Developable Surface Unit 5

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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 surface
Solution
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
  1. 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.
  2. 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.
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

  1. 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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