A Work-Kinetic energy proof that is mathematically sound?

Question

I was looking at proofs of $W=\Delta \text{KE}$ along curves online and they all seem sloppy to me. I'm wondering if this specific proof is mathematically sound, or if anyone has seen a mathematically sound proof:

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Let $v=\frac{dr}{dt}$ denote the velocity vector, as let $r$ denote the position vector. Let $F=m\frac{dv}{dt}$ denote the force at a specified position. Let the mass $m$ be constant. Then the work done along a curve $C$ is defined as,

$$W=\int_{C} F \cdot dr=m \int_{C} \frac{dv}{dt} \cdot dr$$

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The next steps seem sloppy to me.

$$=m \int_{v_i}^{v_f} \frac{dr}{dt} \cdot dv$$

$$=m \int_{v_i}^{v_f} v \cdot dv$$

$$=m \int_{|v_i|}^{|v_f|} |v| \,d|v|$$

$$=\frac{1}{2}m|v_f|^2-\frac{1}{2}m|v_i|^2$$

$$=\Delta \text{KE}$$

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Three things seems weird to me: The treating of differential vectors as if they were just vectors, and the limits of the integration as we change variables of integration. Also even so, assuming we can treat them as vectors, how is $|dv|=d|v|$?

If it's sloppy mathematically, does anyone know how to correct it, I know for example in single variable calculus crossing out differentials can usually be made rigorous by just quoting the chain rule. What about here?

Answer 1

for the physical curves $C$ with a path between $t_1$ and $t_2$ we have from the fundamental theorem of calculus that

$$\int_C\nabla f\cdot \mathrm{d}\mathbf{r}=f(\mathbf{r}(t_2))-f(\mathbf{r}(t_2))$$

and from this your desired result follows (use after change of integration variable to $\mathbf{v}$ to get kinetic energy)

EDIT:

\begin{align*} W&=\int_C \mathbf{F}\cdot\mathrm{d}\mathbf{r} \\ &=m\int_C \mathbf{v}\cdot\mathrm{d}\mathbf{v} \\ &=m\int_C \nabla_v\left( \frac{\mathbf{v}^2}{2}\right)\cdot\mathrm{d}\mathbf{v} \\ &=\frac{1}{2}m\mathbf{v}(t_2)^2-\frac{1}{2}m\mathbf{v}(t_1)^2 \\ &=\Delta\mathrm{KE} \end{align*}

Answer 2

The work of the net force (which is mathematically just stating that the equation $\mathbf{F} \circ \mathbf{c}=m\mathbf{\ddot{c}}$ holds) is given by the variation of kinectic enery, and this is just an application of the fundamental theorem of calculus and a simple computation: that of the derivative of the real function $$E:\mathbb{R} \to \mathbb{R}$$ $$t \mapsto \frac{m}{2}\langle \mathbf{\dot{c}}(t), \mathbf{\dot{c}}(t)\rangle.$$ We have that $E'(t)=\langle m\mathbf{\ddot{c}}(t),\mathbf{\dot{c}}(t) \rangle$ (this computation does not need Newton's second law, and is mathematical only). However, the fact that $E'(t)=\langle \mathbf{F}(\mathbf{c}(t)),\mathbf{\dot{c}}(t) \rangle$ needs it.

By the fundamental theorem of calculus, we have that \begin{align*} E(b)-E(a)=\int_a^bE'=\int_a^b\langle \mathbf{F}(\mathbf{c}(t)),\mathbf{\dot{c}}(t)\rangle dt, \end{align*} and this last one is the definition of the work of $\mathbf{F}$ along $\mathbf{c}$.