In Taylor's Classical Mechanics, one of the problems is as follows:
(1.9) If $\vec{r}$ and $\vec{s}$ are vectors that depend on time, prove that the product rule for differentiating products applies to $\vec{r}$ $\cdot$ $\vec{s}$, that is, that:
$\frac{d}{dt}(\vec{r}$ $\cdot$ $\vec{s}) = \vec{r}$ $\cdot$ $\frac{d\vec{s}}{dt} + \vec{s}$ $\cdot$ $\frac{d\vec{r}}{dt}$
I'm not totally certain that my solution is correct, so if people could give me a hand in checking my work, I'd really appreciate it!
$\vec{r}$ $\cdot$ $\vec{s} = r_xs_x + r_ys_y$
$\frac{d}{dt}(r_xs_x+r_ys_y) = \frac{d}{dt}(r_xs_x) + \frac{d}{dt}(r_ys_y)$
$\frac{d}{dt}(\vec{r}$ $\cdot$ $\vec{s}) = (r_x\frac{d}{dt}s_x + r_y\frac{d}{dt}s_y) + (s_x\frac{d}{dt}r_x + s_y\frac{d}{dt}r_y)$
$\frac{d}{dt}(\vec{r}$ $\cdot$ $\vec{s}) = (r_x+r_y)\frac{d}{dt}(s_x+s_y)+(s_x+s_y)\frac{d}{dt}(r_x+r_y)$
$\frac{d}{dt}(\vec{r}$ $\cdot$ $\vec{s}) = \vec{r}$ $\cdot$ $\frac{d\vec{s}}{dt} + \vec{s}$ $\cdot$ $\frac{d\vec{r}}{dt}$
Best Answer
The penultimate line doesn't look right, and I'm not sure your grouping on the second line is sensible immediately. I would go, in your notation, $$ \begin{align} \frac{d}{dt}(r_x s_x + r_y s_y) &= \frac{d}{dt}(r_x s_x) + \frac{d}{dt}(r_y s_y) &&\text{(Linearity)}\\ &= \left(r_x\frac{d}{dt}s_x+ s_x\frac{d}{dt}r_x \right) + \left(r_y \frac{d}{dt}s_y+ s_y\frac{d}{dt}r_y \right) &&\text{(Product rule)} \\ &= \left(r_x\frac{d}{dt}s_x+ r_y \frac{d}{dt}s_y \right) + \left(s_x\frac{d}{dt}r_x+ s_y\frac{d}{dt}r_y \right) &&\text{(Regrouping)} \\ &= \vec{r} \cdot \frac{d\vec{s}}{dt} + \vec{s} \cdot \frac{d\vec{r}}{dt} \end{align} $$ (using that the component of the derivative is the derivative of the component, and the definition of the dot product).