Dot product and derivative of a vector values function

Question

I'm working of the following questions for a homework.

Q1 : Let $r(t)$ be a vector values function, show that $\frac{d}{dt}\vert r'(t)\vert = \frac{1}{\vert r'(t)\vert} r'(t) \cdot r''(t).$

Q2: If $u = r(t)\cdot( r'(t)\times r''(t))$. Show that $u'(t) = r'(t)\cdot(r'(t)\times r'''(t))$.

I managed to solve them by assuming that $r(t) = f(t)i+g(t)j+h(t)k$ and calculating both sides of the equality and see that they are indeed the equal.

Is there a way to prove this without assuming $r(t)$ is in 3 dimensions?

Answer 1

Q1. On the one hand, $$\dfrac{d}{dt} \bigl[ |\mathbf{r}'(t)|^2 \bigr] = \dfrac{d}{dt} \bigl[ \mathbf{r}'(t) \cdot \mathbf{r}'(t) \bigr] = \mathbf{r}''(t) \cdot \mathbf{r}'(t) + \mathbf{r}'(t) \cdot \mathbf{r}''(t) = 2 \bigl( \mathbf{r}'(t) \cdot \mathbf{r}''(t) \bigr).$$ On the other hand, $$\dfrac{d}{dt} \bigl[ |\mathbf{r}'(t)|^2 \bigr] = 2 |\mathbf{r}'(t)| \, \dfrac{d}{dt} \bigl[ |\mathbf{r}'(t)| \bigr].$$ Compare the right sides of these two lines.

Q2. We have $$\begin{aligned}[t]\mathbf{u}'(t) &= \dfrac{d}{dt} \bigl[ \mathbf{r}(t) \bigr] \cdot [\mathbf{r}'(t) \times \mathbf{r}''(t)] + \mathbf{r}(t) \cdot \dfrac{d}{dt} \bigl[ \mathbf{r}'(t) \times \mathbf{r}''(t)\bigr] \\ &= \mathbf{r}'(t) \cdot [\mathbf{r}'(t) \times \mathbf{r}''(t)] + \mathbf{r}(t) \cdot \biggl[ \dfrac{d}{dt} \bigl[ \mathbf{r}'(t) \bigr] \times \mathbf{r}''(t) + \mathbf{r}'(t) \times \dfrac{d}{dt} \bigl[ \mathbf{r}''(t) \bigr] \biggr]\\ &= 0 + \mathbf{r}(t) \cdot \bigl[ \mathbf{r}''(t) \times \mathbf{r}''(t) + \mathbf{r}'(t) \times \mathbf{r}'''(t) \bigr] \\ &= \mathbf{r}(t) \cdot [\mathbf{r}'(t) \times \mathbf{r}'''(t) ].\end{aligned}$$

Answer 2

Chain rule.

Write $\displaystyle \|x\| = \sqrt{\sum_{k = 1}^d \mathrm{pr}_k^2 x},$ so $\|x\|' \cdot h = \dfrac{(x \mid h)}{\|x\|},$ then $\|r'(t)\|' = \dfrac{(r'(t)\mid r''(t))}{\|r'(t)\|}.$

By Leibniz rule: if $\mathrm{B}$ is a bilinear form, then $\mathrm{B}'(x, y) \cdot (h, k) = \mathrm{B}(x, k) + \mathrm{B}(h, y).$ Then, $u'(t) = r'(t) \cdot r'(t) \times r''(t) + r(t) \cdot (r'(t) \times r''(t))' = r(t) \cdot (r''(t) \times r''(t) + r'(t) \times r'''(t)) = r(t) \cdot r'(t) \times r'''(t).$