First I will write what is written in the book and then I will ask the question:
Suppose $f:\mathbb{R}^3 \rightarrow \mathbb{R}$ is a real-valued function. Let $v$ and $x \in \mathbb{R}^3$ be fixet vectors and consider the function from $\mathbb{R}$ to $\mathbb{R}$ defined by $t \mapsto f(x+tv)$. The set of points of the form $x+tv, t\in \mathbb{R}$ is the line $L$ through the point $x$ parallel to the vector $v$.
The function $t \mapsto f(x+tv)$ represents the function $f$ restricted to the line $L$. For example, is a bird flies along this line with velocity $v$ so that $x+tv$ is its position at time $t$, and if $f$ represents the temperature as a function of position, then $f(x+tv)$ is the temperature at time $t$. We may ask: How fast are the values of $f$ changing along the line $L$ at the point $x$ ? Because the rate of change of a function is given by a derivative, we could say that the answer to this question is the value of the derivative of this function of $t$ at $t=0$ (when $t=0$, $x+tv$ reduce to $x$). This would be the derivative of $f$ at the point $x$ in the direction of $L$, that is, of $v$.
What is written in bold is my issue. I don't understand the question very well and I can't translate the answer from the text correctly. Why is the answer to that question the value of the derivative of that function of $t$ at $t=0$. Why $t=0$ and not $t=1$?
Thanks :)