Prove $φ(R^2)$ where $φ(u,v)=(vcos u,vsin u,bu)$ is the smooth function.

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite












Prove $φ(R^2)$ is a smooth surface and $φ(u,v)=(vcos u,vsin u,bu)$ $R^2rightarrow R^3$ and b>0 constant. Its rank $Dφ$ is 2, so I'm ok with that part. Only thing to prove is that it has an continuous inverse so $φ$ can be an acceptable parametrization. Can i just argue the topology definition . I know the existence of the inverse hence if the pre image of the inverse of open sets of$ R^2$ are open in $R^3$ then the inverse is continuous.But still that needs proof






calculus differential-geometry surfaces







share|cite|improve this question























  • It makes little sense to say "where $varphi$ is a smooth function" and then to define it explicitly.
    – amsmath
    Sep 3 at 14:08










  • thats how the problem was presented . I know its a redundancy.
    – Manolis Lyviakis
    Sep 3 at 14:13










  • What's $b$, a non-zero constant?
    – Mees de Vries
    Sep 3 at 14:15






  • 1




    @ManolisLyviakis Well, first it is a redundancy, but it's also not quite correct language-wise, because the enlish article "a" means "some". So, if you write "where $varphi$ is a smooth function", it can be any smooth function. Then defining it explicitly is actually a contradiction in itself. It is correct to write "where $varphi$ is the smooth function $varphi(u,v) = (vcos u,vsin u, bu)$".
    – amsmath
    Sep 3 at 14:20











  • @amsmath tahnk you for the correction.
    – Manolis Lyviakis
    Sep 3 at 14:21














up vote
0
down vote

favorite












Prove $φ(R^2)$ is a smooth surface and $φ(u,v)=(vcos u,vsin u,bu)$ $R^2rightarrow R^3$ and b>0 constant. Its rank $Dφ$ is 2, so I'm ok with that part. Only thing to prove is that it has an continuous inverse so $φ$ can be an acceptable parametrization. Can i just argue the topology definition . I know the existence of the inverse hence if the pre image of the inverse of open sets of$ R^2$ are open in $R^3$ then the inverse is continuous.But still that needs proof






calculus differential-geometry surfaces







share|cite|improve this question























  • It makes little sense to say "where $varphi$ is a smooth function" and then to define it explicitly.
    – amsmath
    Sep 3 at 14:08










  • thats how the problem was presented . I know its a redundancy.
    – Manolis Lyviakis
    Sep 3 at 14:13










  • What's $b$, a non-zero constant?
    – Mees de Vries
    Sep 3 at 14:15






  • 1




    @ManolisLyviakis Well, first it is a redundancy, but it's also not quite correct language-wise, because the enlish article "a" means "some". So, if you write "where $varphi$ is a smooth function", it can be any smooth function. Then defining it explicitly is actually a contradiction in itself. It is correct to write "where $varphi$ is the smooth function $varphi(u,v) = (vcos u,vsin u, bu)$".
    – amsmath
    Sep 3 at 14:20











  • @amsmath tahnk you for the correction.
    – Manolis Lyviakis
    Sep 3 at 14:21












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Prove $φ(R^2)$ is a smooth surface and $φ(u,v)=(vcos u,vsin u,bu)$ $R^2rightarrow R^3$ and b>0 constant. Its rank $Dφ$ is 2, so I'm ok with that part. Only thing to prove is that it has an continuous inverse so $φ$ can be an acceptable parametrization. Can i just argue the topology definition . I know the existence of the inverse hence if the pre image of the inverse of open sets of$ R^2$ are open in $R^3$ then the inverse is continuous.But still that needs proof






calculus differential-geometry surfaces







share|cite|improve this question















Prove $φ(R^2)$ is a smooth surface and $φ(u,v)=(vcos u,vsin u,bu)$ $R^2rightarrow R^3$ and b>0 constant. Its rank $Dφ$ is 2, so I'm ok with that part. Only thing to prove is that it has an continuous inverse so $φ$ can be an acceptable parametrization. Can i just argue the topology definition . I know the existence of the inverse hence if the pre image of the inverse of open sets of$ R^2$ are open in $R^3$ then the inverse is continuous.But still that needs proof






calculus differential-geometry surfaces




calculus differential-geometry surfaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 3 at 14:49

























asked Sep 3 at 14:05









Manolis Lyviakis

1,307625




1,307625











  • It makes little sense to say "where $varphi$ is a smooth function" and then to define it explicitly.
    – amsmath
    Sep 3 at 14:08










