Showing $ m\int \frac{d\textbf{v}}{dt} \dot \normalsize \textbf{v}dt = \frac{m}{2}\int \frac{d(v^2)}{dt}{}dt$

Question

Can someone please explain how this equation is valid, using intermediate steps if available?

$$ m\int \frac{d\textbf{v}}{dt} \dot \normalsize \textbf{v}dt = \frac{m}{2}\int \frac{d(v^2)}{dt}{}dt$$

And what does the right side mean, if the $dt$ cancels with the $\frac {1}{dt}$?

Answer 1

Using

$v^2 = \vec v \cdot \vec v$

and

$\dfrac{d}{dt} (f g) = f \dfrac{dg}{dt} + \dfrac{df}{dt} g $

write:

$\dfrac{d}{dt}(v^2) = \dfrac{d}{dt}(\vec v \cdot \vec v) = \vec v \cdot \dfrac{d\vec v}{dt} + \dfrac{d\vec v}{dt} \cdot \vec v = 2 (\dfrac{d\vec v}{dt} \cdot \vec v)$

The fact that the $dt$ and $1/dt$ "cancel" in the RHS integral means that you're integrating a differential.

$\int \dfrac{dx}{dt} dt = \int dx = x$

Answer 2

Let $v(t)$ be velocity vector and define the scalar function

$$ f(t)~:=~\langle v(t),v(t)\rangle, $$

which is the absolute value of velocity (speed) squared. The time derivative is

$${d \over dt}f(t)~=~ \left\langle {d \over dt}v(t),v(t)\right\rangle + \left\langle v(t),{d \over dt} v(t) \right\rangle ~=~ 2\ \left\langle{d \over dt}v(t),v(t)\right\rangle. $$

Your equation is just integration of this written in physics language.