# surface is regular if curvature is non zero

• Mar 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

• Mar 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

Is it really alpha' (and not alpha'') in the equation? And could you please precise what U is? Thank you.
• Mar 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 $\displaystyle chart (x,U)$where

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

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

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

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

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

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

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

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

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