Problem 47 - Applied Calculus

Question

In 2009 , the population, \(P\), of India was \(1.166\) billion and growing at \(1.5 \%\) annually. (a) Give a formula for \(P\) in terms of time, \(t\), measured in years since 2009 . (b) Find \(\frac{d P}{d t},\left.\frac{d P}{d t}\right|_{t=0}\), and \(\left.\frac{d P}{d t}\right|_{t=25}\). What do each of these represent in practical terms?

Step-by-Step Solution

Answer

The population formula is \( P(t) = 1.166 \times 1.015^t \). The initial growth rate is \( 1.166 \times \ln(1.015) \), and the growth rate at \( t=25 \) is \( 1.166 \times 1.015^{25} \times \ln(1.015) \).

Step 1: Formula for Population Over Time

The population of India grows at a constant rate of 1.5% annually. We use the exponential growth formula to express this population growth. The formula for the population at any year \( t \) years since 2009 is given by:\[ P(t) = P_0 \times (1 + r)^t \]Where:- \( P_0 = 1.166 \) billion (initial population in 2009).- \( r = 0.015 \) (growth rate as a decimal).Thus, the formula becomes:\[ P(t) = 1.166 \times (1.015)^t \]

Step 2: Differentiate the Population Formula

We need to find \( \frac{dP}{dt} \) to determine the rate of change of the population with respect to time.The derivative of an exponential function \( a^t \) is \( a^t \ln(a) \) by chain rule. Thus:\[ \frac{dP}{dt} = \frac{d}{dt}(1.166 \times 1.015^t) = 1.166 \times 1.015^t \times \ln(1.015) \]

Step 3: Evaluate Derivative at \( t = 0 \)

Substitute \( t = 0 \) into \( \frac{dP}{dt} \) to evaluate the initial rate of population growth:\[ \left.\frac{dP}{dt}\right|_{t=0} = 1.166 \times 1.015^0 \times \ln(1.015) \]Simplifying, \( 1.015^0 = 1 \), so: \[ \left.\frac{dP}{dt}\right|_{t=0} = 1.166 \times \ln(1.015) \]

Step 4: Evaluate Derivative at \( t = 25 \)

Substitute \( t = 25 \) into \( \frac{dP}{dt} \) for the rate after 25 years:\[ \left.\frac{dP}{dt}\right|_{t=25} = 1.166 \times 1.015^{25} \times \ln(1.015) \]

Step 5: Interpretation of the Results

- \( \frac{dP}{dt} \) gives the population's growth rate over time.- \( \left.\frac{dP}{dt}\right|_{t=0} \) represents the initial growth rate in 2009.- \( \left.\frac{dP}{dt}\right|_{t=25} \) represents the growth rate after 25 years, in 2034.

Key Concepts

Population Modeling

Population modeling is a key concept in understanding how populations change over time and predicting future growth. It involves using mathematical formulas to describe the dynamics of population growth, accounting for birth rates, death rates, immigration, and emigration. In our scenario with India, we're modeling the population from the base year 2009 onwards, using exponential growth principles. This approach helps us predict how the population increases given a constant annual growth rate. We represent the population at any time, \( t \), with the equation:
\[ P(t) = P_0 \times (1 + r)^t \]

• \( P_0 \): Initial population at the start year (1.166 billion in 2009).
• \( r \): Growth rate (0.015 for the given 1.5% annual growth).
• \( t \): Time in years since 2009.

This formula caters specifically to scenarios where populations grow at a constant percentage each year, making it ideal for long-term predictions when conditions are stable.

Differentiation

Differentiation is a powerful mathematical tool used to find rates of change. It is essential when we need to understand how quickly a variable, such as population, is changing over time. In the context of our exercise, differentiating the exponential population formula gives us \( \frac{dP}{dt} \), the rate of population growth at any point in time.
The derivative of an exponential function like \( 1.015^t \) is found through the chain rule, leading to the expression:
\[ \frac{dP}{dt} = 1.166 \times 1.015^t \times \ln(1.015) \]This function allows us to calculate the instantaneous rate at which India's population is growing, given any year since 2009. Differentiation, thereby, connects static population figures to dynamic growth scenarios, providing insights into population trends and speeds at various intervals.

Growth Rate Analysis

Growth rate analysis gives us a clearer picture of how a population's size changes over time beyond raw numbers. By calculating the derivative \( \frac{dP}{dt} \), we determine how fast the population is growing each year. This analysis is crucial in making informed predictions and planning resources accordingly.
In practical terms:

• The initial growth rate, \( \left.\frac{dP}{dt}\right|_{t=0} \), tells us how rapidly the population was expected to grow in 2009. It is found by setting \( t = 0 \) in our derivative formula.
• \( \left.\frac{dP}{dt}\right|_{t=25} \) gives the population's growth rate 25 years later, in 2034, showing the rate then based on continuous growth.

By understanding these growth rates, one can assess whether the population will rapidly increase, stabilize, or potentially decline over time, guiding policies and resource allocation effectively.

Exponential Functions

Exponential functions are central to modeling growth processes where change occurs at a rate proportional to the current state. They are expressed in the form \( a^t \), where \( a \) is a positive constant. In population modeling, exponential functions capture the essence of growth at a steady percentage rate per time period, such as years.
Our formula for India's population, \( P(t) = 1.166 \times 1.015^t \), uses an exponential function to model growth, illustrating how populations 'double' or increase over time.

• The base \( 1.015 \) represents the 1.5% annual growth.
• Exponential growth assumes potential unlimited increase, reflecting how populations could grow under optimal conditions.

Understanding exponential functions allows for insight into how small, constant growth rates can lead to significant changes over time, which can be surprising without a grasp of these mathematics fundamentals. This is why exponential functions are a core part of ecological and economic studies, including population dynamics.