Derivatives in an Atwood Machine

[henryd]

Homework Statement

I have the professor's solutions for a homework we handed in. There is a part that is confusing me. We have the following equation:

$$E = \frac{1}{2}(m_1 + m_2)\dot{x}^2-(m_1-m_2)gx$$

Homework Equations

We want to find: $$dE/dt = 0$$

The Attempt at a Solution

The solution says the correct answer is:

$$dE/dt = 0 = (m_1 + m_2)\dot{x}_1\ddot{x}_1 - g(m_1-m_2)\dot{x}_1$$

Why does it contain [itex]\dot{x}\ddot{x}[/itex] instead of just [itex]\ddot{x}[/itex]?

Is it because of the chain rule?

Thanks!

 

[Nugatory]

For any function U(t), what is [itex]\frac{dU^2}{dt}[/itex]?

[itex]\dot{x}[/itex] is a function of t.

 

[henryd]

So then it's just

$$ \frac{dU^2}{dU}\frac{dU}{dt} = 2U\dot{U}$$ ?

 

[Nugatory]

So then it's just

$$ \frac{dU^2}{dU}\frac{dU}{dt} = 2U\dot{U}$$ ?

That doesn't quite work, because [itex]\dot{U} \equiv \frac{dU}{dt}[/itex]

Try [itex]\frac{dU^n}{dt} = nU^{n-1}\frac{dU}{dt}[/itex]

 

[Nickie]

Yes, the use of $\dot{x}\ddot{x}$ instead of just $\ddot{x}$ is due to the chain rule. In this case, the energy equation contains a term with both $\dot{x}$ and $x$, so when taking the derivative with respect to time, we must use the chain rule to account for the change in both variables. This results in the use of $\dot{x}\ddot{x}$ in the final expression.