has exactly three distinct tangent lines to γ passing through it. The same is true if {{nowrap|''y'' > ''x''<sup>3</sup>}} and {{nowrap|''y'' < 0}}. If {{nowrap|1=''y'' = ''x''<sup>3</sup>}} and {{nowrap|''y'' ≠ 0}} then each point (''x'',''y'') has exactly two tangent lines to γ passing through it (this corresponds to the cubic having one ordinary root and one repeated root). The same is true if {{nowrap|''y'' ≠ ''x''<sup>3</sup>}} and {{nowrap|1=''y'' = 0}}. If {{nowrap|1=''y'' = ''x''<sup>3</sup>}} and {{nowrap|1=''x'' = 0}}, i.e., {{nowrap|1=''x'' = ''y'' = 0}}, then this point has a single tangent line to γ passing through it (this corresponds to the cubic having one real root of multiplicity 3). It follows that
:<math>E_3 = \varnothing. </math>

===Example 2===
[[File:Envelope_string_art.svg|thumb|This plot gives the envelope of the family of lines connecting points (''t'',0), (0,''k'' - ''t''), in which ''k'' takes the value 1.]]
In [[string art]] it is common to cross-connect two lines of equally spaced pins. What curve is formed?

For simplicity, set the pins on the ''x''- and ''y''-axes; a non-[[orthogonal]] layout is a [[Coordinate rotation|rotation]] and [[scaling (geometry)|scaling]] away. A general straight-line thread connects the two points (0, ''k''&minus;''t'') and (''t'', 0), where ''k'' is an arbitrary scaling constant, and the family of lines is generated by varying the parameter ''t''. From simple geometry, the equation of this straight line is ''y'' = &minus;(''k''&nbsp;&minus;&nbsp;''t'')''x''/''t'' + ''k''&nbsp;&minus;&nbsp;''t''. Rearranging and casting in the form ''F''(''x'',''y'',''t'') = 0 gives:

{{NumBlk|:|<math>F(x,y,t)=-\frac{kx}{t} - t  + x + k -y = 0\,</math>|{{EquationRef|1}}}}

Now differentiate ''F''(''x'',''y'',''t'') with respect to ''t'' and set the result equal to zero, to get

{{NumBlk|:|<math>\frac{\partial F(x,y,t)}{\partial t}= \frac{kx}{t^2} - 1 = 0\,</math>|{{EquationRef|2}}}}

These two equations jointly define the equation of the envelope. From (2) we have:
: <math>t = \sqrt{kx} \,</math>
Substituting this value of ''t'' into (1) and simplifying gives an equation for the envelope:

{{NumBlk|:|<math>y=(\sqrt{x}-\sqrt{k})^2\,</math>|{{EquationRef|3}}}}

Or, rearranging into a more elegant form that shows the symmetry between x and y:

{{NumBlk|:|<math>\sqrt{x}+\sqrt{y}=\sqrt{k}</math>|{{EquationRef|4}}}}

We can take a rotation of the axes where the ''b'' axis is the line ''y=x'' oriented northeast and the ''a'' axis is the line ''y''=−''x'' oriented southeast. These new axes are related to the original ''x-y'' axes by {{math|1=''x''=(''b''+''a'')/{{sqrt|2}}}} and {{math|1=''y''=(''b''−''a'')/{{sqrt|2}}}} . We obtain, after substitution into (4) and expansion and simplification,

{{NumBlk|:|<math>b = \frac{1}{k\sqrt{2}} a^2 + \frac{k}{2\sqrt{2}},</math>|{{EquationRef|5}}}}

which is apparently the equation for a parabola with axis along ''a''=0, or ''y''=''x''.

