Editing Kinetic energy (section)

=====With vector calculus=====
The work done in accelerating a particle with mass ''m'' during the infinitesimal time interval ''dt'' is given by the dot product of ''force'' '''F''' and the infinitesimal ''displacement'' ''d'''''x'''

<math display="block">\mathbf{F} \cdot d \mathbf{x} = \mathbf{F} \cdot \mathbf{v} d t = \frac{d \mathbf{p}}{d t} \cdot \mathbf{v} d t = \mathbf{v} \cdot d \mathbf{p} = \mathbf{v} \cdot d (m \mathbf{v})\,,</math>

where we have assumed the relationship '''p'''&nbsp;=&nbsp;''m''&nbsp;'''v''' and the validity of [[Newton's second law]]. (However, also see the special relativistic derivation [[#Relativistic kinetic energy of rigid bodies|below]].)

Applying the [[product rule]] we see that:

<math display="block">d(\mathbf{v} \cdot \mathbf{v}) = (d \mathbf{v}) \cdot \mathbf{v} + \mathbf{v} \cdot (d \mathbf{v}) =  2(\mathbf{v} \cdot d\mathbf{v}).</math>

Therefore (assuming constant mass so that ''dm'' = 0), we have

<math display="block">\mathbf{v} \cdot d (m \mathbf{v}) = \frac{m}{2} d (\mathbf{v} \cdot \mathbf{v}) = \frac{m}{2} d v^2  = d \left(\frac{m v^2}{2}\right).</math>

Since this is a [[total differential]] (that is, it only depends on the final state, not how the particle got there), we can integrate it and call the result kinetic energy:

<math display="block">E_\text{k} = \int_{v_1}^{v_2}\mathbf{p}d\mathbf{v} =  \int_{v_1}^{v_2}m\mathbf{v}d\mathbf{v} = {mv^2\over 2}\bigg\vert_{v_1}^{v_2} = {1\over 2}m(v_2^2-v_1^2).</math>

This equation states that the kinetic energy (''E''<sub>k</sub>) is equal to the [[integral]] of the [[dot product]] of the [[momentum]] ('''p''') of a body and the [[infinitesimal]] change of the [[velocity]] ('''v''') of the body. It is assumed that the body starts with no kinetic energy when it is at rest (motionless).