Question

The population of a community is known to increase at a rate proportional to the number of people present at time $\mathit{t}$. If an initial population${\mathit{P}}_{\mathbf{0}}$has doubled in 5years, how long will it take to triple? To quadruple?

Short answer

The time it takes to triple is 7.92 years, and to quadruple is 10 years.

Step-by-step solution

Define growth and decay.

The initial-value problem,$\frac{dx}{dt}=kx,x(t_0)\; =x_0$ where $k$is a constant of proportionality, serves as a model for diverse phenomena involving either growth or decay. This is in the form of a first-order reaction (i.e.) a reaction whose rate, or velocity, $dx/dt$is directly proportional to the amount $x$of a substance that is unconverted or remaining at time $t$.

Solve for first order growth and decay equation.

Let the linear equation with the population of a community as be,

$\frac{dP}{dt}=kP$… (1)

And with the conditions,$P(t=\; 0\mathrm{ye}ars)\; =P_0$ and $P(t=\; 5\mathrm{ye}ars)\; =\; 2P_0$. As the equation (1) is linear and separable, so integrate the equation and separate the variables.

$\begin{array}{rcl}\frac{dP}{P}& =& \mathrm{kd}t\\ \int \frac{1}{P}dP& =& k\int dt\\ lnP& =& \mathrm{kt}+c_1\\ & & e^{lnP}=e^{\left(kt+c_1\right)}\end{array}$

Then, the equation becomes,

role="math" localid="1663834161832" $$\begin{gathered} {\text{P =e}}^{kt}{\mathrm{e}}^{{\mathrm{c}}_{\mathrm{1}}} \\ =c{e}^{kt} \end{gathered}$$… (2)

Obtain the values of constants.

To find the values of constants, apply the point$(P,t)\; =\; (P_0,0)$ in the equation (2), then

$$\begin{gathered} P_0={\mathrm{ce}}^0 \\ c=P_0 \end{gathered}$$

Substitute the value of c in the equation (2).

$P=P_0{e}^{kt}$… (3)

Again, apply the other point $\text{(P,t) =}\left(2P_0,5\right.years)$in the equation (3).

$$\begin{gathered} 2P_0=P_0{e}^{5k} \\ 2=e^{5k} \\ ln(2)\; =\; 5k \end{gathered}$$

$$\begin{gathered} k=\frac{ln(2)}{5} \\ =\frac{0.693}{5} \\ =\; 0.1386 \end{gathered}$$

Substitute the value of in the equation (3).

$P=P_0{e}^{0.1386t}$… (4)

Obtain the time at which the number of people is tripled.

Substitute the value$P_0=\; 3P_0$ into the equation (4).

$\begin{array}{rcl}\backslash \mathrm{begingathered}3P_0& =& P_0{e}^{0.1386t}\\ 3& =& e^{0.1386t}\\ ln\left(3\right)& =& 0.1386t\\ t& =& \frac{ln\left(3\right)}{0.1386}\\ & =& \frac{1.0986}{0.1386}\\ t& =& 7.92\mathrm{years}\end{array}$

Obtain the time at which the number of people is quadruple.

Substitute the value $P_0=4P_0$into the equation (4).

$\begin{array}{rcl}{P}_0& =& P_0{e}^{0.1386t}\\ 4& =& e^{0.1386t}\\ ln\left(4\right)& =& 0.1386t\\ t& =& \frac{ln\left(4\right)}{0.1386}\\ & =& \frac{1.386}{0.1386}\\ t& =& 10\mathrm{years}\end{array}$