# Cauchy Mean Value Theorem/Example

## Example of Use of Cauchy Mean Value Theorem

In the 2012 Olympics Usain Bolt won the 100 metres gold medal with a time of $9.63$ seconds.

By definition, his average speed was the total distance travelled divided by the total time it took:

\(\displaystyle V_a\) | \(=\) | \(\displaystyle \frac {d \left({t_2}\right) - d \left({t_1}\right)} {t_2 - t_1}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {100 \ \mathrm m} {9.63 \ \mathrm s}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 10.384 \ \mathrm {m/s}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 37.38 \ \mathrm {km/h}\) | $\quad$ | $\quad$ |

The Mean Value Theorem gives:

- $f' \left({c}\right) = \dfrac {f \left({b}\right) - f \left({a}\right)} {b - a}$

Hence, at some point Bolt was *actually running* at the average speed of $37.38 \ \mathrm {km/h}$

Asafa Powell was participating in that same race.

He achieved a time of $11.99 \ \mathrm s = 1.245 \times 9.63 \ \mathrm s$.

So Bolt's average speed was $1.245$ times the average speed of Powell.

The Cauchy Mean Value Theorem gives:

\(\displaystyle \frac {f' \left({c}\right)} {g' \left({c}\right)}\) | \(=\) | \(\displaystyle \frac {f \left({b}\right) - f \left({a}\right)} {g \left({b}\right) - g \left({a}\right)}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {\frac {f \left({b}\right) - f \left({a}\right)} {b - a} } {\frac {g \left({b}\right) - g \left({a}\right)} {b - a} }\) | $\quad$ | $\quad$ |

Hence, at some point, Bolt was *actually running* at a speed exactly $1.245$ times that of Powell's.

## Sources

- Mathematics.StackExchange: Post 296194, revision 11