=== Example 3 ===

Let ''I'' ⊂ '''R''' be an open interval and let γ : ''I'' → '''R'''<sup>2</sup> be a smooth plane curve parametrised by [[arc length]]. Consider the one-parameter family of normal lines to γ(''I''). A line is normal to γ at γ(''t'') if it passes through γ(''t'') and is perpendicular to the [[Differential geometry of curves#Tangent vector|tangent vector]] to γ at γ(''t''). Let '''T''' denote the unit tangent vector to γ and let '''N''' denote the unit [[Differential geometry of curves#Normal or curvature vector|normal vector]]. Using a dot to denote the [[dot product]], the generating family for the one-parameter family of normal lines is given by {{nowrap|1=''F'' : ''I'' &times; '''R'''<sup>2</sup> → '''R'''}} where
:<math> F(t,{\mathbf x}) = ({\mathbf x} - \gamma(t)) \cdot {\mathbf T}(t) \ . </math>
Clearly ('''x''' &minus; γ)·'''T''' = 0 if and only if '''x''' &minus; γ is perpendicular to '''T''', or equivalently, if and only if '''x''' &minus; γ is [[Parallel (geometry)|parallel]] to '''N''', or equivalently, if and only if '''x''' = γ + λ'''N''' for some λ ∈ '''R'''. It follows that
:<math> L_{t_0} := \{ {\mathbf x} \in \R^2 : F(t_0,{\mathbf x}) = 0 \} </math>
is exactly the normal line to γ at γ(''t''<sub>0</sub>). To find the discriminant of ''F'' we need to compute its partial derivative with respect to ''t'':
:<math>  \frac{\partial F}{\partial t}(t,{\mathbf x}) = \kappa (t) ({\mathbf x}-\gamma(t))\cdot {\mathbf N}(t) - 1 \ , </math>
where κ is the [[Curvature#Curvature of plane curves|plane curve curvature]] of γ. It has been seen that ''F'' = 0 if and only if '''x''' - γ = λ'''N''' for some λ ∈ '''R'''. Assuming that ''F'' = 0 gives
:<math> \frac{\partial F}{\partial t} = \lambda \kappa(t) - 1 \ . </math>
Assuming that κ ≠ 0 it follows that λ = 1/κ and so
:<math> \mathcal{D} = \gamma(t) + \frac{1}{\kappa(t)}{\mathbf N}(t) \ . </math>
This is exactly the [[evolute]] of the curve γ.

=== 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]].

===Example 5===
[[File:Envelope cast.svg|thumb|The orbits' envelope of the projectiles (with constant initial speed) is a concave parabola. The initial speed is 10 m/s. We take ''g'' = 10 m/s<sup>2</sup>.]]
We consider the following example of envelope in motion. Suppose at initial height 0, one casts a [[Trajectory of a projectile|projectile]] into the air with constant initial velocity ''v'' but different elevation angles θ. Let ''x'' be the horizontal axis in the motion surface, and let ''y'' denote the vertical axis. Then the motion gives the following differential [[dynamical system]]:
:<math>\frac{d^2 y}{dt^2} = -g,\; \frac{d^2 x}{dt^2} = 0, </math>
which satisfies four [[initial condition]]s:
:<math>\frac{dx}{dt}\bigg|_{t=0} = v \cos \theta,\; \frac{dy}{dt}\bigg|_{t=0} = v \sin \theta,\; x\bigg|_{t=0} = y\bigg|_{t=0} = 0.</math>
Here ''t'' denotes motion time, θ is elevation angle, ''g'' denotes [[gravitational acceleration]], and ''v'' is the constant initial speed (not [[velocity]]). The solution of the above system can take an [[implicit function|implicit form]]:
:<math>F(x,y,\theta) = x\tan \theta - \frac{gx^2}{2v^2 \cos^2 \theta} - y = 0.</math>
To find its envelope equation, one may compute the desired derivative:
:<math>\frac{\partial F}{\partial \theta} = \frac{x}{\cos^2 \theta} - \frac{gx^2 \tan \theta}{v^2 \cos^2 \theta} = 0.</math>
By eliminating θ, one may reach the following envelope equation:
:<math>y = \frac{v^2}{2g} - \frac{g}{2v^2}x^2.</math>
Clearly the resulted envelope is also a [[concave function|concave]] [[parabola]].