Alternative Formulation of Arc Length
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If $gamma : [a,b] to Bbb{R}^n$ continuously differentiable, show that the arc length $arc(gamma)$ equals $sup { sum_{i=1}^n ||gamma (t_i)-gamma (t_{i-1})|| mid a = t_0 < t_1 < ... < t_n = b }$.
Recall that $arc(gamma ) = int_{a}^{b} ||gamma '(t)||dt$. I'm trying to solve the above problem. I've already shown that a sequence of points in ${ sum_{i=1}^n ||gamma (t_i)-gamma (t_{i-1})|| mid a = t_0 < t_1 < ... < t_n = b }$ converges to $arc (gamma)$, so I just need to show that $arc (gamma)$ is an upper bound of the set. I've reduced this to showing that $int_{a}^{b} ||gamma ' (t)|| dt ge ||gamma (b) - gamma (a)||$ holds. Intuitively/geometrically it's obvious, but I having trouble mathematizing my intuition. I could use some help.
real-analysis integration multivariable-calculus arc-length
asked Nov 27 at 13:04
user193319
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up vote
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down vote
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If $gamma : [a,b] to Bbb{R}^n$ continuously differentiable, show that the arc length $arc(gamma)$ equals $sup { sum_{i=1}^n ||gamma (t_i)-gamma (t_{i-1})|| mid a = t_0 < t_1 < ... < t_n = b }$.
Recall that $arc(gamma ) = int_{a}^{b} ||gamma '(t)||dt$. I'm trying to solve the above problem. I've already shown that a sequence of points in ${ sum_{i=1}^n ||gamma (t_i)-gamma (t_{i-1})|| mid a = t_0 < t_1 < ... < t_n = b }$ converges to $arc (gamma)$, so I just need to show that $arc (gamma)$ is an upper bound of the set. I've reduced this to showing that $int_{a}^{b} ||gamma ' (t)|| dt ge ||gamma (b) - gamma (a)||$ holds. Intuitively/geometrically it's obvious, but I having trouble mathematizing my intuition. I could use some help.
real-analysis integration multivariable-calculus arc-length
asked Nov 27 at 13:04
user193319
2,3122923
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $gamma : [a,b] to Bbb{R}^n$ continuously differentiable, show that the arc length $arc(gamma)$ equals $sup { sum_{i=1}^n ||gamma (t_i)-gamma (t_{i-1})|| mid a = t_0 < t_1 < ... < t_n = b }$.
Recall that $arc(gamma ) = int_{a}^{b} ||gamma '(t)||dt$. I'm trying to solve the above problem. I've already shown that a sequence of points in ${ sum_{i=1}^n ||gamma (t_i)-gamma (t_{i-1})|| mid a = t_0 < t_1 < ... < t_n = b }$ converges to $arc (gamma)$, so I just need to show that $arc (gamma)$ is an upper bound of the set. I've reduced this to showing that $int_{a}^{b} ||gamma ' (t)|| dt ge ||gamma (b) - gamma (a)||$ holds. Intuitively/geometrically it's obvious, but I having trouble mathematizing my intuition. I could use some help.
real-analysis integration multivariable-calculus arc-length
asked Nov 27 at 13:04
user193319
2,3122923
If $gamma : [a,b] to Bbb{R}^n$ continuously differentiable, show that the arc length $arc(gamma)$ equals $sup { sum_{i=1}^n ||gamma (t_i)-gamma (t_{i-1})|| mid a = t_0 < t_1 < ... < t_n = b }$.
Recall that $arc(gamma ) = int_{a}^{b} ||gamma '(t)||dt$. I'm trying to solve the above problem. I've already shown that a sequence of points in ${ sum_{i=1}^n ||gamma (t_i)-gamma (t_{i-1})|| mid a = t_0 < t_1 < ... < t_n = b }$ converges to $arc (gamma)$, so I just need to show that $arc (gamma)$ is an upper bound of the set. I've reduced this to showing that $int_{a}^{b} ||gamma ' (t)|| dt ge ||gamma (b) - gamma (a)||$ holds. Intuitively/geometrically it's obvious, but I having trouble mathematizing my intuition. I could use some help.
