Problem 14 - Applied 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 \(-100,000 \cdot \ln(2)\) barrels/year. (b) The well will have 50,000 barrels remaining in \(t = - \frac{6 \ln(0.05)}{\ln(2)}\) years.

Step 1: Define the Differential Equation

Let the amount of oil at time \( t \) years be \( Q(t) \). The rate of decrease of oil is proportional to \( Q \), so the differential equation is \( \frac{dQ}{dt} = -kQ \), where \( k \) is a proportionality constant.

Step 2: Solve the Differential Equation

We solve the differential equation \( \frac{dQ}{dt} = -kQ \) by separating variables: \( \frac{1}{Q} \frac{dQ}{dt} = -k \). Integrate both sides to get \( \ln(Q) = -kt + C \). Exponentiate to solve for \( Q \): \( Q(t) = Q_0 e^{-kt} \), where \( Q_0 \) is the initial amount of oil.

Step 3: Determine the Constant \( k \)

Initially, \( Q(0) = 1,000,000 \) barrels. After 6 years, \( Q(6) = 500,000 \). Substituting into the equation \( Q(t) = 1,000,000 e^{-kt} \), we get \( 500,000 = 1,000,000 e^{-6k} \). Solving for \( k \), we get \( e^{-6k} = \frac{1}{2} \), or \( k = \frac{\ln(2)}{6} \).

Step 4: Solve Part (a) - Find \( \frac{dQ}{dt} \) for 600,000 Barrels

The rate \( \frac{dQ}{dt} \) when there are 600,000 barrels is given by \( \frac{dQ}{dt} = -kQ \). We know \( k = \frac{\ln(2)}{6} \) and \( Q = 600,000 \), so \( \frac{dQ}{dt} = -\frac{\ln(2)}{6} \cdot 600,000 = -100,000 \cdot \ln(2) \).

Step 5: Solve Part (b) - Calculate Time for 50,000 Barrels Remaining

We use the equation \( Q(t) = 1,000,000 e^{-kt} \) with \( Q(t) = 50,000 \). Solve \( 50,000 = 1,000,000 e^{-kt} \) to get \( e^{-kt} = \frac{1}{20} \). Substitute \( k = \frac{\ln(2)}{6} \) and solve for \( t \): \( -kt = \ln\left(\frac{1}{20}\right) \), thus \( t = - \frac{6 \ln(0.05)}{\ln(2)} \).

Key Concepts

Exponential Decay

Exponential decay is a process where a quantity decreases at a rate proportional to its current value. In the exercise, the amount of oil in the well is an example of exponential decay. The formula that represents this situation is usually written as: \[ Q(t) = Q_0 e^{-kt} \] Where:

• \( Q(t) \) is the quantity of oil at time \( t \)
• \( Q_0 \) is the initial amount of oil
• \( k \) is the decay constant
• \( t \) is the time

The negative sign in the exponent indicates a decrease or decay over time.
In the context of the exercise, initially, there were 1 million barrels of oil. Over time, this amount decreases according to the exponential decay formula. The key to solving these types of problems is understanding how the rate of decay is directly related to the quantity of oil at any moment.

Proportionality Constant

The proportionality constant, represented by \( k \) in the differential equation, plays an essential role in defining the rate of change. It essentially dictates how quickly the quantity decreases. This constant is generally determined by the conditions of the problem.
In our oil well example, knowing that after 6 years the amount of oil halved, we solved for \( k \) using the condition \( 500,000 = 1,000,000 e^{-6k} \). This calculation shows that:\[ k = \frac{\ln(2)}{6} \] Here, \( \ln(2) \) is the natural logarithm of 2, which arises because the quantity halved.
The constant \( k \) defines the rapidity of exponential decay. A larger \( k \) would indicate a quicker depletion of resources, whereas a smaller \( k \) would indicate a slower decrease.

Separation of Variables

Separation of variables is a method used to solve differential equations, particularly when they are not easily solvable by simpler algebraic methods. The key is rearranging the equation so that each variable is on opposite sides, making it easier to integrate.
In our case, the differential equation \( \frac{dQ}{dt} = -kQ \) can be separated by:\[ \frac{1}{Q} dQ = -k dt \] Next, we integrate both sides separately:

• On the left, integrating \( \frac{1}{Q} \) gives \( \ln(Q) \)
• On the right, integrating \(-k \) gives \(-kt \)

Thus, we get:\[ \ln(Q) = -kt + C \] This helps to solve for \( Q \) by exponentiating both sides, leading to the general solution of the differential equation.

Integration

Integration is the process of finding the integral of a function, which is essentially the reverse of differentiation. It is a key component in solving differential equations, allowing us to find functions from their rates of change.
In our exercise, after separation of variables, integration was performed on both sides of the equation.

• When integrating \( \frac{1}{Q} \), we obtain \( \ln(Q) \)
• When integrating \(-k\), it results in \(-kt\) where t is the variable

This integration step bridges the gap from knowing the rate of change of oil with respect to time, to finding an explicit expression for the amount of oil over time. Understanding how to integrate and apply integration to differential equations is crucial for solving problems involving rates of change, such as exponential decay.