# Distance Traveled

If we wish to find the distance traveled (not the distance from the origin) by a particle along a path ##C## defined by ##y = f(t)## and ##x = g(t)## we would use this integral:

$$L = \int_C ds = \int_{t_1}^{t_2} \sqrt{({\frac{dy}{dt}})^2 + ({\frac{dx}{dt}})^2} dt$$

My question is, does this give a nonzero answer if the particle were at the same position at ##t = t_1## and at ##t = t_2##?

Also, is this a line integral? I don't know much about line integrals and vector calculus, all I know is work done is a good example of a line integral.

Last edited:

ShayanJ
Gold Member
You can simply understand it yourself by calculating this integral for $x=\cos t \ , \ y=\sin t$ from $t=0 \ to \ 2\pi$.

DarthMatter, HallsofIvy and PFuser1232
mathman
If we wish to find the distance traveled (not the distance from the origin) by a particle along a path ##C## defined by ##y = f(t)## and ##x = g(t)## we would use this integral:

$$L = \int_C ds = \int_{t_1}^{t_2} \sqrt{({\frac{dy}{dt}})^2 + ({\frac{dx}{dt}})^2} dt$$

My question is, does this give a nonzero answer if the particle were at the same position at ##t = t_1## and at ##t = t_2##?

Also, is this a line integral? I don't know much about line integrals and vector calculus, all I know is work done is a good example of a line integral.
As long as the particle is moving, the integrand is always positive even if the particle ends up where it started.

DarthMatter and PFuser1232
HallsofIvy
$$L = \int_C ds = \int_{t_1}^{t_2} \sqrt{({\frac{dy}{dt}})^2 + ({\frac{dx}{dt}})^2} dt$$