In-1790-the-population-of-the-United-States-was-3-93-million-and-in-1890-it-was-62-98-million-Using-the-Malthusian-model-estimate-the-U-S-population-as-a-function-of-time-

Question Number 131481 by bemath last updated on 05/Feb/21

$$\mathrm{In}\:\mathrm{1790}\:\mathrm{the}\:\mathrm{population}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{United}\:\mathrm{States}\:\mathrm{was}\:\mathrm{3}.\mathrm{93}\:\mathrm{million} \\ $$$$\mathrm{and}\:\mathrm{in}\:\mathrm{1890}\:\mathrm{it}\:\mathrm{was}\:\mathrm{62}.\mathrm{98}\:\mathrm{million} \\ $$$$\mathrm{Using}\:\mathrm{the}\:\mathrm{Malthusian}\:\mathrm{model}\:, \\ $$$$\mathrm{estimate}\:\mathrm{the}\:\mathrm{U}.\mathrm{S}\:\mathrm{population}\:\mathrm{as} \\ $$$$\mathrm{a}\:\mathrm{function}\:\mathrm{of}\:\mathrm{time} \\ $$

Answered by EDWIN88 last updated on 05/Feb/21

we set t=0 to be year 1790 then we have p(t) = 3.93×e^(kt) ; where p(t) is population in millions . and p(100)=62.98 = 3.93×e^(100k) we find k = ((ln (62.98)−ln (3.98))/(100)) ≈ 0.027742 so p(t)=3.93×e^(0.027742t)

$$\mathrm{we}\:\mathrm{set}\:\mathrm{t}=\mathrm{0}\:\mathrm{to}\:\mathrm{be}\:\mathrm{year}\:\mathrm{1790}\:\mathrm{then}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{p}\left(\mathrm{t}\right)\:=\:\mathrm{3}.\mathrm{93}×\mathrm{e}^{\mathrm{kt}} \:;\:\mathrm{where}\:\mathrm{p}\left(\mathrm{t}\right)\:\mathrm{is}\:\mathrm{population} \\ $$$$\mathrm{in}\:\mathrm{millions}\:.\:\mathrm{and}\:\mathrm{p}\left(\mathrm{100}\right)=\mathrm{62}.\mathrm{98}\:=\:\mathrm{3}.\mathrm{93}×\mathrm{e}^{\mathrm{100k}} \\ $$$$\mathrm{we}\:\mathrm{find}\:\mathrm{k}\:=\:\frac{\mathrm{ln}\:\left(\mathrm{62}.\mathrm{98}\right)−\mathrm{ln}\:\left(\mathrm{3}.\mathrm{98}\right)}{\mathrm{100}}\:\approx\:\mathrm{0}.\mathrm{027742} \\ $$$$\mathrm{so}\:\mathrm{p}\left(\mathrm{t}\right)=\mathrm{3}.\mathrm{93}×\mathrm{e}^{\mathrm{0}.\mathrm{027742t}} \\ $$

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