The mathematical problem is this: given a curve with distance coordinates that are parametric functions of time (duration), find the reparametrization of the curve with duration (time) coordinates that are parametric functions of distance. Symbolically, given the regular curve α(t) = (a1(t), …, an(t)), find β(s) = (b1(s), …, bn(s)) such that bi(s) = t(ai) for i = 1, …, n, where t is time (duration) and s is distance. (Greek letters are used for vectors, Roman letters for scalars.)
The solution is to invert each coordinate function and express them in terms of a common parameter. That is, set each ai(t) = s and solve for t to get t = ai-1(s) = bi(s) for the inverse coordinates in parametric form.
For example, consider a projectile fired from height h with velocity v at angle θ. The path of the projectile is represented by a parametric equation
α(t) = (a1(t), a2(t)) = (vt cos(θ), h + vt sin(θ) – ½gt²),
where g is the acceleration of gravity. Setting s = vt cos(θ) and s = h + vt sin(θ) – ½gt²), then solving for t results in the inverse coordinates, which are in two parts:
β(s) =(s/(v cos(θ)), (v sin(θ) + sqrt(2gh – 2gs + v² sin²(θ)))/g) going up, and
β(s) =(s/(v cos(θ)), (v sin(θ) – sqrt(2gh – 2gs + v² sin²(θ)))/g) coming down.
The spatial position vector α(t) corresponds to a temporal position vector β(s). As there are multiple dimensions of space, so there are multiple dimensions of time. The time in multidimensional space is a scalar, but the space with multiple dimensions of time is a scalar.