Problem 23 - Calculus

Question

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were 1 million barrels of oil in the well; six years later 500,000 barrels remain. (a) At what rate was the amount of oil in the well decreasing when there were 600,000 barrels remaining? (b) When will there be 50,000 barrels remaining?

Step-by-Step Solution

Answer

(a) The rate of decrease is about 69,314 barrels per year. (b) 50,000 barrels remain after about 25.9 years.

Step 1: Understand the Differential Equation

Since the oil is pumped at a rate proportional to the amount remaining, we can express this as \( \frac{dQ}{dt} = -kQ \), where \( Q(t) \) is the quantity of oil in the well at time \( t \), and \( k \) is a positive constant of proportionality.

Step 2: Solve the Differential Equation

To solve the differential equation \( \frac{dQ}{dt} = -kQ \), we separate variables: \( \frac{dQ}{Q} = -k \, dt \). Integrating, we get \( \ln |Q| = -kt + C \). Exponentiating both sides, \( Q = Ce^{-kt} \).

Step 3: Determine the Constant C

Initially, \( Q(0) = 1,000,000 \) barrels. Plugging this into the equation \( Q = Ce^{-kt} \), we get \( 1,000,000 = C \cdot e^{0} \). Thus, \( C = 1,000,000 \). So, \( Q(t) = 1,000,000e^{-kt} \).

Step 4: Determine the Constant k

We know that \( Q(6) = 500,000 \). Substituting into the equation \( 500,000 = 1,000,000e^{-6k} \), we solve for \( k \): \( e^{-6k} = \frac{1}{2} \). Taking logarithms, \( -6k = \ln \frac{1}{2} \), which gives \( k = \frac{-\ln(0.5)}{6} \).

Step 5: Solve (a) - Rate of Decrease When Q = 600,000

The rate of decrease is given by \( \frac{dQ}{dt} = -kQ \). Plugging \( Q = 600,000 \) and our expression for \( k \) gives \( \frac{dQ}{dt} = -\frac{-\ln(0.5)}{6} \times 600,000 \), which simplifies to about \( 69,314 \) barrels per year.

Step 6: Solve (b) - Time When Q = 50,000

Set \( Q(t) = 50,000 \). Using the equation \( 50,000 = 1,000,000e^{-kt} \) and substitute \( k = \frac{-\ln(0.5)}{6} \), solve for \( t \): \( e^{-kt} = 0.05 \), thus \( -kt = \ln(0.05) \). Solving, \( t = \frac{\ln(0.05)}{-ln(0.5)/6} \), which is approximately 25.9 years.

Key Concepts

Exponential Decay

Exponential decay describes a process where the quantity of a substance decreases at a rate proportional to its current value. This concept is essential for understanding how substances diminish over time, such as the oil in our exercise. The concept is governed by a simple differential equation:

• \( \frac{dQ}{dt} = -kQ \)

Here, \( Q \) represents the quantity remaining, \( t \) is time, and \( k \) is a constant of proportionality.

Due to the negative sign, the equation signifies a decay, not growth, giving rise to the 'exponential decay' term. The core characteristic of this process is the constant percentage reduction over equal increments of time, which is reflective in the exponential term \( e^{-kt} \).

Understanding this can be helpful in various fields such as economics, chemistry, and environmental science, where resource depletion or substance decay is analyzed.

Separation of Variables

The separation of variables is a common method for solving differential equations like the one given in our exercise. This technique involves rearranging the terms of the equation so that each side depends only on a single variable. The equation:

• \( \frac{dQ}{dt} = -kQ \)

is rearranged by moving all terms containing \( Q \) to one side and all terms involving \( t \) to the other.

This gives us:

• \( \frac{dQ}{Q} = -k \, dt \)

Integrating both sides results in the natural logarithm form, \( \ln |Q| = -kt + C \), where \( C \) is the integration constant.

Exponentiating both sides resolves the equation to \( Q = Ce^{-kt} \). This method is a critical tool in differential calculus, facilitating the solution of complex real-life problems involving rates of change.

Constant of Proportionality

The constant of proportionality, \( k \), in our differential equation signifies the rate at which decay occurs in exponential decay problems. It acts as a bridge between the rate of change of a variable and the variable itself. The relationship is direct and predictable, allowing us to express the rate of decrease of oil with respect to the quantity left in the well:

• \( \frac{dQ}{dt} = -kQ \)

The magnitude of \( k \) determines how quickly the substance depletes. In the given exercise, it's calculated using known quantities at specific times. By measuring the rate six years after starting the pumping, the equation is refined to find:

• \( k = \frac{-\ln(0.5)}{6} \)

A larger \( k \) would indicate faster depletion, while a smaller \( k \) suggests a slower decrease. Understanding \( k \) is vital in accurately predicting future amounts and timescales for resource management and usage.