Suggested languages for you:

Americas

Europe

Q35E

Expert-verified
Found in: Page 216

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# 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.)

At time $$t$$ both runners have the same speed $${g^\prime }(t) = {h^\prime }(t)$$ is proved.

See the step by step solution

## Step 1: Given data

The $$f(t) = g(t) - h(t)$$ where $$g$$ and $$h$$ are the position functions of the two runners.

## Step 2: Mean value theorem

"If $$f$$ be a function that satisfies the following hypothesis:

1. $$f$$ is continuous on the closed interval $$(a,\;b)$$.

2. $$f$$ is differentiable on the open interval $$(a,\;b)$$.

Then, there is a number $$c$$ in $$(a,\;b)$$ such that $${f^\prime }(c) = \frac{{f(b) - f(a)}}{{b - a}}$$.

Or, equivalently, $$f(b) - f(a) = {f^\prime }(c)(b - a)$$”.

## Step 3: Take the derivative of the given equation

From the given condition, it is observed that the difference between the position of the two runners is $$f(t) = g(t) - h(t).$$

Since both the runners start the race at the same time, initial time is same for both. Similarly, since the runners finish the game at tie, final ending time of the race is also same for both of them.

Let $${t_0}$$ denotes the starting time and $${t_1}$$ denotes the ending time.

At starting and ending point during the race the difference in velocity between the runners was zero. Then,

\begin{aligned}{c}f\left( {{t_0}} \right) &= g\left( {{t_0}} \right) - h\left( {{t_0}} \right)\\ &= 0\\g\left( {{t_0}} \right) &= h\left( {{t_0}} \right)\end{aligned}

And similarly at $${t_1}$$ as follows:

\begin{aligned}{c}f\left( {{t_1}} \right) &= g\left( {{t_1}} \right) - h\left( {{t_1}} \right)\\ &= 0\\g\left( {{t_1}} \right)& = h\left( {{t_1}} \right)\end{aligned}

Take the derivative of the above equation as follows:

$${f^\prime }(t) = {g^\prime }(t) - {h^\prime }(t)$$Step 3: Take the derivative of the given equation

## Step 4: Apply Mean value theorem

Apply Mean value theorem there exists a time $$t$$ with $${t_0} < t < {t_2}$$ as follows:

\begin{aligned}{c}{f^\prime }(t) &= \frac{{f\left( {{t_1}} \right) - f\left( {{t_0}} \right)}}{{{t_1} - {t_0}}}\\ &= \frac{{f\left( {{t_1}} \right)}}{{{t_1} - {t_0}}} - \frac{{f\left( {{t_0}} \right)}}{{{t_1} - {t_0}}}\end{aligned}

Since, $$f\left( {{t_1}} \right) = f\left( {{t_0}} \right) = 0$$, the above equation will be $${f^\prime }(t) = 0$$.

Since $${f^\prime }(t) = 0$$, the above equation will be as follows:

\begin{aligned}{c}{f^\prime }(t) &= {g^\prime }(t) - {h^\prime }(t)\\ &= 0\\{g^\prime }(t) &= {h^\prime }(t)\end{aligned}

Thus, $${g^\prime }(t) = {h^\prime }(t)$$.

Hence, it can be concluded that at time $$t$$ both runners have the same speed $${g^\prime }(t) = {h^\prime }(t)$$.

Therefore, the required statement is showed.