# How does Rolle's theorem apply here?

The derivation below was taken from a book on Classical Differential Geometry. It uses Rolle's theorem to find the characteristic line of a family of planes, but I don't see how it applies.

Given is a family of planes,

$$\mathbf{x} \cdot \mathbf{a}(u) + p(u) = 0$$

with normal vector $\mathbf{a}$ and "distance to origin" $p$ both depending on a parameter $u$ (the planes are supposed not to be parallel). The objective is to find the equation of the characteristic line, which (in my understanding) is the line of intersection of two planes whose $u$'s differ only infinitesimally. Now, the book says

The planes determined by the parameters $u_1$ and $u_2$ (with $u_1 < u_2$) intersect in a straight line, which also lies in the plane $$\mathbf{x} \cdot (\mathbf{a}(u_1)- \mathbf{a}(u_2)) + p(u_1) - p(u_2) = 0$$

or, applying Rolle's theorem,

$$x_1a'_1(v_1) + x_2a'_2(v_2) + x_3a'_3(v_3) + p'(w) = 0$$

with $u_1 \le v_i \le u_2$ and $u_1 \le w \le u_2$.

I don't understand how Rolle's theorem is used here. The theorem says that, for a function $f$ that is continuous on an interval $[a, b]$ with $f(a) = f(b)$, the interval contains a point $q$ at which $f'(q) = 0$.

The interval seems to be $[u_1, u_2]$ in the proof above, but where is $f$ for which $f(a)=f(b)$?

• The author meant the mean value theorem: If $f$ is a ${\it real valued}$ differentiable function then you can write any difference $f(t_2)-f(t_1)$ in the form $f'(\tau)(t_2-t_1)$ where $\tau$ is an unknown point in the interval $[t_1,t_2]$ – Christian Blatter Apr 26 '11 at 11:29
• Neither the answer nor the comment seems to explain why the $v_i$ and $w$ aren't all the same? – joriki Apr 26 '11 at 11:47
• Why would they be? – Plop Apr 26 '11 at 11:58
• @Plop: Both your answer and Christian's comment cite theorems referring to real functions of a single real variable and guaranteeing the existence of a certain value of its derivative at a single intermediate point. If I apply either of those theorems to the function $f(u)=\mathbf x\cdot \mathbf a(u)+p(u)$, I get exactly what I get when I apply the special case of the theorems the OP originally cited, namely $f'(v)=\mathbf x\cdot\mathbf a'(v)+p'(v)=0$ with $u_1\le v\le u_2$. I don't see how either of the generalizations changes the fact that there's only a single intermediate point $v$. – joriki Apr 26 '11 at 12:19
• @joriki: I am asking myself the same thing... maybe it is because $f(u)=\mathbf x\cdot \mathbf a(u)+p(u)$ (other than the $a_i(u)$) still depends on $\mathbf x$? Any small change in $u$ would throw a given $\mathbf x$ off the plane... that confuses me a bit. – koletenbert Apr 26 '11 at 15:36

A variation of Rolle's theorem is the following: If $f$ is realvalued, continuous on $[a,b]$ and differentiable on $]a,b[$, there exists $c \in ]a,b[$ such that $f'(c)= \frac{f(b)-f(a)}{b-a}$.
This is a consequence of Rolle's theorem by considering the function $g : x \mapsto f(x) - \frac{f(b)-f(a)}{b-a} (x-a)$, which satisfies $g(a)=g(b)=f(a)$ and $g'(x) = f'(x) - \frac{f(b)-f(a)}{b-a}$ for all $x \in ]a,b[$.