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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 : http://imgur.com/GeRlOEf

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    Yes, this surface has negative Gaussian curvature at each point (so each point is hyperbolic).2017-01-03

0 Answers 0