Work as integral of change in kinetic energy In the book I'm reading, the work done by a force acting on a particle is given as:
$$W= m\int_{t1}^{t2}\dot{\textbf{v}}\cdot \textbf{v} \,dt = \frac{m}{2} \int_{t1}^{t2}\frac{d}{dt}v^{2}\,dt$$
I don't understand the second step, why can we say that $\dot{\textbf{v}}\cdot \textbf{v} = v \,\dot{v}$ ? Wouldn't that only be the case if $\dot{\textbf{v}}$ and $\textbf{v}$ were pointing in the same direction?
 A: This step is indeed non-trivial; this is what Wikipedia says about it (using the notation $\mathbf{a}$ instead of $\dot{\mathbf{v}}$):

The identity $\textstyle \mathbf{a} \cdot \mathbf{v} = \frac{1}{2} \frac{d v^2}{dt}$ requires some algebra.
From the identity $\textstyle v^2 = \mathbf{v} \cdot \mathbf{v}$ and definition $\textstyle \mathbf{a} = \frac{d \mathbf{v}}{dt}$
it follows: $\frac{d v^2}{dt} = \frac{d (\mathbf{v} \cdot \mathbf{v})}{dt} = \frac{d \mathbf{v}}{dt} \cdot \mathbf{v} + \mathbf{v} \cdot \frac{d \mathbf{v}}{dt} = 2 \frac{d \mathbf{v}}{dt} \cdot \mathbf{v} = 2 \mathbf{a} \cdot \mathbf{v}$.

A: The general proof is already given by @Glorfindel, Here the geometrical intuision behind it which can be found in elementary text :
Consider a vector $\mathbf{A}(t)$. From the definition of derivatives :
$$\frac{d\mathbf{A}}{dt}=\lim_{\Delta t\rightarrow 0}\frac{\mathbf{A}(t+\Delta t)-\mathbf{A}(t)}{\Delta t}$$

In the figure, it's a general  scenario. Note that In the limit $\Delta A\rightarrow 0$, as when we take the derivative, $\Delta A_{||}$  changes the magnitude of $\mathbf{A}$ but not its direction, while $\Delta A_{\perp}$ changes the direction of $\mathbf{A}$ but not its magnitude.

Now let's go back to out main concern:
$$\mathbf{v}\cdot\dot{\mathbf{v}}$$
Here dot product with $\mathbf{v}$ is saying that we are concern with only the parallel component or the component which change the magnitude that  is $dv^2/dt$. So that directly proves :
$$\mathbf{v}\cdot\dot{\mathbf{v}}=\frac{dv^2}{dt}$$
