Question

Let ${\mathbf{r}}_1\left(t\right)$and ${\mathbf{r}}_2\left(t\right)$ both be differentiable three-component vector functions. Prove that

$\frac{d}{dt}\left({\mathbf{r}}_1\left(t\right)\times {\mathbf{r}}_2\left(t\right)\right)={\mathbf{r}}_1^{\text{'}}\left(t\right)\times {\mathbf{r}}_2\left(t\right)+{\mathbf{r}}_1\left(t\right)\times {\mathbf{r}}_2^{\text{'}}\left(t\right)$ (This is Theorem 11.11 (d).)

Short answer

Ans:

Step-by-step solution

Step 1. Given information: 

$r_1\left(t\right)$and$r_2\left(t\right)$ are differentiable vector functions with three components each.

Step 2. Proving :

Let and

Then and

$=r_1{}^{\text{'}}(t)\times r_2(t)+r_1(t)\times r_2{}^{\text{'}}(t)$

Thus

$\frac{d}{dt}\left(r_1\left(t\right)\times r_2\left(t\right)\right)=r_1{}^{\text{'}}(t)\times r_2(t)+r_1(t)\times r_2{}^{\text{'}}(t)$