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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 $\overline{AB}$.

I've been stuck on this question for a while now without any idea on how to get started. Any whatsoever help is appreciated!

Mimi
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1 Answers1

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Define $g : [a,b] \to \Bbb R$ by the equation

$$g(x) = u(x)(v(b) - v(a)) - v(x)(u(b) - u(a))$$

Then $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Show that $g(a) = g(b)$. By the mean value theorem, $g'(c) = 0$ for some $c\in (a,b)$. This implies $u'(c)(v(b) - v(a)) = v'(c)(u(b) - u(a))$. Use this equation to show that $\overline{AB}$ and the tangent line to $C$ at $t = c$ have the same slope.

kobe
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