Question Number 74301 by ~blr237~ last updated on 21/Nov/19
$${Prove}\:{that}\:\:{S}=\left\{\left({x},{y},{z}\right)\in\mathbb{R}^{\mathrm{3}} \backslash\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={z}^{\mathrm{2}} \:\right\}\:{is}\:{a}\:{surface}\: \\ $$$${and}\:{find}\:{out}\:{if}\:{possible}\:{the}\:{tangent}\:{plan}\:{in}\:{O}\left(\mathrm{0},\mathrm{0},\mathrm{0}\right). \\ $$
Answered by mind is power last updated on 21/Nov/19
$${S}−\left\{\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\right\}\:{is}\:{a}\:{surface}\: \\ $$$${let}\:{f}\mathbb{R}^{\mathrm{3}} \rightarrow\mathbb{R} \\ $$$${f}\left({x},{y},{z}\right)={x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{z}^{\mathrm{2}} \\ $$$$\Rightarrow{S}={f}^{−} \left(\mathrm{0}\right) \\ $$$${df}=\left(\mathrm{2}{x},\mathrm{2}{y},−\mathrm{2}{z}\right)\:\:\neq\left(\mathrm{0},\mathrm{0},\mathrm{0}\:\forall\left({x},{y},{z}\right)\in\mathbb{R}^{\mathrm{3}} −\left\{\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\right\}\right. \\ $$$$\Rightarrow{f}\:{is}\:\:{Emersion}\:{in}\:{any}\:{point}\:{IR}^{\mathrm{3}} −\left\{\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\right\} \\ $$$$\Rightarrow{S}−\left\{\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\right\}\:{is}\:{smothe}\:{manifold}\:{of}\:\:{order}\:\mathrm{2}\: \\ $$$${topoligacl}\:{definition}\:{of}\:{surfaces} \\ $$$${in}\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\:{gradf}=\overset{\rightarrow} {\mathrm{0}} \\ $$$$ \\ $$