# Distance Traveled

• PFuser1232

#### PFuser1232

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.

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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
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
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.
Yes, that is a line integral and gives the length of the line. If you go around a circle with circumference 100 meters, you are right back where you started but have walked 100 meters, not 0!