What function provides $d(t)$ distance traveled by time $t$, given $x(t)$ is position of a particle of periodic travel in an interval $[a,b]$ containing $0$?
For example, imagine one particle traces a circle centered at the origin meanwhile another particle travels the $x$-axis. They both begin at $(r, 0)$, meet at $(-r, 0)$, and end at the same place $(r, 0)$ given they traveled at some related pace. The related pace is achieved because, in theory, the one dimensional particle simply pauses when the circular particle undergoes vertical displacements.
Given the vertical displacements, the circular particle got the greater amount of $d(t) = 2\pi r$ worth of steps in while the one dimensional particle got only $d(t) = 4r$ steps in. What can I apply to the position function of the one-dimensional particle, $x(t) = r\cos(\theta(t))$ for $\theta(t) \in [0, 2\pi]$, in order to achieve this intuitive $d(t) = 4r$?
In general case, suppose you have a position given in parametric form: $$\vec r(t)=\left(x(t),y(t),z(t)\right)$$ The velocity is then $$\vec v(t)=\left(x'(t),y'(t),z'(t)\right)$$ Then $$\vec r(t)-\vec r(0)=\int_0^t\vec v(\tau)d\tau$$ If you want to find out the total distance traveled, you need to integrate the speed, which is the absolute value of velocity: $$d(t)=\int_0^t|\vec v(\tau)|d\tau=\int_0^t\sqrt{x'^2(\tau)+y'^2(\tau)+z'^2(\tau)}d\tau$$ In your case you have only two dimensions, so $z'(t)=0$. Then for moving along the circle, $$x(t)=r\cos\theta\\y(t)=r\sin\theta$$ This yields $$x'(t)=-r\sin\theta\\y'(t)=r\cos\theta$$ Then $$d_c=\int_0^{2\pi}\sqrt{r^2\sin^2t+r^2\cos^2t}dt=r\int_0^{2\pi}dt=2\pi r$$ If you move along the $x$ axis, $x$ and $x'$ have the same expression, but $y(t)=y'(t)=0$. Then $$d_d=r\int_0^{2\pi}|\sin(t)|dt=2r\int_0^\pi\sin t dt$$ I've used here that $\sin(\pi+x)=-\sin(x)$. Then $$\int_0^\pi\sin t dt=-\cos(t)|_0^\pi=2$$ so $$d_d=4r$$