Let there be a *displacement* vector **r** that is a parametric function of arc time *t* so that **r** = **r**(*t*). Then define *s* as the arc length of **r** so that

*s* = *s*(*t*) = ∫ || **r**′(*τ*) || *dτ*,

where the integral is from 0 to *t*. Let us further assume that *s* is bijective so that the inverse function is *t*.

Let there also be a *distimement* vector **w** that is a parametric function of arc length *s* so that **w** = **w**(*s*). Then *t* is the arc time of **w** if

*t* = *t*(*s*) = ∫ || **w**′(*σ*) || *dσ*,

where the integral is from 0 to *s*. Now the arc time derivative of **r**, that is, the derivative of *s* with respect to *t* is

*s′* = *ds*/*dt* = || **r**′(*t*) ||.

And the arc length derivative of **w**, that is, the derivative of *t* with respect to *s* is

*t′* = *dt*/*ds* = || **w**′(*s*) ||.

From the inverse function theorem we have that

*t′* = *dt*/*ds* = 1/|| **r**′(*t*) || = 1/(*ds*/*dt*).

And also that

*s′* = *ds*/*dt* = 1/|| **w**′(*s*) || = 1/(*dt*/*ds*).

Putting these together we find

*s′* = *ds*/*dt* = || **r**′(*t*) || = 1/|| **w**′(*s*) || = 1/(*dt*/*ds*).

And also that

*t′* = *dt*/*ds* = || **w**′(*s*) || = 1/|| **r**′(*t*) || = 1/(*ds*/*dt*).

We then have

*s* = *s*(*t*) = ∫ || **r**′(*τ*) || *dτ* = ∫ 1/|| **w**′(*σ*) || d*σ*,

where the first integral is from 0 to *t* and the second integral is from 0 to *s*. And also that

*t* = *t*(*s*) = **∫** || **w**′(*σ*) || *dσ* = **∫** 1/|| **r**′(*τ*) || *dτ*,

where the first integral is from 0 to *s* and the second integral is from 0 to *t*.

Because of the difficulty of inverting *s*(*t*), this shows a bypass is available. That is, from **r**(*t*) we find

*t*(*s*) = ∫ 1/|| **r**′(*τ*) || *dτ*,

and from **w**(*s*) we find

*s*(*t*) = ∫ 1/|| **w**′(*σ*) || *dσ*.

The interpretation is that the space vector, **r**, is a position vector function of the arc time, *t*, and the time vector, **w**, is a time vector function of the arc length, *s*.