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Need help with a Mean Value Theorem Problem

  1. Jun 13, 2005 #1
    Hi all,
    I've been thinking on this a lot but couldn't come up with an answer so I need your help. I've seen this in Thomas' Calculus 10th edition. Anyway here goes the problem

    A marathoner ran the 26.2 mi New York City Marathon in 2.2 h. Show that at least twice the marathoner was running at exactly 11 mph.

    Any help is appreciated.

    Many thanks in advance,
  2. jcsd
  3. Jun 13, 2005 #2

    Doc Al

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    Staff: Mentor

    What does the mean value theorem tell you? (What's the average speed of the runner?)
  4. Jun 13, 2005 #3


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    And what was runner's speed at the beginning and end of the race?
  5. Jun 13, 2005 #4
    The mean value theorem tells me that somewhere between 0 and 2.2 the runner's speed was exactly 11.909090... His initial speed is 0 and I cannot tell his final speed because I do not know the displacement formula.
    I thought a lot about this but obviously I'm missing something.
  6. Jun 13, 2005 #5

    Doc Al

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    OK. So you know he started from rest and at some point reached 11.9 m/h. So what can you conclude about his having a speed of exactly 11 m/h?

    Regarding his final speed: Assume he stopped running when he finished the race. :smile:
  7. Jun 13, 2005 #6
    well, actually I cannot conclude anything about his running at 11 mph. As to the second question :) speed is 0. Actually I could think this way about 11mph problem. In order for a function to be the derivative of another one it has to take on every value on the interval it is defined. So we know that the v(t) function is defined between 0 and 2.2. By the intermediate value theorem somewhere between 11.9 and 0 the function v(t) must take the value 11. Is my reasoning correct?

    Correct or not I still have the problem of at least two times of 11 mph. I don't know how i will make it.
  8. Jun 13, 2005 #7
    The mean value theorem:
    If a continuous function f(x) exists between the closed interval a and b, then theres a point c between a and b such that

    [tex] f'(c) = \frac{f(b)-f(a)}{b-a} [/tex]

    a = 0, b = 2.2, f(a) = 0, f(b) = 26.2.

    Its pretty much saying that if its continuously increasing from a to b, then it must take on every value between "a" and "b" during its course.
  9. Jun 13, 2005 #8

    Doc Al

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    Exactly! The mean value theorem tells you that the runner's speed must have been 11.9 m/h at some point. And the intermediate value theorem tells you that sometime between the time t=0 and the point where his speed was 11.9, that the speed must have hit exactly 11 m/h.

    Use the same reasoning for the latter part of the trip. Between the point where his speed was 11.9 m/h and the end of the race (speed = 0), once again his speed must have hit exactly 11 m/h at least once.
  10. Jun 13, 2005 #9
    Gosh, I should have seen that! :) Thanks a million. I've gotta work more. Anyway I still have the university to learn all these by profs. :) Thanks again to all of you...
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