Proof for Derivative of Dot Product

Question

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!

$\vec{r}$ $\cdot$ $\vec{s} = r_xs_x + r_ys_y$

$\frac{d}{dt}(r_xs_x+r_ys_y) = \frac{d}{dt}(r_xs_x) + \frac{d}{dt}(r_ys_y)$

$\frac{d}{dt}(\vec{r}$ $\cdot$ $\vec{s}) = (r_x\frac{d}{dt}s_x + r_y\frac{d}{dt}s_y) + (s_x\frac{d}{dt}r_x + s_y\frac{d}{dt}r_y)$

$\frac{d}{dt}(\vec{r}$ $\cdot$ $\vec{s}) = (r_x+r_y)\frac{d}{dt}(s_x+s_y)+(s_x+s_y)\frac{d}{dt}(r_x+r_y)$

$\frac{d}{dt}(\vec{r}$ $\cdot$ $\vec{s}) = \vec{r}$ $\cdot$ $\frac{d\vec{s}}{dt} + \vec{s}$ $\cdot$ $\frac{d\vec{r}}{dt}$

Answer 1

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).

Answer 2

You're assuming the dot product is $x*y = x_0 y_0 + x_1 y_1 + x_2 y_2 + ... + x_n y_n + ...$

The proof can be extended to any kind of dot product defined over any vector space.

$$\dfrac{d}{dt}(\vec{x}*\vec{y})=\lim_{\Delta t\to0}\dfrac{\vec{x}(t+\Delta t)*\vec{y}(t+\Delta t)-\vec{x}*\vec{y}}{\Delta t}=\\ \lim_{\Delta t\to0}\dfrac{\vec{x}(t+\Delta t)*\vec{y}(t+\Delta t)-\vec{x}(t+\Delta t)*\vec{y}(t)+\vec{x}(t+\Delta t)*\vec{y}(t)-\vec{x}*\vec{y}}{\Delta t}=\\\lim_{\Delta t\to0}\vec{x}(t+\Delta t)*\dfrac{\vec{y}(t+\Delta t)-\vec{y}(t)}{\Delta t}+\lim_{\Delta t\to0}\dfrac{\vec{x}(t+\Delta t)-\vec{x}}{\Delta t}*\vec{y}=\\ \vec{x}(t)*\lim_{\Delta t\to0}\dfrac{\vec{y}(t+\Delta t)-\vec{y}(t)}{\Delta t}+\left(\lim_{\Delta t\to0}\dfrac{\vec{x}(t+\Delta t)-\vec{x}}{\Delta t}\right)*\vec{y}=\\ \vec{x}*\dfrac{d\vec{y}}{dt}+\dfrac{d\vec{x}}{dt}*\vec{y}$$

Answer 3

$$\left(\sum r_is_i\right)'=\sum(r_is_i)'=\sum(r_i's_i+r_is_i')=\sum r_i's_i+\sum r_is_i'$$ $$\mathbf r\cdot \mathbf s=\mathbf{r'}\cdot \mathbf s+ \mathbf r\cdot{\mathbf s'}$$ is enough.