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Chapter 3: Problem 35

The number \(N\) of farms in the United States has declined continually since \(1950 .\) In \(1950,\) there were 5,650,000 farms, and in \(2012,\) that number had decreased to 2,170,000 . (Sources: U.S. Department of Agriculture; National Agricultural Statistics Service.) Assuming the number of farms decreased according to the exponential decay model: a) Find the value of \(k,\) and write an exponential function that describes the number of farms after time \(t,\) where \(t\) is the number of years since \(1950 .\) b) Estimate the number of farms in 2016 and in 2020 c) At this decay rate, when will only 1,000,000 farms remain?

Short Answer

Expert verified
In 2016, there were approximately 2,081,096 farms. In 2020, about 1,996,506 farms. Only 1,000,000 farms will remain by 2078.

Step by step solution

01

Understand the Exponential Decay Model

The exponential decay model is given by the formula \( N(t) = N_0 e^{-kt} \), where:- \( N(t) \) is the number of farms at time \( t \),- \( N_0 \) is the initial number of farms,- \( k \) is the decay constant,- \( t \) is the number of years since 1950.
02

Find the Decay Constant k

We know \( N_0 = 5,650,000 \) and \( N(62) = 2,170,000 \) for \( t = 62 \) (from 1950 to 2012). Substitute these values into the formula:\[ 2,170,000 = 5,650,000 \, e^{-62k} \]Divide both sides by \( 5,650,000 \), then take the natural log to solve for \( k \):\[ e^{-62k} = \frac{2,170,000}{5,650,000} = 0.384 \\ln(0.384) = -62k \k = -\frac{\ln(0.384)}{62} \approx 0.0153 \]
03

Write the Exponential Function

With \( k = 0.0153 \), the function describing the number of farms over time \( t \) is:\[ N(t) = 5,650,000 \, e^{-0.0153t} \]
04

Estimate Farms in 2016

Substitute \( t = 66 \) (2016 is 66 years after 1950) into the function:\[ N(66) = 5,650,000 \, e^{-0.0153 \times 66} \]Calculate:\[ N(66) \approx 5,650,000 \, e^{-1.0098} \approx 2,081,096 \]
05

Estimate Farms in 2020

Substitute \( t = 70 \) (2020 is 70 years after 1950) into the function:\[ N(70) = 5,650,000 \, e^{-0.0153 \times 70} \]Calculate:\[ N(70) \approx 5,650,000 \, e^{-1.071} \approx 1,996,506 \]
06

Determine When Farms Will Reduce to 1,000,000

Set \( N(t) = 1,000,000 \) and solve for \( t \):\[ 1,000,000 = 5,650,000 \, e^{-0.0153t} \]Divide by \( 5,650,000 \) and solve for \( t \):\[ e^{-0.0153t} = \frac{1,000,000}{5,650,000} \]\\[ \ln(0.177) = -0.0153t \]\[ t = -\frac{\ln(0.177)}{0.0153} \approx 128 \]Since \( t \) is 128, adding 128 to 1950 gives the year 2078.

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Key Concepts

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

Decay Constant
In the context of exponential decay, the decay constant, represented by the symbol \( k \), is a critical value that defines the rate at which something reduces over time. For scenarios involving exponential decay, like the decline in the number of farms, \( k \) quantifies how quickly the quantity decreases. The decay constant is a positive value, indicating an exponential decrease rather than growth.
In our problem, we found \( k \) by using the equation \( 2,170,000 = 5,650,000 \, e^{-62k} \), where the initial number of farms \( N_0 \) was 5,650,000 in 1950 and \( N(62) = 2,170,000 \) was the number in 2012. Solving mathematically, the decay constant was calculated as \( k \approx 0.0153 \).
This number tells us how quickly the farms are declining each year. A larger decay constant would mean a faster rate of reduction.
Exponential Function
An exponential function is a mathematical expression in the form of \( N(t) = N_0 e^{-kt} \). This function shows how a quantity changes at a rate proportional to its current value. Here, \( e \) is the base of natural logarithms, approximately equal to 2.718. It is used in calculations involving exponential growth or decay.
The use of an exponential function is crucial for modeling situations where there's a consistent percentage change over time, like the decline of farms from 1950 to 2012.
  • \( N(t) \) is the number of farms at time \( t \)
  • \( N_0 \) represents the initial amount, which starts at 5,650,000 farms
  • \( k \) is the decay constant
  • \( t \) stands for the time elapsed in years since the starting point, 1950
Plugging these values into the function creates a powerful tool to predict future outcomes or determine past states.
Natural Logarithm
A natural logarithm, expressed as \( \ln \), is the logarithm to the base \( e \) (where \( e \approx 2.718 \)) and is used to solve equations involving exponential functions. In the context of exponential decay, the natural logarithm helps to isolate variables and solve for unknowns, such as finding the decay constant \( k \).
To solve for \( k \) in the equation \( e^{-62k} = 0.384 \), we took the natural log of both sides to simplify: \( \ln(0.384) = -62k \). This transformation is necessary because it linearizes the exponential growth or decay equation, making it solvable using basic algebraic methods.
The resulting expressions allow researchers or students to manipulate and understand the relationships between quantities in real-world exponential models.

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