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Arc length of a regular parametrized curve

by tuggler
Tags: curve, length, parametrized, regular
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tuggler
#1
Aug31-13, 07:22 PM
P: 45
Given [tex]t\in I[/tex]the arc length of a regular parametrized curve [tex]\alpha : I \to \mathbb{R}^3[/tex] from the point [tex]t_0[/tex] is by definition [tex]s(t) = \int^t_{t_0}|\alpha'(t)|dt[/tex] where [tex]|\alpha'(t)| = \sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}[/tex] is the length of the vector [tex]\alpha'(t).[/tex] Since [tex]\alpha'(t) \ne 0[/tex] the arc length [tex]s[/tex] is a differentiable function of and [tex]ds/dt = |\alpha'(t)|.[/tex]

This is where I get confused.

It can happen that the parameter [tex]t[/tex]is already the arc length measured from some point. In this case, [itex]ds/dt = 1 =|\alpha'(t)|[/tex]. Conversely, if [tex]|\alpha'(t)| = 1[/tex] then [tex]s = \int_{t_0}^t dt = t - t_0.[/tex]

How did they get that it equals 1? I am not sure what they are saying?
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tuggler
#2
Aug31-13, 07:23 PM
P: 45
Opps, I am in the wrong thread. How can I delete this?
mathman
#3
Sep1-13, 03:19 PM
Sci Advisor
P: 6,076
Quote Quote by tuggler View Post
Given [tex]t\in I[/tex]the arc length of a regular parametrized curve [tex]\alpha : I \to \mathbb{R}^3[/tex] from the point [tex]t_0[/tex] is by definition [tex]s(t) = \int^t_{t_0}|\alpha'(t)|dt[/tex] where [tex]|\alpha'(t)| = \sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}[/tex] is the length of the vector [tex]\alpha'(t).[/tex] Since [tex]\alpha'(t) \ne 0[/tex] the arc length [tex]s[/tex] is a differentiable function of and [tex]ds/dt = |\alpha'(t)|.[/tex]

This is where I get confused.

It can happen that the parameter [tex]t[/tex]is already the arc length measured from some point. In this case, [itex]ds/dt = 1 =|\alpha'(t)|[/tex]. Conversely, if [tex]|\alpha'(t)| = 1[/tex] then [tex]s = \int_{t_0}^t dt = t - t_0.[/tex]

How did they get that it equals 1? I am not sure what they are saying?
If t is arc length (that is: s = t), then ds/dt = 1. If this doesn't answer your question you need to elaborate.


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