Q. 73 - Calculus

Question

 Another model for population growth is what is called super growth. It assumes that the rate of change in a population is proportional to a higher power of P than P¹. For example, suppose the rate of change of the world’s human population P(t) is proportional to P^1.1.

 (a) Set up a differential equation that describes $\frac{dP}{dt}$ , and solve it to get an equation for the world population P(t). Your answer will involve two unsolved constants.

(b)  Given that the world population was estimated to be 6.451 billion in 2005 and 6.775 billion in 2010, what is the value of the proportionality constant for this super growth model? 

(c) According to your model, when will the world population be 7 billion?

Step-by-Step Solution

Answer

(a) $P\left(t\right)={\left[-0.1t+C\right]}^{-10}$

(b)$P\left(t\right)={\left[-0.1\left(0.0082\right)t+0.8299\right]}^{-10}$

(c) $$\begin{gathered} \mathrm{Hence}, \mathrm{for} 8.54 \mathrm{world}\text{'}\mathrm{s} \mathrm{population} \mathrm{will} \mathrm{become} 7 \mathrm{billion}. \mathrm{Since}-\mathrm{gcorresponds} \mathrm{to} \mathrm{the} \mathrm{year} \\ 2005 \mathrm{it} \mathrm{implies} \mathrm{that} \mathrm{according} \mathrm{to} \mathrm{the} \mathrm{super} \mathrm{growth} \mathrm{model} \mathrm{described} \mathrm{in} \mathrm{the} \mathrm{problem} \\ \mathrm{the} \mathrm{world}\text{'}\mathrm{s} \mathrm{population} \mathrm{will} \mathrm{become} 7 \mathrm{billion} \mathrm{by} \mathrm{the} \mathrm{middle} \mathrm{of} \mathrm{year} 2013. \end{gathered}$$

Step 1. Given

 Another model for population growth is what is called super growth. It assumes that the rate of change in a population is proportional to a higher power of P than P¹. For example, suppose the rate of change of the world’s human population P(t) is proportional to P^1.1.

Part(a) Step 2. Finding the differential equation

Note that the problem is of exponential growth model; and the rate of growth of the population is proportional to p^1.1. Hence, the differential equation describing the problem is

$$\begin{gathered} \frac{dP}{dt}\alpha P^{1.1} \\ \frac{dP}{dt}=k{P}^{1.1} \end{gathered}$$

Now, proceed to solve the differential equation. Observe that the differential equation does not contain the independent variable at all, so solve it by using anti differentiating method

$$\begin{gathered} \frac{dP}{dt}\alpha P^{1.1} \\ \frac{dP}{dt}=k{P}^{1.1} \\ \int \frac{dP}{P^{1.1}}=k\int dt \\ \frac{P^{-0.1}}{-0.1}=kt+C_1 \\ P\left(t\right)={\left[-0.1t+C\right]}^{-10} \end{gathered}$$

Part(b) Step 3. Finding solution to the super growth model

(b) Suppose that the year 2005 corresponds to l = 0 , then the year 2010 will correspond to r = 5 . Now, it is given that world population was 6.451 billion in the year 2005. So, take P(1)=6.451 and t = 0 in equation (1) and solve for C

$$\begin{gathered} 6.451={\left[0+C\right]}^{-10} \\ C={\left(6.451\right)}^{-1/10} \\ =0.8299 \\ \mathrm{Substitute} \mathrm{this} \mathrm{value} \mathrm{of} \mathrm{C} \mathrm{in} \mathrm{equation} \left(1\right) \mathrm{to} \mathrm{find} \mathrm{the} \mathrm{solution} \mathrm{of} \mathrm{the} \mathrm{super} \mathrm{growth} \mathrm{model} \mathrm{as} \\ P\left(t\right)={[-0.1kt+0.8299]}^{-10} \end{gathered}$$

Next, for finding the constant of proportionality k, use the other given data, namely, the population was 6.775 billion in 2010. So, take P(r)=6.775 and t = 5 in equation (2) and solve

for k

$$\begin{gathered} 6.775={\left[-0.1k.\left(5\right)+0.8299\right]}^{-10} \\ {\left(6.775\right)}^{-0.1}=-0.5k+0.8299 \\ 0.5k=0.8299-{\left(6.775\right)}^{-0.1} \\ k=0.0082 \\ \mathrm{Hence}, \mathrm{the} \mathrm{solution} \mathrm{to} \mathrm{the} \mathrm{super} \mathrm{growth} \mathrm{model} \mathrm{is} \\ P\left(t\right)={\left[-0.1\left(0.0082\right)t+0.8299\right]}^{-10} \end{gathered}$$

Part(c) Step 4. Finding time

$$\begin{gathered} \mathrm{In} \mathrm{order} \mathrm{to} \mathrm{find} \mathrm{the} \mathrm{time} \mathrm{when} \mathrm{world} \mathrm{population} \mathrm{will} \mathrm{become} 7 \mathrm{billion}, \mathrm{take} P\left(t\right) = 7 in equation \left(3\right) and solve for t \\ 7={[-0.1(0.0082)t+0.8299]}^{-10} (7{)}^{-0.1}=0.00082t+0.8299 \\ t=\frac{1}{0.00082} [0.8299-{\left(7\right)}^{-0.1}] \\ =8.54 \\ \mathrm{Hence}, \mathrm{for} 8.54 \mathrm{world}\text{'}\mathrm{s} \mathrm{population} \mathrm{will} \mathrm{become} 7 \mathrm{billion}. \mathrm{Since}-\mathrm{gcorresponds} \mathrm{to} \mathrm{the} \mathrm{year} \\ 2005 \mathrm{it} \mathrm{implies} \mathrm{that} \mathrm{according} \mathrm{to} \mathrm{the} \mathrm{super} \mathrm{growth} \mathrm{model} \mathrm{described} \mathrm{in} \mathrm{the} \mathrm{problem} \\ \mathrm{the} \mathrm{world}\text{'}\mathrm{s} \mathrm{population} \mathrm{will} \mathrm{become} 7 \mathrm{billion} \mathrm{by} \mathrm{the} \mathrm{middle} \mathrm{of} \mathrm{year} 2013. \end{gathered}$$