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Thread: tortoise and the hare problem

  1. #1
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    tortoise and the hare problem

    Hi,

    I'm working on a problem right now that involves the story about the race between the tortoise and the hare, and I'm having some trouble.

    The race is 2000m. The tortoise travels 20m per minute, and the hare travels 1000m in the first minute and half the remaining distance per minute after that. Right away this question seems odd because by definition, the hare will never finish the race if it is always going half the remaining distance. But anyway, I'm asked to write the series for both the tortoise and the hare, and I'm told specifically "each term of the series represents the distance the tortoise/hare travels in that minute."

    So by what they're asking, the series for the tortoise would be:

    20 + 20 + 20 + 20....

    a = 20 d = 0

    and the hare:

    1000 + 500 + 250 + 125....

    a = 1000 r = 0.5

    I am then asked to find formulas for Sn for both the tortoise and the hare, where Sn represents the total distance travelled after n minutes.

    For the tortoise, it doesn't seem to fit the equation we have learned for arithmetic series': Sn = n/2[2a + (n - 1)d]/2 it seems the most fitting equation is just: Sn = 20n

    For the hare, the equation Sn = [1000(1 - 0.5^n)]/0.5 seems to work just fine


    Could somebody please help me with this? I'm not sure if I'm doing something wrong, or if my equations are wrong, or anything. The strangeness of the question just has my mind all mixed up. I know by definition the hare cannot finish the race, and the tortoise finishes after 100 minutes (simple math), but this all can't be that easy. Can it?

    Thanks very much
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    Re: tortoise and the hare problem

    Quote Originally Posted by jpompey View Post
    The race is 2000m. The tortoise travels 20m per minute, and the hare travels 1000m in the first minute and half the remaining distance per minute after that. Right away this question seems odd because by definition, the hare will never finish the race if it is always going half the remaining distance. But anyway, I'm asked to write the series for both the tortoise and the hare, and I'm told specifically "each term of the series represents the distance the tortoise/hare travels in that minute."
    I am then asked to find formulas for Sn for both the tortoise and the hare, where Sn represents the total distance travelled after n minutes.
    $\text{Tortoise :}{T_n} = \sum\limits_{k = 1}^n {20 \cdot k}$

    $\text{Hare :}{H_n} = \sum\limits_{k = 1}^n {\frac{{2000}}{{{2^k}}}}$
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    Re: tortoise and the hare problem

    Quote Originally Posted by Plato View Post
    $\text{Tortoise :}{T_n} = \sum\limits_{k = 1}^n {20 \cdot k}$

    $\text{Hare :}{H_n} = \sum\limits_{k = 1}^n {\frac{{2000}}{{{2^k}}}}$
    Thank you, but I'm afraid that doesn't help me at all.
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    Re: tortoise and the hare problem

    Quote Originally Posted by jpompey View Post
    that doesn't help me at all.
    In that case your next step is to have a live tutor help you.
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    Re: tortoise and the hare problem

    your equation for the tortoise is the same as what i found, but your equation for the hare gives the value of the term depending on n, or k in your case. If k=5, 2000/2^k gives me 62.5, which is the fifth term. The equation i need is supposed to find the total distance traveled after n amount of minutes. And i believe i found it. I was just hoping you could let me know if im on the right path or not, and if not perhaps point me in the right direction. Thank you for your response anyway.
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    Re: tortoise and the hare problem

    Quote Originally Posted by jpompey View Post
    your equation for the tortoise is the same as what i found, but your equation for the hare gives the value of the term depending on n, or k in your case. If k=5, 2000/2^k gives me 62.5,
    That is not true You need to understand what you are talking about before you what an absurdity. Look at this sum
    Do you see that $T_5=1937.5$ feet?
    On that webpage change the five to 25, the click the $~\boxed{=}~$ at the end of the input window.
    see that the distance is now 1999.999940395355224609375ft.
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    Re: tortoise and the hare problem

    I can't even begin to understand how that works. All I see is 2000 / 2^5 = 62.5 ... which it is. Regardless, this has nothing to do with what I'm doing. I appreciated your response, helpful or not, but now I just find you rude. I was polite to you, I'm not sure where it came from. I'm here for clarification, but you don't seem to want to want to clarify anything. A couple of equations with no explanation of any kind isn't helpful to anyone. What good is an answer if I don't understand how it was found?
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  8. #8
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    Re: tortoise and the hare problem

    Don't mind Plato, he gets cranky sometimes.

