Everybody knows this one-dimensional kinematic equation from high school physics: $$2a\,\Delta x = v^2 - v_0^2 \quad (\label(1)$$ which is essentially (adopting the usual abuses of notation found in undergraduate mechanics textbooks) $$\int_{x_0}^x a \, dx = \int_{v_0}^v v \, dv $$ in the case when the acceleration $a$ is constant. Now this is justified rigorously as follows (proof is my own, all corrections/suggestions welcome):
- Suppose we have the injective, twice-continously differentiable function $x : \Omega \to \mathbb R$, $\Omega \subseteq \mathbb R$ open, indicating the position of our particle, and the two functions $$V : x(\Omega) \to \mathbb R, \qquad A : x(\Omega) \to \mathbb R $$ that satisfy $$\dot x(\tau) = V(x(\tau)), \qquad \ddot x(\tau) = A(x(\tau)) \qquad \forall \tau \in \Omega$$
- Then – for the single-variable chain rule – we must have, for some $t_0 \in \Omega$, that $$A(x(t_0)) = \frac{dV}{dy}(x(t_0))\dot x(t_0) = \frac{dV}{dy}(x(t_0))V(x(t_0)) $$
- We now choose $t \in \Omega$ such that $t > t_0$, integrate both sides and use the theorem of integration by substitution over the interval $x([t_0,t]) = [x(t_0),x(t)]$: $$\int_{x(t_0)}^{x(t)} A(\xi) \, d\xi = \int_{x(t_0)}^{x(t)} V(\xi) \frac{dV}{dy}(\xi) \, d\xi = \int_{V(x(t_0))}^{V(x(t))} \lambda \, d\lambda = \left[\frac 1 2\lambda^2\right]_{V(x(t_0))}^{V(x(t))}$$
My question would be: is there an equivalent $(1)$ in space? That would mean we'd be integrating vector-valued functions over some (curvelike) region in $\mathbb R^3$. I think one would have to use the multivariable version of integral substitution (also mentioned in the link above), but I'm not sure of all the details of such a derivation. I'm also guessing, since multiplication of two vectors is not well defined in Euclidean space, that we'd have dot products or square moduli in lieu of the squares on the right-hand side of $(1)$.