How would you derive these properties using component form?

Question

I'm currently looking over lecture notes and am looking at the following properties:

Can anyone explain how they might be derived using component form?

Answer 1

I'll show how some of them can be done, first let $e_1=(1,0,0),\ e_2=(0,1,0),\ e_3=(0,0,1)$, then if we let $\underline a=\sum_{n=1}^3 a_n(t)e_n$ and $\underline b=\sum_{n=1}^3 b_n(t)e_n$ we can show that

$$\begin{align} \frac d{dt}(\underline a+\underline b) &= \frac d{dt}\{\sum_{n=1}^3 (a_n(t)+b_n(t))e_n\} \\ &= \frac d{dt}(\sum_{n=1}^3 a_n(t)e_n+\sum_{n=1}^3 b_n(t)e_n) \\ &= \frac d{dt}\sum_{n=1}^3 a_n(t)e_n+\frac d{dt}\sum_{n=1}^3 b_n(t)e_n \\ &= \frac {d\underline a}{dt}+\frac {d\underline b}{dt} \end{align}$$

For the second equality, we have $$\begin{align} \frac d{dt}(\lambda(t)\underline a) &= \frac d{dt}\sum_{n=1}^3(\lambda(t)a_n(t))e_n \\ &= \sum_{n=1}^3\frac d{dt}(\lambda(t)a_n(t))e_n \\ &= \sum_{n=1}^3\left((\frac d{dt}\lambda(t))a_n(t)+ \lambda(t)(\frac d{dt}a_n(t))\right)e_n \\ &= \sum_{n=1}^3(\frac d{dt}\lambda(t))a_n(t)e_n+\sum_{n=1}^3 \lambda(t)(\frac d{dt}a_n(t))e_n \\ &=(\frac d{dt}\lambda(t))\sum_{n=1}^3a_n(t)e_n+\lambda(t)\sum_{n=1}^3 (\frac d{dt}a_n(t))e_n \\ &=\frac {d\lambda(t)}{dt}\underline a+\lambda(t)\frac {d\underline a}{dt} \end{align}$$

The rest are similar, just expand the sum and use the product law. You should be able to do it yourself :)