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~~> A particle is subjected to the potential V (x) = −F x, where F is a constant. The~~
~~particle travels from x = 0 to x = a in a time interval t0 . Assume the motion of the~~
~~particle can be expressed in the form $x(t) = A + B t + C t^2$ . Find the values of A, B,~~
~~and C such that the action is a minimum.~~
I was thinking it can solved using Lagrangian rather than Hamilton. There's no frictional force.
$$L=\frac{1}{2}m\dot{x}^2+Fx$$
$$\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})-\frac{\partial L}{\partial x}=0$$
$$m\ddot{x}=F$$
$$\ddot{x}=\frac{F}{m}$$
Differentiate $x(t)$ twice. $$2C=\frac{F}{m}=>C=\frac{F}{2m}$$
For finding B I was thinking to integrate $\ddot{x}$ once. $$\dot{x}=\int \ddot{x} \mathrm dt$$
$$=\ddot{x}t$$
initial position is 0 so, not writing constant.
$$\dot{x}=\frac{F}{m}$$
Differentiate $x(t)$ once.
$$B+2Ct=\frac{F}{m}$$
$$B=\frac{F}{m}-\frac{2Ft}{2m}$$
$$=-\frac{Ft}{2m}$$
Again, going to integrate $\ddot{x}$ twice.
$$x=\int \int \ddot{x} dt dt$$
$$=\frac{\ddot{x}t^2}{2}$$
initial velocity and initial position is 0.
$$x=\frac{Ft^2}{2m}$$
$$A+Bt+Ct^2=\frac{Ft^2}{2m}$$
$$A=\frac{Ft^2+Ft-F}{2m}$$
According to my, I think that C is the minimum (I think B is cause, B is negative; negative is less than positive). And, A is maximum.
A person were saying that `It asked you to minimise the action; it told you the particle moved from $0$ to $a$ in time $t_0$; it gave you the equation of the trajectory.`
In my work where should I put the interval?

> A particle is subjected to the potential V (x) = −F x, where F is a constant. The particle travels from x = 0 to x = a in a time interval t0 . Assume the motion of the particle can be expressed in the form $x(t) = A + B t + C t^2$ . Find the values of A, B, and C such that the action is a minimum.
I was thinking it can solved using Lagrangian rather than Hamilton. There's no frictional force.
$$L=\frac{1}{2}m\dot{x}^2+Fx$$
$$\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})-\frac{\partial L}{\partial x}=0$$
$$m\ddot{x}=F$$
$$\ddot{x}=\frac{F}{m}$$
Differentiate $x(t)$ twice. $$2C=\frac{F}{m}=>C=\frac{F}{2m}$$
For finding B I was thinking to integrate $\ddot{x}$ once. $$\dot{x}=\int \ddot{x} \mathrm dt$$
$$=\ddot{x}t$$
initial position is 0 so, not writing constant.
$$\dot{x}=\frac{F}{m}$$
Differentiate $x(t)$ once.
$$B+2Ct=\frac{F}{m}$$
$$B=\frac{F}{m}-\frac{2Ft}{2m}$$
$$=-\frac{Ft}{2m}$$
Again, going to integrate $\ddot{x}$ twice.
$$x=\int \int \ddot{x} dt dt$$
$$=\frac{\ddot{x}t^2}{2}$$
initial velocity and initial position is 0.
$$x=\frac{Ft^2}{2m}$$
$$A+Bt+Ct^2=\frac{Ft^2}{2m}$$
$$A=\frac{Ft^2+Ft-F}{2m}$$
According to my, I think that C is the minimum (I think B is cause, B is negative; negative is less than positive). And, A is maximum.
A person were saying that `It asked you to minimise the action; it told you the particle moved from $0$ to $a$ in time $t_0$; it gave you the equation of the trajectory.`
In my work where should I put the interval?

lagrangian-formalism