On texts of multivariable calculus and real analysis I have always seen the work made by $\mathbf{F}$ along the path $\gamma$ defined as the integral$$\int_\gamma\mathbf{F}\cdot\text{d}s:=\int_a^b\mathbf{F}(t)\cdot\mathbf{r}'(t)\text{d}t$$where $\mathbf{r}:[a,b]\to\mathbb{R}^3$ is a piecewise smooth curve whose support is $\mathbf{r}([a,b])=\gamma$. By piecewise smooth it is intended in the text I have read that $\mathbf{r}$ is the juxtaposition of a finite number of smooth curves $\mathbf{r}_i:[a_i,b_i]\to\mathbb{R}^3$, where smooth means that $$\mathbf{r}_i\in C^1[a_i,b_i]\quad\text{ and }\quad\forall t\in[a_i,b_i]\quad\|\mathbf{r}_i'(t)\|\ne 0.$$This implies that $\forall t\in[a,b]\quad\|\mathbf{r}'(t)\|\ne 0$ and that $\mathbf{r}$ has finite left and right derivative for any $t\in[a,b]$.
All the proofs of the work-energy theorem that I have found in books of physics use the velocity $\mathbf{v}$ of a point, upon which the force $\mathbf{F}$ acts, to parametrise its path $\gamma$ and get, by taking into account Newton's second law $\mathbf{F}=m\frac{\text{d}\mathbf{v}}{\text{d}t}$, $$\int_\gamma\mathbf{F}\cdot\text{d}s=\int_a^b m\frac{\text{d}\mathbf{v}(t)}{\text{d}t}\cdot\mathbf{v}(t)\text{d}t=m\int_a^b\frac{1}{2}\frac{\text{d}\|\mathbf{v}(t)\|^2}{\text{d}t}\text{d}t=\frac{1}{2}m\|\mathbf{v}(b)\|^2-\frac{1}{2}m\|\mathbf{v}(a)\|^2.$$Anyhow, when examples and exercise are presented in texts of physics, $\|\mathbf{v}(a)\|$ or $\|\mathbf{v}(b)\|$ may well be zero, which is, to my eyes, a contradiction of the fact that the parametrisation of $\gamma$ must have a non-null derivative $\|\mathbf{r}'(t)\|=\|\mathbf{v}(t)\|\ne 0$ for all $t\in[a,b]$. What am I missing or misunderstanding? I $\infty$-ly thank you for any answer!
Why would you need to have $||{\bf r}' (t)|| \neq 0$? this constraint is not necessary at all. You have pointed out yourself that you are using Newton's second law, which states ${\bf F} = m \frac{d {\bf v}}{dt}$, if ${\bf F}$ was to be directed in oppoiste direction to ${\bf v}$ and you started with a given value of ${\bf v} = {\bf v}_0$ then certainly $||{\bf r}' (t)||=0$ at some point. There is no contradiction. An example of this is throwing an object vertically in the earth's gravitational field.