# Thread: Points on the surface

1. ## Points on the surface

Points on a surface can be: elliptic, hyperbolic, parabolic, (flat or planar)
I have the surface
$\vec{r}(u,v)=( sh(u) , ch(u)*cos(v) ,ch(u) *sin(v) )$
$sh(u)=sinh(u), ch(u)=cosh(u)$
$\vec{r'_{u}}(u,v)=( ch(u) , sh(u)*cos(v) ,sh(u) *sin(v) )$
$\vec{r'_{v}}(u,v)=( 0 , -ch(u)*sin(v) ,ch(u) *cos(v) )$
$\vec{r''_{u^2}}(u,v)=( sh(u) , ch(u)*cos(v) ,ch(u) *sin(v) )$
$\vec{r''_{u*v}}(u,v)=( 0 , -sh(u)*sin(v) ,sh(u) *cos(v) )$
$\vec{r''_{v^2}}(u,v)=( 0 , -ch(u)*cos(v) ,-ch(u)*sin(v) )$
$E=ch^2(u)+sh^2(u)*cos^2(v)+sh^2(u)*sin^2(v)=ch^2(u )+sh^2(u);$
$F=0;G=ch^2(u);H=ch(u)*{(ch^2(u)+sh^2(u))}^{(1/2)}$
$D=(\frac{1}{H})*(-ch(u)) ; D'=0 ; D''=(\frac{1}{H})*{(ch^3(u))};$
$Nature: D*D''-D'^2=(\frac{(-ch^4(u))}{(H^2)})=(\frac{-ch^2(u)}{(ch^2(u)+sh^2(u))})<0\forall{u}$
So all points are hyperbolic?
Here is a graph : Imgur: The most awesome images on the Internet

2. ## Re: Points on the surface

All points are hyperbolic? Yes or No. An answer is greatly appreciated.