Problem 20 - Calculus Early Transcendentals

Question

Starting in \(2010(t=0),\) the rate at which oil is consumed by a small country increases at a rate of \(1.5 \% / \mathrm{yr},\) starting with an initial rate of 1.2 million barrels/yr. a. How much oil is consumed over the course of the year 2010 (between \(t=0\) and \(t=1\) )? b. Find the function that gives the amount of oil consumed between \(t=0\) and any future time \(t\). c. How many years after 2010 will the amount of oil consumed since 2010 reach 10 million barrels?

Step-by-Step Solution

Answer

#Answer# 31 years

Step 1: Define the consumption rate function

Let's define the consumption rate function. The problem states that the rate of consumption increases by 1.5% per year, starting at 1.2 million barrels/year. This can be modeled by an exponential function: \(r(t) = 1.2(1 + (1.5/100))^t\) This function represents the oil consumption rate (in million barrels/year) at any time t (in years).

Step 2: Compute the integral of the consumption rate function

To find the amount of oil consumed between t=0 and any time t, we need to obtain the integral function of the rate function: \(A(t) = \int{1.2(1 + (1.5/100))^t dt}\)

Step 3: Solve the integral for the consumption function

Solving the integral yields the oil consumption function: \(A(t) = \frac{1.2}{(1.5/100)}(1 + (1.5/100))^t + C\) Where C is the constant of integration. Since at t=0, no oil has been consumed, we have: \(A(0) = \frac{1.2}{(1.5/100)}(1 + (1.5/100))^0 + C = 0\) Solving for C, we have C=0. The oil consumption function, A(t), is then: \(A(t) = \frac{1.2}{(1.5/100)}(1 + (1.5/100))^t\)

Step 4: Calculate oil consumption between t=0 and t=1

For part a, we must calculate the oil consumed between t=0 and t=1. We can do this by plugging t=1 into the consumption function obtained in step 3: \(A(1) = \frac{1.2}{(1.5/100)}(1 + (1.5/100))^1 \approx 1.218\) million barrels Therefore, about 1.218 million barrels of oil are consumed in the year 2010.

Step 5: Determine the number of years to reach 10 million barrels

For part c, we need to find when the total oil consumption reaches 10 million barrels. To do this, we'll solve A(t) = 10 for t: \(\frac{1.2}{(1.5/100)}(1 + (1.5/100))^t = 10\) To solve for t, take the natural logarithm of both sides: \(t = \frac{\ln(\frac{10(1.5/100)}{1.2})}{\ln(1+(1.5/100))} \approx 30.98\) years Therefore, it takes about 31 years after 2010 for the amount of oil consumed since 2010 to reach 10 million barrels.

Key Concepts

Exponential Growth

Exponential growth refers to the increase in quantity at a rate proportional to its current value. In this context, the oil consumption rate of a country is an example of exponential growth. The initial rate of oil consumption is 1.2 million barrels per year and it increases by 1.5% annually.

The formula for exponential growth is often given as:

• \( r(t) = r_0(1 + r)^t \)

Where:

• \( r(t) \) is the rate at time \( t \).
• \( r_0 \) is the initial rate.
• \( r \) is the growth rate as a decimal.

This formula helps calculate how the consumption rate changes over time. Such growth leads to a rapid increase over the years, which can be effectively captured by this mathematical representation.

Understanding exponential growth is crucial because it helps predict future trends and assess sustainability issues like oil supply.

Integration

Integration is a fundamental concept in calculus that helps calculate the area under a curve. It's particularly useful here for finding the total oil consumed over a period. The consumption rate function represents the rate of oil used per year, and integrating this function over time gives the total amount of oil consumed.

In this problem, the integral of the consumption rate function \( r(t) \) from 0 to \( t \) gives the total oil consumption, \( A(t) \). The integration process involves finding the antiderivative or primitive of the function:

• \( A(t) = \int{1.2(1.015)^t \, dt} \)

Integration here requires knowledge of handling exponential functions. Solving this integral provides a function that indicates cumulative consumption up to any year \( t \) since 2010.

This process turns a dynamic rate into a tangible quantity, which is essential for answering questions about total resources used over time.

Logarithms

Logarithms are the inverse operations to exponentiation, essential for solving equations involving exponential functions. In this exercise, logarithms help determine how long it takes to consume a specific amount of oil.

Once you have an equation of the form:

• \( 1.2(1.015)^t = x \)

Taking the logarithm (typically natural log, \( \ln \)) of both sides helps isolate the variable \( t \):

• \( t = \frac{\ln(\frac{x}{1.2})}{\ln(1.015)} \)

This method effectively handles exponentiation by utilizing logarithmic properties to transform the equation into a solvable format.

Understanding how to apply logarithms simplifies solving real-world problems involving exponential growth, enabling the calculation of time spans or growth rates efficiently.

Consumption Rate Function

The consumption rate function is a mathematical representation of how a quantity, such as oil consumption, changes over time. This function is particularly useful for modeling real-world scenarios where consumption is not constant.

In this exercise, the consumption rate function is given by:

• \( r(t) = 1.2(1 + 0.015)^t \)

This expression captures the increase in oil consumption over time, starting from an initial rate of 1.2 million barrels per year, growing at 1.5% annually.

The function is essential for accurately calculating both instantaneous consumption at any given time and total consumption over a period through integration. By understanding this function, one can assess how consumption patterns might evolve and plan accordingly for resource management.

This approach is fundamental in fields such as environmental science, economics, and engineering, where resource usage needs efficient monitoring and forecasting.