Why must the total time derivative only be a linear function of velocity? [duplicate]

Question

I'm hung up on page 7 of Landau & Lifshitz Course on Mechanics. They claim,

$$L(v'^2) = L(v^2)+\frac{\partial L}{\partial v^2}2\textbf v\cdot \epsilon \tag{p.7}$$ The second term on the right of this equation is a total time derivative only if it is a linear function of the velocity $\textbf v$.

I can understand the Taylor expansion to get to this equation, but I do not understand why this must be a linear function of $\textbf v$.

If I understand correctly, we need to find a function$f(q, t)$ such that:

$$\frac{d}{dt}(f(q,t)) = \frac{\partial L}{\partial v^2}2\textbf v\cdot \epsilon$$

If $f(q,t) = q^2$, would the total time derivative not include a nonlinear function of $\textbf v$? Is there some restriction imposed upon $f$ such that it must also be a linear function of $q$?

Answer 1

The Taylor expansion is:

$f(x+\delta)=f(x)+f'(x)\delta+...$

Here $x=v^2$, and you know that

$v^2+\delta=(\vec{v}+\vec{\epsilon})^2=v^2+2\vec{v}.\vec{\epsilon}+\epsilon^2$

thus,

$\delta=2\vec{v}.\vec{\epsilon}+\epsilon^2$

discarding terms of order $\epsilon^2$, then you get the expression in the book.

Answer 2

It comes down to how the differentiation works. The function I mentioned, $f(q, t)=q^2$ would indeed be a linear function once differentiated:

$$\frac{d}{dt}(q^2)=\frac{d}{dt}(q\cdot q)$$

Which, when used with the product rule, would be:

$$\frac{dq}{dt} \cdot q + q \frac{dq}{dt}= 2q\cdot v$$

Where $v$ only appears in its first order still. It turns out that for higher degrees, i.e. $ f=q^n$, the solution would be:

$$n \cdot q(t)^{n-1}\cdot v$$

So, regardless of how you look at it, the function must be a linear function of $v$.