Problem 8 - Active Calculus

Question

According to the U.S. census, the population of the city of Grand Rapids, MI, was 181,843 in \(1980 ; 189,126\) in 1990 ; and 197,800 in 2000 . a. Between 1980 and 2000 , by how many people did the population of Grand Rapids grow? b. In an average year between 1980 and 2000 , by how many people did the population of Grand Rapids grow? c. Just like we can find the average velocity of a moving body by computing change in position over change in time, we can compute the average rate of change of any function \(f\). In particular, the average rate of change of a function \(f\) over an interval \([a, b]\) is the quotient $$\frac{f(b)-f(a)}{b-a}$$ What does the quantity \(\frac{f(b)-f(a)}{b-a}\) measure on the graph of \(y=f(x)\) over the interval \([a, b] ?\) d. Let \(P(t)\) represent the population of Grand Rapids at time \(t,\) where \(t\) is measured in years from January \(1,1980 .\) What is the average rate of change of \(P\) on the interval \(t=0\) to \(t=20 ?\) What are the units on this quantity? e. If we assume the population of Grand Rapids is growing at a rate of approximately \(4 \%\) per decade, we can model the population function with the formula $$P(t)=181843(1.04)^{t / 10}$$ Use this formula to compute the average rate of change of the population on the intervals \([5,10],[5,9],[5,8],[5,7],\) and [5,6] f. How fast do you think the population of Grand Rapids was changing on January 1 , 1985 ? Said differently, at what rate do you think people were being added to the population of Grand Rapids as of January \(1,1985 ?\) How many additional people should the city have expected in the following year? Why?

Step-by-Step Solution

Answer

a) 15,957 people; b) 797.85 people/year; c) Slope of the secant line; d) 797.85 people/year; e) Apply the given formula for each interval; f) Estimate using average rate in intervals.

Step 1: Calculate the total population growth from 1980 to 2000

Find the difference between the population in 2000 and the population in 1980 to determine the total growth. Population in 1980: 181,843 Population in 2000: 197,800 Total growth = 197,800 - 181,843

Step 2: Calculate the average annual population growth

Divide the total population growth by the number of years (2000 - 1980 = 20 years) to find the average annual growth. Total growth = 15,957 Years = 20 Average annual growth = 15,957 / 20

Step 3: Understand the average rate of change formula

The formula for the average rate of change of a function f over an interval [a, b] is \[ \frac{f(b) - f(a)}{b - a} \]. This measures the slope of the secant line connecting the points (a, f(a)) and (b, f(b)) on the graph of y = f(x).

Step 4: Calculate the average rate of change for population

Using the population at t = 0 (1980) and t = 20 (2000), apply the formula \[ \frac{P(20) - P(0)}{20 - 0} \]. Population in 1980: 181,843 Population in 2000: 197,800 Average rate of change = \frac{197,800 - 181,843}{20}

Step 5: Determine average rate of change using the model

Use the formula \[ P(t) = 181843(1.04)^{t / 10} \] to compute the average rate of change for the specified intervals. For each interval [5, 10], [5, 9], [5, 8], [5, 7], and [5, 6], apply the formula \[ \frac{P(b) - P(a)}{b - a} \].

Step 6: Evaluate how fast the population was changing in 1985

Given the model \[ P(t) = 181843(1.04)^{t / 10} \], calculate P(5) for t = 5 (0 for 1980, 5 for 1985). P(5) measures the instantaneous population rate change at t = 5 (1985). Use the difference quotient for t = 4 and t = 6.

Step 7: Conclusion on population change in 1985

By evaluating the average rate of change near January 1985, estimate how many people were being added per year. Multiply this rate by 1 year to estimate additional people in the following year.

Key Concepts

population growth

The growth in a population over a specified time period is a measure of how the number of individuals in that population changes. In this exercise, we are examining how the populace of Grand Rapids, MI, increased over 20 years from 1980 to 2000.

- Start by noting the population at the beginning and end of the period. In this case, the population in 1980 was 181,843 and in 2000 it was 197,800.
- Population growth is calculated by subtracting the initial population from the final population, so: \(197,800 - 181,843 = 15,957\). That means Grand Rapids grew by 15,957 people in 20 years.
- To find the average yearly increase, you divide the total growth by the number of years, here that would be: \(\frac{15,957}{20} = 797.85\). So on average, the population grew by approximately 798 people per year.

secant line

A secant line is a straight line that connects two points on a curve. In the context of population growth, if we plot time on the x-axis and population on the y-axis, a secant line can be used to demonstrate the average rate of change between two points in time.

- To find the slope of the secant line, we use the formula for the average rate of change: \(\frac{f(b) - f(a)}{b - a}\).
- For our exercise, this would be: \(\frac{197,800 - 181,843}{20} = 797.85\).
- This slope represents the average increase in population per year.

In summary, the secant line here shows how the population of Grand Rapids changes on average from 1980 to 2000.

difference quotient

The difference quotient is a formula that provides the average rate of change of a function over an interval. It’s crucial for understanding how a function changes between two points and is the backbone for calculating derivatives.

- In our exercise, the difference quotient for population growth is given by: \(\frac{P(20) - P(0)}{20 - 0} = \frac{197,800 - 181,843}{20} = 797.85\).
- This represents the average number of people added to the Grand Rapids population per year over 20 years.
- The difference quotient can also be applied to smaller intervals to understand changes over shorter periods.

In essence, the difference quotient helps us quantify how rapidly a function is changing, which in our case is the population over time.