# Find the values of A, B, and C such that the action is a minimum

• Istiak
Istiak
Homework Statement
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.
Relevant Equations
Lagrangian
> 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. Did I do any mistake?

• PeroK

Homework Helper
Gold Member
2022 Award
Where to begin?

First, you were supposed to compute the action and minimise it.

Everything you did from integrating ##\ddot x## got pretty wild. You definitely cannot integrate ##\ddot x## as though it were constant.

##A, B, C## were supposed to be constants, not functions of ##t##.

Istiak
Where to begin?

First, you were supposed to compute the action and minimise it.

Everything you did from integrating ##\ddot x## got pretty wild. You definitely cannot integrate ##\ddot x## as though it were constant.

##A, B, C## were supposed to be constants, not functions of ##t##.
Umm, I had found ##F=m\ddot{x}## 🤔. couldn't get you... then started differentiating ##x## function.