Question

Suppose it is known that the population of the communityin Problem 1is 10,000after 3years. What was the initialpopulation${\mathit{P}}_{\mathbf{0}}$? What will be the population in$\mathbf{10}$years? Howfast is the population growing at$\mathit{t}\mathbf{=}\mathbf{10}$?

Short answer

The initial population is 6597.54, the population in ten years is 26930, and the rate of growth of the population is 3659 people/year.

Step-by-step solution

Step 1:Define growth and decay.

The initial-value problem, $\frac{dx}{dt}=\mathrm{kx},\mathrm{x(}{t}_0)=x_0$where 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 be,

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

And with the conditions,${\text{P(t = 0 years ) =P}}_{\text{0}}$ and ${\text{P(t = 5 years ) = 2P}}_{\text{0}}$. As the equation (1) is linear and separable, so integrate the equation and separate the variables.

$$\begin{gathered} \frac{\text{dP}}{\text{P}}\text{= kdt} \\ \int \frac{\text{1}}{\text{P}}\text{dP = k}\int \text{dt} \\ {\text{lnP = kt +c}}_{\text{1}} \\ {\text{e}}^{\text{lnP}}{\text{=e}}^{\left({\text{kt +c}}_{\text{1}}\right)} \end{gathered}$$

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

$$\begin{gathered} {\text{P}}_{\text{0}}{\text{= ce}}^{\text{0}} \\ {\text{c =P}}_{\text{0}} \end{gathered}$$

Substitute the value of in the equation (2).

${\text{P =P}}_{\text{0}}{\text{e}}^{\text{kt}}$… (3)

Again, apply the other point$\text{(P,t) =}\left({\text{2P}}_{\text{0}}\text{,5}\right.\text{years)}$ in the equation (3).

$$\begin{gathered} {\text{2P}}_{\text{0}}{\text{=P}}_{\text{0}}{\text{e}}^{\text{5k}} \\ {\text{2 =e}}^{\text{5k}} \\ \text{ln(2) = 5k} \end{gathered}$$

$$\begin{gathered} k=\frac{ln\left(2\right)}{5} \\ =\frac{ln2}{5}y{r}^{-1} \end{gathered}$$

Substitute the value of in the equation (3).

${\text{P =P}}_{\text{0}}{\text{e}}^{\frac{\text{ln2}}{\text{5}}\text{t}}$… (4)

Obtain the initial population and the population after ten years.

Substitute the value$P=10,000$and$t=3$ into the equation (4).

role="math" localid="1663835854821" $\begin{array}{rcl}{P}_0& =& 10000\times e^{\frac{l{m}^{3}3}{3}}\\ & \approx & 6597.54\end{array}$

When $t=10$, then the population after ten years is given by,

role="math" localid="1663835848941" $\begin{array}{rcl}P\left(10\right)& =& 6597.5e^{\frac{ln2}{6}10}\\ & \approx & 26390\end{array}$

Obtain the growth of the population at ten years.

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

$\begin{array}{rcl}P\text{'}\left(10\right)& =& kP\left(10\right)\\ & =& \frac{ln2}{5}26930\\ & =& 3658.45\\ & \approx & 3659\end{array}$