real-analysis integration multivariable-calculus arc-length
real-analysis integration multivariable-calculus arc-length
asked Nov 27 at 13:04
user193319
2,3122923
asked Nov 27 at 13:04
user193319
2,3122923
asked Nov 27 at 13:04
user193319
2,3122923
asked Nov 27 at 13:04
user193319
2,3122923
asked Nov 27 at 13:04
user193319
2,3122923
2,3122923
add a comment |
add a comment |
1 Answer
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You have essentially solved this already. $$ |gamma(b)-gamma(a)| = left| int_a^b gamma'(t) , dt right| le int_a^b |gamma'(t)| , dt $$
by the triangle inequality. This is another version of the Mean Value Theorem, summarisable as: "global growth $le$ local growth".
answered Nov 27 at 13:12
Richard Martin
1,63918
I thought that the mean value theorem didn't hold for vector valued functions.
– user193319
Nov 27 at 13:14
I think you only need it to hold component by component. Anyway the above argument based on the triangle inequality is correct, vector or not.
– Richard Martin
Nov 27 at 13:19
I don't see how component by component will help, since we'll get a different value of $t$ in the mean value function for each coordinate function. As a matter of fact, the mean value theorem doesn't hold: $gamma : [0,1] to Bbb{R}^2$ defined by $gamma (t) = (t,t^2)$ is a counterexample,
– user193319
Nov 27 at 13:21
Oh, I see what you're doing. You're using FOTC, which does hold for vector-valued functions. Sorry about the confusion.
– user193319
Nov 27 at 13:39
No probs, I didn't do a very good job of explaining it, and I could have omitted my comment about the MVT, which might have been better.
– Richard Martin
Nov 27 at 13:43
|
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You have essentially solved this already. $$ |gamma(b)-gamma(a)| = left| int_a^b gamma'(t) , dt right| le int_a^b |gamma'(t)| , dt $$
by the triangle inequality. This is another version of the Mean Value Theorem, summarisable as: "global growth $le$ local growth".
answered Nov 27 at 13:12
Richard Martin
1,63918
I thought that the mean value theorem didn't hold for vector valued functions.
– user193319
Nov 27 at 13:14
I think you only need it to hold component by component. Anyway the above argument based on the triangle inequality is correct, vector or not.
– Richard Martin
Nov 27 at 13:19
I don't see how component by component will help, since we'll get a different value of $t$ in the mean value function for each coordinate function. As a matter of fact, the mean value theorem doesn't hold: $gamma : [0,1] to Bbb{R}^2$ defined by $gamma (t) = (t,t^2)$ is a counterexample,
– user193319
Nov 27 at 13:21
Oh, I see what you're doing. You're using FOTC, which does hold for vector-valued functions. Sorry about the confusion.
– user193319
Nov 27 at 13:39
No probs, I didn't do a very good job of explaining it, and I could have omitted my comment about the MVT, which might have been better.
– Richard Martin
Nov 27 at 13:43
|
show 1 more comment
up vote
1
down vote
accepted
You have essentially solved this already. $$ |gamma(b)-gamma(a)| = left| int_a^b gamma'(t) , dt right| le int_a^b |gamma'(t)| , dt $$
by the triangle inequality. This is another version of the Mean Value Theorem, summarisable as: "global growth $le$ local growth".
answered Nov 27 at 13:12
Richard Martin
1,63918
I thought that the mean value theorem didn't hold for vector valued functions.
– user193319
Nov 27 at 13:14
I think you only need it to hold component by component. Anyway the above argument based on the triangle inequality is correct, vector or not.
– Richard Martin
Nov 27 at 13:19
I don't see how component by component will help, since we'll get a different value of $t$ in the mean value function for each coordinate function. As a matter of fact, the mean value theorem doesn't hold: $gamma : [0,1] to Bbb{R}^2$ defined by $gamma (t) = (t,t^2)$ is a counterexample,
– user193319
Nov 27 at 13:21
Oh, I see what you're doing. You're using FOTC, which does hold for vector-valued functions. Sorry about the confusion.
– user193319
Nov 27 at 13:39
No probs, I didn't do a very good job of explaining it, and I could have omitted my comment about the MVT, which might have been better.
