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Quick math question

  1. Jul 12, 2005 #1
    Hi guys...can someone help me out with this problem...

    The population of Newfoundland and Labrador was 533,800 in 2001. At that time the population of Newfoundland was decreasing at a rate of 0.6% per year. The population of PEI was 138,500 in 2001. At that time the population was increasing at a rate of 0.2% per year. In what year will the population of PEI be more than Newfoundland? Explain?


    I have this so far: y=#of years

    533,800(1-0.006)^y

    138,500(1-0.002)^y


    Now what?
     
  2. jcsd
  3. Jul 12, 2005 #2
    the equation of pop. growth is y(t) = y0 * e ^(kt)

    where y0 is begining pop
    k is the growth rate
    and t is the time

    plug in the values and make them equal to each other and solve for t
     
  4. Jul 13, 2005 #3
    we havent done that yet..any other way...
     
  5. Jul 13, 2005 #4
    138,500*(1+0.002)^y = 533,800*(1-0.006)^y
    138,500*(1.002)^y = 533,800*(0.994)^y
    ln(138,500*(1.002)^y) = ln(533,800*(0.994)^y)
    ln(138,500) + ln((1.002)^y) = ln(533,800) + ln((0.994)^y)
    ln(138,500) + y*ln(1.002) = ln(533,800) + y*ln(0.994)
    y*ln(1.002) - y*ln(0.994) = ln(533,800) - ln(138,500)
    y = ln(533,800/138,500) / ln(1.002/0.994) = 168.3

    2001 + 168.3 = 2169.3

    ... theoretical
     
  6. Jul 13, 2005 #5
    Thx faruk but wats ln?
     
  7. Jul 13, 2005 #6
    Natural log
     
  8. Jul 13, 2005 #7
    i haven't done logs yet...this is suppose to be grade 11 math...
     
    Last edited: Jul 13, 2005
  9. Jul 13, 2005 #8
    It's been many years since I've done stuff like this but isn't it the same sort of calculation as working out compound interests?

    Create a graph with the decreasing population of Newfoundland represented using one colour and then do the same for Labrador. Where the line cross is where they equal each other.
     
  10. Jul 13, 2005 #9
    *joejo, logarithms are used to find exponents, which would be the years in your problem. For example, solving for [itex] c [/itex] in the equation
    [tex] a = b^c ,\left( {a,b,c} \right) \in \mathbb{R} [/tex]
    will give you
    [tex] c = \log _b a \Leftrightarrow \frac{{\log a}}{{\log b}} \Leftrightarrow \frac{{\ln a}}{{\ln b}} [/tex]
    *What's left is just plugging in values (*hint:smile:*)
    ---------------------------------------------------------------------
    *P.S., I am a high school junior too (Grade 11), and the mathematics
    at my school can seem at times quite slow as well :shy:;
    [tex] \downarrow [/tex] For example, I just finished second-semester calculus this year, but I
    could have studied it as a sophomore last year (i.e., in 10th grade :rolleyes:~)
     
    Last edited: Jul 13, 2005
  11. Jul 13, 2005 #10
    11th grade: that's 16yrs old? Is that the equivalent of O levels (GCSE's)
     
  12. Jul 13, 2005 #11
    No-one calls them O Levels any more :smile:. Edit: A Levels still remain A Levels, though!
     
  13. Jul 13, 2005 #12
    You can also do it by the trial and error method

    For Newfoundland : 533800*(1-0.006)^t => 533800*(0.994)^t
    For PEI: 138500*(1+0.002)^t => 138500*(1.002)^t

    Start with t+0 (t equals the number of years elapsed) and start jumping:

    when t=20 Newfoundland=476123.8 & PEI=143858.8 which is no good

    t=120 gives Newfoundland=259265.5 & PEI=176025.8 which is closer

    t=150 gives Newfoundland=216439.4 & PEI=186899.4 which is closer still

    t=170 gives Newfoundland=191895.1 & PEI=194519.2 has overshot a little

    t=168 gives Newfoundland=194218.7 & PEI=193743.4 nearly there

    t=169 gives Newfoundland=193053.4 & PEI=194130.9 which is the number you want

    2001+169=2170 which is the year PEI overtakes Newfoundland :)
     
  14. Jul 13, 2005 #13
    Hah, I'm old. Leave me alone :cry:

    Mmmm, the crossover occurs in the later part of 168 so the year would be 2169 (I think)
     
    Last edited: Jul 13, 2005
  15. Jul 13, 2005 #14

    BobG

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    I would think they would have at least introduced common logs (base 10). Or maybe I'm old, too. When I was in high school, you had to have some understanding of logs or you wouldn't understand your calculator. But then, our calculators were three pieces of bamboo lashed together.
     
  16. Jul 13, 2005 #15
    Actually, in Belgium logs are introduced in grade 12, but that may not be a great reference... :rolleyes:
     
  17. Jul 13, 2005 #16
    thanks guys.....you all helped out a bit...Damincs answer was the way we did it in the book...Thanks Daminc!
     
  18. Jul 13, 2005 #17

    saltydog

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    This is what I have JoeJo:

    Consider differential equations for both places:

    [tex]\frac{dN}{dt}=-0.006N[/tex]

    [tex]\frac{dP}{dt}=0.002P[/tex]

    Solving them with the initial conditions given and equating them for when the populations just become equal we get:

    [tex]533800e^{-0.006t}=138500e^{-0.002t}[/tex]

    little of this, little of that, we get t=169 so in 2170 the populations become equal as PEI starts to overtake the other one.

    Edit: I'm such a slow-poke. :yuck:
     
    Last edited: Jul 13, 2005
  19. Jul 14, 2005 #18
    When I was younger I think we were introduced to logs by having a small blue book filled with numbers. Can anyone else remember that book?

    p.s. I'm sure we did logs when we were about 12-13 or so.
     
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