Arc length does not change with a reparametrizationShow: $int_{mathbb{R}^n}mbox{div }a(x), d^nx=0$A curve...
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Arc length does not change with a reparametrization
Show: $int_{mathbb{R}^n}mbox{div }a(x), d^nx=0$A curve parametrized by arc lengthParametrized curve in $mathbb{R}^n$Arclength does not change with reparametrizationWhat is the Convention in Arc Length Parametrization?The change of parameter of a regular curve is a diffeomorphism, and preserves the lengthProof that any unit-speed-reparametrization of a curve preserves orientation and is an inverse of an arc length function based at some $t_0$.Any unit speed reparametrization $beta=alpha(h)$ of $alpha$ is reparametrized by an arc length function.Why is a curve parameterized by arc length necessarily a unit speed curve?Is the unit tangent with any parametrisation the same as the tangent with arc-length parametrisation?
$begingroup$
Show that the arc length of a curve doesn't change with a reparametrization.
Let $sigma: [a,b]tomathbb{R}^{3}, ;g: [c,d]to[a,b]$ and $alpha: [c,d]tomathbb{R}^{3}$. $alpha(s)=(sigma(g(s))$.
The length of the curve $sigma$ is $Lsigma =int_{a}^{b} leftlVert sigma'(t) rightrVert dt$, if $t=g(s)$ then $dt=g'(s)ds$
So $Lsigma=int_{c}^{d} leftlVertsigma'(g(s)) rightrVert g'(s)ds$, if $g'(s)gt 0$, then
$Lsigma=int_{c}^{d} leftlVertsigma'(g(s)) g'(s)rightrVert ds = int_{c}^{d} leftlVertalpha' (s)rightrVert ds= Lalpha$
But i don't know how to proceed if $g'(s)lt 0$
multivariable-calculus
asked Mar 19 at 6:17
Robert ArnoldRobert Arnold
31
$endgroup$
add a comment |
$begingroup$
Show that the arc length of a curve doesn't change with a reparametrization.
Let $sigma: [a,b]tomathbb{R}^{3}, ;g: [c,d]to[a,b]$ and $alpha: [c,d]tomathbb{R}^{3}$. $alpha(s)=(sigma(g(s))$.
The length of the curve $sigma$ is $Lsigma =int_{a}^{b} leftlVert sigma'(t) rightrVert dt$, if $t=g(s)$ then $dt=g'(s)ds$
So $Lsigma=int_{c}^{d} leftlVertsigma'(g(s)) rightrVert g'(s)ds$, if $g'(s)gt 0$, then
$Lsigma=int_{c}^{d} leftlVertsigma'(g(s)) g'(s)rightrVert ds = int_{c}^{d} leftlVertalpha' (s)rightrVert ds= Lalpha$
But i don't know how to proceed if $g'(s)lt 0$
multivariable-calculus
asked Mar 19 at 6:17
Robert ArnoldRobert Arnold
31
$endgroup$
add a comment |
$begingroup$
Show that the arc length of a curve doesn't change with a reparametrization.
Let $sigma: [a,b]tomathbb{R}^{3}, ;g: [c,d]to[a,b]$ and $alpha: [c,d]tomathbb{R}^{3}$. $alpha(s)=(sigma(g(s))$.
The length of the curve $sigma$ is $Lsigma =int_{a}^{b} leftlVert sigma'(t) rightrVert dt$, if $t=g(s)$ then $dt=g'(s)ds$
So $Lsigma=int_{c}^{d} leftlVertsigma'(g(s)) rightrVert g'(s)ds$, if $g'(s)gt 0$, then
$Lsigma=int_{c}^{d} leftlVertsigma'(g(s)) g'(s)rightrVert ds = int_{c}^{d} leftlVertalpha' (s)rightrVert ds= Lalpha$
But i don't know how to proceed if $g'(s)lt 0$
multivariable-calculus
asked Mar 19 at 6:17
Robert ArnoldRobert Arnold
31
$endgroup$
Show that the arc length of a curve doesn't change with a reparametrization.
Let $sigma: [a,b]tomathbb{R}^{3}, ;g: [c,d]to[a,b]$ and $alpha: [c,d]tomathbb{R}^{3}$. $alpha(s)=(sigma(g(s))$.
The length of the curve $sigma$ is $Lsigma =int_{a}^{b} leftlVert sigma'(t) rightrVert dt$, if $t=g(s)$ then $dt=g'(s)ds$
So $Lsigma=int_{c}^{d} leftlVertsigma'(g(s)) rightrVert g'(s)ds$, if $g'(s)gt 0$, then
$Lsigma=int_{c}^{d} leftlVertsigma'(g(s)) g'(s)rightrVert ds = int_{c}^{d} leftlVertalpha' (s)rightrVert ds= Lalpha$
But i don't know how to proceed if $g'(s)lt 0$
multivariable-calculus
multivariable-calculus
asked Mar 19 at 6:17
Robert ArnoldRobert Arnold
31
asked Mar 19 at 6:17
Robert ArnoldRobert Arnold
31
asked Mar 19 at 6:17
Robert ArnoldRobert Arnold
31
asked Mar 19 at 6:17
Robert ArnoldRobert Arnold
31
asked Mar 19 at 6:17
Robert ArnoldRobert Arnold
31
31
add a comment |
add a comment |
1 Answer
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votes
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You are not applying the change of variables properly. You get $L_{sigma} =int_c^{d} |sigma'(g(s))||g'(s)|, ds$. Also $|sigma'(g(s))g'(s)|=|sigma'(g(s))| |g'(s)|$.
answered Mar 19 at 6:22
Kavi Rama MurthyKavi Rama Murthy
72.3k53170
$endgroup$
$begingroup$
So $frac{dg(s)}{ds} = leftlvert g'(s)rightrvert $ ?