    Your formula for S_n for the tortoise looks very strange. I think you have an extra divisor 2 in there.
    S_n= \frac12{n\big(2a+(n-1)\big)}
    looks better to me. It also gives the correct formula that you suggested. Your formula for the hare is also correct.

    Your series are OK too. Plato just used the more advanced sigma notation (but his formula for the hare is wrong (as far as I can tell). The sigma notation involves writing a formula the generates each term of the series separately, and then using the \sum sign and its subscript and superscript to show which particular elements of the series we are summing.

    Your formulas just show the distance travelled after n minutes (as required). You are correct that the way in which the question is framed means that the hare never gets to the end of the race. It's an example of Zeno's paradox, except that Zeno used Achilles and a tortoise for his thought experiment.
    Thanks from jpompey
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    Re: tortoise and the hare problem

    Wonderful thank you very much.
    I see what you mean about the formula, in my lesson text they wrote it two different ways: n/2(2a+(n-1)) and [n(2a+(n-1))]/2 I must have jumbled them together a bit.
    One thing that still confuses me, is the formula for distance traveled after n minutes for the hare works well, but this one here for the tortoise doesn't seem to.

    When I plug in the numbers I can't seem to make it work:
    a = 20
    d = 0
    n = 100

    Sn = n/2(2a+(n-1))
    Sn = 100/2[2(20)+(100-1)]
    Sn = 50(40+99)
    Sn = 50(139)
    Sn = 6950

    Am I doing something wrong here? When n=100 the distance should be at 2000m.

    I can only seem to get 2000m when Sn=20n

    Thanks again
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    Re: tortoise and the hare problem

    Quote Originally Posted by jpompey View Post
    a = 20
    d = 0
    n = 100

    Sn = n/2(2a+(n-1)) *******************
    Sn = 100/2[2(20)+(100-1)]
    Sn = 50(40+99)
    Sn = 50(139)
    Sn = 6950

    Am I doing something wrong here? When n=100 the distance should be at 2000m.

    I can only seem to get 2000m when Sn=20n
    *******************whadda heck you doin' neighbor?
    That should be: Sn = n/2(2a + d(n-1))

    That'll give you 2000.

    Btw, I see no need for the arithmetic sum formula here;
    simply use distance = speed * time
    Thanks from jpompey
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    Re: tortoise and the hare problem

    Quote Originally Posted by DenisB View Post
    That should be: Sn = n/2(2a + d(n-1))

    That'll give you 2000.
    I see my mistake now, I was placing 0 as d but I wasn't doing 0(n-1) properly. I forgot that equals 0, not (n-1).

    Thanks for your help
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    Re: tortoise and the hare problem

    Quote Originally Posted by Archie View Post
    Don't mind Plato, he gets cranky sometimes.
    Your series are OK too. Plato just used the more advanced sigma notation (but his formula for the hare is wrong (as far as I can tell).
    Tell us how how could you tell? This is not mathematics as a second language, is it? English as a second language is your gig is it not? Here learn something before embarrassing of yourself.
    The track is 2000m. The Hare covers half of it, 1000m, in the first minute. Because that means 1000m is left, the here covers 500m in the second minute, he covers 250m in the third minute, 125m in the fourth minute, and 62.5m in the fifth minute. So in the first five minutes the Hare has covered (1000+500+250+125+62.5)m total.
    So that is ${T_5} = \sum\limits_{k = 1}^5 {\frac{{2000}}{{{2^k}}}}=1937.5M$ SEE HERE
    But in the same time period, the tortoise runs 100m.
    However in 100 min. the tortoise runs 2000M the total distance but the hare only runs 1999.99999999999m
    See Here

    Quote Originally Posted by jpompey View Post
    Wonderful thank you very much.
    I see what you mean about the formula, in my lesson text they wrote it two different ways: n/2(2a+(n-1)) and [n(2a+(n-1))]/2 I must have jumbled them together a bit.
    One thing that still confuses me, is the formula for distance traveled after n minutes for the hare works well, but this one here for the tortoise doesn't seem to.
    You really do need a live tutor.
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  13. #13
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    Re: tortoise and the hare problem

    I meant the tortoise, actually. But whatever.
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    Re: tortoise and the hare problem

    Quote Originally Posted by Plato View Post
    You really do need a live tutor.
    No, with the help of these two good people I was able to find what I did right and wrong. I see now that my mistakes were simple ones. Funny how a little bit of explanation can help clear things up eh?
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  15. #15
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    Re: tortoise and the hare problem

    Worry none 'bout Plato: his mother-in-law was visiting...
    Thanks from jonah
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