If all the influences on a [point particle] are represented as [works|work], the net work done by the forces produces a change in the kinetic energy of the particle according to:

{latex}\begin{large}\[ \Delta K = W_{\rm net}\]\end{large}{latex}

From [Newton's 2nd Law|Newton's Second Law] for a point particle, we know

{latex}\begin{large}\[ \vec{F}_{\rm net} = m\frac{d\vec{v}}{dt}\]\end{large}{latex}

Now suppose that the particle undergoes an infinitesimal displacement _dr_.  Since we want to bring the left side of the equation into line with the form of the expression for work, we take the dot product of each side with the displacement:

{latex}\begin{large}\[ \vec{F}_{\rm net}\cdot d\vec{r} = m\frac{d\vec{v}}{dt} \cdot d\vec{r} \]\end{large}{latex}

Before we can integrate, we make a substitution.  Since _v_ is the velocity of the particle, we can re-express the infinitesimal displacement as:

{latex}\begin{large}\[ d\vec{r} = \vec{v}dt\]\end{large}{latex}

Making this substitution on the right hand side of the equation, we have:

{latex}\begin{large}\[ \vec{F}_{\rm net}\cdot d\vec{r} = m\frac{d\vec{v}}{dt}\cdot \vec{v}\:dt = m\vec{v}\cdot d\vec{v} = m(v_{x}\;dv_{x} + v_{y}\;dv_{y}+v_{z}\;dv_{z})\]\end{large}{latex}

We can now integrate over the path:

{latex}\begin{large}\[ \int_{\rm path} \vec{F}_{\rm net} \cdot d\vec{r} = \frac{1}{2}m(v_{x,f}^{2}-v_{x,i}^{2} + v_{y,f}^{2} - v_{y,i}^{2} + v_{z,f}^{2} - v_{z,i}^{2}) = \frac{1}{2}m(v_{f}^{2}-v_{i}^{2})\]\end{large}{latex}

which is equivalent to the Work-Kinetic Energy Theorem.