Prove the relation of curves

Question

Let $\overrightarrow{\sigma}(t)$ a curve, $\overrightarrow{v}(t)$ the vector velocity, and $\overrightarrow{a}(t)$ thre acceleration. We suppose that $\overrightarrow{F}$ is a vector function $\mathbb{R}^3\rightarrow \mathbb{R}^3$ which is $C^1$, $m>0$ and $\overrightarrow{F}(\overrightarrow{\sigma}(t))=m\overrightarrow{a}(t)$. Prove that $$\frac{d}{dt}[m\overrightarrow{\sigma}(t)\times \overrightarrow{v}(t)]=\overrightarrow{\sigma}(t)\times \overrightarrow{F}(\overrightarrow{\sigma}(t))$$

Could you give me some hints how we could prove this??

Answer 1

It is

$$\frac{d}{dt}[m\overrightarrow{\sigma}(t)\times \overrightarrow{v}(t)]=\frac{d}{dt}[m\overrightarrow{\sigma}(t)]\times \overrightarrow{v}(t)+m\overrightarrow{\sigma}(t)\times \frac{d}{dt}[\overrightarrow{v}(t)].$$

Now,

$$\frac{d}{dt}[m\overrightarrow{\sigma}(t)]\times \overrightarrow{v}(t)=\vec{0}$$ and

$$m\overrightarrow{\sigma}(t)\times \frac{d}{dt}[\overrightarrow{v}(t)]=m\overrightarrow{\sigma}(t)\times \overrightarrow{a}(t)=\overrightarrow{\sigma}(t)\times \overrightarrow m{a}(t).$$ Thus, we are done.