homework and exercises - Use of the term first order dependency

In a question I am doing it says:

Show explicitly that the function

$$y(t)=\frac{-gt^2}{2}+\epsilon t(t-1)$$ yields an action that has no first order dependency on $\epsilon$.

Also my textbook says that

[...] if a certain function $x_0(t)$ yields a stationary value of $S$ then any other function very close to $x_0(t)$ (with the same endpoint values) yields essentially the same $S$, up to first order of any deviations.

I am confused about the first order bit? In the first case does it mean that $\frac{\partial S}{\partial \epsilon}=0$ or that it does not depend of $\epsilon$ but may take some other constant value. In the second case does it mean likewise or something different, please explain?

Answer

Hints:

1. 
   

   The action is

   $$\tag{A} S[y]:=\int_0^1 \! dt ~L(y,\dot{y}), \qquad L(y,\dot{y})~:=~\frac{m}{2}\dot{y}^2 -mgy,$$ with Dirichlet boundary conditions $$\tag{B} y(0)~=~0 \quad\text{and}\quad y(1)~=~-\frac{g}{2}.$$
   
   


2. 
   

   Calculate explicitly the composed function

   $$\tag{C} s(\epsilon)~:=~ S[y_{\epsilon}] ,$$ where $$\tag{D} y_{\epsilon}(t)~:=~-\frac{gt^2}{2}+\epsilon t(t-1).$$
   
   


3. 
   

   Check that the virtual paths (D) satisfy the Dirichlet boundary conditions (B). Why do we need to check that?

   
   
   


4. 
   

   Show explicitly that the function $s(\epsilon)$ has no first order dependence on $\epsilon$. What is the physical significance of this fact?

   
   

References:

1. David Morin, The Lagrangian Method, Chap 6, Lecture notes, 2007; Exercise 6.30.