Problem 37 - Calculus and its applications

Question

Annual consumption of beef per person was about \(64.6 \mathrm{lb}\) in 2000 and about \(61.2 \mathrm{lb}\) in 2008 . Assuming that \(B(t),\) the annual beef consumption \(t\) years after \(2000,\) is decreasing according to the exponential decay model a) Find the value of \(k,\) and write the equation. b) Estimate the consumption of beef in 2015 . c) In what year (theoretically) will the consumption of beef be 20 lb per person?

Step-by-Step Solution

Answer

a) \( k \approx 0.00674; B(t) = 64.6 e^{-0.00674t} \), b) 2015 estimate: 58.3 lb, c) Year: 2173.

Step 1: Set Up the Exponential Decay Model

To model the situation, we use the exponential decay equation: \[ B(t) = B_0 e^{-kt} \]where \( B_0 \) is the initial beef consumption at \( t = 0 \) (in 2000), \( k \) is the decay constant, and \( t \) is the number of years since 2000. Given \( B_0 = 64.6 \text{ lb} \).

Step 2: Use Data from 2008 to Find k

In 2008, \( t = 8 \) and \( B(t) = 61.2 \text{ lb} \). Substitute these values into the decay model:\[61.2 = 64.6 e^{-8k}\]Solve for \( k \) by first isolating \( e^{-8k} \):\[e^{-8k} = \frac{61.2}{64.6}\]Calculate the division:\[e^{-8k} \approx 0.94727\]Take the natural logarithm on both sides:\[-8k = \ln(0.94727)\]Solve for \( k \):\[k = -\frac{\ln(0.94727)}{8} \approx 0.00674\]Thus, the decay constant \( k \approx 0.00674 \).

Step 3: Write the Exponential Decay Equation

Substitute \( k \approx 0.00674 \) into the exponential decay model:\[B(t) = 64.6 e^{-0.00674t}\]This is the equation that models the annual beef consumption over time.

Step 4: Estimate Beef Consumption in 2015

For 2015, \( t = 15 \). Substitute \( t \) into the decay equation:\[B(15) = 64.6 e^{-0.00674 \times 15}\]Calculate the exponent:\[e^{-0.00674 \times 15} \approx 0.90260\]Now compute \( B(15) \):\[B(15) \approx 64.6 \times 0.90260 \approx 58.3 \text{ lb}\]Thus, the estimated beef consumption in 2015 is approximately 58.3 lb per person.

Step 5: Find the Year When Beef Consumption is 20 lb per Person

Set \( B(t) = 20 \text{ lb} \) and solve for \( t \): \[20 = 64.6 e^{-0.00674t}\]Isolate \( e^{-0.00674t} \):\[e^{-0.00674t} = \frac{20}{64.6} \approx 0.30991\]Take the natural logarithm:\[-0.00674t = \ln(0.30991)\]Solve for \( t \):\[t = -\frac{\ln(0.30991)}{0.00674} \approx 172.64\]Since \( t \) is the number of years after 2000, add this to 2000 to find the year:\[2000 + 172.64 \approx 2173\]Thus, in the year 2173, the beef consumption will theoretically be 20 lb per person.

Key Concepts

Mathematical Modeling

Mathematical modeling is a powerful tool used to represent real-world scenarios using mathematical expressions. With the help of equations, it allows us to predict future outcomes based on past data. In the case of the decreasing beef consumption, the exponential decay model is employed. This model helps to understand how the consumption decreases over time.

To construct a reliable mathematical model, it requires the identification of key variables and constants. In this case, the model is expressed as:

• \( B(t) = B_0 e^{-kt} \)

where:

• \( B(t) \) is the beef consumption at time \( t \)
• \( B_0 \) is the initial amount of beef consumed
• \( k \) is the decay constant, representing the rate at which beef consumption decreases
• \( t \) is the time in years since 2000

The mathematical model not only helps in analyzing the past data but also in forecasting future values, such as predicting when consumption will reach a certain level. By adjusting parameters like the decay constant, we can explore different scenarios and their potential outcomes.

Decay Constant

The decay constant \( k \) is a crucial factor in the exponential decay model. It quantifies the rate at which a quantity decreases over time. In simpler terms, it tells us how fast the consumption of beef is declining year after year.

The decay constant is found by using known data points and substituting them into the decay model. For example, knowing the beef consumption in 2000 and 2008, allows us to rearrange the equation \( B(t) = B_0 e^{-kt} \) and solve for \( k \).

Essentially, the decay constant provides valuable insight into the dynamics of the decay process. A larger value of \( k \) would indicate a faster rate of decline, whereas a smaller \( k \) means a more gradual decrease. Understanding \( k \) equips us with the ability to make more accurate predictions about the future trends of beef consumption and better prepares us for planning and adjustments needed in response to these trends.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a fundamental concept within the exponential decay model. It serves as the inverse operation of taking an exponent of Euler's number, \( e \), which is approximately 2.718.

When solving problems involving exponential decay, the natural logarithm is often used to determine the decay constant \( k \). For instance, taking the natural logarithm of both sides of the equation helps us isolate \( k \).

Here’s a step-by-step on how it’s utilized:

• First, you isolate the exponential term, such as \( e^{-kt} \).
• Then, by taking the natural logarithm of both sides, you convert the exponential equation into a linear form.
• This transformation makes it simple to solve for the unknown variable, \( k \), as it allows for easy manipulation of the equation to isolate it.

The use of \( \ln \) is key to turning exponential equations, which can be complex, into manageable algebraic calculations. Its application simplifies the process of finding essential parameters within the exponential decay model, offering a clear path from data to prediction.