In 3-D space, a particle with mass m\,\!, velocity \mathbf{v} has
kinetic energy T T =\frac{1}{2}m v^2 . Velocity is the derivative of position r with respect to time t\,\!. Use
chain rule for several variables: \mathbf{v} = \frac{d\mathbf{r}}{dt} = \sum_i\ \frac{\partial \mathbf{r}}{\partial q_i} \dot{q}_i + \frac{\partial \mathbf{r}}{\partial t} . where q_i are
generalized coordinates. Therefore, T = \frac{1}{2} m \left(\sum_i\ \frac{\partial \mathbf{r}}{\partial q_i}\dot{q}_i+\frac{\partial \mathbf{r}}{\partial t}\right)^2 . Rearranging the terms carefully, \begin{align} T &= T_0 + T_1 + T_2 : \\[1ex] T_0 &= \frac{1}{2} m \left(\frac{\partial \mathbf{r}}{\partial t}\right)^2 , \\ T_1 &= \sum_i\ m\frac{\partial \mathbf{r}}{\partial t}\cdot \frac{\partial \mathbf{r}}{\partial q_i}\dot{q}_i\,\!, \\ T_2 &= \sum_{i,j}\ \frac{1}{2}m\frac{\partial \mathbf{r}}{\partial q_i}\cdot \frac{\partial \mathbf{r}}{\partial q_j}\dot{q}_i\dot{q}_j, \end{align} where T_0\,\!, T_1\,\!, T_2 are respectively
homogeneous functions of degree 0, 1, and 2 in generalized velocities. If this system is scleronomous, then the position does not depend explicitly with time: \frac{\partial \mathbf{r}}{\partial t}=0. Therefore, only term T_2 does not vanish: T = T_2. Kinetic energy is a homogeneous function of degree 2 in generalized velocities. ==Example: pendulum==