My question concerns the following problem from Spivak's "Calculus" (Problem 9 of the Appendix to Chapter 12).
Problem
Let $u$ and $v$ be continuous on $[a, b]$ and differentiable on $(a, b)$; then, $u$ and $v$ give a parametric representation of a curve from $P = (u(a), v(a))$ to $Q = (u(b), v(b))$. Geometrically, it seems clear that at some point on the curve the tangent line is parallel to the line segment from $P$ to $Q$. Prove this analytically. Assume that we don't have $u'(x) = v'(x) = 0$ for any $x$ in $(a, b)$.
Relevant theorems
Theorem 1. Assume that $x = u(t),\ y = v(t)$ is a parametric representation of a curve, and that $u$ is one-to-one on some interval. If $u$ is differentiable on this interval and $u'(t) \neq 0$, at the point $x = u(t)$ we have
$$f'(x) = \dfrac{v'(t)}{u'(t)}$$
Attempted solution
Assume that $P$ and $Q$ are not the same point.
By the Cauchy Mean Value Theorem, there is a $c$ in $(a, b)$ such that:
$$[u(a) - u(b)]v'(c) = [v(a) - v(b)]u'(c) \tag1$$
There are three cases:
- $u(a) = u(b)$;
- $v(a) = v(b)$;
- $u(a) \neq u(b)$ and $v(a) \neq v(b)$.
To keep this question shorter, I'm only going to show case 3.
Assume that $u(a) \neq u(b)$ and $v(a) \neq v(b)$. Note that $u'(c) \neq 0$, otherwise $(1)$ would imply that $v'(c) = 0$. So:
$$\dfrac{v'(c)}{u'(c)} = \dfrac{v(a) - v(b)}{u(a) - u(b)}$$
At this point, I would like to be able to use Theorem 1 (see Relevant theorems section) to reach the desired conclusion. However, in order to be able to apply Theorem 1, I must find an interval around $c$ where $u$ is one-to-one.
If $u$ is one-to-one in an interval around $c$, then let $x = u(c)$. So, the slope of $f$ at $x$ is given by:
$$f'(x) = \dfrac{v'(c)}{u'(c)}$$
This implies the desired conclusion:
$$f'(x) = \dfrac{v(a) - v(b)}{u(a) - u(b)}$$
The problem remains to show that $u$ actually is one-to-one is an interval around $c$.
In the solutions manual, the author states that "since $u'(x) \neq 0$, $u$ is one-to-one in an interval containing $x$", but I'm not seeing how that is necessarily true.
Any hint?
The author of the solutions manual is wrong. Here is a counterexample if $u$ is not assumed to be continuously differentiable. So, we cannot rely on your "relevant theorem" (where, btw, you did not specify $f$).
Happily, we don't need it! Spivak's hypothesis that the vector $\overrightarrow T(t):=(u'(t),v'(t))$ is non zero for every $t\in(a,b)$ is sufficient for this vector to give the direction of the tangent line to the curve at any point $M(t):=(u(t),v(t)).$ We don't need either to distinguish your three cases.
The vector $\overrightarrow{PQ}:=(u(b)-u(a),v(b)-v(a))$ has the same direction as $\overrightarrow T(c)$, i.e. $[PQ]$ is parallel to the tangent line at $M(c),$ iff $$\begin{vmatrix}u(b)-u(a)&u'(c)\\ v(b)-v(a)&v'(c)\end{vmatrix}=0,$$ and you already found such a $c$ by Cauchy's mean value theorem.