Problem 52 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

PropucnoN The production of oil (in millions of barrels per day) extracted from oil sands in Canada is projected to grow according to the function $$ P(t)=\frac{4.76}{1+4.11 e^{-0.22 r}} \quad(0 \leq t \leq 15) $$ where \(t\) is measured in years, with \(t=0\) corresponding to \(2005 .\) What is the total production of oil from oil sands over the years from 2005 until \(2020(t=15)\) ?

Step-by-Step Solution

Answer

The total production of oil from oil sands over the years from 2005 until 2020 is approximately 19.46 million barrels per day.

Step 1: Rewrite the given function

Rewrite the given function in a more suitable form for integration: \(P(t) = \frac{4.76}{1 + 4.11e^{-0.22t}}\) Now, let \(u = 1 + 4.11e^{-0.22t}\), and use substitution to simplify the integration process.

Step 2: Calculate du/dt

Next, we need to find the derivative of \(u\) with respect to \(t\), denoted as \(\frac{du}{dt}\): \(\frac{du}{dt} = -0.22 \cdot 4.11e^{-0.22t}\)

Step 3: Find the integral of P(t)dt using substitution

Substitute \(u\) into the original integral and calculate the integral: \(\int_{0}^{15} P(t)dt = \int_{1}^{1+4.11e^{-0.22 \cdot 15}}\frac{4.76}{u} \frac{1}{-0.22 \cdot 4.11e^{-0.22t}} du\) After canceling out the constant terms, we end up with: \(=-\frac{4.76}{0.22 \cdot 4.11} \int_{1}^{1+4.11e^{-0.22 \cdot 15}}\frac{1}{u} du\) Then we can apply the natural logarithm rule when integrating, which is: \(\int \frac{1}{u} du = \ln|u|\) This gives us: \(-\frac{4.76}{0.22 \cdot 4.11} (\ln|1 + 4.11e^{-0.22 \cdot 15}|- \ln|1|)\)

Step 4: Evaluate the integral and find the total production

Now, we can evaluate this expression and find the total production of oil from 2005 to 2020: \(-\frac{4.76}{0.22 \cdot 4.11} (\ln|1 + 4.11e^{-0.22 \cdot 15}|- \ln|1|) \approx 19.46\) So the total production of oil from oil sands over the years from 2005 until 2020 is approximately 19.46 million barrels per day.

Key Concepts

Natural Logarithm

The natural logarithm, often written as \( \ln \), plays a significant role in integration, particularly for functions of the form \( \frac{1}{u} \). In calculus, the integral of \( \frac{1}{u} \) is \( \ln|u| \). This relationship is fundamental and simplifies many integrals, especially when they need to be evaluated over a range.
Understanding the concept of natural logarithms is crucial for integrating functions like \( P(t) \). It helps identify patterns and simplify expressions involving exponential growth or decay. When dealing with real-world applications involving growth, such as the production of oil, recognizing logarithmic forms can streamline calculations.

• Natural logarithm is the inverse of the exponential function with base \( e \).
• It is especially useful for dealing with continuous growth or processes with a constant rate of decay.

Substitution Method

Integration can sometimes be complicated, but the substitution method simplifies the process. This technique is essentially a change of variables that makes an integral more manageable. It's a powerful tool for solving integrals that involve composite functions. In our exercise, we used substitution to simplify the integration of \( P(t) \).
The core idea of substitution is to replace a complicated part of the function with a single variable, \( u \). After substitution, the integral takes a simpler form, making it easier to solve.

• Choose a substitution that simplifies the integral, such as letting \( u = 1 + 4.11e^{-0.22t} \).
• Find the derivative \( \frac{du}{dt} \) and ensure that all parts of the integral are expressed in terms of \( u \).
• After integrating with respect to \( u \), revert back to the original variable to find the solution.

Exponential Function

Exponential functions, like \( e^{-0.22t} \), are crucial in modeling continuous growth and decay. They appear frequently in real-world applications, from finance to natural sciences. In our original exercise, the exponential function models the growth of oil production.
When an exponential function is used in calculations of growth or decay, it represents a process where the rate of change is proportional to the current value, leading to rapid changes over time. This characteristic is why exponential functions are popular in scenarios like population growth or radioactive decay.

• Exponential functions have the form \( f(t) = ae^{kt} \), where \( a \) is the initial amount and \( k \) is the growth or decay rate.
• These functions describe situations where change accelerates over time, which is frequent in processes involving resource production.
• Understanding exponential growth is key to predicting how quickly a resource like oil could develop in situations described by the exercise.