  • thats how the problem was presented . I know its a redundancy.
    – Manolis Lyviakis
    Sep 3 at 14:13










  • What's $b$, a non-zero constant?
    – Mees de Vries
    Sep 3 at 14:15






  • 1




    @ManolisLyviakis Well, first it is a redundancy, but it's also not quite correct language-wise, because the enlish article "a" means "some". So, if you write "where $varphi$ is a smooth function", it can be any smooth function. Then defining it explicitly is actually a contradiction in itself. It is correct to write "where $varphi$ is the smooth function $varphi(u,v) = (vcos u,vsin u, bu)$".
    – amsmath
    Sep 3 at 14:20











  • @amsmath tahnk you for the correction.
    – Manolis Lyviakis
    Sep 3 at 14:21
















  • It makes little sense to say "where $varphi$ is a smooth function" and then to define it explicitly.
    – amsmath
    Sep 3 at 14:08










  • thats how the problem was presented . I know its a redundancy.
    – Manolis Lyviakis
    Sep 3 at 14:13










  • What's $b$, a non-zero constant?
    – Mees de Vries
    Sep 3 at 14:15






  • 1




    @ManolisLyviakis Well, first it is a redundancy, but it's also not quite correct language-wise, because the enlish article "a" means "some". So, if you write "where $varphi$ is a smooth function", it can be any smooth function. Then defining it explicitly is actually a contradiction in itself. It is correct to write "where $varphi$ is the smooth function $varphi(u,v) = (vcos u,vsin u, bu)$".
    – amsmath
    Sep 3 at 14:20











  • @amsmath tahnk you for the correction.
    – Manolis Lyviakis
    Sep 3 at 14:21















It makes little sense to say "where $varphi$ is a smooth function" and then to define it explicitly.
– amsmath
Sep 3 at 14:08




It makes little sense to say "where $varphi$ is a smooth function" and then to define it explicitly.
– amsmath
Sep 3 at 14:08












thats how the problem was presented . I know its a redundancy.
– Manolis Lyviakis
Sep 3 at 14:13




thats how the problem was presented . I know its a redundancy.
– Manolis Lyviakis
Sep 3 at 14:13












What's $b$, a non-zero constant?
– Mees de Vries
Sep 3 at 14:15




What's $b$, a non-zero constant?
– Mees de Vries
Sep 3 at 14:15




1




1




@ManolisLyviakis Well, first it is a redundancy, but it's also not quite correct language-wise, because the enlish article "a" means "some". So, if you write "where $varphi$ is a smooth function", it can be any smooth function. Then defining it explicitly is actually a contradiction in itself. It is correct to write "where $varphi$ is the smooth function $varphi(u,v) = (vcos u,vsin u, bu)$".
– amsmath
Sep 3 at 14:20





@ManolisLyviakis Well, first it is a redundancy, but it's also not quite correct language-wise, because the enlish article "a" means "some". So, if you write "where $varphi$ is a smooth function", it can be any smooth function. Then defining it explicitly is actually a contradiction in itself. It is correct to write "where $varphi$ is the smooth function $varphi(u,v) = (vcos u,vsin u, bu)$".
– amsmath
Sep 3 at 14:20













