Question

In 2012, the population of a city was 6.13 million. The exponential growth rate was 3.68% per year.

Answer 1

Step 1

State the formula for exponential growth

$P(t)=P_o(1+r)^t$

where;

$\begin{gathered} P(t)=\text{ Population at year t} \\ P_o=The\text{ population at t= 0 = 6.13 million} \\ r=\text{Exponential growth rate= }\frac{3.68}{100}=0.0368 \\ t=\text{ time} \end{gathered}$

Step 2

A) Find the exponential growth function

$\begin{gathered} P(t)=6.13_{}(1+0.0368)^t \\ P(t)=6.13(1.0368)^t \\ \text{Where P(t) is in millions of people} \end{gathered}$

Step 3

B) Estimate the population of the city in 2018

$\begin{gathered} For\text{ 2018, t = 6} \\ \text{Hence,} \\ P(t)=6.13(1.0368)^6 \\ P(t)\text{ }\approx7.61\text{ million people} \end{gathered}$

Step 4

C) When will the population of the city be 10million

$\begin{gathered} P(t)=6.13(1.0368)^t \\ 10=6.13(1.0368)^t \\ \frac{10}{6.13}=(1.0368)^t \\ find\text{ t} \\ \ln (\frac{10}{6.13})=t\ln (1.0368) \\ \frac{0.489390343}{\ln(1.0368)}=\frac{t\ln(1.0368)}{\ln(1.0368)} \\ t=13.54187199\text{ } \\ \text{This corresponds to 13.5 years after 2012 to 1 decimal place} \end{gathered}$

Step 5

Find the doubling time

$\begin{gathered} We\text{ want the time where } \\ \frac{P(t)}{P_o}=\frac{P(t)}{6.13}=2 \end{gathered}$$\begin{gathered} But\text{ }\frac{P(t)}{P_o}=(1.0368)^t \\ (1.0368)^t=\frac{P(t)}{6.13} \end{gathered}$$\begin{gathered} \text{But we want }\frac{P(t)}{6.13}=2 \\ \text{Therefore} \\ (1.0368)^t=2 \\ t\ln (1.0368)=\ln (2) \\ \frac{t\ln(1.0368)}{\ln(1.0368)}=\frac{\ln(2)}{\ln(1.0368)} \\ t=19.18000737 \\ t\approx19.2\text{ years to 1 decimal place} \end{gathered}$

The doubling time approximately to 1 decimal place = 19.2 years