# I Euler-Lagrange query

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1. Dec 20, 2016

### volican

2. Dec 20, 2016

### BvU

Hello Volican,

It comes from the time derivative of $\partial {\mathcal L}\over \partial \dot\theta$ $${\partial {\mathcal L}\over \partial \dot\theta }= mr^2\dot \theta$$

3. Dec 20, 2016

### volican

Thanks for writting back, much appreciated. I can see where that comes from and also the -mgrsinθ but cant see how they got the 2mr˙r˙θ. Any ideas? I am thinking that I can just completley ignore the dr/dt as it is not in the euler lagrange equation? Is this wrong?

4. Dec 20, 2016

### BvU

$${d\over dt} \left ( {\partial {\mathcal L}\over \partial \dot\theta } \right ) = {d\over dt} mr^2\dot \theta = m {d\over dt} \left ( r^2\right ) \dot \theta + mr^2 {d\over dt} \dot \theta = 2mr\dot r\dot\theta + mr^2\ddot\theta$$

5. Dec 20, 2016

### volican

Aewsome! to differentiate r^2 did you use the chain rule?

6. Dec 20, 2016

### BvU

Yes, the chain rule is awesome $${d\over dt} r^2 = {d\over dt} \left ( r \right ) r + r {d\over dt} \left ( r \right ) = 2 r\dot r$$

Make sure you speak this jargon fluently when pursuing further studies !

Last edited: Jan 11, 2017