<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\dy}{\frac{dy}{dx}}
\parindent=0pt
\begin{document}

{\bf Question}

$(*)$ Assume that the population of the earth changes at a rate
proportional to the current population.
\begin{enumerate}
\item Write the ODE satisfied by the population

\item In 1650 the population is estimated to have been about 600
million, and in 1950 about 2,800 million. By solving the ODE and
fitting this data estimate the population at time $t$ (years AD).

\item Using this fitted solution, and assuming that the greatest
population the earth can support is $2.5\times10^{10}$ people, in
what year will this limit be reached?
\end{enumerate}




\vspace{0.25in}

{\bf Answer}

$\ds N(t)=$population
\begin{itemize}
\item[a)]
$\ds\frac{dN}{dt}=kN$

\item[b)]
$\ds t=1650,\,\,\,\, N=600*10^6$

$\ds t=1950,\,\,\,\, N=2800*10^6$

from a) \,\,$\ds N=Ae^{kt}$

$\ds \begin{array}{ll} N(1650) &amp; \Rightarrow 600*10^6=Ae^{k1650}
\Rightarrow A=6*10^8e^{-k1650}\\ N(1950) &amp; \Rightarrow
2800*10^6=Ae^{k1950}=6*10^8e^{k(1950-1650)}\\ &amp; \Rightarrow
2800*10^6=6*10^8e^{300k} \end{array}$

$\ds k=\frac{1}{300}\ln\left(\frac{14}{3}\right)\approx0.0015$ per
year.

\item[c)]
$\ds 2.5*10^{10}=6*10^8e^{k(t-1650)}$

$\ds \Rightarrow
\frac{\ln\left(\frac{2.5*10^2}{6}\right)}{k}+1650=t$

$\ds 726+1650\approx t$

$\ds 2376\approx t$
\end{itemize}


\end{document}
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