Arc length of a regular parametrized curve

[tuggler]

Given t\in Ithe arc length of a regular parametrized curve \alpha : I \to \mathbb{R}^3 from the point t_0 is by definition s(t) = \int^t_{t_0}|\alpha'(t)|dt where |\alpha'(t)| = \sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2} is the length of the vector \alpha'(t). Since \alpha'(t) \ne 0 the arc length s is a differentiable function of and ds/dt = |\alpha'(t)|.

This is where I get confused.

It can happen that the parameter tis already the arc length measured from some point. In this case, ds/dt = 1 =|\alpha&#039;(t)|[/tex]. Conversely, if |\alpha&amp;#039;(t)| = 1 then s = \int_{t_0}^t dt = t - t_0.<br /> <br /> How did they get that it equals 1? I am not sure what they are saying?

 

[tuggler]

Opps, I am in the wrong thread. How can I delete this?

 

[mathman]

Given t\in Ithe arc length of a regular parametrized curve \alpha : I \to \mathbb{R}^3 from the point t_0 is by definition s(t) = \int^t_{t_0}|\alpha&#039;(t)|dt where |\alpha&#039;(t)| = \sqrt{(x&#039;(t))^2+(y&#039;(t))^2+(z&#039;(t))^2} is the length of the vector \alpha&#039;(t). Since \alpha&#039;(t) \ne 0 the arc length s is a differentiable function of and ds/dt = |\alpha&#039;(t)|.

This is where I get confused.

It can happen that the parameter tis already the arc length measured from some point. In this case, ds/dt = 1 =|\alpha&#039;(t)|[/tex]. Conversely, if |\alpha&amp;#039;(t)| = 1 then s = \int_{t_0}^t dt = t - t_0.<br /> <br /> How did they get that it equals 1? I am not sure what they are saying?

<br /> <br /> If t is arc length (that is: s = t), then ds/dt = 1. If this doesn&#039;t answer your question you need to elaborate.