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Population Growth

  1. Apr 14, 2003 #1
    The following question is from a text book..and I can't seem to solve it. Can someone help.
    --==============================
    The population of a midwestern city follows the exponential law. If the population decreased from 900,000 to 800,000 from 1993 to 1995, what will the population be in 1997?
     
  2. jcsd
  3. Apr 14, 2003 #2
    600000 ?
     
  4. Apr 14, 2003 #3
    I'll solve this problem from scratch, so that you can see how the mathematical model is formulated:

    We assume that population change is proportional to the current population size. (This assumption holds until the population grows large enough that competition for resources occurs). We thus have a differential equations as follows:

    dN(t)/dt = kN(t) , t is measured in years

    with boundary conditions:

    N(0) = 900,000 ; N(2) = 800,000

    We can solve this by seperating the variables:

    ∫ dN(t)/N(t) = ∫ kdt
    ln|N(t)| = kt +A , A is an arbitrary constant
    N(t) = Bekt

    Now we use the boundary conditions:

    N(0) = A = 900,000
    N(2) = 900,000e2k = 800,000
    => k = -0.058891518
    => N(t) = 900,000e-0.058891518t

    We now use the equation to find the population at time t = 4 (1997):

    N(4) ~ 711,111

    Now, if you haven't done any work on differential equations, then all of the above may as well have been written in French. So I'll solve the problem using the info given:

    You were told that the population grows exponentially, and the most general form for an exponential equations is:

    N(t) = Aekt , A and k are arbitrary constants

    So we will use this equation and solve for A and k, as we did above:

    N(0) = A = 900,000
    N(2) = 900,000e2k = 800,000
    => k = -0.058891518
    => N(t) = 900,000e-0.058891518t

    We now use the equation to find the population at time t = 4 (1997):

    N(4) ~ 711,111
     
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