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The differential equations for the motion of a particle under any forces when we use rectangular coordinates are known from Newston's laws of motion
where
 are the components of the actual forces on the particle resolved parallel to each of the fixed rectangular axes, or rather their equivalents
 , are called the effective forces on the particle. They are of course a set of forces mechanically equivalent to the actual forces acting on the particle.
The equations of motion of the particle in terms of any other system of coordinates are easily obtained.
Let
 , be the coordinates in question. The appropriate formulas for transformation of coordinates express  in terms of
 .
For the component velocity  we have
and
 are explicit functions of
 linear and homogeneous in terms of
 .
We may note in passing that it follows from this fact that
 are homogeneous quadratic functions of
 .
Obviously
and since
and
Let us now find an expression for the work
 done by the effective forces when the coordinate  is changed by an infinitesimal amount
 without changing  or  . If
 are changes thus produced in  , obviously from the definition of work
if expressed in rectangular coordinates. We need, however, to express
 in terms of our coordinates
 .
Now
but from earlier definitions
Hence
and therefore
![$\displaystyle \delta_{q_1} W = \left [ \frac{d}{dt} \frac{\partial T}{\partial \dot{q_1}} - \frac{\partial T}{\partial q_1} \right ] \delta q_1$](http://images.physicslibrary.org/cache/objects/286/l2h/img35.png "$\displaystyle \delta_{q_1} W = \left [ \frac{d}{dt} \frac{\partial T}{\partial \dot{q_1}} - \frac{\partial T}{\partial q_1} \right ] \delta q_1$") |
(1) |
where
and is the kinetic energy of the particle.
To get our differential equation we have only to write the second member of (1) equal to the work done by the actual forces when  is changed by
 .
If we represent the work in question by
 , our equation is
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(2) |
and of course we get such an equation for every coordinate. Even though we derived this differential equation for a single particle in free motion, it is the same for a systems of particles, except the kinetic energy is for all the particles in the system, which brings us to Lagrange's equations
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(3) |
In any concrete problem,  must be expressed in terms of
 , and their time derivatives before we can form the expression for the work done by the effective forces.
 , the work done by the actual forces, must be obtained from direct examination of the problem.
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![$\displaystyle \delta_{q_1}W = m\left [ \ddot{x} \delta x + \ddot{y} \delta y + \ddot{z} \delta z \right ]$](http://images.physicslibrary.org/cache/objects/286/l2h/img28.png "$\displaystyle \delta_{q_1}W = m\left [ \ddot{x} \delta x + \ddot{y} \delta y + \ddot{z} \delta z \right ]$")
![$\displaystyle \delta_{q_1}W = m\left [ \ddot{x} \frac{\partial x}{\partial q_1}... ... y}{\partial q_1} + \ddot{z} \frac{\partial z}{\partial q_1} \right ]\delta q_1$](http://images.physicslibrary.org/cache/objects/286/l2h/img31.png "$\displaystyle \delta_{q_1}W = m\left [ \ddot{x} \frac{\partial x}{\partial q_1}... ... y}{\partial q_1} + \ddot{z} \frac{\partial z}{\partial q_1} \right ]\delta q_1$")
![$\displaystyle T = \frac{m}{2} \left [ \dot{x}^2 + \dot{y}^2 + \dot{z}^2 \right ] $](http://images.physicslibrary.org/cache/objects/286/l2h/img36.png "$\displaystyle T = \frac{m}{2} \left [ \dot{x}^2 + \dot{y}^2 + \dot{z}^2 \right ] $")