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Chapter 5: Problem 10

The per capita (per person) consumption of milk was 27.6 gallons in 1980 and has been steadily decreasing by an annual decay factor of 0.99 a. Form an exponential function for per capita milk consumption \(M(t)\) for year \(t\) after 1980 . b. According to your function, what was the per capita consumption of milk in \(2000 ?\) If available, use the Internet to check your predictions.

Short Answer

Expert verified
The exponential function is \(M(t) = 27.6 \times 0.99^t\). In 2000, the per capita milk consumption was approximately 22.55 gallons.

Step by step solution

01

- Understand the problem

Given the initial per capita consumption of milk in 1980 was 27.6 gallons and the annual decay factor is 0.99, this indicates each year the consumption is 99% of the previous year's consumption.
02

- Form the exponential function

The general form of an exponential decay function is given by \[ M(t) = M_0 \times (decay\text{ factor})^t \]where \(M_0\) is the initial amount, and \(t\) is the number of years after the initial year (1980 in this case). Substituting the known values:\[ M(t) = 27.6 \times 0.99^t \]
03

- Calculate per capita consumption in 2000

The year 2000 is 20 years after 1980. Substitute \(t = 20\) into the exponential function:\[ M(20) = 27.6 \times 0.99^{20} \]Calculate the result:\[ M(20) = 27.6 \times 0.8171 \]\[ M(20) \approx 22.55 \text{gallons} \]

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Key Concepts

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

Milk Consumption
Let's talk about milk consumption over time. In 1980, the average person drank 27.6 gallons of milk per year. But as the years went by, people started drinking less and less milk each year. We use a special kind of mathematical function to describe this decrease. This brings us to our next concept: exponential functions.
Exponential Functions
Exponential functions are a type of mathematical function that explain how quantities grow or shrink at a consistent rate over time. In our case, we鈥檙e looking at an exponential decay because milk consumption decreases each year. An exponential decay function can be written as:

\[ M(t) = M_0 \times (decay\text{ factor})^t \]
\(M_0\) is the initial amount (27.6 gallons in 1980). The decay factor here is 0.99, meaning milk consumption decreases by 1% each year.

Using these values, we derived the function:
\[ M(t) = 27.6 \times 0.99^t \]

This function helps us model how milk consumption changes over time.
Mathematical Modeling
Mathematical modeling is a way to represent real-world scenarios using mathematical formulas. Think of it as a way to predict the future or understand the past through numbers. In our exercise, we used a model to predict milk consumption in the year 2000. Here鈥檚 how:
  • We found the formula: \[ M(t) = 27.6 \times 0.99^t \]
  • Then we calculated for 20 years after 1980: \[ t = 2000 - 1980 = 20 \]
  • By substituting \( t = 20 \) into the function, we found that:
\[ M(20) = 27.6 \times 0.99^{20} \approx 22.55 \text{gallons} \]
This shows how much milk an average person consumed in 2000 according to our model.
Annual Decay Factor
The annual decay factor is a key piece of the puzzle in understanding how exponential decay works. It represents the percentage by which something decreases each year. For milk consumption, the annual decay factor is 0.99. This means milk consumption in any given year is 99% of what it was the previous year.

Here鈥檚 how it looks mathematically:
  • Initial consumption in 1980: 27.6 gallons
  • Consumption in 1981: \( 27.6 \times 0.99 \)
  • Consumption in 1982: \( 27.6 \times 0.99^2 \)
  • And so on...
Each year, you simply multiply the previous year's consumption by 0.99. This is how the exponential function captures the idea of steady decrease over time.

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Most popular questions from this chapter

Match the statements (a) through (d) with the correct exponential function in (e) through (h). Assume time \(t\) is measured in the unit indicated. a. Radon-222 decays by \(50 \%\) every \(t\) days. b. Money in a savings account increases by \(2.5 \%\) per year. c. The population increases by \(25 \%\) per decade. d. The pollution in a stream decreases by \(25 \%\) every year. e. \(A=1000(1.025)^{t}\) f. \(A=1000(0.75)^{t}\) g. \(A=1000\left(\frac{1}{2}\right)^{t}\) h. \(A=1000(1.25)^{t}\)

A bank compounds interest annually at \(4 \%\). a. Write an equation for the value \(V\) of \(\$ 100\) in \(t\) years. b. Write an equation for the value \(V\) of \(\$ 1000\) in \(t\) years. c. After 20 years will the total interest earned on \(\$ 1000\) be ten times the total interest earned on \(\$ 100 ?\) Why or why not?

(Graphing program recommended.) The infant mortality rate (the number of deaths per 1000 live births) fell in the United States from 7.2 in 1996 to 6.4 in 2006 . a. Assume that the infant mortality rate is declining linearly over time. Construct an equation modeling the relationship between infant mortality rate and time, where time is measured in years since \(1996 .\) Make sure you have clearly identified your variables. b. Assuming that the infant mortality rate is declining exponentially over time, construct an equation modeling the relationship, where time is measured in years since \(1996 .\) c. Graph both of your models on the same grid. d. What would each of your models predict for the infant mortality rate in \(2010 ?\)

Between 1970 and 2000 , the United States grew from about 200 million to 280 million, an increase of approximately \(40 \%\) in this 30 -year period. If the population continues to expand by \(40 \%\) every 30 years, what will the U.S. population be in the year \(2030 ?\) In \(2060 ?\) Use technology to estimate when the population of the United States will reach I billion.

Identify the doubling time or half-life of each of the following exponential functions. Assume \(t\) is in years. [Hint: What value of \(t\) would give you a growth (or decay) factor of 2 (or \(1 / 2\) )?] a. \(Q=70(2)^{t}\) b. \(Q=1000(2)^{t / 50}\) c. \(Q=300\left(\frac{1}{2}\right)^{t}\) d. \(Q=100\left(\frac{1}{2}\right)^{t / 250}\) e. \(N=550(2)^{t / 10}\) f. \(N=50\left(\frac{1}{2}\right)^{t / 20}\)
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