Derivative of Action Integral Equals Generalized Momentum?

[shinobi20]

Homework Statement

I need to find the partial derivative of the action S with respect to the generalized coordinate q(tf) and according to my textbook, it should equal the generalized momentum p(tf). How can I derive this?

Homework Equations

S = integral of L dt, with boundary ti to tf. (ti and tf are initial and final times)
qi = 0 (initial generalized coordinate)

The Attempt at a Solution

I took the partial derivative of both sides with respect to the generalized coordinate q, so the right side will have partial L partial q (which is equal to (p dot)) so I'm left with integral of (p dot) dt, but this is simple p(tf) - p(ti). Since qi = 0, p(ti) = 0. Hence partial S partial q(tf) = p(tf).

Is this a possible solution? I have also uploaded my solution. Thanks.

 

[andrewkirk]

I think you can stop and declare victory as soon as you get to $p(t_f)-p(t_i)$. That's a momentum quantity, which is all that's needed.

It is not necessarily the case that $p(t_i)=0$ or that $q(t_i)=0$. Even if they were, a change of coordinates such as a shift of origin and/or a velocity boost could render them nonzero.