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.






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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







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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







      share|cite|improve this question














      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






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      asked Nov 27 at 13:04









      user193319

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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".






          share|cite|improve this answer





















          • 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
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          active

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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          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".






          share|cite|improve this answer





















          • 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















          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".






          share|cite|improve this answer





















          • 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













          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".






          share|cite|improve this answer












          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".







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • 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


















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