Physics Question Math Related:

Question

When do Physicists ever use the following expression in whatever context: \begin{align} \frac{d}{dt}[r_1(t)\ \cdot \ r_2(t)] = r_1 \ \cdot \ \frac{dr_2}{dt} + \frac{dr_1}{dt} \ \cdot \ r_2 \\ \frac{d}{dt}[r_1(t)\ \times \ r_2(t)] = r_1 \ \times \ \frac{dr_2}{dt} + \frac{dr_1}{dt} \ \times \ r_2 \end{align}

For an answer type I could be visualizing one from linear momentum, or any where applicable. Any thoughts would be helpful, or even a migration to Physics.SE I posted here since it was about the math component.

Answer 1

2) Angular momentum of a solid object is defined as $\vec{L} = \vec{r} \times \vec{p}$, where $\vec{p}$ is regular momentum of that object, and $\vec{r}$ is the distance vector from the point w.r.t which the angular momentum is calculated. Thus, change in angular momentum can come either from changing the linear momentum or the change of length of the "arm" . Taking the derivative of $\vec{L}$ shows that the result is the sum of these two effects

1) Your factory produces $\vec{r}_1 = (3/5 bottles, 4/5 cans)$ per minute. You should be producing $\vec{r}_2 = (4/5 bottles, 3/5 cans)$ per minute. Thus your efficiency is $\vec{r}_1 \cdot \vec{r}_2$. Your efficiency can change due to two factors - you yourself produce different ratio of bottles and cans, or your goal changes. Again, the net effect is sum of both

Answer 2

An example:

Particle moving in a circle of radius $a$:

Then

1) $\vec r \cdot \vec r =a^2$ (constant).

$\frac{d}{dt} (\vec r\cdot \vec r)=2 (\frac{d}{dt} \vec r )\cdot \vec r =0$;

$\vec v =(\frac{d}{dt} \vec r )$ is perpendicular to $\vec r$.

2) Assume particle is moving in a circle with constant speed.

Similarly: The acceleration vector is perpendicular to the velocity.

Does this sound familiar?