Developable Surface

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Developable surface

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.

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 K is zero
or K=0
or \frac{LN-M^2}{EG-F^2}=0
or LN-M^2=0

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
H \vec{N}=-vκ\vec{b} (A)
Taking magnitude we get
H=vκ
And substituting H in (A), we get
\vec{N}=\frac{-vκ \vec{b}}{H}
Now, fundamental coefficients are
L=\vec{R}_{11}.\vec{N}={κ\vec{n}+vκ(τ\vec{b}-κ\vec{t})+vκ'\vec{n}}.\frac{-vκ \vec{b}}{H} =\frac{-v^2 κ^2 τ}{H}
M=\vec{R}_{12}\vec{N}=κ\vec{n}.\frac{-vκ \vec{b}}{H}=0
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.
i.e.K=\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×\vec{r}_2 ).(\vec{N}_1×\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 we proceed similar in case 1.

Therefore the surface is developable.

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