Suppose $u$ and $v$ are continuous on $[a,b]$ and differentiable on $(a,b)$, and assume that for all $t\in(a,b)$, at least one of $u'(t)$ and $v'(t)$ is nonzero. Let $C$ be the curve given by $(u(t),v(t))$ for $t \in [a,b]$. Let $A = (u(a),v(a))$ and $B = (u(b),v(b))$ be the endpoints of the curve, and assume $A\ne B$. Show that there is some point $c \in (a,b)$ such that the tangent line to $C$ at $(u(c),v(c))$ is parallel to $\overline{AB}$.
How do I go about creating a parameterization that proves that the lines are parallel for a value of $c$?
Let $k=\frac{u(b)-u(a)}{v(b)-v(a)}$, and define $f(t)=u(t)-kv(t)$. If $v(b)=v(a)$, then the condition $A\ne B$ gives us that $u(b)\ne u(a)$, so we reverse the roles of $u$ and $v$. Now we have $$ f(b)-f(a)=u(b)-u(a)-\left(\frac{u(b)-u(a)}{v(b)-v(a)}\right)(v(b)-v(a))=0. $$ So $f(b)=f(a)$. As the hypotheses guarantee that $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, we get by Rolle's Theorem that there exists $c\in(a,b)$ with $f'(c)=0$. So $$ 0=f'(c)=u'(c)-kv'(c). $$ That is, $$\tag1 \frac{v(b)-v(a)}{v'(c)}=\frac{u(b)-u(a)}{u'(c)}. $$ Note that if $v'(c)=0$ then the hypotheses give us that $u'(c)\ne0$ and we can reverse the roles. If we name $\alpha$ the constant in $(1)$, we have obtained that $$ (u'(c),v'(c))=\frac1\alpha\,(u(b)-u(a),v(b)-v(a)). $$