In the arc length formula, where do $x_1$, $x_2$ and $x_3$ represent? The length of a regular arc $x=x(t)$ is given by $$\int^b _a \sqrt{ \bigg(\frac{dx_1}{dt}\bigg)^2 + \bigg(\frac{dx_2}{dt}\bigg)^2 + \bigg(\frac{dx_3}{dt}\bigg)^2} dt$$
What do $x_1$, $x_2$ and $x_3$ represent?
 A: The vector $x = \langle x_1,\,x_2,\,x_3\rangle$ is a vector in $\mathbb{R}^3$ with three components. 
If $x$ is written as a function of $t$, we have
$$x(t) = \langle x_1(t),\,x_2(t),\,x_3(t)\rangle$$
A: $(x_1,x_2,x_3)$ is the point on the curve corresponding to the parameter value $t$.
$(dx_1,dx_2,dx_3)$ may by intuitively thought of as an infinitely small change in $(x_1,x_2,x_3)$ corresponding to an infinitely small change $dt$ in the value of $t$. 
The corresponding infinitely small distance that the point $(x_1,x_2,x_3)$ moves along the curve is $\sqrt{(dx_1)^2+(dx_2)^2+(dx_3)^2}$, given by the Pythagorean theorem.  The whole arc length is the sum of these infinitely many infinitely small distances; thus it is
$$
\int_{t\,:=\,a}^{t\,:=\,b} \sqrt{(dx_1)^2+(dx_2)^2+(dx_3)^2}.
$$
This is of course the same as
$$
\int^b_a \sqrt{ \left(\frac{dx_1}{dt}\right)^2 + \left(\frac{dx_2}{dt}\right)^2 + \left(\frac{dx_3}{dt}\right)^2} \, dt.
$$
This last form is used when you know $(x_1,x_2,x_3)$ as a function of $t$, thus reducing the problem to that of evaluating a definite integral with respect to a scalar variable $t$.
A: They are the three components of a point, $(x_1, x_2, x_3)$, on the curve.  You could also write the point as (x, y, z) and the arclength of the curve as $\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2+ \left(\frac{dz}{dt}\right)^2} dt$.
