# Developable Surface #### 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

1. 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.
2. Case2
If then we proceed similar in case 1.

Therefore the surface is developable.