Let the vector r be a displacement vector of motion K that is a parametric function of its arc time t so that r = r(t). Then define s as the arc length of K such 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 its inverse function is t(s).
Let the vector w be a dischronment vector of K 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′(t) = ds/dt = || r′(t) || = || v(t) ||,
where v(t) is the speed function of K. And the arc length derivative of w, that is, the derivative of t with respect to s is
t′(s) = dt/ds = || w′(s) || = || u(s) ||,
where u(s) is the pace function of K.
From the inverse function theorem we have that
s′(t) = 1/|| w′(t) || = 1/|| u(t) || and
t′(s) = 1/|| r′(s) || = 1/|| v(s) ||.
Putting these together we find
s′(t) = || r′(t) || = || v(t) || = 1/|| u(t) || = 1/|| w′(t) || and
t′(s) = || w′(s) || = || u(s) || = 1/|| v(s) || = 1/|| r′(s) ||.
We then have
s = s(t) = ∫ || r′(τ) || dτ = ∫ || v(τ) || dτ = ∫ 1/|| u(σ) || dσ = ∫ 1/|| w′(σ) || dσ,
where the integrals are from 0 to t. And also that
t = t(s) = ∫ || w′(σ) || dσ = ∫ || u(σ) || dσ = ∫ 1/|| v(τ) || dτ = ∫ 1/|| r′(τ) || dτ,
where the integrals are from 0 to s.
Because of the difficulty of inverting s(t) and s(t) this shows a bypass is available. That is,
t(s) = ∫ 1/|| r′(τ) || dτ = ∫ 1/|| v(τ) || dτ.
and
s(t) = ∫ 1/|| w′(σ) || dσ = ∫ 1/|| u(σ) || dσ.