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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?

Short Answer

Expert verified
(a) Rate of decrease is approximately 69,300 barrels per year. (b) 50,000 barrels will remain after about 25.9 years.

Step by step solution

01

Model the Situation

Since the rate of change is proportional to the amount of oil left, we can model this situation using a differential equation. Let's denote the amount of oil left at time \( t \) as \( Q(t) \). The differential equation is \( \frac{dQ}{dt} = -kQ \), where \( k \) is the proportionality constant.
02

Solve the Differential Equation

To solve the differential equation \( \frac{dQ}{dt} = -kQ \), separate variables to get \( \frac{1}{Q} dQ = -k dt \). Integrating both sides gives \( \ln|Q| = -kt + C \). Solving for \( Q \), we get \( Q(t) = Ce^{-kt} \), where \( C \) is the integration constant.
03

Determine Initial Conditions

We know that initially there were 1 million barrels, so \( Q(0) = 1000000 \). Substitute this into the equation to find \( C \: 1000000 = Ce^0 \Rightarrow C = 1000000 \). So, \( Q(t) = 1000000 e^{-kt} \).
04

Find the Proportionality Constant

Using the information that 500,000 barrels remain after 6 years, substitute into the equation: \( 500000 = 1000000 e^{-6k} \). Solving this equation gives \( e^{-6k} = 0.5 \), thus \(-6k = \ln(0.5)\). Therefore, \( k = -\frac{\ln(0.5)}{6} \approx 0.1155 \).
05

Step 5a: Calculate the Rate of Decrease When 600,000 Barrels Remain

Substitute \( Q = 600000 \) into the rate equation \( \frac{dQ}{dt} = -kQ \). With \( k \approx 0.1155 \), we have \( \frac{dQ}{dt} = -(0.1155) \times 600000 \approx -69300 \). Therefore, the rate of decrease is approximately 69,300 barrels per year.
06

Step 5b: Find Time When 50,000 Barrels Remain

To find when there are 50,000 barrels remaining, set \( Q(t) = 50000 \) and solve for \( t \): \( 50000 = 1000000 e^{-kt} \). Simplify to get \( e^{-kt} = 0.05 \), so \( -kt = \ln(0.05) \). Using \( k \approx 0.1155 \), solve that \( t = \frac{-\ln(0.05)}{0.1155} \approx 25.9 \). This means after approximately 25.9 years.

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Key Concepts

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

Exponential Decay
Exponential decay is a concept where the quantity of something decreases at a rate proportional to its current value. In our oil well scenario, this concept is important because the rate at which the oil is pumped out decreases as there is less oil left. The smaller the amount left, the slower the rate at which it reduces.
This is mathematically represented by the differential equation \( \frac{dQ}{dt} = -kQ \), where \( Q \) is the amount of oil left, \( t \) is time, and \( k \) is a positive constant known as the proportionality constant. The minus sign indicates a decrease or decay in the amount.
Once you break down this equation, it leads us to a solution of the form: \( Q(t) = Ce^{-kt} \). Here, \( C \) is determined using initial conditions, and it serves as a starting point for modeling exponential decay.
Initial Conditions
Initial conditions are the known values at the start of the scenario that are crucial for finding particular solutions to differential equations. They provide the specific circumstances of the problem at \( t = 0 \).
In our exercise, the initial condition is that there were 1 million barrels of oil in the well initially. This helps us find the constant \( C \) in the equation \( Q(t) = Ce^{-kt} \). By substituting \( t = 0 \) and \( Q(0) = 1000000 \), we calculate \( C = 1000000 \).
With this initial condition, we have the specific exponential decay model \( Q(t) = 1000000 e^{-kt} \), giving a precise formula that includes all the known conditions at the beginning.
Proportionality Constant
The proportionality constant \( k \) is a pivotal element in differential equations that depicts how fast or slow the exponential decay proceeds. It essentially determines the steepness of the exponential decay curve.
In the example, we know that 500,000 barrels remain after 6 years, which allows us to determine \( k \). By substituting \( Q(6) = 500000 \) into the equation \( Q(t) = 1000000 e^{-kt} \), we find \( e^{-6k} = 0.5 \). Solving this gives \( k = -\frac{\ln(0.5)}{6} \).
Using this calculated value of \( k \), we can determine important facts like the rate of decrease when there are 600,000 barrels remaining and the time duration before only 50,000 barrels are left. This constant is crucial for understanding and predicting how rapidly a given process or phenomenon will change over time.

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