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.
73.6k43175
asked Jan 9 at 23:31
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
327
$endgroup$
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.
73.6k43175
asked Jan 9 at 23:31
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
327
$endgroup$
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.
73.6k43175
asked Jan 9 at 23:31
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
327
$endgroup$
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.
73.6k43175
asked Jan 9 at 23:31
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
327
edited Jan 10 at 1:15
zhw.
73.6k43175
asked Jan 9 at 23:31
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
327
edited Jan 10 at 1:15
zhw.
73.6k43175
edited Jan 10 at 1:15
zhw.
73.6k43175
edited Jan 10 at 1:15
zhw.
73.6k43175
73.6k43175
asked Jan 9 at 23:31
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
327
asked Jan 9 at 23:31
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
327
asked Jan 9 at 23:31
Alberto Torrejon ValenzuelaAlberto Torrejon Valenzuela
327
327
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 |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068064%2fprove-the-injectivity-and-differentiability-of-xu-v-u-cosv-u-sinv%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068064%2fprove-the-injectivity-and-differentiability-of-xu-v-u-cosv-u-sinv%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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