Question

Find the envelope of the family of spheres with centre (l,m,o) and radius one . Also obtain the equation of the characteristic curve of the family when l=m .

Answer 1

Answer on Question #40389 – Math - Differential Calculus

We have $c = (\ell, m, o)$ and the radius $r = 1$. The equation is

$$(x - \ell)^2 + (y - m)^2 + (z - o)^2 - 1^2 = 0$$

Let $\ell$ be a parameter. Then we have

$$F(x, y, z, \ell) = 0.$$

The envelope of the family of spheres is determined from the equations

$$F(x, y, z, \ell) = 0, \quad \frac{\partial F}{\partial \ell}(x, y, z, \ell) = 0.$$

Since

$$\frac{\partial F}{\partial \ell} = -2(x - \ell) = 0,$$

we find $\ell = x$.

The envelope of the family of spheres is given by

$$F[x, y, z, \ell(x)] = (y - m)^2 + (z - o)^2 - 1 = 0.$$

This is the equation of a circular cylinder of radius 1 whose axis coincides with the $x$-axis.

Consider another case. Let $m$ be a parameter. Similarly

$$\frac{\partial F}{\partial m} = -2(y - m) = 0,$$

then

$$m = y.$$

The envelope of the family of spheres is given by

$$F[x, y, z, m(y)] = (x - \ell)^2 + (z - o)^2 - 1 = 0.$$

This is the equation of a circular cylinder of radius 1 whose axis coincides with the $y$-axis.

And the third case. Let $o$ be a parameter. Similarly

$$\frac{\partial F}{\partial o} = -2(z - o) = 0,$$

then

$$o = z.$$

The envelope of the family of spheres is given by

$$F[x, y, z, o(z)] = (x - \ell)^2 + (y - m)^2 - 1 = 0.$$

This is the equation of a circular cylinder of radius 1 whose axis coincides with the $z$-axis.

So we have three equations, that give us the envelope of the family of spheres.

**Answer:**

$$(y - m)^2 + (z - o)^2 - 1 = 0,$$ $$(x - \ell)^2 + (z - o)^2 - 1 = 0,$$ $$(x - \ell)^2 + (y - m)^2 - 1 = 0.$$