# The time derivative of kinetic energy

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• LagrangeEuler
Therefore, ##\frac{dp}{dt} = \vec{F}## and using the chain rule, ##\frac{dT}{dt} = \frac{1}{2m}2p\frac{dp}{dt} = \frac{1}{2m}2p\vec{F} = \vec{v}\cdot \vec{F}##. In summary, we can see that ##\frac{dT}{dt}=\vec{v}\cdot \vec{F}## from considering the equations for kinetic energy and Newton's second law.

#### 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 $$\vec{v}\cdot \vec{F}$$?

LagrangeEuler said:
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 $$\vec{v}\cdot \vec{F}$$?
Well, ##\vec {p} = m \vec {v}## and ##\vec F = \frac{d\vec p}{dt}## is Newton's second law.

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