Alexandre Magueresse
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Envelope of a family of curves

Definition (Envelope) The envelope of a family of planar curves is a curve that is tangent, at each of its points, to a member of the family.

Suppose we are given a family of curves described by

Ct={(x,y)∈R2,ft(x,y)=0}C_{t} = \{(x, y) \in \mathbb{ R}^{2}, f_{t}(x, y) = 0\}

for all t∈Rt \in \mathbb{ R}. For convenience, consider the function F:(t,x,y)↦ft(x,y)F: (t, x, y) \mapsto f_{t}(x, y) and assume FF is differentiable. A point (x,y)∈R2(x, y) \in \mathbb{ R}^{2} belongs to the envelope of the family (Ct)(C_{t}), if and only if there exists t∈Rt \in \mathbb{ R} such that

F(t,x,y)=∂tF(t,x,y)=0F(t, x, y) = \partial_{t} F(t, x, y) = 0

simultaneously.

An important and easy special case is when the family of curves are lines. Suppose ft(x,y)=a(t)x+b(t)y−c(t)f_{t}(x, y) = a(t) x + b(t) y - c(t) for some differentiable functions a,b,c:R→Ra, b, c: \mathbb{ R} \to \mathbb{ R}. A point (x,y)∈R2(x, y) \in \mathbb{ R}^{2} belongs to the envelope if there exists t∈Rt \in \mathbb{ R} such that

a(t)x+b(t)y=c(t),a′(t)x+b′(t)y=c′(t).a(t) x + b(t) y = c(t), \qquad a'(t) x + b'(t) y = c'(t).

This is a linear system in (x,y)(x, y), which we can solve if it is not singular, that is, if a(t)b′(t)≠a′(t)b(t)a(t) b'(t) \neq a'(t) b(t). In that case, we find

(x,y)=(b′c−bc′ab′−a′b,a′c−ac′a′b−ab′),(x, y) = \left(\frac{b' c - b c'}{a b' - a' b}, \frac{a' c - a c'}{a' b - a b'}\right),

where we omitted the parameter tt for cleaner notations.

Example Let k≥1k \geq 1 and consider connecting the lines connecting the complex points eite^{i t} and ekite^{k i t}, for all t∈[0,2π)t \in \left[ 0, 2\pi\right). The equation of these lines correspond to the coefficients a(t)=sin⁡(kt)−sin⁡(t)a(t) = \sin(k t) - \sin(t), b(t)=cos⁡(t)−cos⁡(kt)b(t) = \cos(t) - \cos(k t) and c(t)=sin⁡((k−1)t)c(t) = \sin((k - 1) t). Noticing that aa and bb can be factored by sin⁡([(k−1)/2]t)\sin([(k - 1) / 2] t), we are left with new coefficients A(t)=cos⁡([(k+1)/2]t)A(t) = \cos([(k+1)/2] t), B(t)=sin⁡([(k+1)/2]t)B(t) = \sin([(k+1)/2] t) and C(t)=cos⁡([(k−1)/2]t)C(t) = \cos([(k-1)/2] t) after simplifying. Applying the parameterisation of the envelope established above, we obtain

x(t)=cos⁡(k+12t)cos⁡(k−12t)+k−1k+1sin⁡(k+12t)sin⁡(k−12t),y(t)=sin⁡(k+12t)cos⁡(k−12t)−k−1k+1cos⁡(k+12t)sin⁡(k−12t),\begin{align*} x(t) &= \cos\left(\frac{k+1}{2}t\right) \cos\left(\frac{k-1}{2} t\right) + \frac{k-1}{k+1} \sin\left(\frac{k+1}{2} t\right) \sin\left(\frac{k-1}{2}t\right), \\ y(t) &= \sin\left(\frac{k+1}{2}t\right) \cos\left(\frac{k-1}{2} t\right) - \frac{k-1}{k+1} \cos\left(\frac{k+1}{2} t\right) \sin\left(\frac{k-1}{2}t\right), \end{align*}

which simplifies into

x(t)=1k+1[kcos⁡(t)+cos⁡(kt)],y(t)=1k+1[ksin⁡(t)+sin⁡(kt)].\begin{align*} x(t) &= \frac{1}{k+1} [k \cos(t) + \cos(kt)], \\ y(t) &= \frac{1}{k+1} [k \sin(t) + \sin(kt)]. \end{align*}

This is the equation of an epicycloid; it can also be obtained by rolling (without slipping) a small circle of radius 1/(k+1)1/(k+1) around a circle of radius 1−2/(k+1)1 - 2/(k+1).

CC BY-SA 4.0 Alexandre Magueresse. Last modified: March 04, 2026.
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