Two questions on Do Carmo’s definition of variation of a curve

Clash Royale CLAN TAG#URR8PPP

up vote
0
down vote

favorite

On page 345 of Differential Geometry of Curves and Surfaces by Manfredo Do Carmo there is the following definition of variation of a curve:

Let $alpha colon [0, l] to S$ be a regular parametrised curve, where the parameter $s$ is the arc length. A variation of $alpha$ is a differentiable map $h colon [0, l] times (-varepsilon, varepsilon) subseteq mathbbR^2 to S$ such that $h(s, 0) = alpha(s)$ for all $s in (0, l].$

I have two questions regarding this defintion:

1. Shouldn’t $h(s, 0) = alpha(s)$ also hold for $s = 0$?


2. What does it mean for $h$ to be differentiable on $[0, l] times (-varepsilon, varepsilon),$ which is not an open set?

Any help will be appreciated.

differential-geometry calculus-of-variations

share|cite|improve this question

asked Jul 31 at 16:00

Verísimo

14

add a comment | 

up vote
0
down vote

favorite

On page 345 of Differential Geometry of Curves and Surfaces by Manfredo Do Carmo there is the following definition of variation of a curve:

Let $alpha colon [0, l] to S$ be a regular parametrised curve, where the parameter $s$ is the arc length. A variation of $alpha$ is a differentiable map $h colon [0, l] times (-varepsilon, varepsilon) subseteq mathbbR^2 to S$ such that $h(s, 0) = alpha(s)$ for all $s in (0, l].$

I have two questions regarding this defintion:

1. Shouldn’t $h(s, 0) = alpha(s)$ also hold for $s = 0$?


2. What does it mean for $h$ to be differentiable on $[0, l] times (-varepsilon, varepsilon),$ which is not an open set?

Any help will be appreciated.

differential-geometry calculus-of-variations

share|cite|improve this question

asked Jul 31 at 16:00

Verísimo

14

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

On page 345 of Differential Geometry of Curves and Surfaces by Manfredo Do Carmo there is the following definition of variation of a curve:

Let $alpha colon [0, l] to S$ be a regular parametrised curve, where the parameter $s$ is the arc length. A variation of $alpha$ is a differentiable map $h colon [0, l] times (-varepsilon, varepsilon) subseteq mathbbR^2 to S$ such that $h(s, 0) = alpha(s)$ for all $s in (0, l].$

I have two questions regarding this defintion:

1. Shouldn’t $h(s, 0) = alpha(s)$ also hold for $s = 0$?


2. What does it mean for $h$ to be differentiable on $[0, l] times (-varepsilon, varepsilon),$ which is not an open set?

Any help will be appreciated.

differential-geometry calculus-of-variations

share|cite|improve this question

asked Jul 31 at 16:00

Verísimo

14

On page 345 of Differential Geometry of Curves and Surfaces by Manfredo Do Carmo there is the following definition of variation of a curve:

Let $alpha colon [0, l] to S$ be a regular parametrised curve, where the parameter $s$ is the arc length. A variation of $alpha$ is a differentiable map $h colon [0, l] times (-varepsilon, varepsilon) subseteq mathbbR^2 to S$ such that $h(s, 0) = alpha(s)$ for all $s in (0, l].$

I have two questions regarding this defintion:

1. Shouldn’t $h(s, 0) = alpha(s)$ also hold for $s = 0$?


2. What does it mean for $h$ to be differentiable on $[0, l] times (-varepsilon, varepsilon),$ which is not an open set?

Any help will be appreciated.

differential-geometry calculus-of-variations

share|cite|improve this question

asked Jul 31 at 16:00

Verísimo

14

share|cite|improve this question

share|cite|improve this question

asked Jul 31 at 16:00

Verísimo

14

asked Jul 31 at 16:00

Verísimo

14

asked Jul 31 at 16:00

Verísimo

14

14

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
0
down vote

1. No, because the idea is that this concept is defined for every continuous curve. If we imposed that the equality holds when $s=0$, that would mean that $alpha$ would be differentiable.