– Richard Martin
Nov 27 at 13:43
|
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You have essentially solved this already. $$ |gamma(b)-gamma(a)| = left| int_a^b gamma'(t) , dt right| le int_a^b |gamma'(t)| , dt $$
by the triangle inequality. This is another version of the Mean Value Theorem, summarisable as: "global growth $le$ local growth".
answered Nov 27 at 13:12
Richard Martin
1,63918
You have essentially solved this already. $$ |gamma(b)-gamma(a)| = left| int_a^b gamma'(t) , dt right| le int_a^b |gamma'(t)| , dt $$
by the triangle inequality. This is another version of the Mean Value Theorem, summarisable as: "global growth $le$ local growth".
answered Nov 27 at 13:12
Richard Martin
1,63918
answered Nov 27 at 13:12
Richard Martin
1,63918
answered Nov 27 at 13:12
Richard Martin
1,63918
answered Nov 27 at 13:12
Richard Martin
1,63918
1,63918
I thought that the mean value theorem didn't hold for vector valued functions.
– user193319
Nov 27 at 13:14
I think you only need it to hold component by component. Anyway the above argument based on the triangle inequality is correct, vector or not.
– Richard Martin
Nov 27 at 13:19
I don't see how component by component will help, since we'll get a different value of $t$ in the mean value function for each coordinate function. As a matter of fact, the mean value theorem doesn't hold: $gamma : [0,1] to Bbb{R}^2$ defined by $gamma (t) = (t,t^2)$ is a counterexample,
– user193319
Nov 27 at 13:21
Oh, I see what you're doing. You're using FOTC, which does hold for vector-valued functions. Sorry about the confusion.
– user193319
Nov 27 at 13:39
No probs, I didn't do a very good job of explaining it, and I could have omitted my comment about the MVT, which might have been better.
– Richard Martin
Nov 27 at 13:43
|
show 1 more comment
I thought that the mean value theorem didn't hold for vector valued functions.
– user193319
Nov 27 at 13:14
I think you only need it to hold component by component. Anyway the above argument based on the triangle inequality is correct, vector or not.
– Richard Martin
Nov 27 at 13:19
I don't see how component by component will help, since we'll get a different value of $t$ in the mean value function for each coordinate function. As a matter of fact, the mean value theorem doesn't hold: $gamma : [0,1] to Bbb{R}^2$ defined by $gamma (t) = (t,t^2)$ is a counterexample,
– user193319
Nov 27 at 13:21
Oh, I see what you're doing. You're using FOTC, which does hold for vector-valued functions. Sorry about the confusion.
– user193319
Nov 27 at 13:39
No probs, I didn't do a very good job of explaining it, and I could have omitted my comment about the MVT, which might have been better.
– Richard Martin
Nov 27 at 13:43
I thought that the mean value theorem didn't hold for vector valued functions.
– user193319
Nov 27 at 13:14
I thought that the mean value theorem didn't hold for vector valued functions.
– user193319
Nov 27 at 13:14
I think you only need it to hold component by component. Anyway the above argument based on the triangle inequality is correct, vector or not.
– Richard Martin
Nov 27 at 13:19
I think you only need it to hold component by component. Anyway the above argument based on the triangle inequality is correct, vector or not.
– Richard Martin
Nov 27 at 13:19
I don't see how component by component will help, since we'll get a different value of $t$ in the mean value function for each coordinate function. As a matter of fact, the mean value theorem doesn't hold: $gamma : [0,1] to Bbb{R}^2$ defined by $gamma (t) = (t,t^2)$ is a counterexample,
– user193319
Nov 27 at 13:21
I don't see how component by component will help, since we'll get a different value of $t$ in the mean value function for each coordinate function. As a matter of fact, the mean value theorem doesn't hold: $gamma : [0,1] to Bbb{R}^2$ defined by $gamma (t) = (t,t^2)$ is a counterexample,
– user193319
Nov 27 at 13:21
Oh, I see what you're doing. You're using FOTC, which does hold for vector-valued functions. Sorry about the confusion.
– user193319
Nov 27 at 13:39
Oh, I see what you're doing. You're using FOTC, which does hold for vector-valued functions. Sorry about the confusion.
– user193319
Nov 27 at 13:39
No probs, I didn't do a very good job of explaining it, and I could have omitted my comment about the MVT, which might have been better.
– Richard Martin
Nov 27 at 13:43
No probs, I didn't do a very good job of explaining it, and I could have omitted my comment about the MVT, which might have been better.
– Richard Martin
Nov 27 at 13:43
|
show 1 more comment
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