# The time derivative of kinetic energy

LagrangeEuler
Lets consider $$T(\vec{p})=\frac{\vec{p}^2}{2m}=\frac{\vec{p}\cdot \vec{p}}{2m}$$. Then $$\frac{dT}{dt}=\vec{v}\cdot \vec{F}$$.
And if we consider
$$T=\frac{p^2}{2m}$$ than $$\frac{dT}{dt}=\frac{1}{2m}2p\frac{dp}{dt}$$
Could I see from that somehow that this is $$\vec{v}\cdot \vec{F}$$?

Lets consider $$T(\vec{p})=\frac{\vec{p}^2}{2m}=\frac{\vec{p}\cdot \vec{p}}{2m}$$. Then $$\frac{dT}{dt}=\vec{v}\cdot \vec{F}$$.
$$T=\frac{p^2}{2m}$$ than $$\frac{dT}{dt}=\frac{1}{2m}2p\frac{dp}{dt}$$
Could I see from that somehow that this is $$\vec{v}\cdot \vec{F}$$?