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Chapter 2: Problem 90

Accumulation Years People from their mid- 40 s to their mid-50s are in the prime investing years. Demographic studies of this type are of particular importance to financial institutions. The function $$ N(t)=34.4(1+0.32125 t)^{0.15} \quad 0 \leq t \leq 12 $$ gives the projected number of people in this age group in the United States (in millions) in year \(t\), where \(t=0\) corresponds to the beginning of 1996 . a How large was this segment of the population projected to be at the beginning of 2005 ? b. How fast was this segment of the population growing at the beginning of 2005 ? Source: U.S. Census Bureau.

Short Answer

Expert verified
At the beginning of 2005, the population segment in the prime investing years was approximately 39.98 million people. The growth rate of this segment at that time was approximately 0.3954 million people per year.

Step by step solution

01

Calculate the value of t for the beginning of 2005

Given that t = 0 corresponds to the beginning of 1996, we need to find the value of t for the beginning of 2005. As there are 9 years between the beginning of 1996 and the beginning of 2005, then t = 9 for the beginning of 2005.
02

Find the size of the population segment at the beginning of 2005

We need to plug in t = 9 into the function N(t) to find the size of the population segment at the beginning of 2005. $$ N(t) = 34.4(1+0.32125t)^{0.15} $$ $$ N(9) = 34.4(1+0.32125(9))^{0.15} $$ Now, compute the value of N(9): $$ N(9) \approx 39.98 $$ So, at the beginning of 2005, there were approximately 39.98 million people in the prime investing years.
03

Find the derivative of N(t)

To find the rate of growth of the population segment at the beginning of 2005, we need to find the derivative of N(t) which represents the growth rate function. Let's find the derivative of N(t) with respect to t: $$ \dfrac{dN(t)}{dt}=\dfrac{d}{dt}(34.4(1+0.32125t)^{0.15}) $$ Using chain rule, we have: $$ \dfrac{dN(t)}{dt}=34.4\cdot0.15(1+0.32125t)^{-0.85}\cdot0.32125 $$
04

Find the rate of growth of the population segment at the beginning of 2005

Now, we need to plug in t = 9 (the beginning of 2005) into the derived growth rate function to find the rate of growth of the population segment: $$ \dfrac{dN(t)}{dt}=34.4\cdot0.15(1+0.32125(9))^{-0.85}\cdot0.32125 $$ Compute the value of the growth rate at t = 9: $$ \dfrac{dN(9)}{dt}\approx 0.3954 $$ Hence, at the beginning of 2005, the population segment in the prime investing years was growing at a rate of approximately 0.3954 million people per year.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Demographic Studies
Demographic studies are a vital component in understanding population dynamics. They involve the examination of statistical data relating to the structure and distribution of populations, which in turn informs economic planning, policy development, and social services.

For instance, demographic studies can reveal trends in age groups that are entering their prime investing years. This information is crucial for financial institutions. These trends can be represented through mathematical models that project populations at specific times. In our exercise, the function given, \( N(t) \), represents such a model for the U.S. population.
Exponential Functions
Exponential functions are mathematical representations where a constant base is raised to a variable exponent. In the realm of population growth, exponential functions such as \( N(t) = 34.4(1+0.32125 t)^{0.15} \) model how populations change over time. These functions help to forecast future values based on current trends, assuming that the growth rate remains constant.

In the given exercise, we see how demographic changes can be modeled using this function, which gives financial institutions an insight into potential future markets. By substituting different values of \( t \) for various years, we can calculate projected population sizes using this exponential relationship.
Rate of change
The rate of change in a function, represented mathematically as its derivative, is a concept that tells us how quickly a quantity is increasing or decreasing at a particular point in time.

In our example, finding the derivative of \( N(t) \) gives us the rate at which the segment of the population in their prime investing years is growing per year. Calculating this rate of change at \( t = 9 \) (the beginning of 2005), we find it as approximately 0.3954 million people per year. This rate is critical for anticipating growth trends, allocating resources, and planning for future demands.

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