2. So what? Isn't the map $fcolon[-1,1]longrightarrowmathbb R$ defined by $f(x)=x$ differentiable?

share|cite|improve this answer

answered Jul 31 at 16:05

José Carlos Santos

112k1696172

• 
  
  
  

  
  

  
  
  
  
  
  

  
  1. $alpha$ is supposed to be $mathcalC^infty,$ so requiring that $h(0, 0) = alpha(0)$ wouldn’t change anything right?
  – Verísimo
  Jul 31 at 16:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  2. I have only seen a definition of differentiability on non-open sets for one-variable functions.
  – Verísimo
  Jul 31 at 16:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  1. Where is it that Do Carmo says that $alpha$ is $C^infty$? I have a copy of the book before my eyes and I didn't sse that. 2. My guess then is that, for him, a function definid on an arbitrary set is differentiable if it is the restriction to that set of a differentiable function defined on an open set.
  – José Carlos Santos
  Jul 31 at 16:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  1. In the first paragraph of section 1-2.
  – Verísimo
  Jul 31 at 17:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  2. Okay, thanks.
  – Verísimo
  Jul 31 at 17:19
  

  
  
  
  

 | 
show 1 more comment

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

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2868209%2ftwo-questions-on-do-carmo-s-definition-of-variation-of-a-curve%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

1. No, because the idea is that this concept is defined for every continuous curve. If we imposed that the equality holds when $s=0$, that would mean that $alpha$ would be differentiable.


2. So what? Isn't the map $fcolon[-1,1]longrightarrowmathbb R$ defined by $f(x)=x$ differentiable?

share|cite|improve this answer

answered Jul 31 at 16:05

José Carlos Santos

112k1696172

• 
  
  
  

  
  

  
  
  
  
  
  

  
  1. $alpha$ is supposed to be $mathcalC^infty,$ so requiring that $h(0, 0) = alpha(0)$ wouldn’t change anything right?
  – Verísimo
  Jul 31 at 16:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  2. I have only seen a definition of differentiability on non-open sets for one-variable functions.
  – Verísimo
  Jul 31 at 16:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  1. Where is it that Do Carmo says that $alpha$ is $C^infty$? I have a copy of the book before my eyes and I didn't sse that. 2. My guess then is that, for him, a function definid on an arbitrary set is differentiable if it is the restriction to that set of a differentiable function defined on an open set.
  – José Carlos Santos
  Jul 31 at 16:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  1. In the first paragraph of section 1-2.
  – Verísimo
  Jul 31 at 17:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  2. Okay, thanks.
  – Verísimo
  Jul 31 at 17:19
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

1. No, because the idea is that this concept is defined for every continuous curve. If we imposed that the equality holds when $s=0$, that would mean that $alpha$ would be differentiable.


2. So what? Isn't the map $fcolon[-1,1]longrightarrowmathbb R$ defined by $f(x)=x$ differentiable?

share|cite|improve this answer

answered Jul 31 at 16:05

José Carlos Santos

112k1696172

• 
  
  
  

  
  

  
  
  
  
  
  

  
  1. $alpha$ is supposed to be $mathcalC^infty,$ so requiring that $h(0, 0) = alpha(0)$ wouldn’t change anything right?
  – Verísimo
  Jul 31 at 16:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  2. I have only seen a definition of differentiability on non-open sets for one-variable functions.
  – Verísimo
  Jul 31 at 16:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  1. Where is it that Do Carmo says that $alpha$ is $C^infty$? I have a copy of the book before my eyes and I didn't sse that. 2. My guess then is that, for him, a function definid on an arbitrary set is differentiable if it is the restriction to that set of a differentiable function defined on an open set.
  – José Carlos Santos
  Jul 31 at 16:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  1. In the first paragraph of section 1-2.
  – Verísimo
  Jul 31 at 17:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  2. Okay, thanks.
  – Verísimo
  Jul 31 at 17:19
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

up vote
0
down vote

1. No, because the idea is that this concept is defined for every continuous curve. If we imposed that the equality holds when $s=0$, that would mean that $alpha$ would be differentiable.


2. So what? Isn't the map $fcolon[-1,1]longrightarrowmathbb R$ defined by $f(x)=x$ differentiable?

share|cite|improve this answer

answered Jul 31 at 16:05

José Carlos Santos

112k1696172

1. No, because the idea is that this concept is defined for every continuous curve. If we imposed that the equality holds when $s=0$, that would mean that $alpha$ would be differentiable.


