Problem 53 - Applied Calculus

Question

The number of barrels of oil produced from North Dakota oil wells is estimated to be \(B=21.88 e^{0.059 t}\) million barrels, where \(t\) is in months since Sept \(2012 .^{20}\) Estimate the relative rate of change of oil production in December 2012 using (a) \(\quad \Delta t=1\) (b) \(\Delta t=0.1\) (c) \(\Delta t=0.01\)

Step-by-Step Solution

Answer

The relative rate of change is constant at 0.059 for December 2012, regardless of \( \Delta t \).

Step 1: Understand the Function and Define the Variables

The function for the oil production is given by \( B = 21.88 e^{0.059 t} \). Here, \( B \) represents the oil production in million barrels and \( t \) is the time in months since September 2012. We need to find the relative rate of change at December 2012, which is \( t = 3 \).

Step 2: Find Expression for Relative Rate of Change

The relative rate of change of the function \( B(t) \) is given by the derivative \( \frac{B'(t)}{B(t)} \). Calculate \( B'(t) \):\[ B'(t) = \frac{d}{dt}(21.88 e^{0.059 t}) = 21.88 \times 0.059 e^{0.059 t} \].Thus, \( \frac{B'(t)}{B(t)} = \frac{21.88 \times 0.059 e^{0.059 t}}{21.88 e^{0.059 t}} = 0.059 \). The relative rate of change is constant at \( 0.059 \).

Step 3: Estimate the Relative Change Over Interval for Δt=1

To find the relative change over a time interval \( \Delta t \), calculate \( \frac{B(t + \Delta t) - B(t)}{B(t) \Delta t} \). For \( \Delta t = 1 \), compute:\[ \frac{B(4) - B(3)}{B(3) \cdot 1} = \frac{21.88 e^{0.059 \cdot 4} - 21.88 e^{0.059 \cdot 3}}{21.88 e^{0.059 \cdot 3}} \].Simplify this expression. Since the exponential terms cancel in relative terms, it approximates \( 0.059 \) over such intervals.

Step 4: Estimate the Relative Change Over Interval for Δt=0.1

Use the same process as in Step 3 for \( \Delta t = 0.1 \):\[ \frac{B(3.1) - B(3)}{B(3) \cdot 0.1} = \frac{21.88 e^{0.059 \cdot 3.1} - 21.88 e^{0.059 \cdot 3}}{21.88 e^{0.059 \cdot 3} \cdot 0.1} \].Consider the approximate nature: this should converge closely to \( 0.059 \).

Step 5: Estimate the Relative Change Over Interval for Δt=0.01

Repeat with \( \Delta t = 0.01 \):\[ \frac{B(3.01) - B(3)}{B(3) \cdot 0.01} = \frac{21.88 e^{0.059 \cdot 3.01} - 21.88 e^{0.059 \cdot 3}}{21.88 e^{0.059 \cdot 3} \cdot 0.01} \].The result should be increasingly accurate at approaching \( 0.059 \) as \( \Delta t \) decreases.

Key Concepts

Exponential Growth

Exponential growth describes a process where the quantity increases rapidly over time according to a mathematical function. In the context of oil production, this is characterized by the formula \( B = 21.88 e^{0.059 t} \). Here, \( B \) is the number of barrels produced, \( t \) is the time in months since September 2012, and \( e^{0.059 t} \) models the exponential increase.
Understanding exponential growth is essential in modeling scenarios where growth happens at a rate proportional to the current quantity. Such a growth rate is consistent, like compounding interest in finance or population growth under ideal conditions. In our exercise, the constant \( 0.059 \) signifies the growth rate, meaning that the production increases by approximately \( 5.9\% \) every month.
Exponential functions often model real-life situations effectively because they capture the essence of continuous growth. This process is 'exponential' because the rate of increase in the quantity becomes quicker over time, which is seen in this oil production formula.

Derivative Calculation

The derivative is a fundamental concept in calculus that measures how a function changes as its input changes. In this exercise, we need to calculate the derivative of the function \( B(t) \) to understand the rate of change in oil production.
The given function is \( B = 21.88 e^{0.059 t} \). To find its derivative, \( B'(t) \), we apply the general rule for the derivative of an exponential function, \( \frac{d}{dt}(e^{kt}) = k e^{kt} \). By multiplying by the constant coefficient, the derivative becomes:

• \( B'(t) = 21.88 \times 0.059 e^{0.059 t} \)

Once the derivative is calculated, the relative rate of change is obtained by dividing the derivative by the original function, \( B(t) \):

• \( \frac{B'(t)}{B(t)} = \frac{21.88 \times 0.059 e^{0.059 t}}{21.88 e^{0.059 t}} = 0.059 \)

This simple step highlights how derivatives are used to measure the speed of growth. In this exercise, calculating the derivative reveals that the relative rate of change remains constant at \( 0.059 \), indicating consistent exponential growth in oil production.

Oil Production Modeling

Modeling oil production involves crafting mathematical functions that accurately describe the trend of oil extraction over time. By understanding these models, we can assess and predict future oil production, assisting in planning and resource management.
In our exercise, the production function is given by \( B = 21.88 e^{0.059 t} \), showcasing exponential growth properties. This model helps decision-makers forecast production values by analyzing how production scales with time. It's especially useful in predicting the supply available to meet energy demands.
To evaluate the model's prediction accuracy and refine it, we calculate the relative rate of change over different intervals like \( \Delta t = 1, 0.1, \) and \( 0.01 \). This gives insight into how production changes month by month and over smaller periods. Understanding these variations helps financiers and resource planners make informed decisions about extraction strategies and investments.
Through such models, stakeholders can estimate future trends, adapting strategies accordingly and managing resources with greater foresight. Such analysis underscores the broader implications of mathematical modeling in realistic, industrial scenarios.