Let's break down the problem step by step: --- ## **Step 1: Write the new Lagrangian** Given: \[ \tilde{L}(q, \dot{q}, t) = L(q, \dot{q}, t) + \frac{d}{dt} f(q, t) \] where \( f(q, t) \) is any arbitrary function of \( q \) and \( t \). --- ## **Step 2: Compute the Euler-Lagrange Equation for \( \tilde{L} \)** The Euler-Lagrange equation for a Lagrangian \( L \) is: \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} = \] Apply this to \( \tilde{L} \): \[ \frac{d}{dt} \left( \frac{\partial \tilde{L}}{\partial \dot{q}} \right) - \frac{\partial \tilde{L}}{\partial q} = \] --- ## **Step 3: Calculate the Partial Derivatives** ### **a) Compute \(\frac{\partial \tilde{L}}{\partial \dot{q}}\):** \[ \tilde{L}(q, \dot{q}, t) = L(q, \dot{q}, t) + \frac{d}{dt} f(q, t) \] Note: \[ \frac{d}{dt} f(q, t) = \frac{\partial f}{\partial q} \dot{q} + \frac{\partial f}{\partial t} \] So: \[ \tilde{L}(q, \dot{q}, t) = L(q, \dot{q}, t) + \frac{\partial f}{\partial q} \dot{q} + \frac{\partial f}{\partial t} \] Now, \[ \frac{\partial \tilde{L}}{\partial \dot{q}} = \frac{\partial L}{\partial \dot{q}} + \frac{\partial f}{\partial q} \] --- ### **b) Compute \(\frac{\partial \tilde{L}}{\partial q}\):** \[ \frac{\partial \tilde{L}}{\partial q} = \frac{\partial L}{\partial q} + \frac{\partial}{\partial q} \left( \frac{\partial f}{\partial q} \dot{q} + \frac{\partial f}{\partial t} \right) \] \[ = \frac{\partial L}{\partial q} + \frac{\partial^2 f}{\partial q^2} \dot{q} + \frac{\partial^2 f}{\partial q \partial t} \] --- ### **c) Compute \(\frac{d}{dt} \left( \frac{\partial \tilde{L}}{\partial \dot{q}} \right)\):** \[ \frac{d}{dt} \left( \frac{\partial \tilde{L}}{\partial \dot{q}} \right) = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} + \frac{\partial f}{\partial q} \right) \] \[ = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) + \frac{d}{dt} \left( \frac{\partial f}{\partial q} \right) \] \[ = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) + \frac{\partial^2 f}{\partial q^2} \dot{q} + \frac{\partial^2 f}{\partial q \partial t} \] --- ## **Step 4: Plug into Euler-Lagrange Equation** \[ \frac{d}{dt} \left( \frac{\partial \tilde{L}}{\partial \dot{q}} \right) - \frac{\partial \tilde{L}}{\partial q} \] \[ = \left[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) + \frac{\partial^2 f}{\partial q^2} \dot{q} + \frac{\partial^2 f}{\partial q \partial t} \right] - \left[ \frac{\partial L}{\partial q} + \frac{\partial^2 f}{\partial q^2} \dot{q} + \frac{\partial^2 f}{\partial q \partial t} \right] \] \[ = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} \] --- ## **Step 5: Conclusion** The extra terms from \( f(q, t) \) **cancel out** exactly. Therefore, the equations of motion for \( \tilde{L} \) are **identical** to those for \( L \). --- ## **Final Answer** **Adding a total time derivative of any function \( f(q,t) \) to the Lagrangian does not change the equations of motion.** Mathematically, \[ \frac{d}{dt} \left( \frac{\partial \tilde{L}}{\partial \dot{q}} \right) - \frac{\partial \tilde{L}}{\partial q} = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} \] So the equations of motion for \( \tilde{L} \) and \( L \) are **identical**.