In differential geometry, a developable surface is a special type of surface that can be flattened onto a plane without stretching, tearing, or distorting. This means that a developable surface can be unrolled or unfolded into a flat sheet, like a piece of paper, without any changes in the distances between points on the surface.It is defined as below.

Let \( S:\vec{r}=\vec{r}(u,v)\) be a surface. Now, S is called developable surface if Gaussian curvature K is zero. In such surface,

\(LN-M^2=0\) at all points.

#### Example

Show that \( \vec{r}=(a \cos u,a \sin u,v)\) is developable surfaceSolution

The equation of the surface is

\( \vec{r}=(a \cos u,a \sin u,v)\) (i)

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

\( E=a^2,F=0,G=1,H=a,L=-a,M=0,N=0 \)

Here

\(LN-M^2=0 \)

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 \( S:\vec{r}=\vec{r}(u,v)\) be a surface.

Necessary condition

Assume that S is developable, then its equation is

\( R=\vec{r}(s)+v\vec{t}(s) \) (i)

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

\( R_{1}=\vec{t}+v \kappa \vec{n} \)

\( R_{2}=\vec{t}\)

\( R_{1} \times R_{2} =(\vec{t}+v \kappa \vec{n}) \times (\vec{t})\)

\( H \vec{N}=-v \kappa \vec{b}\) (A)

Taking magnitude we get

\(H=v\kappa\)

And substituting H in (A), we get

\(\vec{N}=\frac{-v\kappa \vec{b}}{H}\)

Again

\( R_{11}=\kappa \vec{n}+v \kappa (\tau \vec{b}-\kappa \vec{t}) \)

\( R_{12}=\kappa \vec{n}\)

\( R_{22}=0\)

Now, fundamental coefficients are

\(M=\vec{R}_{12}\vec{N}\)

or\(M=\kappa \vec{n}.\frac{-v\kappa \vec{b}}{H}\)

or\(M=0 \)

Again

\(N=\vec{R}_{22}.\vec{N}=0 \)

Now, Gaussian curvature of the surface is

\( K=\frac{LN-M^2}{EG-F^2 }=0 \)

Sufficient condition

Let Gaussian curvature of S is zero.

\( K=0\)

\( \frac{LN-M^2}{EG-F^2 }=0 \)

or \( LN-M^2=0 \)

or \( (\vec{r}_1.\vec{N}_1 )(\vec{r}_2.\vec{N}_2 )-(\vec{r}_1.\vec{N}_2 )(\vec{r}_2.\vec{N}_1 )=0 \)

or \( (\vec{r}_1 \times \vec{r}_2 ).(\vec{N}_1 \times \vec{N}_2 )=0 \)

This implies either \( \vec{N}_1\) or \( \vec{N}_2\) is zero

- Case1

If \( \vec{N}_1=0\) then, equation of tangent plane to the surface S is

\( (R-\vec{r} ).\vec{N}=0 \) (ii)

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

\( \vec{r}_1.\vec{N}+(R-\vec{r} ).\vec{N}_1=0 \)

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 \( \vec{N}_2=0\) then equation of tangent plane to the surface S is

\( (R-\vec{r} ).\vec{N}=0 \) (iii)

Differentiating (iii) w. r. to. v, we get

\( \vec{r}_2.\vec{N}+(R-\vec{r} ).\vec{N}_2=0 \)

It shows that, tangent plane to the surface is independent from parameter \(v\), thus surface is envelope of single parameter family of planes, and therefore the surface is developable.

#### Theorem

Necessary and sufficient condition for a surface to be developable is \(LN-M^2=0 \)Proof

Let \( S:\vec{r}=\vec{r}(u,v)\) be a surface.

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

Gaussian curvature is zero

or\(K=0\)

or\( \frac{LN-M^2}{EG-F^2}=0 \)

or\(LN-M^2=0 \)

This completes the proof.

#### Exercise

- Show following surface are developable.

(a) \(\vec{r}=(\cos u, \sin u ,v)\)

(b) \(\vec{r}=(v \cos u, v \sin u ,v)\)

(c) \(\vec{r}=(u,v,u+v)\)

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