Problem 31 - Applied Calculus

Question

The population of the world can be represented by \(P=\) \(6.4(1.0126)^{t}\), where \(P\) is in billions of people and \(t\) is years since 2004 . Find a formula for the population of the world using a continuous growth rate.

Step-by-Step Solution

Answer

The continuous population model is \(P = 6.4e^{0.0125t}\).

Step 1: Understanding the Problem

The given formula for the population is based on a discrete growth rate. To convert this to a continuous growth rate, we need to rewrite the exponential function accordingly.

Step 2: Identifying the Relationship

In the formula \(P = 6.4(1.0126)^t\), the factor \(1.0126\) represents the annual growth rate from a discrete perspective. Our goal is to express this growth continuously, using the base \(e\), the natural exponential base.

Step 3: Using the Continuous Growth Formula

The continuous exponential model can be expressed as \(P = P_0 e^{rt}\), where \(P_0\) is the initial population, \(r\) is the continuous growth rate, and \(t\) is the time in years. We know \(P_0 = 6.4\), the initial population in billions.

Step 4: Equating and Solving for \(r\)

To find the continuous growth rate \(r\), equate the expressions: \((1.0126)^t = e^{rt}\). Taking the natural logarithm of both sides, we have: \(\ln(1.0126) = rt\). Solve for \(r\) by recognizing \(t = 1\): \(r = \ln(1.0126)\).

Step 5: Calculating \(r\)

Calculate \(r\) using a calculator: \(r = \ln(1.0126) \approx 0.0125\). This represents the continuous growth rate.

Step 6: Writing the Continuous Model Formula

Substitute \(r = 0.0125\) and \(P_0 = 6.4\) into the continuous model: \(P = 6.4e^{0.0125t}\). This is the formula for the population using a continuous growth rate.

Key Concepts

Exponential Functions

Exponential functions play a major role when studying growth and decay processes. An exponential function is typically represented as \( f(x) = a \, b^x \), where \( a \) is the initial amount, \( b \) is the growth factor, and \( x \) represents time. In our context, this means \( a \) is the starting population, and \( b \) is the growth factor.It's important to distinguish between continuous and discrete growth:

• Discrete growth happens in regular, separate intervals (e.g., annually).
• Continuous growth means the quantity grows uninterrupted over time.

Understanding these differences helps us transition between modeling types, as seen in our original exercise when we convert to continuous growth using the base \( e \), the natural exponential base.

Population Growth

Population growth can be modeled effectively using exponential functions. Typically, a population grows by a certain percentage over a specific time frame, such as annually. This is referred to as the growth rate. In our exercise, the world's population was modeled with a discrete growth rate using the factor \( 1.0126 \), meaning the population grows 1.26% each year.In cases where growth is smoother and happens continuously, we use the mathematical constant \( e \) for modeling. This transformation is essential in providing a more precise representation of growth over time.Continuous growth provides flexibility in examining trends over any time frame. Whether looking at yearly, daily, or minute-by-minute changes, modeling populations with continuous growth gives more refined insights into future trends.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm with base \( e \). In our exercise, it was a crucial tool to switch from a discrete growth rate to a continuous growth rate. To do this, we started with the equation \( (1.0126)^t = e^{rt} \). By applying the natural logarithm to both sides to eliminate the exponent \( t \), we find:\[ \ln(1.0126) = rt \]This equation allowed us to solve for the continuous growth rate \( r \), showing that \( r = \ln(1.0126) \approx 0.0125 \). This is a small, but meaningful change that gives a complete picture of continuous population growth. Natural logarithms are pivotal in calculus and growth rate problems because they simplify exponential equations, making them easier to analyze and solve.