Population Growth

Let $P$ denote the population of the country at a given time, and $t$ the time elapsed (in years). Suppose that at $t = 0$ the population is $P_{ 0 }$ , and that $t = t_{ 2 } = 50$ is the time at which the population doubles to $2 P _{ 0 }$ . Then we have to find $t_{ 3 }$ , the time at which the population triples to $3P_{ 0 } :$

$\underbrace{ \text{rate of increase in the number of inhabitants } } _{ \huge{ \frac{ dP }{ dt } } } \underbrace{ \text{ is} } _{ \huge{ = } } \underbrace{ \text{ proportional to} } _{ \huge{ k } } \underbrace{ \text{ the number of inhabitants.} } _{ \huge{ P } }$

Now, $\frac{ dP } { P } = k \ dt\implies \int \frac{ dP } { P } = \int k \ dt \implies \ln P = kt + C.$ We use the initial values to solve two linear equations in two unknowns and find their values. Since $P = P_{ 0 }$ at $t = 0$ and $P=2P_0$ , we have

$\begin{cases} -\ln P _{ 0 } & = k \times 0 + C \\ - \ln 2P_ { 0 } & = k \times 50 + C, \end{cases}$

where the solution is $C = \ln P _{ 0 }$ and $k = \frac{ \ln 2 }{ 50 }$ . Hence we obtain the particular solution

$\begin{aligned} \ln P &= \frac{ \ln 2 }{ 50 } t + \ln P_{ 0 } \\ P &= P_{ 0 } 2 ^ { t / 50 }. \end{aligned}$

Now, we want to find the times taken for the population to triple, i.e. $P = 3 P _{ 0 }:$

$3 P _{ 0 } = P_{ 0 } 2 ^ { t_{ 3 } / 50 }.$

Solving for $t _ { 3 },$ the time taken (in years) is $50 \log _{ 2 } 3\approx 79.$

We note that population $P$ grows proportionally to its size at time $t$ , that is:

$\begin{aligned} \frac {dP}{dt} & \propto P \\ \implies \frac {dP}{dt} & = \color{#3D99F6}{k}P \quad \quad \small \color{#3D99F6}{k \text{ is a constant}} \\ \frac {dP}{P} & = k \ dt \\ \int \frac {dP}{P} & = \int k \ dt \\ \ln P & = kt + C \quad \quad \small \color{#3D99F6}{k \text{ is a constant}} \\ P & = P_0 e^{kt} \quad \quad \small \color{#3D99F6}{P_0 = e^C \text{ the initial population, when } t = 0} \end{aligned}$

We know that $P = 2P_0$ , when $t=50$ years, therefore,

$\begin{aligned} 2P_0 & = P_0 e^{50k} \\ \implies k & = \frac {\ln 2}{50} \end{aligned}$

For $P = 3 P_0$ , we have:

$\begin{aligned} 3P_0 & = P_0 e^{\frac {\ln 2}{50}t} \\ \implies t & = \frac {50 \ln 3}{\ln 2} \\ & \approx \boxed{79} \end{aligned}$