Geometric intuition of the mean value theorem of several variables

Question

Mean value theorem

Let $f:U\to \mathbb R$ be defined in the open set $U\subset \mathbb R^n$. Suppose the segment $[a,a+v]$ be contained in $U$ and the restriction $f|_{[a,a+v]}$ be continous and there exists the directional derivative $\frac{\partial f}{\partial v}(x)$ for every $x\in (a,a+v)$. Then there exists $\theta\in (0,1)$ such that $f(a+v)-f(a)=\frac{\partial f}{\partial v}(a+\theta v)$.

Remark: $v$ is not necessarily unitary

I know intuitively what is $\frac{\partial f}{\partial v}(x)$. However, what the geometric meaning of $f(a+v)-f(a)$ and $f(a+v)-f(a)$ be equal to $\frac{\partial f}{\partial v}(x)$?

In another words in the single variable case if $f[a,b]\to \mathbb R$ is continuous and derivable in $(a,b)$, then there exists $c\in (a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$.

Can I have an similar intuition in the several variables case?

Thanks

Answer 1

All the action here is taking place on the line segment $[a, a+v]$. If you restrict your attention to this line segment, you're just in the single-variable version of the Mean Value Theorem. That is, if $g(t) = f(a + t v)$, then $g'(t) = \dfrac{\partial f}{\partial v}(a+tv)$, and the result is just saying $ g(1) - g(0) = g'(\theta)$ for some $\theta$, $0 < \theta < 1$.

EDIT: The picture you have drawn for the one-variable case is exactly the picture here, where (in the notation I used above) you have $t$ instead of $x$ and $g(t)$ instead of $f(x)$. You can think of it as a cross-section of the graph of $y = f(x)$, cut along the line $x = a + t v$.

Answer 2

Basically, the geometry is a more general version of the single variable case.

Consider a two variable function $z=f(x_{1},x_{2})$. Let $\mathbf{a}=(a_{1},a_{2})$ and $\mathbf{b}=(b_{1},b_{2})$. The mean value theorem says that there exists a $\mathbf{w}=(w_{1},w_{2})=\lambda\mathbf{a}+(1-\lambda)\mathbf{b}$ (i.e., it is on $\mathbf{a}-\mathbf{b})$ such that $$ f(\mathbf{a})-f(\mathbf{b})=\nabla f(\mathbf{w})(\mathbf{a}-\mathbf{b}). $$ If we draw a tangent line (tangent to $f$) passing through $f(\mathbf{w})$ in the direction of $\mathbf{a}-\mathbf{b}$, then this tangent line (call it $T_{\mathbf{w}}$) is parallel to the line segment joining $f(\mathbf{a})$ and $f(\mathbf{b})$. (call the line segment $\mathbf{f}_{\mathbf{a-\mathbf{b}}}$.)

Define $h(x_1,x_2,z)=f(x_1,x_2)-z$. Then $h=0$ is the level surface which is the graph of $f$ in the three dimensional space. Let $\mathbf{x}=(x_{1},x_{2})$ be a point on $x_{1}x_{2}$-plane. A point on the graph of $f$ is denoted by $(\mathbf{x},f(\mathbf{x}))=(x_{1},x_{2},f(x_{1},x_{2}))$.

We would like to show that on the tangent plane, the vector passing through $w$ in the direction of $\mathbf{a}-\mathbf{b}$ shares the same direction as $\mathbf{f}_{\mathbf{a}-\mathbf{b}}=(\mathbf{a}-\mathbf{b},f(\mathbf{a})-f(\mathbf{b}))$.

Consider point $(\mathbf{w},f(\mathbf{w}))$. For function $h$, we have $$ \nabla h(\mathbf{w})=(f_{1}'(\mathbf{w}),f_{2}'(\mathbf{w}),-1)=(\nabla f(\mathbf{w}),-1). $$

We know $\nabla h(\mathbf{w})$ is orthogonal to the tangent plane. Thus, it is orthogonal to any vector on the plane. The tangent line that we draw, $T_{\mathbf{w}}$, is on the tangent plane. Thus, $\nabla h(\mathbf{w})$ is also orthogonal to it.

It turns out that $\nabla h(\mathbf{w})$ is also orthogonal to $\mathbf{f}_{\mathbf{a}-\mathbf{b}}$. \begin{align*} \nabla h(\mathbf{w})\mathbf{f}_{\mathbf{a}-\mathbf{b}} & =(\nabla f(\mathbf{w}),-1)(\mathbf{a}-\mathbf{b},f(\mathbf{a})-f(\mathbf{b}))\\ & =\nabla f(\mathbf{w})(\mathbf{a}-\mathbf{b})-(f(\mathbf{a})-f(\mathbf{b}))\\ & =0 \end{align*} Since $T_{\mathbf{w}}$ and $\mathbf{f}_{\mathbf{a}-\mathbf{b}}$ are on the same vertical plane, they are parallel to each other.