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

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 $\frac {1}{dt}$?

-
Joebevo, in words, what does this equation express? –  Alfred Centauri Jul 20 '12 at 12:12
Let's see.. the left hand side (inside the integral) is equivalent to the acceleration dotted with an infinitesimal displacement. I'm not quite sure how to interpret the part inside the integral on the right hand side.. the differential of $v^2$.. I haven't seen this very often. I know that a $dx$ is an infinitesimal displacement along the $x$-axis, but what is a $d(v^2)$? –  Joebevo Jul 20 '12 at 13:09
OK then, forget the integrals (take the time derivative of both sides) and then you have: $m \vec a \cdot \vec v = \dfrac{d}{dt}(\frac{1}{2}m v^2)$. What does this express? –  Alfred Centauri Jul 20 '12 at 13:13
Ah.. work done equals rate of change of KE. :) –  Joebevo Jul 20 '12 at 13:19
Close! But not quite there. See: link –  Alfred Centauri Jul 20 '12 at 13:36

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.

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

-

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.

-
I've never seen this $<v(t),v(t)>$ operator (or whatever you call it). Could you tell me more about it? What is it called? Any references? Which part of math does it come from? Thanks. –  Joebevo Jul 20 '12 at 10:56
It's just the notation for inner product –  straups Jul 20 '12 at 14:30
I loved the answer but, in this context, I think the notation over-complexes the OP's problem, given his apparent level (he's trying to understand classical mechanics!). I know, however, that this notation is really normal to the people that have had a QM course or some Lagrangian mechanics, but I think that, like when trying to learn a new language, one has to be careful with the words in order to be more clear :-). –  Néstor Jul 20 '12 at 23:25