Editing Envelope (mathematics) (section)

=== 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.