Why sign changed when I had put derivative respect to time before another value?

Question

In the answer, he wrote that

$$\frac{\partial L}{\partial \dot{q}}\frac{d}{dt}\delta q=-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}\delta q$$

My question is why the sign changed when he had put derivative respect to time before Lagrangian "formalism". Even someone in the comment told me the same thing. But I couldn't understand. I think I am missing something in Differentiation,ain't I?

score 1 · Answer 1 · answered Oct 07 '21 at 16:10

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He is partially integrating $\int_{0}^{1} \frac{\partial L}{\partial \dot{q}}\delta \dot{q} dt$ and assume that $\frac{\partial L}{\partial \dot{q}}\delta q$ is $0$ on the bounds.

answered Oct 07 '21 at 16:10

Gábor Pálovics

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It wasn't my question. My question was about sign not this one.. – Unknown Oct 08 '21 at 04:01
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When you partiallyintegrate $\int_{0}^{1} \frac{\partial L}{\partial \dot{q}}\delta \dot{q} dt = \left[\frac{\partial L}{\partial q } \delta q\right]0^1 - \int{0}^{1} \delta q\frac{d}{dt} \frac{\partial L}{\partial \dot{q}} dt $ – Gábor Pálovics Oct 08 '21 at 05:22
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The minus sign comes from the partial integration. – Gábor Pálovics Oct 08 '21 at 05:38

Why sign changed when I had put derivative respect to time before another value?

1 Answers1