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I had read about the envelope of the family of the curve. It is defined as a curve which is tangent to each member of the family at a single point and it is union of all such points.

To find envelope they had provided condition:

The envelope is the simultaneous solution of following

$F(x,y,t)=0$ and $\frac{\partial}{\partial t}F(x,y,t)=0$

I do not understand how the definition and above conditions are equivalent.

Please Help me.

Any Help will be appreciated.

cqfd
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Curious student
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1 Answers1

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Consider the curve $t\mapsto (x(t),y(t))$ and let's see what are the necessary conditions it must fullfil to be an envelope.

I had read about the envelope of the family of the curve. It is defined as a curve which is tangent to each member of the family at a single point and it is union of all such points.

  1. First of all, the curve $t\mapsto (x(t),y(t))$ must have a point of contact with the one-parameter $t$ family of curves implicitly defined by $F(x,y,t)=0$. So you must have $F(x(t),y(t),t)=0$ which is your first equation.

  2. Then the curve $t\mapsto (x(t),y(t))$ must be tangent to the curves implicitly defined by $F(x,y,t)=0$.

    On one side, the vector $\left(\begin{array}{c} x'(t^\star) \\ y'(t^\star) \end{array}\right)$ is the tangente direction at $t^\star$ of the curve $t\mapsto (x(t),y(t))$.

    On the other side, for a fixed $t=t^\star$ the vector $\nabla F=\left(\begin{array}{c} \partial_x F \\ \partial_y F \end{array}\right)$ is a vector orthogonal to the tangent of the level-set of $F(x,y,t^\star)=0 $ (if this last point is not clear, you can read this)

    Like we want to have an identical tangent direction at point $(x(t^\star),y(t^\star))$ we must impose a zero scalar product $$ \left(\begin{array}{c} x'(t^\star) \\ y'(t^\star) \end{array}\right).\nabla F = 0 = \partial_x F(x(t^\star),y(t^\star),t^\star) x(t^\star)+\partial_y F(x(t^\star),y(t^\star),t^\star) y(t^\star)\ \ \ \ \ (1) $$ However, from point 1/, we also have $F(x(t),y(t),t)=0$, thus $\frac{d}{dt}F(x(t),y(t),t)=0$ which is $$ \partial_x F(x(t^\star),y(t^\star),t^\star) x(t^\star)+\partial_y F(x(t^\star),y(t^\star),t^\star)y(t^\star)+\partial_tF(x(t^\star),y(t^\star),t^\star) = 0\ \ \ \ \ (2) $$ Equations (1) & (2) implies that $\partial_tF(x(t^\star),y(t^\star),t^\star) = 0$

To recap, a necessary condition for $(x(t),y(t))$ being your envelope is that:

  • $F(x(t),y(t),t)=0$

  • $\partial_t F(x(t),y(t),t)=0$

which is the expected result.

Picaud Vincent
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