The following table gives estimates of the world population, in millions, from 1750 to 2000 |...

Question:

The following table gives estimates of the world population, in millions, from 1750 to 2000

Year Population
1750 680
1800 860
1850 1200
1900 1590
1950 2580
2000 5890

Use the exponential model and the population figures for 1750 and 1800 to predict the world population in 1900.

Exponential Growth Function

Exponential growth function given by {eq}f(t) = f_0 e^{kt} {/eq} where {eq}f_0 {/eq} is the initial value of the function, {eq}f(t) {/eq}, and k is the growth constant which is related to how fast or how slow the function is increasing over time. This function is used to model the increase in the number of individuals in a population. Aside from the exponential function, there are other models that describe population dynamics such as agent-based modeling or probabilistic population models.

Answer and Explanation:


The population can be modeled by the function

{eq}P(t) = P_0 e^{kt} {/eq}

where {eq}P_0 {/eq} is the initial population. We need to solve for k using the data given from the table using the values in 1750 to 1800.


We set {eq}t = 0 {/eq}, for the year 1750 and {eq}t = 50 {/eq} for the year 1800.


With this we have the following equations:

{eq}\displaystyle \begin{align*} P(0) &= P_0 e^{k(0)}\\ 680 &= P_0 e^0\\ P_0 &= 680\\ \end{align*} {/eq}

Now, that we have {eq}P_0 = 680 {/eq}, we can plug this into the equation that represents the population in the year 1800.


For the year 1800, we have {eq}t = 50 {/eq}.


{eq}\displaystyle \begin{align*} P(t) &= P_0 e^{kt} \\ P(50) &= 680 e^{k(50)}\\ 860 &= 680 e^{50k}\\ e^{50k} &= \frac{860}{680} \\ e^{50k} &= \frac{43}{34} \end{align*} {/eq}

Taking the natural logarithm of the equation to solve for k.


{eq}\displaystyle \begin{align*} \ln \bigg( e^{50k} &= \frac{43}{34}\bigg) \\ 50k &= \ln \frac{43}{34}\\ k &= \frac{1}{50} \ln \frac{43}{34}\\ &\approx 0.004697 \end{align*} {/eq}


If we want to calculate the population in the year 1900, we can use the equation {eq}\displaystyle P(t) = 680 e^{kt} {/eq} and set {eq}t = 150 {/eq}, where P(t) is in millions.


{eq}\displaystyle \begin{align*} P(t) &= 680 e^{kt} \\ &= 680 e^{\bigg( \frac{1}{50} \ln \frac{43}{34} \bigg) t} \\ &= 680 e^{\bigg( \frac{1}{50} \ln \frac{43}{34} \bigg) (150)} \\ &= 1380\ \rm{million} \end{align*} {/eq}


Based on thepopulation values in the year 1750 and 1800, the expected population by the year 1900 will be {eq}1380 {/eq} million. The value derived from the exponential function underestimated the population. This means that a simple exponential growth function is not an accurate model for the population of the world.


Learn more about this topic:

Loading...
Exponential Growth: Definition & Examples

from High School Algebra I: Help and Review

Chapter 6 / Lesson 10
99K

Related to this Question

Explore our homework questions and answers library