$endgroup$
– Robert Arnold
Mar 19 at 6:42
$begingroup$
$frac {dg(s)} {ds} =g'(s)$ but in the change of variable formula you have to multiply by $|frac {dg(s)} {ds}|$, not $frac {dg(s)} {ds}$.
$endgroup$
– Kavi Rama Murthy
Mar 19 at 6:45
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You are not applying the change of variables properly. You get $L_{sigma} =int_c^{d} |sigma'(g(s))||g'(s)|, ds$. Also $|sigma'(g(s))g'(s)|=|sigma'(g(s))| |g'(s)|$.
answered Mar 19 at 6:22
Kavi Rama MurthyKavi Rama Murthy
72.3k53170
$endgroup$
$begingroup$
So $frac{dg(s)}{ds} = leftlvert g'(s)rightrvert $ ?
$endgroup$
– Robert Arnold
Mar 19 at 6:42
$begingroup$
$frac {dg(s)} {ds} =g'(s)$ but in the change of variable formula you have to multiply by $|frac {dg(s)} {ds}|$, not $frac {dg(s)} {ds}$.
$endgroup$
– Kavi Rama Murthy
Mar 19 at 6:45
add a comment |
$begingroup$
You are not applying the change of variables properly. You get $L_{sigma} =int_c^{d} |sigma'(g(s))||g'(s)|, ds$. Also $|sigma'(g(s))g'(s)|=|sigma'(g(s))| |g'(s)|$.
answered Mar 19 at 6:22
Kavi Rama MurthyKavi Rama Murthy
72.3k53170
$endgroup$
$begingroup$
So $frac{dg(s)}{ds} = leftlvert g'(s)rightrvert $ ?
$endgroup$
– Robert Arnold
Mar 19 at 6:42
$begingroup$
$frac {dg(s)} {ds} =g'(s)$ but in the change of variable formula you have to multiply by $|frac {dg(s)} {ds}|$, not $frac {dg(s)} {ds}$.
$endgroup$
– Kavi Rama Murthy
Mar 19 at 6:45
add a comment |
$begingroup$
You are not applying the change of variables properly. You get $L_{sigma} =int_c^{d} |sigma'(g(s))||g'(s)|, ds$. Also $|sigma'(g(s))g'(s)|=|sigma'(g(s))| |g'(s)|$.
answered Mar 19 at 6:22
Kavi Rama MurthyKavi Rama Murthy
72.3k53170
$endgroup$
You are not applying the change of variables properly. You get $L_{sigma} =int_c^{d} |sigma'(g(s))||g'(s)|, ds$. Also $|sigma'(g(s))g'(s)|=|sigma'(g(s))| |g'(s)|$.
answered Mar 19 at 6:22
Kavi Rama MurthyKavi Rama Murthy
72.3k53170
answered Mar 19 at 6:22
Kavi Rama MurthyKavi Rama Murthy
72.3k53170
answered Mar 19 at 6:22
Kavi Rama MurthyKavi Rama Murthy
72.3k53170
answered Mar 19 at 6:22
Kavi Rama MurthyKavi Rama Murthy
72.3k53170
72.3k53170
$begingroup$
So $frac{dg(s)}{ds} = leftlvert g'(s)rightrvert $ ?
$endgroup$
– Robert Arnold
Mar 19 at 6:42
$begingroup$
$frac {dg(s)} {ds} =g'(s)$ but in the change of variable formula you have to multiply by $|frac {dg(s)} {ds}|$, not $frac {dg(s)} {ds}$.
$endgroup$
– Kavi Rama Murthy
Mar 19 at 6:45
add a comment |
$begingroup$
So $frac{dg(s)}{ds} = leftlvert g'(s)rightrvert $ ?
$endgroup$
– Robert Arnold
Mar 19 at 6:42
$begingroup$
$frac {dg(s)} {ds} =g'(s)$ but in the change of variable formula you have to multiply by $|frac {dg(s)} {ds}|$, not $frac {dg(s)} {ds}$.
$endgroup$
– Kavi Rama Murthy
Mar 19 at 6:45
$begingroup$
So $frac{dg(s)}{ds} = leftlvert g'(s)rightrvert $ ?
$endgroup$
– Robert Arnold
Mar 19 at 6:42
$begingroup$
So $frac{dg(s)}{ds} = leftlvert g'(s)rightrvert $ ?
$endgroup$
– Robert Arnold
Mar 19 at 6:42
$begingroup$
$frac {dg(s)} {ds} =g'(s)$ but in the change of variable formula you have to multiply by $|frac {dg(s)} {ds}|$, not $frac {dg(s)} {ds}$.
$endgroup$
– Kavi Rama Murthy
Mar 19 at 6:45
$begingroup$
$frac {dg(s)} {ds} =g'(s)$ but in the change of variable formula you have to multiply by $|frac {dg(s)} {ds}|$, not $frac {dg(s)} {ds}$.
$endgroup$
– Kavi Rama Murthy
Mar 19 at 6:45
add a comment |
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