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

In summary, the conversation discusses finding the values of A, B, and C in the equation ##x(t) = A + B t + C t^2## in order to minimize the action of a particle subjected to a potential function. The participant suggests using Lagrangian rather than Hamilton and assumes there is no frictional force. However, their approach of integrating ##\ddot{x}## and differentiating ##x(t)## does not follow the given problem parameters and does not result in the correct values for A, B, and C. The correct approach should involve computing the action and minimizing it, taking into account the given conditions of the particle's motion.
  • #1
Istiak
158
12
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?
 
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  • #2
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##.
 
  • #3
PeroK said:
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.
 
  • #4
Istiakshovon said:
Umm, I had found ##F=m\ddot{x}## 🤔. couldn't get you... then started differentiating ##x## function.
You ignored most things in the question:

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.

You didn't do the problem that was asked.
 

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

1. How do I find the values of A, B, and C to minimize the action?

To find the values of A, B, and C that minimize the action, you will need to use a mathematical optimization method such as calculus. This involves taking the derivative of the action with respect to each variable and setting them equal to 0 to solve for the minimum values.

2. What is the significance of finding the minimum value of the action?

Finding the minimum value of the action is important in many scientific fields, as it allows us to determine the most efficient or optimal solution to a problem. In physics, for example, it can help us find the path of least resistance or the configuration that requires the least energy.

3. Can the values of A, B, and C be negative when the action is minimized?

Yes, the values of A, B, and C can be negative when the action is minimized. The minimum value of the action is not dependent on the sign of these variables, but rather on their relative values to each other.

4. Are there any limitations to using calculus to find the minimum value of the action?

While calculus is a powerful tool for finding the minimum value of the action, it may not always be applicable. In some cases, the action may be too complex or have too many variables for calculus to be effective. In these situations, alternative optimization methods may need to be used.

5. How can I verify that the values of A, B, and C I have found are the true minimum?

To verify that the values of A, B, and C you have found are the true minimum, you can use a technique called the second derivative test. This involves taking the second derivative of the action with respect to each variable and evaluating it at the minimum values. If the second derivative is positive, then the values are confirmed to be the minimum.

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