@amsmath tahnk you for the correction.
– Manolis Lyviakis
Sep 3 at 14:21




@amsmath tahnk you for the correction.
– Manolis Lyviakis
Sep 3 at 14:21










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










Let $M := varphi(Bbb R^2)$. Then $varphi : Bbb R^2to M$ is obviously onto and you proved that it is also injective. Furthermore, the rank of the Jacobian is $2$ in each point (which you also proved). Now, the inverse of $varphi$ is given by
$$
psi(x,y,z) = left(frac z b,,,xcosfrac z b+ysinfrac z bright),quad (x,y,z)in M.
$$
This is easily verifyable by computing $psi(varphi(u,v))$. So, $varphi^-1$ is obviously continuous.






share|cite|improve this answer




















  • how did ou come up with it? any sort tricks?
    – Manolis Lyviakis
    Sep 3 at 14:57






  • 1




    When you write $varphi(u,v) = (x,y,z)$, then $u = z/b$ and $vbinomcos z/bsin z/b = binom x y$. Taking the scalar product with $binomcos z/bsin z/b$ gives $v = langlebinom x y,binomcos z/bsin z/brangle = xcos z/b + ysin z/b$.
    – amsmath
    Sep 3 at 15:03










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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2903896%2fprove-%25cf%2586r2-where-%25cf%2586u-v-v-cos-u-v-sin-u-bu-is-the-smooth-function%23new-answer', 'question_page');

);

Post as a guest

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










Let $M := varphi(Bbb R^2)$. Then $varphi : Bbb R^2to M$ is obviously onto and you proved that it is also injective. Furthermore, the rank of the Jacobian is $2$ in each point (which you also proved). Now, the inverse of $varphi$ is given by
$$
psi(x,y,z) = left(frac z b,,,xcosfrac z b+ysinfrac z bright),quad (x,y,z)in M.
$$
This is easily verifyable by computing $psi(varphi(u,v))$. So, $varphi^-1$ is obviously continuous.






share|cite|improve this answer




















  • how did ou come up with it? any sort tricks?
    – Manolis Lyviakis
    Sep 3 at 14:57






  • 1




    When you write $varphi(u,v) = (x,y,z)$, then $u = z/b$ and $vbinomcos z/bsin z/b = binom x y$. Taking the scalar product with $binomcos z/bsin z/b$ gives $v = langlebinom x y,binomcos z/bsin z/brangle = xcos z/b + ysin z/b$.
    – amsmath
    Sep 3 at 15:03














up vote
2
down vote



accepted










Let $M := varphi(Bbb R^2)$. Then $varphi : Bbb R^2to M$ is obviously onto and you proved that it is also injective. Furthermore, the rank of the Jacobian is $2$ in each point (which you also proved). Now, the inverse of $varphi$ is given by
$$
psi(x,y,z) = left(frac z b,,,xcosfrac z b+ysinfrac z bright),quad (x,y,z)in M.
$$
This is easily verifyable by computing $psi(varphi(u,v))$. So, $varphi^-1$ is obviously continuous.






share|cite|improve this answer




















  • how did ou come up with it? any sort tricks?
    – Manolis Lyviakis
    Sep 3 at 14:57






  • 1




    When you write $varphi(u,v) = (x,y,z)$, then $u = z/b$ and $vbinomcos z/bsin z/b = binom x y$. Taking the scalar product with $binomcos z/bsin z/b$ gives $v = langlebinom x y,binomcos z/bsin z/brangle = xcos z/b + ysin z/b$.
    – amsmath
    Sep 3 at 15:03












up vote
2
down vote



accepted







up vote
2
down vote



accepted






Let $M := varphi(Bbb R^2)$. Then $varphi : Bbb R^2to M$ is obviously onto and you proved that it is also injective. Furthermore, the rank of the Jacobian is $2$ in each point (which you also proved). Now, the inverse of $varphi$ is given by
$$
psi(x,y,z) = left(frac z b,,,xcosfrac z b+ysinfrac z bright),quad (x,y,z)in M.
$$
This is easily verifyable by computing $psi(varphi(u,v))$. So, $varphi^-1$ is obviously continuous.






