Problem 15 - Calculus

Question

The population of Cairo grew from 5 million to 10 million in 20 years. From \(y^{\prime}=c y\) find \(c .\) When was \(y=8\) million?

Step-by-Step Solution

Answer

\(c \approx 0.03465\) per year; \(y = 8\) million at \(t \approx 11.7\) years.

Step 1: Understand the Growth Model

The differential equation given is \[\frac{dy}{dt} = c y\]where \(y\) represents the population and \(c\) is the constant growth rate we need to find.

Step 2: Find the General Solution

The general solution of the differential equation can be expressed as:\[y(t) = y_0 e^{ct}\]where \(y_0\) is the initial population at time \(t = 0\). We know that \(y_0 = 5\) million.

Step 3: Use Given Data to Find \(c\)

We have \(y(20) = 10\) million. Plug these values into the general solution:\[10 = 5 e^{20c}\]Solving for \(c\), we get:\[2 = e^{20c} \\ln 2 = 20c \c = \frac{\ln 2}{20}\]Therefore, \(c \approx 0.03465\) per year.

Step 4: Solve for When Population Reaches 8 Million

We need to find \(t\) such that \(y(t) = 8\). Using the general solution:\[8 = 5 e^{ct}\]Substitute \(c = \frac{\ln 2}{20}\):\[\frac{8}{5} = e^{\ln 2 \cdot \frac{t}{20}} \\ln\left(\frac{8}{5}\right) = \frac{t \ln 2}{20} \t = \frac{20 \cdot \ln\left(\frac{8}{5}\right)}{\ln 2}\]This simplifies to \(t \approx 11.702\) years after the initial time.

Key Concepts

Differential Equation

A differential equation is a type of mathematical equation that relates a function with its derivatives. In other words, it shows how a rate of change in one variable is related to the variable itself. For population growth, this is incredibly useful, as it allows us to model how populations increase or decrease over time.

The exercise involves the differential equation \( \frac{dy}{dt} = c y \), which represents an exponential growth model. Here:

• \( y \): Represents the size of the population at a given time \( t \).
• \( \frac{dy}{dt} \): Is the rate of change of the population with respect to time.
• \( c \): Is a constant that represents the growth rate.

By solving the differential equation, you can find out how populations grow, how fast they grow, and predict future population sizes under certain conditions.

Population Growth

Population growth describes how the number of individuals in a population increase over time. This is often modeled using exponential functions, especially if the conditions are ideal and resources are unlimited. In our exercise, we model the population growth of Cairo from 5 million to 10 million in 20 years using an exponential growth model.

The key components involved in understanding this growth are:

• Initial population size (\( y_0 \)), which was 5 million in our case.
• Final population size, which increased to 10 million over 20 years.
• Exponential growth formula: \( y(t) = y_0 e^{ct} \).

The growth rate \( c \) found in the exercise (\( c \approx 0.03465 \)) helps us understand the speed of this growth. This model can also predict when the population would reach a certain size, like 8 million.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a special mathematical function that is the inverse of the exponential function. It is used to solve equations involving exponentials, like those in exponential growth models.

When solving for the growth constant \( c \) in the exercise, we use the natural logarithm to simplify equations and solve for unknowns. Here's how it is used:

• The equation \( 2 = e^{20c} \) becomes \( \ln 2 = 20c \) after applying \( \ln \) to both sides.
• This allows us to solve for \( c \) by dividing both sides by 20, yielding \( c = \frac{\ln 2}{20} \).

Similarly, to find when the population reaches 8 million, we use the natural logarithm again to solve the equation \( \ln\left(\frac{8}{5}\right) = \frac{t \ln 2}{20} \). The natural logarithm helps in transforming multiplicative growth into a more manageable additive form, making problem-solving easier.