Q:

Kinetic energy derivation: Why is $\frac{d \mathbf v}{dt} \cdot \mathbf v= \frac 12 \frac{d}{dt}(v^2)~?$

In Goldstein's Classical Mechanics 3rd edition, page 3, the Kinetic energy is derived by considering the work done on a particle by an external force $\mathbf F$ from point $1$ to point $2$ $$W_{12}=\int_1^2\mathbf F \cdot d\mathbf s.$$ It is then argued that for constant mass, the integral reduces to $$W_{12}=m\int \frac{d \mathbf v}{dt} \cdot \mathbf v dt=\frac m2 \int \frac{d}{dt}(v^2)dt$$ from which you get the usual formula $W_{12}=\frac m2 (v_2^2-v_1^2)$. I'm confused however on how the last integral above was obtained. Why must $$\frac{d \mathbf v}{dt} \cdot \mathbf v= \frac 12 \frac{d}{dt}(v^2)$$ when it may not be the case that $\frac {d \mathbf v}{dt}$ and $\mathbf v$ are parallel? Shouldn't this depend on the angle between the velocity and acceleration vectors?

A:

I'm confused however on how the last integral above was obtained.

$$v^2 = \mathbf{v}\cdot\mathbf{v}$$
$$\frac{d}{dt}(v^2) =  \frac{d}{dt}(\mathbf{v}\cdot\mathbf{v}) = \frac{d\mathbf{v}}{dt}\cdot\mathbf{v} + \mathbf{v}\cdot\frac{d\mathbf{v}}{dt} = 2\left(\frac{d\mathbf{v}}{dt}\cdot\mathbf{v}\right)$$
$$\Rightarrow \frac{d\mathbf{v}}{dt}\cdot\mathbf{v} = \frac{1}{2}\frac{d}{dt}(v^2) $$

when it may not be the case that $\frac{d\mathbf{v}}{dt}$ and $\mathbf{v}$ are parallel?

Consider, for example, that case that $\frac{d\mathbf{v}}{dt}$ is perpendicular to $\mathbf{v}$ and then the left-hand side must be zero.
But, as we know, for uniform circular motion, the acceleration is always perpendicular to the velocity and that the speed $v = \sqrt{v^2}$ is then constant and thus, the right-hand side is zero too which is consistent.  

