Mixing
Prove the injectivity and differentiability of $x(u,v) = (u cos(v), u sin(v), frac{u^2}{2}).$
$begingroup$
Prove the injectivity and differentiability of
$$x(u,v) = (u cos(v), u sin(v), frac{u^2}{2})$$
where $u in (0,infty),$ $v in (0,2pi).$
What I really want to prove is that (U,x) is a simple surface.
where U=(0,inf)x(0,2$pi$)
What I have:
For the injectivity Let (u1,v1),(u2,v2) $epsilon$ U
Got to prove x(u1,v1)=x(u2,v2) $Rightarrow$ (u1,v1)=(u2,v2)
$frac{u1^2}{2}$ = $frac{u2^2}{2}$ $Rightarrow$ u1=$pm$u2
u1*cosv1 = u2*cosv2 $Rightarrow$ ?
u1*senv1 = u2*senv2 $Rightarrow$ ?
And I don't have any clue about how to prove differentiability
differential-geometry surfaces
edited Jan 10 at 1:15
zhw.
72.7k43175
asked Jan 9 at 23:31
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
277
$endgroup$
add a comment |
$begingroup$
Prove the injectivity and differentiability of
$$x(u,v) = (u cos(v), u sin(v), frac{u^2}{2})$$
where $u in (0,infty),$ $v in (0,2pi).$
What I really want to prove is that (U,x) is a simple surface.
where U=(0,inf)x(0,2$pi$)
What I have:
For the injectivity Let (u1,v1),(u2,v2) $epsilon$ U
Got to prove x(u1,v1)=x(u2,v2) $Rightarrow$ (u1,v1)=(u2,v2)
$frac{u1^2}{2}$ = $frac{u2^2}{2}$ $Rightarrow$ u1=$pm$u2
u1*cosv1 = u2*cosv2 $Rightarrow$ ?
u1*senv1 = u2*senv2 $Rightarrow$ ?
And I don't have any clue about how to prove differentiability
differential-geometry surfaces
edited Jan 10 at 1:15
zhw.
72.7k43175
asked Jan 9 at 23:31
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
277
$endgroup$
2
$begingroup$
Please show us your work and show us specifically where you got stuck.
$endgroup$
– Ted Shifrin
Jan 9 at 23:33
$begingroup$
Haven't got too much
$endgroup$
– Alberto Torrejon Valenzuela
Jan 9 at 23:43
$begingroup$
click on edit to see a few of the tricks in MathJax.
$endgroup$
– zhw.
Jan 10 at 1:17
$begingroup$
Observe that $u_1, u_2>0$. Thus $frac{u1^2}{2}=frac{u2^2}{2}$ implies $u1=u2$.
$endgroup$
– Jens Schwaiger
Jan 11 at 3:58
add a comment |
$begingroup$
Prove the injectivity and differentiability of
$$x(u,v) = (u cos(v), u sin(v), frac{u^2}{2})$$
where $u in (0,infty),$ $v in (0,2pi).$
What I really want to prove is that (U,x) is a simple surface.
where U=(0,inf)x(0,2$pi$)
What I have:
For the injectivity Let (u1,v1),(u2,v2) $epsilon$ U
Got to prove x(u1,v1)=x(u2,v2) $Rightarrow$ (u1,v1)=(u2,v2)
$frac{u1^2}{2}$ = $frac{u2^2}{2}$ $Rightarrow$ u1=$pm$u2
u1*cosv1 = u2*cosv2 $Rightarrow$ ?
u1*senv1 = u2*senv2 $Rightarrow$ ?
And I don't have any clue about how to prove differentiability
differential-geometry surfaces
edited Jan 10 at 1:15
zhw.
72.7k43175
asked Jan 9 at 23:31
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
277
$endgroup$
Prove the injectivity and differentiability of
$$x(u,v) = (u cos(v), u sin(v), frac{u^2}{2})$$
where $u in (0,infty),$ $v in (0,2pi).$
What I really want to prove is that (U,x) is a simple surface.
where U=(0,inf)x(0,2$pi$)
What I have:
For the injectivity Let (u1,v1),(u2,v2) $epsilon$ U
Got to prove x(u1,v1)=x(u2,v2) $Rightarrow$ (u1,v1)=(u2,v2)
$frac{u1^2}{2}$ = $frac{u2^2}{2}$ $Rightarrow$ u1=$pm$u2
u1*cosv1 = u2*cosv2 $Rightarrow$ ?
u1*senv1 = u2*senv2 $Rightarrow$ ?
