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Q35E

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Essential Calculus: Early Transcendentals

Found in: Page 216

Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

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 by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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.

Most popular questions for Math Textbooks

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