Suggested Problem: Hamilton's principle of least action applies given the missile trajectory is a parabola which motion is made in least time, i.e. for which the action S is a minimum:
S = ò t2 t1 [½ m dx(t) 2/ dt 2 - mg x(t)] dt
Or: Letting y = x = x(t) to obtain the least time context:
S = ò t2 t1 [½ m dx 2/ dt 2 - mgx] dt
Say between time t = t1 and t = t2. The path taken is always such that we have a least difference between kinetic energy and potential energy on average. Hence,
ò t2 t1 [T - V] dt = min
I.e. the action is a minimum for the (true, i.e. x) path that satisfies the differential equation:
[- m dx 2/ dt 2 - V' x] = 0
We see at missile launch the KE is high then becomes zero at apex, while PE is zero initially then becomes a maximum at apex. Throughout the trajectory the balance of PE (V) and KE (T) changes but the average remains a minimum to comport with least time, i.e. from launch point to end point.
If y = A sin x defines the true (x) path then another expression representing a variation on it might be: y = 3A sin x or in the least time coordinates: e.g. x(t) = 3 A sin wt. Given a small variation h then the variation in the action would be:
dS = ò t2 t1 [½ m (dx / dt) (d h/ dt) - h V' x] dt
where x defines the true or natural path.
The Lagrangian for the natural path (x) is:
L = ½ m dx 2/ dt 2 - mg x
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