Problem 45 - Calculus and its applications

Question

In \(2013,\) annual world demand for crude oil was approximately 33.3 billion barrels, and it was projected to increase by \(1.5 \%\) per year. a) Assuming an exponential growth model, predict the demand in 2020 . b) World reserves of crude oil in 2013 were approximately 1635 billion barrels. Assuming that no new oil reserves are found, when will the reserves be depleted? The model $$ \int_{0}^{T} P e^{-k t} d t=\frac{P}{k}\left(1-e^{-k T}\right) $$ can be applied to calculate the buildup of a radioactive material that is being released into the atmosphere at a constant annual rate. Some of the material decays, but more continues to be released. The amount present at time \(\mathrm{T}\) is given by the integral above, where \(P\) is the amount released per year and \(k\) is the half-life.

Step-by-Step Solution

Answer

By 2020, demand is approximately 36.98 billion barrels; reserves deplete in 2274.

Step 1: Understand the Exponential Growth Model

The exponential growth model can be expressed as \( D(t) = D_0(1 + r)^t \), where \( D_0 \) is the initial demand, \( r \) is the growth rate, and \( t \) is the time in years from the initial point. Given \( D_0 = 33.3 \text{ billion barrels} \), and \( r = 0.015 \) (1.5\%), we need to find the demand after 7 years (from 2013 to 2020).

Step 2: Calculate Demand in 2020

Using the formula, \( D(7) = 33.3 \times (1 + 0.015)^7 \). Calculating this gives:\[ D(7) = 33.3 \times (1.015)^7 \approx 33.3 \times 1.1107 \approx 36.98 \text{ billion barrels} \]. This is the projected demand for 2020.

Step 3: Determine When Reserves Will Be Depleted

To estimate when the reserves will deplete, set the demand equation equal to the total reserves. So, we solve \( 1635 = 33.3(1.015)^t \). Solving for \( t \), first take the natural logarithm of both sides: \( \ln(1635) = \ln(33.3) + t \ln(1.015) \).

Step 4: Solve for Time \( t \)

Solving for \( t \), rearrange it as follows:\[ t \ln(1.015) = \ln(1635) - \ln(33.3) \]. Calculate:\( \ln(1635) \approx 7.399 \) and \( \ln(33.3) \approx 3.505 \), so \( t = \frac{7.399 - 3.505}{\ln(1.015)} \approx \frac{3.894}{0.0149} \approx 261.21 \). Therefore, the reserves will be depleted in about 261 years from 2013.

Key Concepts

Crude Oil Demand

Crude oil demand refers to the amount of crude oil that stakeholders, including industries and governments, require to meet their energy and production needs. Over time, this demand often follows an exponential growth pattern, particularly when the market is healthy and industries are expanding. Exponential growth is characterized by a constant percentage increase over time. In our exercise, the demand for crude oil was expected to rise at a rate of 1.5% per year, starting from a demand of 33.3 billion barrels in 2013.

To predict future demand using exponential growth, one can employ the formula \( D(t) = D_0(1 + r)^t \), where \( D_0 \) is the initial demand, \( r \) is the growth rate, and \( t \) is the number of years from the initial point. This formula allows us to calculate the projected demand in future years. For instance, in 2020, which is 7 years from 2013, the predicted crude oil demand was approximately 36.98 billion barrels. This scenario helps illustrate how small yearly percentage increases can lead to substantial growth over time.

Understanding these concepts can assist students in grasping the potential implications of increasing demand on resources and the importance of sustainable energy practices.

Radioactive Decay

Radioactive decay is a process by which an unstable atomic nucleus loses energy by emitting radiation. This natural process occurs at a constant rate for each radioactive isotope, and it is often modeled using exponential functions. Unlike exponential growth, radioactive decay involves a reduction over time.

The half-life \( k \) of a radioactive substance is a critical concept, referring to the time it takes for half of the radioactive material to decay. In our example, the model \[ \int_{0}^{T} P e^{-k t} d t=\frac{P}{k}(1-e^{-k T}) \] is used to calculate the buildup of radioactive material continuously released into the atmosphere. Here, \( P \) indicates the quantity released per year, while \( k \) defines the rate of decay.

By applying such models, we can understand how much of the material remains at any time \( T \). It provides insight into environmental impacts and helps in planning for prolonged exposures to radiation. Understanding radioactive decay is essential for fields such as nuclear physics, environmental sciences, and medical radiology.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm to the base \( e \), where \( e \) is an irrational number approximately equal to 2.71828. It plays a crucial role in many areas of mathematics and real-world applications, particularly in processes involving exponential growth and decay.

In the context of exponential growth or decay problems, the natural logarithm helps us solve for time \( t \). Once the exponential equation \( a(1 + r)^t = b \) is set up, taking the natural logarithm of both sides simplifies solving for \( t \). For instance, when estimating when crude oil reserves will be depleted by setting initial demand equal to reserves, the equation \( 1635 = 33.3(1.015)^t \) is used. By applying the natural logarithm, this equation is transformed into a linear form \( \ln(1635) = \ln(33.3) + t \ln(1.015) \) which can then be solved algebraically.

These natural logarithm applications, whether predicting future values or uncovering growth rates, are invaluable tools in economic, scientific, and engineering disciplines.