Mathematical identity related to d'Alembert's Principle

In Hand & Finch's book on Analytical Mechanics, I came across this mathematical identity Eq. 1.19 in Chapter 1, page 5, which is related to the description of d'Alembert's principle:

$$\dot{\vec{p}} \cdot \delta \vec{r}=\frac{d(\vec{p} \cdot \delta \vec{r})}{d t}-\vec{p} \cdot \frac{d(\delta \vec{r})}{d t}.\tag{1.19}$$

Does this identity have a name and could someone please explain it to me?

• It sometimes goes by the name of product rule ;) Jokes aside, momentum and position can both be time-dependent. Thus, the time derivative of the product $\vec{p} \cdot \delta \vec{r}$ has got two components. One is just brought to the right-hand side of the equation.
– lmr
May 6 '19 at 10:39
• This is an application of the chain rule in differentiation. Notice that the first term on the RHS is a time derivative of the scalar product of two vectors. Apply chain rule to this term and rearrange to obtain the above result. May 6 '19 at 10:40
• I see it now! Thanks. It is simply a re-arranging of the product rule. May 6 '19 at 11:14
• Oh man, that is written strange. I didn't notice that the first that the first $p$ was actually dotted. I'm sure it makes more sense in context, but showing everything else as $\frac {d}{dt}$ except for that $p$ makes it harder to see that this is just chain rule.
– JMac
May 6 '19 at 11:21
• @JMac yeah It really confused me! I wish authors would describe what they are doing in precise detail instead of saying "here is a mathematical identity". It would save readers (especially dumbasses like me) a lot of time. May 6 '19 at 11:31

Bring over the term on the right, and substitute the notation $$\dot{x}\equiv \frac{dx}{dt}$$ to get

$$\frac{d(\vec{p} \cdot \delta \vec{r})}{d t} = \frac{d\vec{p}}{dt} \cdot \delta \vec{r} + \vec{p} \cdot \frac{d(\delta \vec{r})}{d t},$$

which is an application of the product rule.

Hint: Do you know the Leibniz rule $$(fg)'=f'g + f (g')\,?$$

Yes this identity has a name, it is the variational equivalent of "integration by part". You are probably familiar with the fact from basic calculus that integration by parts is given by $$\int u.dv = (u.v) - \int v\,du$$

In your equation, if you substitute $$v=\vec p$$ (that is, $$dv=d\vec p$$), and $$u=\delta \vec r$$ , you can then follow the integration-by-part rule and write (multiplying up-and-down by $$dt$$ each term in both members): $$\int \frac{d\vec p}{dt}. \delta \vec r dt = \int \frac{d(\vec p . \delta \vec r)}{dt}dt- \int\vec p\frac{\delta \vec r}{dt}dt$$ and therefore, by equaling the left and hand side of integration arguments in the equation, you end up with:

$$\frac{d\vec p}{dt}. \delta \vec r = \frac{d(\vec p . \delta \vec r)}{dt}- \vec p\frac{\delta \vec r}{dt}$$ which is exactly uour equation. As others have said in their comments/answer, this is nothing else than a rearrangement of the product rule (that's really what "integration by parts" actually does)