surface is regular if curvature is non zero

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March 9th 2009, 03:16 PM
dopi

surface is regular if curvature is non zero

alpha(u) is a unit speed space curve, the chart (x,U) where

x(u,v) = alpha(u) + v*(alpha'(u)), (u,v) in U

i want to show that this surface is regular if the curvature k(u): = || alpha''(u)|| of the unit speed curve alpha(u) is non zero

does anyone know how i would go about this?
March 11th 2009, 06:31 AM
Laurent

Quote:

Originally Posted by dopi

alpha(u) is a unit speed space curve, the chart (x,U) where

x(u,v) = alpha(u) + v*(alpha'(u)), (u,v) in U

i want to show that this surface is regular if the curvature k(u): = || alpha''(u)|| of the unit speed curve alpha(u) is non zero

does anyone know how i would go about this?

Is it really alpha' (and not alpha'') in the equation? And could you please precise what U is? Thank you.
March 11th 2009, 03:59 PM
dopi

Quote:

Originally Posted by Laurent

Is it really alpha' (and not alpha'') in the equation? And could you please precise what U is? Thank you.

alpha(u) is a unit speed space curve, and use this curve to construct a tangent developable surface with $chart (x,U)$ where

$x(u,v) = alpha(u) + v*(alpha'(u)), (u,v) in U$

where $U = {(u,v) in R^2 : -infinity < u < infinity, v>0}$

i want to show that this surface is regular if the curvature k(u): = || alpha''(u)|| of the unit speed curve alpha(u) is non zero

so it is really alpha' in the equation. And i have re-written the question to specify what U is. Thank you.

the solution i got was
$Xu (cross-product)Xv = v*alpha''(u) (cross-product)t,$ where i used $t$ as $t = alpha(u)'$ from the frenet serret equations.

i replaced $alpha''(u)$ with the $k$ curvature, wasnt too sure how else to do it

and so as $v>o, k$ must be non zero for the chart to be regular.
March 12th 2009, 01:26 AM
Laurent

Quote:

Originally Posted by dopi

the solution i got was
$Xu (cross-product)Xv = v*alpha''(u) (cross-product)t,$ where i used $t$ as $t = alpha(u)'$ from the frenet serret equations.

i replaced $alpha''(u)$ with the $k$ curvature, wasnt too sure how else to do it

and so as $v>o, k$ must be non zero for the chart to be regular.

In the Serret-Frenet system, there are three vectors, $(T,N,B)$ . We have $T=\alpha'(u)$ , and $\alpha''(u)=\kappa N$ . Therefore, $X_u\times X_v=(T+v\kappa N)\times T=v\kappa N\times T=v\kappa B$ and $\|X_u\times X_v\|=v\kappa\neq 0$ .