A How does time derivative commute from one variable to another?

[irishetalon00]

TL;DR Summary

how does time derivative commute from one variable to another?

I'm reviewing Meirovitch's "Methods of Analytical Dynamics," and I don't understand the commutation of the derivative from r to dr:
$$
\mathbf{F} \cdot d\mathbf{r} = m \ddot{\mathbf{r}} \cdot d\mathbf{r} = m\mathbf{\dot{r}} \cdot d\mathbf{\dot{r}}
$$

 

[PeroK]

TL;DR Summary: how does time derivative commute from one variable to another?

I'm reviewing Meirovitch's "Methods of Analytical Dynamics," and I don't understand the commutation of the derivative from r to dr:
$$
\mathbf{F} \cdot d\mathbf{r} = m \ddot{\mathbf{r}} \cdot d\mathbf{r} = m\mathbf{\dot{r}} \cdot d\mathbf{\dot{r}}
$$

The definition of a differential that I use is:

If $x = f(t)$ then $dx = f'(t)dt$, where $f'(t) = \frac{df}{dt}$.

So, as $\dot x = f'(t)$ we have $d\dot x = f''(t)dt$.

The result follows from this and an extension to the dot product.

 

[irishetalon00]

Thank you PeroK. I think I got it. But to be sure, let me try to reiterate what you said in the context of my original question.

$\ddot{\mathbf{r}}$ , which is the time derivative of $\dot{\mathbf{r}}$, can also be thought of as being equal to $\frac{d\dot{\mathbf{r}}}{dt}$, where $d\dot{\mathbf{r}}$ and $dt$ are differential algebraic variables. after making the substitutions, since dt is a scalar it can be passed through the dot product to $d\mathbf{r}$, after which $\frac{d\mathbf{r}}{dt}$ is just $\dot{\mathbf{r}}$

 

[PeroK]

Thank you PeroK. I think I got it. But to be sure, let me try to reiterate what you said in the context of my original question.

$\ddot{\mathbf{r}}$ , which is the time derivative of $\dot{\mathbf{r}}$, can also be thought of as being equal to $\frac{d\dot{\mathbf{r}}}{dt}$, where $d\dot{\mathbf{r}}$ and $dt$ are differential algebraic variables. after making the substitutions, since dt is a scalar it can be passed through the dot product to $d\mathbf{r}$, after which $\frac{d\mathbf{r}}{dt}$ is just $\dot{\mathbf{r}}$

That's the hand-waving way to do it. If you want to justify it more formally, then you need a sound mathematical basis for what is a differential.

The first point to note is that a time derivative only makes sense if you can write $x$ as a function of $t$. I'll use one dimension as it's easier on my phone.

If we have $x = f(t)$, then:
$$(\ddot x )dx = f''(t)f'(t)dt$$And:
$$(\dot x)d\dot x = f'(t)f''(t)dt$$And we can see without any hand waving that the two expressions are equal.

That said, physicists wave their hands a lot, so you could just say it looks right and let the maths students worry about proving it!

 

[wrobel]

In order for simplicity sake some authors make the narration harder.
By definition the kinetic energy of a particle is
$T=\frac{m}{2}|\boldsymbol {\dot r}|^2=\frac{m}{2}(\boldsymbol {\dot r},\boldsymbol {\dot r}).$
Thus we have
$\dot T=m(\boldsymbol {\dot r},\boldsymbol {\ddot r})=(\boldsymbol {\dot r},\boldsymbol {F}).$
The term from the right is called a power.
Integrating the last formula we get
$T\mid_{t=t_2}-T\mid_{t=t_1}=\int_{t_1}^{t_2}(\boldsymbol {\dot r},\boldsymbol {F})dt.$
The term from the right is called a work done.

If $\boldsymbol F=\boldsymbol F(\boldsymbol r)$ then the integral is an integral of a differential form $(\boldsymbol F, d\boldsymbol r)$ and thus it does not depend on a parametrization of the given trajectory.

 

[Svein]

The impulse of an object is given by $$ m\cdot v.$$ Derive this with respect to time: $$ \frac{d}{dt}(m\cdot v) = \frac{d}{dt}(m)\cdot v+m\cdot\frac{d}{dt}(v). $$ In most cases m is constant...