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Two runners start a race at the same time and finish in a tie. Prove that at some time during the race they have the same speed. [Hint: Consider $ f(t) = g(t) - h(t) $, where $ g $ and $ h $ are the position functions of the two runners.]

Let $g(t)$ and $h(t)$ be the position functions of the two runners and let $f(t)=g(t)-h(t) .$ By hypothesis,$f(0)=g(0)-h(0)=0$ and $f(b)=g(b)-h(b)=0,$ where $b$ is the finishing time. Then by the Mean Value Theorem,there is a time $c,$ with $0<c<b,$ such that $f^{\prime}(c)=\frac{f(b)-f(0)}{b-0} .$ But $f(b)=f(0)=0,$ so $f^{\prime}(c)=0 .$ since$f^{\prime}(c)=g^{\prime}(c)-h^{\prime}(c)=0,$ we have $g^{\prime}(c)=h^{\prime}(c) .$ So at time $c,$ both runners have the same speed $g^{\prime}(c)=h^{\prime}(c)$

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 2

The Mean Value Theorem

Derivatives

Differentiation

Volume

Campbell University

Oregon State University

Idaho State University

Boston College

Lectures

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Okay, So we are being told at two under starter race at the same time and finishing a type prove that at some point during the race they were They have the same spirit and we're told to consider half of Casey called Jean T minus hp Herget, age of the position from tens of the two run ish. So we're going tow, create some notation. So we're going to let key not be the initial starting point to the initial time. So this is the starting time, and then we're gonna let two you want equals the final time. And since they are running inner brace the you can assume that the grafts are continuous indefensible because you don't suddenly change. You don't have discrete point or not. It has to be continuous in a race so we can assume continuity into French ability, which then allows us to apply the mean value. Put it to soak continuity in different abilities. I about the differential. So then, if if if we let every vehicle GFT managed hft, we also can assume the F prime of Qi is equal to G Prime FT minus h prime ft. And since we are in the interval. T zero to t one. We can apply the mean value there. And here it is on the next page so I can have more space so we can write the G prime of tea minus h Primary T is equal tio to mean very demagogy. GFT one minor GFT night over to you one minus t Hey, tonight I want to bring this over and then minus two same thing for aged So Asia t one minus age off t not over Tea of one minus. Not now, If you recall, we're told that two runners started the same time and finish in a tie. So that means we can assume we can assume that the initial speech starting point is dear for sex. We can assume that the initial on starting conditions is the same as the second runner, and so are the final thirteen points. And if that's true, that means just whole function is equal to zero. And so what that means is that at some point there's a point where both of them have the exact same speed and so by by letting the thie average slope of the two equal each other. We have proven basically that at some point we don't know what point. But during some point at time t during the race they have the same speed because we know that the average show opens equal to zero. Yeah.

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