"Rigorous" derivation of kinetic energy

$\newcommand{\dd}{{\rm d}}$ $\newcommand{\vect}[1]{{\bf #1}}$

First note that

$$ \frac{\dd }{\dd t}(\vect{v}\cdot \vect{v}) =2 \vect{v}\cdot \frac{\dd \vect{v}}{\dd t} $$

Therefore

$$ \vect{F}\cdot \vect{v} = \vect{F}\cdot \frac{\dd \vect{r}}{\dd t} = m\vect{v}\cdot \frac{\dd \vect{v}}{\dd t} = m \left({\frac{1}{2}}\frac{\dd }{\dd t}(\vect{v}\cdot \vect{v})\right) = \frac{1}{2}m\frac{\dd v^2}{\dd t} $$

Integrating you have

$$ \frac{1}{2}m \Delta v^2 = \int_\gamma\dd {\vect r}\cdot \vect F $$

Two years have gone by since I asked this question, and now after some time at uni here's my take on a "rigorous" (whatever I meant with that at the time) derivation (Since now I don't find the other answer very clear, what's the curve integrated over, what's the force field the particle is moving in?):

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Suppose the position of a particle of mass $m > 0$ moving between time $0\in \mathbb{R}$ and time $1 \in \mathbb{R}$ is given by a smooth curve $x: [0,1]\to\mathbb{R^3}$. If the particle moves through a force field $F: \mathbb{R^3} \to \mathbb{R^3}$, then the work done on the particle is given by the line integral $$\Delta W=\int_xds\cdot F(s).$$

We want to find the work needed to accelerate the object from it's initial velocity $\dot{x}(0)$ to it's final velocity $\dot{x}(1)$. Suppose there are no external forces, only the inertia of the particle is "trying to counteract" the acceleration. For the sake of the argument, suppose further that $x$ is injective. Then the force field of the intertia at point $y \in x([0,1])$ is given as follows by Newton's third law: $$F(y):=m\ddot{x}(x^{-1}(y)) \in \mathbb{R^3}.$$ The expression simply assigns every point on the trajectory of the particle the force needed to give the particle it's current acceleration.

The work integral computes as follows: $$\Delta W = \int_x ds\cdot F(s)= \int_0^1 dt \langle F(x(t)) \ | \ \dot{x}(t)\rangle = \int_0^1 dt \langle m\ddot{x}(x^{-1}(x(t))) \ | \ \dot{x}(t)\rangle = m \int_0^1 dt \langle \ddot{x}(t) \ | \ \dot{x}(t)\rangle.$$ One quickly checks that $\langle\ddot{x}(t) | \dot{x}(t)\rangle = \frac{1}{2}\frac{d}{dt} ||\dot{x}(t)||^2$. So by defining $\Delta v^2 := ||\dot{x}(1)||^2-||\dot{x}(0)||^2$, the work integral becomes $$\Delta W = \frac{1}{2}m \int_0^1 dt \frac{d}{dt}||\dot{x}(t)||^2 = \frac{1}{2}m(||\dot{x}(1)||^2-||\dot{x}(0)||^2) = \frac{1}{2}m\Delta v^2,$$ which is the familiar expression of the kinetic energy.