Prove that S={(x,y,z)∈R^3 \ x^2 +y^2 =z^2 } is a surface and find out if possible the tangent plan in O(0,0,0).


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Question Number 74301 by ~blr237~ last updated on 21/Nov/19
$Prove that S={(x,y,z)∈R^3 \ x^2 +y^2 =z^2 } is a surface and find out if possible the tangent plan in O(0,0,0).$
$P r o v e t h a t S = {(x, y, z) \in R^{3} ∖ x^{2} + y^{2} = z^{2}} i s a s u r f a c e$ $a n d f i n d o u t i f p o s s i b l e t h e t a n g e n t p l a n i n O (0, 0, 0) .$
	Answered by mind is power last updated on 21/Nov/19
	$S−{(0,0,0)} is a surface let fR^3 →R f(x,y,z)=x^2 +y^2 −z^2 ⇒S=f^− (0) df=(2x,2y,−2z) ≠(0,0,0 ∀(x,y,z)∈R^3 −{(0,0,0)} ⇒f is Emersion in any point IR^3 −{(0,0,0)} ⇒S−{(0,0,0)} is smothe manifold of order 2 topoligacl definition of surfaces in(0,0,0) gradf=0^→$
	$S - {(0, 0, 0)} i s a s u r f a c e$ $l e t f R^{3} \to R$ $f (x, y, z) = x^{2} + y^{2} - z^{2}$ $\Rightarrow S = f^{-} (0)$ $d f = (2 x, 2 y, - 2 z) \neq (0, 0, 0 \forall (x, y, z) \in R^{3} - {(0, 0, 0)}$ $\Rightarrow f i s E m e r s i o n i n a n y p o i n t {I R}^{3} - {(0, 0, 0)}$ $\Rightarrow S - {(0, 0, 0)} i s s m o t h e m a n i f o l d o f o r d e r 2$ $t o p o l i g a c l d e f i n i t i o n o f s u r f a c e s$ $i n (0, 0, 0) g r a d f = \vec{0}$
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