# Distance Traveled

1. Jan 3, 2015

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: Jan 3, 2015
2. Jan 3, 2015

### ShayanJ

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

3. Jan 3, 2015

### mathman

As long as the particle is moving, the integrand is always positive even if the particle ends up where it started.

4. Jan 5, 2015

### HallsofIvy

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!