Need a Intuitive thinking for the proof of Mean value theorem for scalar field Ok so the proof is laid out on my book but I'm genuinely struggling to have a geometrical/Intuitive thinking to this particular proof. I badly need help with it.
So here is the statement of the proof (PLEASE NOTE THAT $D_u$ in this case implies directional derivative in the direction of the unit vector $u$):
Let $f: S\to \mathbb R$ be a scalar field. Let $a\in S$ and where $S\subseteq \mathbb R^n$
So let $u$ be a unit vector and $u\in\mathbb{R}^n$ and $a + ut \in S$ for all $0\leq t \leq 1.$ Suppose $D_uf(a)$ exists for all $a + ut$ where $0 \leq t \leq 1$ then there exist a $\theta$ such that $0<\theta<1$ and
$$f(a+u)-f(a) = D_uf(z), \text{where }z= a + u\theta$$
Now I'm not gonna write the proof that is written in my book because I fairly understood the steps well. I don't understand where this $z= a + u\theta$ came or what does it even imply or how it looks geometrically.
 A: Let $f: S\to \mathbb R$ be a scalar field. Let $a\in S$ and where $S\subseteq \mathbb R^n$.

So let $u$ be a unit vector and $u\in\mathbb{R}^n$ and $a + ut \in S$ for all $0\leq t \leq 1.$

$S$ is some region in $\mathbb{R^n}$; $a$ is within $S$ such that all points up to $1$ unit away in the $\arg(u)$ direction are also in $S$. We define $A$ as the segment leading from $a$ to $a+ut$; i.e. $A:=\{ a+ut | t\in[0,1]\}$.

Suppose $D_uf(a)$ exists for all $a + ut$ where $0 \leq t \leq 1$ then there exists a $\theta$ such that $0<\theta<1$ and
$$f(a+u)-f(a) = D_uf(z), \text{where }z= a + u\theta$$

$z$ lies in $A$, as $z=a+u\theta$ for some $\theta\in[0,1]$. So this is saying: for some point on this line $A$, there exists a point $z\in A$ such that the derivative of $S$ at $z$ is equal to the difference of the values of $f$ at the endpoints of $A$; the "mean value" of $f$ on $A$.
This is notated a little confusingly; I'm not a fan of using $t$ and $\theta$ to parametrize the same thing. Also $a,u,z$ seem to have no clear pattern.
