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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
edited Jul 15 at 3:28
Andrei
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asked Jul 15 at 3:19
Gene Naden
203
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up vote
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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
edited Jul 15 at 3:28
Andrei
7,5152822
asked Jul 15 at 3:19
Gene Naden
203
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
add a comment |Â
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
edited Jul 15 at 3:28
Andrei
7,5152822
asked Jul 15 at 3:19
Gene Naden
203
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
edited Jul 15 at 3:28
Andrei
7,5152822
asked Jul 15 at 3:19
Gene Naden
203
edited Jul 15 at 3:28
Andrei
7,5152822
edited Jul 15 at 3:28
Andrei
7,5152822
edited Jul 15 at 3:28
Andrei
7,5152822
7,5152822
asked Jul 15 at 3:19
Gene Naden
203
asked Jul 15 at 3:19
Gene Naden
203
asked Jul 15 at 3:19
Gene Naden
203
203
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
add a comment |Â
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
add a comment |Â
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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