This problem is an application of the Intermediate Value Theorem and Mean Value Theorem in calculus. Let's solve it step-by-step.

Problem Recap

We have two runners, starting at the same point and finishing at the same time, meaning their race ends in a tie. We need to show that at some time during the race, they both have the same speed.

Step-by-Step Solution

1. Define Position Functions: Let $g(t)$ and $h(t)$ be the position functions of the two runners, where $t$ represents time. Since they start and end together, we know:

   • $g(0) = h(0)$ (start at the same point),
   • $g(T) = h(T)$ (finish at the same point after time $T$).

2. Define the Difference Function: Let $f(t) = g(t) - h(t)$, which represents the difference in positions between the two runners at any time $t$.

   • At $t = 0$: $f(0) = g(0) - h(0) = 0$.
   • At $t = T$: $f(T) = g(T) - h(T) = 0$.

   So, $f(t)$ starts and ends at 0, meaning $f(0) = f(T) = 0$.

3. Apply the Mean Value Theorem: Since $f(t)$ is continuous on $[0, T]$ and differentiable on $(0, T)$, by the Mean Value Theorem, there exists some point $c$ in $(0, T)$ such that: $f'(c) = \frac{f(T) - f(0)}{T - 0} = \frac{0 - 0}{T} = 0.$

4. Interpret $f'(c) = 0$: Since $f(t) = g(t) - h(t)$, we have: $f'(t) = g'(t) - h'(t).$ Thus, $f'(c) = g'(c) - h'(c) = 0$, which implies: $g'(c) = h'(c).$

Conclusion

There exists a time $c$ in $(0, T)$ where the derivatives $g'(c)$ and $h'(c)$ are equal, meaning the two runners have the same speed at that moment.

This completes the proof.

Would you like further details or have any questions about this explanation?

Related Questions

1. How does the Intermediate Value Theorem relate to this proof?
2. What is the importance of differentiability in applying the Mean Value Theorem here?
3. Can this approach be used if the runners do not finish at the same time?
4. How would this proof change if the runners had different starting points?
5. How could this theorem be applied in other real-life racing scenarios?

Tip

Remember that the Mean Value Theorem can only be applied to continuous and differentiable functions over closed intervals.