share|cite|improve this answer












Let $M := varphi(Bbb R^2)$. Then $varphi : Bbb R^2to M$ is obviously onto and you proved that it is also injective. Furthermore, the rank of the Jacobian is $2$ in each point (which you also proved). Now, the inverse of $varphi$ is given by
$$
psi(x,y,z) = left(frac z b,,,xcosfrac z b+ysinfrac z bright),quad (x,y,z)in M.
$$
This is easily verifyable by computing $psi(varphi(u,v))$. So, $varphi^-1$ is obviously continuous.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 3 at 14:56









amsmath

2,645114




2,645114











  • how did ou come up with it? any sort tricks?
    – Manolis Lyviakis
    Sep 3 at 14:57






  • 1




    When you write $varphi(u,v) = (x,y,z)$, then $u = z/b$ and $vbinomcos z/bsin z/b = binom x y$. Taking the scalar product with $binomcos z/bsin z/b$ gives $v = langlebinom x y,binomcos z/bsin z/brangle = xcos z/b + ysin z/b$.
    – amsmath
    Sep 3 at 15:03
















  • how did ou come up with it? any sort tricks?
    – Manolis Lyviakis
    Sep 3 at 14:57






  • 1




    When you write $varphi(u,v) = (x,y,z)$, then $u = z/b$ and $vbinomcos z/bsin z/b = binom x y$. Taking the scalar product with $binomcos z/bsin z/b$ gives $v = langlebinom x y,binomcos z/bsin z/brangle = xcos z/b + ysin z/b$.
    – amsmath
    Sep 3 at 15:03















how did ou come up with it? any sort tricks?
– Manolis Lyviakis
Sep 3 at 14:57




how did ou come up with it? any sort tricks?
– Manolis Lyviakis
Sep 3 at 14:57




1




1




When you write $varphi(u,v) = (x,y,z)$, then $u = z/b$ and $vbinomcos z/bsin z/b = binom x y$. Taking the scalar product with $binomcos z/bsin z/b$ gives $v = langlebinom x y,binomcos z/bsin z/brangle = xcos z/b + ysin z/b$.
– amsmath
Sep 3 at 15:03




When you write $varphi(u,v) = (x,y,z)$, then $u = z/b$ and $vbinomcos z/bsin z/b = binom x y$. Taking the scalar product with $binomcos z/bsin z/b$ gives $v = langlebinom x y,binomcos z/bsin z/brangle = xcos z/b + ysin z/b$.
– amsmath
Sep 3 at 15:03

















 

draft saved


draft discarded















































 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2903896%2fprove-%25cf%2586r2-where-%25cf%2586u-v-v-cos-u-v-sin-u-bu-is-the-smooth-function%23new-answer', 'question_page');

);

Post as a guest
































































-

這個網誌中的熱門文章

How to combine Bézier curves to a surface?

圖片
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite My aim is to smooth the terrain in a video game. Therefore I contrived an algorithm that makes use of Bézier curves of different orders. But this algorithm is defined in a two dimensional space for now. To shift it into the third dimension I need to somehow combine the Bézier curves. Given two curves in each two dimensions, how can I combine them to build a surface? I thought about something like a curve of a curve. Or maybe I could multiply them somehow. How can I combine them to cause the wanted behavior? Is there a common approach? In the image you can see the input, on the left, and the output, on the right, of the algorithm that works in a two dimensional space. For the real application there is a three dimensional input. The algorithm relays only on the adjacent tiles. Given these it will define the control points of the mentioned Bézier cur...
閱讀完整內容

Mutual Information Always Non-negative

圖片
Clash Royale CLAN TAG #URR8PPP up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 '12 at 15:38 Cameron Buie 83.6k 7 71 154 asked Jun 17 '12 at 15:30 Omri 359 3 13 3 Convexity of the function $tmapsto tlog t$. – Did Jun 17 '12 at 15:33 In addition, the convexity properties require the coefficients in the linear combination sum 1. Then, as p(x,y) is a probability distribution, it fullfits such condition. – Francisco Javier Delgado Ceped Aug 10 at 18:44 add a comment  |  up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 ...
閱讀完整內容

Why am i infinitely getting the same tweet with the Twitter Search API?

圖片
up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia...
閱讀完整內容