Register to reply 
Arc length of a regular parametrized curve 
Share this thread: 
#1
Aug3113, 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? 


#2
Aug3113, 07:23 PM

P: 45

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



#3
Sep113, 03:19 PM

Sci Advisor
P: 6,027




Register to reply 
Related Discussions  
Arc length of a regular parametrized curve  Introductory Physics Homework  1  
Arclength of exp from 0 to 1  Calculus & Beyond Homework  3  
Simpson’s rule to evaluate this integral  Calculus & Beyond Homework  0  
Arclength TI89  General Math  5  
Find the arc length of the given function  Introductory Physics Homework  8 