2. So what? Isn't the map $fcolon[-1,1]longrightarrowmathbb R$ defined by $f(x)=x$ differentiable?

share|cite|improve this answer

answered Jul 31 at 16:05

José Carlos Santos

112k1696172

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 16:05

José Carlos Santos

112k1696172

answered Jul 31 at 16:05

José Carlos Santos

112k1696172

answered Jul 31 at 16:05

José Carlos Santos

112k1696172

112k1696172

• 
  
  
  

  
  

  
  
  
  
  
  

  
  1. $alpha$ is supposed to be $mathcalC^infty,$ so requiring that $h(0, 0) = alpha(0)$ wouldn’t change anything right?
  – Verísimo
  Jul 31 at 16:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  2. I have only seen a definition of differentiability on non-open sets for one-variable functions.
  – Verísimo
  Jul 31 at 16:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  1. Where is it that Do Carmo says that $alpha$ is $C^infty$? I have a copy of the book before my eyes and I didn't sse that. 2. My guess then is that, for him, a function definid on an arbitrary set is differentiable if it is the restriction to that set of a differentiable function defined on an open set.
  – José Carlos Santos
  Jul 31 at 16:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  1. In the first paragraph of section 1-2.
  – Verísimo
  Jul 31 at 17:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  2. Okay, thanks.
  – Verísimo
  Jul 31 at 17:19
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  1. $alpha$ is supposed to be $mathcalC^infty,$ so requiring that $h(0, 0) = alpha(0)$ wouldn’t change anything right?
  – Verísimo
  Jul 31 at 16:17
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  2. I have only seen a definition of differentiability on non-open sets for one-variable functions.
  – Verísimo
  Jul 31 at 16:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  1. Where is it that Do Carmo says that $alpha$ is $C^infty$? I have a copy of the book before my eyes and I didn't sse that. 2. My guess then is that, for him, a function definid on an arbitrary set is differentiable if it is the restriction to that set of a differentiable function defined on an open set.
  – José Carlos Santos
  Jul 31 at 16:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  1. In the first paragraph of section 1-2.
  – Verísimo
  Jul 31 at 17:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  2. Okay, thanks.
  – Verísimo
  Jul 31 at 17:19
  

  
  
  
  

1. $alpha$ is supposed to be $mathcalC^infty,$ so requiring that $h(0, 0) = alpha(0)$ wouldn’t change anything right?
– Verísimo
Jul 31 at 16:17

1. $alpha$ is supposed to be $mathcalC^infty,$ so requiring that $h(0, 0) = alpha(0)$ wouldn’t change anything right?
– Verísimo
Jul 31 at 16:17

2. I have only seen a definition of differentiability on non-open sets for one-variable functions.
– Verísimo
Jul 31 at 16:19

2. I have only seen a definition of differentiability on non-open sets for one-variable functions.
– Verísimo
Jul 31 at 16:19

1. Where is it that Do Carmo says that $alpha$ is $C^infty$? I have a copy of the book before my eyes and I didn't sse that. 2. My guess then is that, for him, a function definid on an arbitrary set is differentiable if it is the restriction to that set of a differentiable function defined on an open set.
– José Carlos Santos
Jul 31 at 16:30

1. Where is it that Do Carmo says that $alpha$ is $C^infty$? I have a copy of the book before my eyes and I didn't sse that. 2. My guess then is that, for him, a function definid on an arbitrary set is differentiable if it is the restriction to that set of a differentiable function defined on an open set.
– José Carlos Santos
Jul 31 at 16:30

1. In the first paragraph of section 1-2.
– Verísimo
Jul 31 at 17:19

1. In the first paragraph of section 1-2.
– Verísimo
Jul 31 at 17:19

2. Okay, thanks.
– Verísimo
Jul 31 at 17:19

2. Okay, thanks.
– Verísimo
Jul 31 at 17:19

 | 
show 1 more comment

 

draft saved

draft discarded

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2868209%2ftwo-questions-on-do-carmo-s-definition-of-variation-of-a-curve%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

Name Email

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

Name Email

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

Name Email

Name Email

Name

Email

Name Email

Name

Email