Calculus Question within Lagrangian mechanics

In summary, the "Calculus Question within Lagrangian mechanics" explores the application of calculus in deriving the equations of motion for a mechanical system using Lagrangian principles. It emphasizes the use of the Lagrangian function, which is the difference between kinetic and potential energy, and involves applying the Euler-Lagrange equation to obtain the dynamics of the system. The discussion may include examples and problem-solving strategies relevant to understanding the behavior of physical systems through variational principles.
  • #1
Hennessy
22
10
TL;DR Summary
Product rules with hidden chain rules
Hi all currently got a lagrangian function which i've found to be :
\begin{equation}\mathcal{L}=\frac{1}{2}m(\dot{x}^2+\dot{y}^2+4x^2\dot{x}^2+4y^2\dot{y}^2+8xy\dot{x}\dot{y})- mg(x^2+y^2)
\end{equation}
Let us first calculate
$$(\frac{\partial L}{\partial \dot{x}})$$ which leads us to $$m\dot{x}+4x^2m\dot{x}+8xy\dot{y}$$ now we also have to differentiate this again with respect to t.
$$\frac{d}{dt}(m\dot{x}+4x^2m\dot{x}+8xy\dot{y}) $$ Now this is where i'm stuck. I'm stuck because of the 2 product rule in the middle of this term and then the triple product rule on the right hand and then within them i know there are chain rules as $$x,y,\dot{x},\dot{y}$$ are all f(t). Im basically asking how to use the product rule again. using $$x,\dot{x}$$ as my uv then the product rule is $$x'\dot{x}+x\dot{x}'$$ but when i calculate the primes i get confused. so for example $$x'\dot{x}+x\dot{x}'$$ does this mean differentiate the entire function wrt x and then multiply it just by $\dot{x}$ or does it mean multiply it by the entire thing? Advice would be appreciated , i know this is more a calculus question but just trying to figure it out apologies if this is in the wrong place. Put it here as it requires knowledge of lagrangian mechanics for $$x,y,\dot{x},\dot{y}$$ as being time derivatives is all. Thank you!
 
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  • #2
Use the chain rule for derivatives. If you have a function ##F(t,f_1(t),f_2(t),\ldots)## then
$$
\frac{dF}{dt} =
\frac{\partial F}{\partial t} + \frac{\partial F}{\partial f_1} \dot f_1 + \frac{\partial F}{\partial f_2} \dot f_2 + \ldots
$$

You may want to redo this derivative:
Hennessy said:
Let us first calculate
$$(\frac{\partial L}{\partial \dot{x}})$$ which leads us to $$m\dot{x}+4x^2m\dot{x}+8xy\dot{y}$$

It is also unclear what you mean by ##x’##. We already use dots to denote time derivatives.
 
  • Like
Likes PhDeezNutz
  • #3
Orodruin said:
Use the chain rule for derivatives. If you have a function ##F(t,f_1(t),f_2(t),\ldots)## then
$$
\frac{dF}{dt} =
\frac{\partial F}{\partial t} + \frac{\partial F}{\partial f_1} \dot f_1 + \frac{\partial F}{\partial f_2} \dot f_2 + \ldots
$$

You may want to redo this derivative:


It is also unclear what you mean by ##x’##. We already use dots to denote time derivatives.
Hi there, apologies for the confusion. I rewrote the product rule from uv' +vu' and I wrote it in terms of my two functions x and $\dot{x}$ i also understand that i originally wrote my product rule as u'v+v'u , but this shouldn't of changed the result if im not mistaken?
 

FAQ: Calculus Question within Lagrangian mechanics

What is the role of calculus in Lagrangian mechanics?

Calculus is fundamental in Lagrangian mechanics as it is used to derive the equations of motion. Specifically, the Euler-Lagrange equations, which are central to this framework, are derived using calculus. These equations involve taking partial derivatives of the Lagrangian function with respect to the generalized coordinates and their time derivatives.

How do you derive the Euler-Lagrange equations?

The Euler-Lagrange equations are derived by applying the principle of stationary action. This involves taking the Lagrangian, which is a function of generalized coordinates, their time derivatives, and time, and calculating the partial derivatives with respect to these variables. Setting the variation of the action integral to zero leads to the Euler-Lagrange equations.

What is the Lagrangian function?

The Lagrangian function, typically denoted as L, is defined as the difference between the kinetic energy (T) and potential energy (V) of a system, L = T - V. It is a function of the generalized coordinates, their time derivatives, and possibly time. The Lagrangian encapsulates the dynamics of the system and is used to derive the equations of motion.

How do you handle constraints in Lagrangian mechanics?

Constraints in Lagrangian mechanics can be handled using the method of Lagrange multipliers. When constraints are present, they are incorporated into the Lagrangian through additional terms involving the Lagrange multipliers. This modifies the Euler-Lagrange equations to account for the constraints, ensuring the resulting equations of motion respect these constraints.

What are generalized coordinates in Lagrangian mechanics?

Generalized coordinates are a set of parameters that uniquely define the configuration of a system relative to some reference configuration. They are not necessarily Cartesian coordinates and can be any convenient set of variables that simplify the problem. In Lagrangian mechanics, the equations of motion are derived with respect to these generalized coordinates.

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