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- Thread starter Ahmed1029
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In summary, if the inertial Cartesian coordinates as functions of the generalized coordinates depend explicitly on time, the kinetic energy in the Lagrangian will also be explicitly time dependent. This is because the time derivative of the position vector includes a term involving the partial derivative with respect to time.

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$$\vec{x}=\vec{x}(q^k,t), \quad k \in \{1,\ldots,f \}$$

the time derivative (Einstein summation convention applies)

$$\dot{\vec{x}}=\dot{q}^k \partial_k \vec{x} + \partial_t \vec{x}$$

and thus

$$T=\frac{m}{2} \dot{\vec{x}}^2 = \frac{m}{2} \left (\dot{q}^k \partial_k \vec{x} + \partial_t \vec{x} \right)^2,$$

which is in general explicitly time dependent.

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