Question

The population of a town grows at a rate proportional to thepopulation present at time t. The initial population of 500increases by 15% in 10years. What will be the population in30 years? How fast is the population growing at$\mathit{t}\mathbf{=}\mathbf{30}$?

Short answer

The population in thirty years is 760 people, and the rate of growth of the population is 11 people/year.

Step-by-step solution

Define growth and decay.

The initial-value problem,$\frac{\text{dx}}{\text{dt}}{\text{= kx, x(t}}_{\text{0}}{\text{) =x}}_{\text{0}}$ 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, $\text{dx/dt}$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 P be,

$\frac{\text{dP}}{\text{dt}}\text{= kP}$… (1)

And with the conditions,$\text{P(t = 0 years) = 500}$ and role="math" localid="1663847230031" $\mathrm{P}(\mathrm{t}=10\mathrm{years})=1.15\times 500=575$. As the equation (1) is linear and separable, so integrate the equation and separate the variables.

role="math" localid="1663847276890" $$\begin{gathered} \frac{\mathrm{dP}}{\mathrm{P}}=\mathrm{kdt} \\ \int \frac{\mathrm{1}}{\mathrm{P}}\mathrm{dP}=\mathrm{k}\int \mathrm{dt} \\ \mathrm{lnP}=\mathrm{kt}+{\mathrm{c}}_{\mathrm{1}} \\ {\mathrm{e}}^{\mathrm{lnP}}={\mathrm{e}}^{\left(\mathrm{kt}+{\mathrm{c}}_{\mathrm{1}}\right)} \end{gathered}$$

Then, the equation becomes,

$$\begin{gathered} {\text{P =e}}^{\text{kt}}{\text{e}}^{{\text{c}}_{\text{1}}} \\ {\text{= ce}}^{\text{kt}} \end{gathered}$$… (2)

Obtain the values of constants.

To find the values of constants, apply the point in the equation (2), then

$\begin{array}{rcl}& & {\text{500 = ce}}^{\text{0}}\\ & & \text{c = 500}\end{array}$

Substitute the value of in the equation (2).

${\text{P = 500e}}^{\text{kt}}$… (3)

Again, apply the other point$\text{(P,t) =}\left(\text{575,10}\right.\text{years)}$ in the equation (3).

role="math" localid="1663847397530" $\begin{array}{rcl}575& =& 500{\mathrm{e}}^{10\mathrm{k}}\\ \frac{575}{500}& =& {\mathrm{e}}^{10\mathrm{k}}\\ \ln (1.15)& =& 10\mathrm{k}\\ \mathrm{k}& =& \frac{\ln (1.15)}{10}\\ & =& \frac{0.14}{10}\\ & =& 0.014\end{array}$

Substitute the value of in the equation (3).

$\mathrm{P}=500{\mathrm{e}}^{0.014\mathrm{t}}$… (4)

Obtain the population after thirty years.

Substitute the value into the equation (4).

$\begin{array}{rcl}\mathrm{P}& =& 500{\mathrm{e}}^{0.014\times 30}\\ & =& 500\times {\mathrm{e}}^{0.4192}\\ & =& 500\times 1.5208\\ & \approx & 760\mathrm{people}\end{array}$

Obtain the growth of the population at thirty years.

Let the rate at which the population grows at thirty years be,

$\begin{array}{rcl}\frac{\mathrm{dP}}{\mathrm{dt}}& =& 0.014\times 760\\ & \approx & 11\mathrm{persons}/\mathrm{year}\end{array}$