#### Developable surface

Let be a surface. Now, S is called developable surface if Gaussian curvature K is zero. In such surface,

at all points.

#### Theorem

Necessary and sufficient condition for a surface to be developable is

Proof

Let be a surface.

Now, necessary and sufficient condition for S to be developable is

Gaussian curvature K is zero

or
K=0

or

or

#### Example

Show that is developable surface.

Solution

The equation of the surface is

(i)

By diff. of (i) w. r. to. u, and v respectively, we get

Here

Thus, the surface is developable.

#### Theorem

The necessary and sufficient condition for a surface to be developable is that its Gaussian curvature is zero.

Proof

Let be a surface.

Necessary condition

Assume that S is developable, then its equation is

(i)

Differentiating (i) w. r. to. s and v respectively, we get

(A)

Taking magnitude we get

H=vκ

And substituting H in (A), we get

Now, fundamental coefficients are

Now, Gaussian curvature of the surface is

Sufficient condition

Let Gaussian curvature of S is zero.

i.e.

or

or

or

This implies either or is zero

- Case1

If then, equation of tangent plane to the surface S is

(ii)

Differentiating (ii) w. r. to. u, we get

It shows that, tangent plane to the surface is independent from parameter u, thus surface is envelope of single parameter family of planes, and therefore the surface is developable. - Case2

If then we proceed similar in case 1.

Therefore the surface is developable.