the world population was 2560 million in 1950 and 6,080 million in 2000. assume the growth rate of the population P'(t) is proportional to the size of the population P(t): P'(t)=kP(t)

1 Answer

  • Let [tex]P(t)[/tex] denote the population at time [tex]t[/tex] in millions.

    The population's growth rate is modeled by the differential equation


    This equation is separable, and we can solve easily:

    [tex]\dfrac{\mathrm dP(t)}{\mathrm dt}=kP(t)\iff\dfrac{\mathrm dP(t)}{P(t)}=k\,\mathrm dt[/tex]
    [tex]\implies\displaystyle\int\frac{\mathrm dP}P=k\int\mathrm dt[/tex]
    [tex]\implies \ln|P(t)|=kt+C[/tex]
    [tex]\implies P(t)=e^{kt+C}=e^{kt}e^C=Ce^{kt}[/tex]

    Let's assume 1950 corresponds to year [tex]t=0[/tex], so that 2000 would correspond to [tex]t=50[/tex]. Then [tex]P(0)=2560[/tex], which means we have


    and [tex]P(50)=6080[/tex], which means

    [tex]6080=2560e^{50k}\implies \dfrac{19}8=e^{50k}[/tex]
    [tex]\implies 50k=\ln\dfrac{19}8[/tex]
    [tex]\implies k=\dfrac1{50}\ln\dfrac{19}8\approx0.0173[/tex]

    So the population is modeled by


    (assuming [tex]t=0[/tex] refers to the year 1950)