Problem 36 - Calculus

Question

The population, \(P(t),\) of China, in billions, can be approximated by \(^{4}\) $$P(t)=1.394(1.006)^{t}$$,where \(t\) is the number of years since the start of 2014 According to this model, how fast was the population growing at the start of 2014 and at the start of \(2015 ?\) Give your answers in millions of people per year.

Step-by-Step Solution

Answer

The population grew at 8.34 million/year in 2014 and 8.39 million/year in 2015.

Step 1: Understand the Population Growth Model

The population model given is \( P(t) = 1.394(1.006)^t \), where \( t \) is the number of years since the start of 2014. This means \( t = 0 \) corresponds to the start of 2014, and \( t = 1 \) corresponds to the start of 2015.

Step 2: Calculate the Growth Rate at Start of 2014

To find the growth rate at the start of 2014, differentiate \( P(t) \) with respect to \( t \): \[ P'(t) = 1.394 \cdot \ln(1.006) \cdot (1.006)^t \] For the growth rate at the start of 2014, substitute \( t = 0 \): \[ P'(0) = 1.394 \cdot \ln(1.006) \cdot (1.006)^0 = 1.394 \cdot \ln(1.006) \] Calculating \( \ln(1.006) \approx 0.00598 \), we find: \[ P'(0) = 1.394 \cdot 0.00598 \approx 0.00834 \text{ billion/year} \] Convert to millions of people: \[ 0.00834 \text{ billion/year} = 8.34 \text{ million/year} \] The population growth rate at the start of 2014 was approximately 8.34 million people per year.

Step 3: Calculate the Growth Rate at Start of 2015

Now, calculate the growth rate at the start of 2015. Substitute \( t = 1 \) into the derivative: \[ P'(1) = 1.394 \cdot \ln(1.006) \cdot (1.006)^1 \] Using \( \ln(1.006) \approx 0.00598 \), this becomes: \[ P'(1) = 1.394 \cdot 0.00598 \cdot 1.006 \approx 0.00839 \text{ billion/year} \] Convert to millions of people: \[ 0.00839 \text{ billion/year} = 8.39 \text{ million/year} \] Thus, the population growth rate at the start of 2015 was approximately 8.39 million people per year.

Key Concepts

Population Growth Model

In modeling how a population changes over time, exponential growth models are frequently used. Specifically, they express the population as a function of time. For example, in the population model of China, the formula is given by \( P(t) = 1.394(1.006)^t \), where \( P(t) \) represents the population in billions, and \( t \) is the number of years since the start of 2014. This means when \( t = 0 \), it corresponds to the beginning of the year 2014, and \( t = 1 \) relates to the start of 2015. The constant term, 1.394, denotes the initial population at \( t = 0 \). The term \( 1.006^t \) indicates the growth factor over time. Every year, the population is multiplied by 1.006. In simpler terms, this suggests that each year, the population is intended to grow by 0.6%.

• The base, 1.006, indicates the annual growth rate.
• \( t \) represents the time in years since 2014.
• This model effectively forecasts how the population will increase over time.

This example illustrates how exponential models can effectively predict trends, allowing us to assess population prospects over the years.

Differentiation

Differentiation is a mathematical operation used to determine how a function changes at any point. When we want to find the growth rate of a population, we use differentiation to understand how fast the population is changing at a specific time. In this case, we have the function \( P(t) = 1.394(1.006)^t \), where we differentiate \( P \) with respect to \( t \) to find \( P'(t) \). Differentiating exponential functions typically involves using the natural logarithm of the base as part of the process:

• Differentiate the given function: \[ P'(t) = 1.394 \cdot \ln(1.006) \cdot (1.006)^t \]
• Substitute \( t = 0 \) to get the initial growth rate, and \( t = 1 \) for the rate at the start of 2015.
• The derivative \( P'(t) \) gives the rate of change or how many millions the population grows per year.

Through differentiation, we get an explicit rate of increase in the population from one year to the next, reflecting that the growth is not just increasing but accelerating over time.

Natural Logarithm

The natural logarithm, often represented as \( \ln \), is a special type of logarithm that uses the base of Euler's number \( e \), approximately 2.718. In the context of exponential growth models, natural logarithms help us manage and manipulate terms that involve powers.For our model, \( \ln(1.006) \) is a crucial component of finding the derivative of the exponential function.

• Natural logarithms allow us to find the instantaneous rate of percentage change at a point.
• In practical terms, \( \ln(1.006) \) helps to translate the base 1.006 into its continuous compounding form, which is vital for differentiation.
• The estimated value \( \ln(1.006) \approx 0.00598 \) is used in calculating the growth rate at specific time points.

Understanding natural logarithms is an essential part of calculus and helps in realistic modeling of growth and decay, particularly with continuous data as seen in population studies or finance.