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Chapter 2: Problem 35

Suppose that a 5 -year projection of population trends suggests that \(t\) years from now, the population of a certain community will be \(P\) thousand, where $$ P(t)=-2 t^{3}+9 t^{2}+8 t+200 $$ a. At what rate will the population be growing 3 years from now? b. At what rate will the rate of population growth be changing with respect to time 3 years from now?

Short Answer

Expert verified
a. 8 b. -18

Step by step solution

01

Define the population function

The population function given is \[ P(t) = -2t^{3} + 9t^{2} + 8t + 200 \]
02

Find the first derivative of P(t)

To find the rate of population growth, compute the first derivative of the population function with respect to time, \( t \): \[ P'(t) = \frac{d}{dt}(-2t^{3} + 9t^{2} + 8t + 200) \] Apply the power rule: \[ P'(t) = -6t^{2} + 18t + 8 \]
03

Evaluate the first derivative at t = 3

Substitute \( t = 3 \) into \( P'(t) \) to find the rate of population growth 3 years from now: \[ P'(3) = -6(3)^{2} + 18(3) + 8 \] Simplify: \[ P'(3) = -6(9) + 54 + 8 \] \[ P'(3) = -54 + 54 + 8 = 8 \] Thus, the population will be growing at 8 thousand people per year 3 years from now.
04

Find the second derivative of P(t)

To determine how the rate of population growth is changing, compute the second derivative of the population function: \[ P''(t) = \frac{d}{dt}(-6t^{2} + 18t + 8) \] Apply the power rule: \[ P''(t) = -12t + 18 \]
05

Evaluate the second derivative at t = 3

Substitute \( t = 3 \) into \( P''(t) \) to find the rate of change of the population growth rate 3 years from now: \[ P''(3) = -12(3) + 18 \] Simplify: \[ P''(3) = -36 + 18 = -18 \] Thus, the rate of population growth will be decreasing by 18 thousand people per year per year 3 years from now.

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Key Concepts

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

First Derivative
The first derivative of a function tells us the rate at which the function's value is changing at any given point. In this exercise, the first derivative of the population function, denoted as \( P'(t) \), informs us about the rate of population growth. The original population function provided is \[ P(t) = -2t^{3} + 9t^{2} + 8t + 200 \]. Using the power rule in calculus, we differentiate each term with respect to \( t \). Thus, \( P'(t) \) becomes \[ P'(t) = -6t^{2} + 18t + 8 \]. This derivative helps us determine how many thousand people the population will increase or decrease per year.
Second Derivative
The second derivative, \( P''(t) \), provides information on how the rate of change of the function's value itself is changing. Essentially, it reflects the acceleration or deceleration of the population growth. After finding \( P'(t) \), we differentiate it once more to get \( P''(t) \): \[ P''(t) = -12t + 18 \]. This second derivative tells us not just if the population is growing, but whether it is growing faster or slower over time. By evaluating \( P''(3) \), we see how rapidly the population growth rate is changing three years from now.
Power Rule in Calculus
The power rule is a fundamental concept in calculus used to find the derivative of functions of the form \( ax^n \). According to the power rule, the derivative of \( ax^n \) is \( anx^{n-1} \). In this exercise, we use the power rule multiple times to differentiate the population function. For each term such as \( -2t^{3} \), \( 9t^{2} \), and \( 8t \), we apply the power rule to find: \[ \frac{d}{dt}(-2t^{3}) = -6t^{2} \] \[ \frac{d}{dt}(9t^{2}) = 18t \] \[ \frac{d}{dt}(8t) = 8 \]. Combining these, we obtain the first derivative \( P'(t) \).
Rate of Change
The rate of change concept is crucial for understanding how a quantity changes over time. It is quantified using the derivative in calculus. In the context of this exercise, the rate of change refers to how quickly the population is growing. By computing the first derivative \( P'(t) \), we find out how many thousands of people per year the population is increasing or decreasing. For example, at \( t = 3 \), \( P'(3) = 8 \) means that the population is growing by 8 thousand people per year 3 years from now. Similarly, the second derivative \( P''(t) \) gives us the rate of change of the rate of population growth.
Applied Calculus in Population Dynamics
Calculus, particularly differentiation, is extensively applied in population dynamics to model and predict changes. In this exercise, differentiation allows us to study how the population of a community changes over time by examining the rates of growth and their accelerations. The first derivative helps determine the growth rate, while the second derivative sheds light on how the growth rate is evolving. These insights are valuable for making projections, planning resources, or anticipating future societal needs. By leveraging calculus, we can make more informed decisions based on how populations grow and change.

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Most popular questions from this chapter

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