What's the relationship between δ and d/dt?

Question

I understand how

$$m\int \frac{d\mathbf{v}}{dt} \cdot \mathbf{v} dt = \frac{m}{2}\int \frac{d}{dt}(\mathbf{v}^2)dt$$

But how was the leap below made?

Equation

Wouldn't that require δv to equal $\frac{dv}{dt}$? How can that be?

Answer 1

In the context of calculus of variations, $\delta$ can be interpreted as a directional derivative (of a function, typically of a function — so calculus of variations texts often refer to this as a functional) in an arbitrary direction (which is again a function). The heuristic your book is using is this: If you look at the variation of $F(f)$ in the direction of $g$, you are taking $$\delta_g F(f) = \frac{d}{dt}\Big|_{t=0}F(f+tg),$$ quite in analogy with directional derivatives in calculus, where you have $$D_v f(p) = \frac d{dt}\Big|_{t=0} f(p+tv).$$ So the usual sum and product rules (and so forth) hold nicely.

This $t$ has nothing to do with the $t$ appearing in your time integrals.