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Population growth help

  1. Mar 25, 2009 #1
    A population grows at a constant relative rate. After 10 months the population grows to 72, and after 18 months the population grows to 96. Find C and k.

    the answer I got was c=50.25 and k=.0359

    can somebody just double check for me? I dont know if the method I was used was correct.
     
  2. jcsd
  3. Mar 26, 2009 #2

    CompuChip

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    If you have doubts about your method, you better post the method :)

    I get (up to two decimals) the same answers, but that might be a co-incidence.

    [Also, please be clear about your notation. I just assumed that C would be the initial population and k the rate of growth.]
     
  4. Mar 26, 2009 #3
    ok heres the method I used, its very long and we haven't had to solve these type of problems this way in class so I was apprehensive about it

    C = Initial Population K = Relative Growth Rate

    P(10) = Ce^(10k) = 72 P(18) = Ce^(18k) = 96
    e^(10k) = 72/C e^(18k) = 96/C
    10k = ln(72/C) 18k = ln(96/C)
    k = (ln(72/C))/10 k = (ln(96/C))/18

    k = k
    (ln(72/C))/10 = (ln(96/C))/18

    18(ln(72/C)) = 10(ln(96/C))
    18ln72 - 18lnC = 10ln96 - 10lnC
    76.98 - 18lnC = 45.64 - 10lnC
    -10lnC + 18lnC = 76.98 - 45.64
    8lnC = 31.34
    lnC = 3.92
    C = 50.4 (I got 50.25 because I didnt round so I could get a more accurate answer)

    then I plugged in C to find k

    (ln(72/50.4))/10 = .0356 = k

    was my method correct? I felt like I was making a simple problem, harder than what it was, but idk

    shouldnt C be a whole number since it is a population?
     
  5. Mar 27, 2009 #4

    CompuChip

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    By the looks of it, your method is correct, although it is not the most clear way to write down what you are doing. In particular, I was wondering where the exponential is coming from on the first line, although I think in the end it doesn't matter because taking the wrong base number (e instead of something else) produces a common factor which drops out when you equate the two.

    Here is what I would expect.
    The formula for growth at a constant relative rate is
    P(n) = C (1 + k)^n
    because for n = 0 the population is C, and for each time step you have to multiply the previous population P(n - 1) by (1 + k).

    Then it is given that
    P(10) = 72 = C (1 + k)^(10)
    P(18) = 96 = C (1 + k)^(18)
    which gives you two equations in two unknowns.

    What you could do is divide them, and get
    [tex]\frac{P(18)}{P(10)} = \frac{96}{72} = \frac{C (1 + k)^{18}}{C (1 + k)^{10}} = (1 + k)^8[/tex]
    and you see that C drops out. So now it's easy to solve for k:
    [tex](1 + k)^8 = 4/3[/tex]
    so
    [tex]1 + k = \sqrt[8]{4/3} = 1,0366...[/tex]
    [tex]\qquad\implies k = 0,0366...[/tex]

    To find C, you only need to plug it back in to one of the equations, for example
    P(10) = 72 = C * (1,0366...)^(10)
    (try not to round, e.g. if you use a calculator try to use it's ANS function to plug in 1 + k) and so
    C = 72 / (1,0366...)^(10) = 50.25....

    Note how the formula one starts with is more intuitive (I see an exponential in your formula, while it says the growth is at constant relative rate; my formula explicitly has this behaviour by multiplying by the growth percentage at every step) and easier to work with (you don't need logarithms, just some root).

    That C and P(n) are in general not integer numbers is common in such exercises. You need to remember that this is merely a model for some "real" process, which makes assumptions (constant growth), tries to describe something discrete (counting the population every month) by a continuous function (you can calculate C (1 + k)^n for any n), and all we want of it is to give us more or less the right numbers at n = 0, 1, 2, ... although we will never be able to use it to predict any exact historic or future data.
     
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