Why is arc length defined on a closed interval?

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My differential geometry text defined a curve as a function from an open interval of $mathbb R$ onto $mathbb R^3$, but it defines arc length as the integral in terms of the velocity along a closed interval. Why "closed"? Would the velocity not be well-defined if we used an open interval?




differential-geometry




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    Consider the path $f: (0,1) to mathbbR^3$ defined by $f(t) = (t, t,1/t)$. What would its arclength be?
    – Randall
    Jul 15 at 3:37










  • Riemann integrals are generally defined for a function defined on a closed interval. This to avoid problems like the on given by @Randall. It’s just a definition.
    – John Mitchell
    Jul 15 at 7:18














up vote
1
down vote

favorite












My differential geometry text defined a curve as a function from an open interval of $mathbb R$ onto $mathbb R^3$, but it defines arc length as the integral in terms of the velocity along a closed interval. Why "closed"? Would the velocity not be well-defined if we used an open interval?




differential-geometry




share|cite|improve this question

















  • 2




    Consider the path $f: (0,1) to mathbbR^3$ defined by $f(t) = (t, t,1/t)$. What would its arclength be?
    – Randall
    Jul 15 at 3:37










  • Riemann integrals are generally defined for a function defined on a closed interval. This to avoid problems like the on given by @Randall. It’s just a definition.
    – John Mitchell
    Jul 15 at 7:18












up vote
1
down vote

favorite









up vote
1
down vote

favorite











My differential geometry text defined a curve as a function from an open interval of $mathbb R$ onto $mathbb R^3$, but it defines arc length as the integral in terms of the velocity along a closed interval. Why "closed"? Would the velocity not be well-defined if we used an open interval?




differential-geometry




share|cite|improve this question













My differential geometry text defined a curve as a function from an open interval of $mathbb R$ onto $mathbb R^3$, but it defines arc length as the integral in terms of the velocity along a closed interval. Why "closed"? Would the velocity not be well-defined if we used an open interval?





differential-geometry





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share|cite|improve this question




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edited Jul 15 at 3:28









Andrei

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asked Jul 15 at 3:19









Gene Naden

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  • 2




    Consider the path $f: (0,1) to mathbbR^3$ defined by $f(t) = (t, t,1/t)$. What would its arclength be?
    – Randall
    Jul 15 at 3:37










  • Riemann integrals are generally defined for a function defined on a closed interval. This to avoid problems like the on given by @Randall. It’s just a definition.
    – John Mitchell
    Jul 15 at 7:18












  • 2




    Consider the path $f: (0,1) to mathbbR^3$ defined by $f(t) = (t, t,1/t)$. What would its arclength be?
    – Randall
    Jul 15 at 3:37










  • Riemann integrals are generally defined for a function defined on a closed interval. This to avoid problems like the on given by @Randall. It’s just a definition.
    – John Mitchell
    Jul 15 at 7:18







2




2




Consider the path $f: (0,1) to mathbbR^3$ defined by $f(t) = (t, t,1/t)$. What would its arclength be?
– Randall
Jul 15 at 3:37




Consider the path $f: (0,1) to mathbbR^3$ defined by $f(t) = (t, t,1/t)$. What would its arclength be?
– Randall
Jul 15 at 3:37












Riemann integrals are generally defined for a function defined on a closed interval. This to avoid problems like the on given by @Randall. It’s just a definition.
– John Mitchell
Jul 15 at 7:18




Riemann integrals are generally defined for a function defined on a closed interval. This to avoid problems like the on given by @Randall. It’s just a definition.
– John Mitchell
Jul 15 at 7:18















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