And I don't have any clue about how to prove differentiability
differential-geometry surfaces
differential-geometry surfaces
edited Jan 10 at 1:15
zhw.
72.7k43175
asked Jan 9 at 23:31
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
277
edited Jan 10 at 1:15
zhw.
72.7k43175
asked Jan 9 at 23:31
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
277
edited Jan 10 at 1:15
zhw.
72.7k43175
edited Jan 10 at 1:15
zhw.
72.7k43175
edited Jan 10 at 1:15
zhw.
72.7k43175
72.7k43175
asked Jan 9 at 23:31
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
277
asked Jan 9 at 23:31
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
277
asked Jan 9 at 23:31
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
277
277
2
$begingroup$
Please show us your work and show us specifically where you got stuck.
$endgroup$
– Ted Shifrin
Jan 9 at 23:33
$begingroup$
Haven't got too much
$endgroup$
– Alberto Torrejon Valenzuela
Jan 9 at 23:43
$begingroup$
click on edit to see a few of the tricks in MathJax.
$endgroup$
– zhw.
Jan 10 at 1:17
$begingroup$
Observe that $u_1, u_2>0$. Thus $frac{u1^2}{2}=frac{u2^2}{2}$ implies $u1=u2$.
$endgroup$
– Jens Schwaiger
Jan 11 at 3:58
add a comment |
2
$begingroup$
Please show us your work and show us specifically where you got stuck.
$endgroup$
– Ted Shifrin
Jan 9 at 23:33
$begingroup$
Haven't got too much
$endgroup$
– Alberto Torrejon Valenzuela
Jan 9 at 23:43
$begingroup$
click on edit to see a few of the tricks in MathJax.
$endgroup$
– zhw.
Jan 10 at 1:17
$begingroup$
Observe that $u_1, u_2>0$. Thus $frac{u1^2}{2}=frac{u2^2}{2}$ implies $u1=u2$.
$endgroup$
– Jens Schwaiger
Jan 11 at 3:58
2
2
$begingroup$
Please show us your work and show us specifically where you got stuck.
$endgroup$
– Ted Shifrin
Jan 9 at 23:33
$begingroup$
Please show us your work and show us specifically where you got stuck.
$endgroup$
– Ted Shifrin
Jan 9 at 23:33
$begingroup$
Haven't got too much
$endgroup$
– Alberto Torrejon Valenzuela
Jan 9 at 23:43
$begingroup$
Haven't got too much
$endgroup$
– Alberto Torrejon Valenzuela
Jan 9 at 23:43
$begingroup$
click on edit to see a few of the tricks in MathJax.
$endgroup$
– zhw.
Jan 10 at 1:17
$begingroup$
click on edit to see a few of the tricks in MathJax.
$endgroup$
– zhw.
Jan 10 at 1:17
$begingroup$
Observe that $u_1, u_2>0$. Thus $frac{u1^2}{2}=frac{u2^2}{2}$ implies $u1=u2$.
$endgroup$
– Jens Schwaiger
Jan 11 at 3:58
$begingroup$
Observe that $u_1, u_2>0$. Thus $frac{u1^2}{2}=frac{u2^2}{2}$ implies $u1=u2$.
$endgroup$
– Jens Schwaiger
Jan 11 at 3:58
add a comment |
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2
$begingroup$
Please show us your work and show us specifically where you got stuck.
$endgroup$
– Ted Shifrin
Jan 9 at 23:33
$begingroup$
Haven't got too much
$endgroup$
– Alberto Torrejon Valenzuela
Jan 9 at 23:43
$begingroup$
click on edit to see a few of the tricks in MathJax.
$endgroup$
– zhw.
Jan 10 at 1:17
$begingroup$
Observe that $u_1, u_2>0$. Thus $frac{u1^2}{2}=frac{u2^2}{2}$ implies $u1=u2$.
$endgroup$
– Jens Schwaiger
Jan 11 at 3:58