Considering what Greg Graviton wrote, I'll write out the derivation and see if I can make sense of it.

$$ S = \int_{t_1}^{t_2} L(q, \dot q, t) dt $$

where S is the action and L the Lagrangian. We vary the path and find the extremum of the action:

$$ \delta S = \int_{t_1}^{t_2} ({\partial L \over \partial q}\delta q + {\partial L \over \partial \dot q}\delta \dot q) dt = 0. $$

Here, q and $\dot q$ are varied independently. But then in the next step we use this identity,

$$ \delta \dot q = {d \over dt} \delta q. $$

And here is where the relationship between q and $\dot q$ enters the picture. I think that what is happening here is that q and $\dot q$ are treated as independent initially, but then the independence is removed by the identity.

$$ \delta S = \int_{t_1}^{t_2} ({\partial L \over \partial q}\delta q + {\partial L \over \partial \dot q}{d \over dt} \delta q) dt = 0 $$

And then follows the rest of the derivation. We integrate the second term by parts:

$$ \delta S = \lbrack {\partial L \over \partial \dot q}\delta q\rbrack_{t_1}^{t_2} + \int_{t_1}^{t_2} ({\partial L \over \partial q} - {d \over dt}{\partial L \over \partial \dot q})\delta q dt = 0, $$

and the bracketed expression is zero because the endpoints are held fixed. And then we can pull out the Euler-Lagrange equation:

$$ {\partial L \over \partial q} - {d \over dt}{\partial L \over \partial \dot q} = 0. $$

Now it makes more sense to me. You start by treating the variables as independent but then remove the independence by imposing a condition during the derivation.

I think that makes sense. I expect in general other problems can be treated the same way.

(I copied the above equations from Mechanics by Landau and Lifshitz.)

Considering what Greg Graviton wrote, I'll write out the derivation and see if I can make sense of it.

$$ S = \int_{t_1}^{t_2} L(q, \dot q, t)\, \mathrm{d}t $$

where S is the action and L the Lagrangian. We vary the path and find the extremum of the action:

$$ \delta S = \int_{t_1}^{t_2} \left({\partial L \over \partial q}\delta q + {\partial L \over \partial \dot q}\delta \dot q\right) \,\mathrm{d}t = 0\,. $$

Here, q and $\dot q$ are varied independently. But then in the next step we use this identity,

$$ \delta \dot q = {\mathrm{d} \over \mathrm{d}t} \delta q. $$

And here is where the relationship between q and $\dot q$ enters the picture. I think that what is happening here is that q and $\dot q$ are treated as independent initially, but then the independence is removed by the identity.

$$ \delta S = \int_{t_1}^{t_2} \left({\partial L \over \partial q}\delta q + {\partial L \over \partial \dot q}{d \over \mathrm{d}t} \delta q\right) \,\mathrm{d}t = 0 $$

And then follows the rest of the derivation. We integrate the second term by parts:

$$ \delta S = \left[ {\partial L \over \partial \dot q}\delta q\right]_{t_1}^{t_2} + \int_{t_1}^{t_2} \left({\partial L \over \partial q} - {d \over dt}{\partial L \over \partial \dot q}\right)\delta q\, \mathrm{d}t = 0\,, $$

and the bracketed expression is zero because the endpoints are held fixed. And then we can pull out the Euler-Lagrange equation:

$$ {\partial L \over \partial q} - {\mathrm{d} \over \mathrm{d}t}{\partial L \over \partial \dot q} = 0\,. $$

Now it makes more sense to me. You start by treating the variables as independent but then remove the independence by imposing a condition during the derivation.

I think that makes sense. I expect in general other problems can be treated the same way.

(I copied the above equations from Mechanics by Landau and Lifshitz.)

Considering what Greg Graviton wrote, I'll write out the derivation and see if I can make sense of it.

$$ S = \int_{t_1}^{t_2} L(q, \dot q, t) dt $$

where S is the action and L the Lagrangian. We vary the path and find the extremum of the action:

$$ \delta S = \int_{t_1}^{t_2} ({\partial L \over \partial q}\delta q + {\partial L \over \partial \dot q}\delta \dot q) dt = 0. $$

Here, q and $\dot q$ are varied independently. But then in the next step we use this identity,

$$ \delta \dot q = {d \over dt} \delta q. $$

And here is where the relationship between q and $\dot q$ enters the picture. I think that what is happening here is that q and $\dot q$ are treated as independent initially, but then the independence is removed by the identity.

And then follows the rest of the derivation. We integrate by parts:

$$ \delta S = \lbrack {\partial L \over \partial \dot q}\delta q\rbrack_{t_1}^{t_2} + \int_{t_1}^{t_2} ({\partial L \over \partial q} - {d \over dt}{\partial L \over \partial \dot q})\delta q dt = 0, $$

and the bracketed expression is zero because the endpoints are held fixed. And then we can pull out the Euler-Lagrange equation:

$$ {\partial L \over \partial q} - {d \over dt}{\partial L \over \partial \dot q} = 0. $$

Now it makes more sense to me. You start by treating the variables as independent but then remove the independence by imposing a condition during the derivation.

I think that makes sense. I expect in general other problems can be treated the same way.

(I copied the above equations out of Mechanics by Landau and Lifshitz.)

Considering what Greg Graviton wrote, I'll write out the derivation and see if I can make sense of it.

$$ S = \int_{t_1}^{t_2} L(q, \dot q, t) dt $$

where S is the action and L the Lagrangian. We vary the path and find the extremum of the action:

$$ \delta S = \int_{t_1}^{t_2} ({\partial L \over \partial q}\delta q + {\partial L \over \partial \dot q}\delta \dot q) dt = 0. $$

Here, q and $\dot q$ are varied independently. But then in the next step we use this identity,

$$ \delta \dot q = {d \over dt} \delta q. $$

And here is where the relationship between q and $\dot q$ enters the picture. I think that what is happening here is that q and $\dot q$ are treated as independent initially, but then the independence is removed by the identity.

$$ \delta S = \int_{t_1}^{t_2} ({\partial L \over \partial q}\delta q + {\partial L \over \partial \dot q}{d \over dt} \delta q) dt = 0 $$

And then follows the rest of the derivation. We integrate the second term by parts:

$$ \delta S = \lbrack {\partial L \over \partial \dot q}\delta q\rbrack_{t_1}^{t_2} + \int_{t_1}^{t_2} ({\partial L \over \partial q} - {d \over dt}{\partial L \over \partial \dot q})\delta q dt = 0, $$

and the bracketed expression is zero because the endpoints are held fixed. And then we can pull out the Euler-Lagrange equation:

$$ {\partial L \over \partial q} - {d \over dt}{\partial L \over \partial \dot q} = 0. $$

Now it makes more sense to me. You start by treating the variables as independent but then remove the independence by imposing a condition during the derivation.

I think that makes sense. I expect in general other problems can be treated the same way.

(I copied the above equations from Mechanics by Landau and Lifshitz.)