Editing Envelope (mathematics) (section)

===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>