# arc length of a regular parametrized curve

by tuggler
Tags: curve, length, parametrized, regular
 P: 45 Given $$t\in I$$the 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 $$t$$is already the arc length measured from some point. In this case, [latex]ds/dt = 1 =|\alpha'(t)|[/tex]. Conversely, if $$|\alpha'(t)| = 1$$ then $$s = \int_{t_0}^t dt = t - t_0.$$ How did they get that it equals 1? I am not sure what they are saying?
 P: 45 Opps, I am in the wrong thread. How can I delete this?
 Quote by tuggler Given $$t\in I$$the 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 $$t$$is already the arc length measured from some point. In this case, [latex]ds/dt = 1 =|\alpha'(t)|[/tex]. Conversely, if $$|\alpha'(t)| = 1$$ then $$s = \int_{t_0}^t dt = t - t_0.$$ How did they get that it equals 1? I am not sure what they are saying?