Problem 48 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Based on a preliminary report by a geological survey team, it is estimated that a newly discovered oil field can be expected to produce oil at the rate of $$ R(t)=\frac{600 t^{2}}{t^{3}+32}+5 \quad(0 \leq t \leq 20) $$ thousand barrels/year, \(t\) yr after production begins. Find the amount of oil that the field can be expected to yield during the first 5 yr of production, assuming that the projection holds true.

Step-by-Step Solution

Answer

The total amount of oil that can be expected to be produced during the first 5 years is approximately: \(O(t) = 200 (\ln(157) - \ln(32)) + 25 \approx 313.98\) thousand barrels.

Step 1: Identify the function and the interval

We are given that the rate of production, \(R(t)\), is defined as: \( R(t)=\frac{600 t^{2}}{t^{3}+32}+5 \quad(0 \leq t \leq 20) \) We are interested in the interval \(0 \leq t \leq 5\), as we want to find the amount of oil produced during the first 5 years.

Step 2: Set up the integral

Since the rate of oil production varies with time, we need to integrate over the time interval to find the total amount produced. We can do this by setting up the following definite integral: \( O(t) = \int_{0}^{5} R(t) dt \)

Step 3: Integrate R(t) with respect to time

We need to perform the integration of R(t) over the interval from 0 to 5: \( O(t) = \int_{0}^{5} \left(\frac{600 t^{2}}{t^{3}+32}+5\right) dt = \int_{0}^{5} \frac{600 t^{2}}{t^{3}+32} dt + \int_{0}^{5} 5 dt \) Here, we can notice that the second integral is trivial to evaluate. For the first integral, we may need to use substitution to find the antiderivative.

Step 4: Evaluate the second integral

We can directly evaluate the integral on the right side: \( \int_{0}^{5} 5 dt = 5 \int_{0}^{5} dt = 5 [t]_{0}^{5} = 5(5-0) = 25 \) For the first integral, we will use substitution.

Step 5: Use substitution for the first integral

Let \(u = t^3 + 32\), then \(\frac{du}{dt} = 3t^2\). So, we change variables as: $$ du = 3t^2 dt \Rightarrow \frac{1}{3} du = t^2 dt $$ The integral becomes: $$ \int_{0}^{5} \frac{600 t^2}{t^3 + 32} dt = 600 \int_{32}^{157} \frac{1}{3u} du = 200 \int_{32}^{157} \frac{1}{u} du $$

Step 6: Evaluate the first integral

Now, we can evaluate the first integral using the antiderivative of \(\frac{1}{u}\): $$ 200 \int_{32}^{157} \frac{1}{u} du = 200 [\ln(u)]_{32}^{157} = 200 (\ln(157) - \ln(32)) $$

Step 7: Compute the total amount of oil produced

Finally, we evaluate the main integral by adding the results from the first and second integrals: $$ O(t) = 200 (\ln(157) - \ln(32)) + 25 $$ Thus, the total amount of oil that can be expected to be produced during the first 5 years is approximately: $$ O(t) = 200 (\ln(157) - \ln(32)) + 25 \approx 313.98 \text{ thousand barrels.} $$

Key Concepts

Oil Production Rate

The oil production rate refers to how quickly oil is extracted from a reservoir over a specific period. It's crucial for understanding the output capacity of an oil field. In this exercise, the production rate is described by the function \( R(t) = \frac{600 t^{2}}{t^{3}+32}+5 \). This equation represents the rate at which oil is produced in thousand barrels per year, measured at different years \( t \) after drilling begins. Understanding the shape of this function over time gives insights into how production might vary. Initially, as time increases, the rate could rise due to increased extraction efficiency. But it may eventually decline as the reservoir depletes. By writing the production rate as an equation, it becomes easier to integrate over time, which allows us to calculate the total oil produced over a set interval.

Definite Integral

A definite integral is a fundamental concept in calculus. It calculates the accumulation of quantities, which in this exercise, represents the total amount of oil produced. A definite integral has limits of integration, which are the time periods we are focused on. In this example, the definite integral is set up over the interval from 0 to 5 years. This means we're adding up small bits of oil produced, from the start of production to the end of the fifth year. The definite integral is expressed as:

• \( O(t) = \int_{0}^{5} R(t) \, dt \)

Here, \( O(t) \) represents the total amount of oil that has been extracted over the time interval. Calculating this requires integrating the production rate function \( R(t) \) between these bounds. This tells us how many thousand barrels of oil will be produced over the first five years.

Substitution Method

The substitution method is a powerful technique in calculus, used to simplify complex integrals for easier calculation. When faced with an integral that seems challenging to solve directly, substitution allows us to transform it into a simpler form. In the exercise, substitution involves letting \( u = t^3 + 32 \). This changes the derivative, allowing for a switch in variables:

• \( du = 3t^2 \, dt \) transforms into \( \frac{1}{3} du = t^2 \, dt \)

By substituting \( u \) for \( t^3 + 32 \), the original complex integral \( \int \frac{600 t^2}{t^3 + 32} \, dt \) reduces to a simpler form: \( 200 \int_{32}^{157} \frac{1}{u} \, du \). This method makes it feasible to find the antiderivative, necessary for calculating the total oil production.

Antiderivative Calculation

Calculating the antiderivative is a key step in solving a definite integral. It involves finding a function whose derivative is the given function, allowing us to evaluate the integral over the desired limits.In the substitution method applied here, we set \( u = t^3 + 32 \) and transformed the integral into:

• \( 200 \int_{32}^{157} \frac{1}{u} \, du \)

The antiderivative of \( \frac{1}{u} \) is \( \ln|u| \). So, calculating this gave: \( 200 [\ln(u)]_{32}^{157} \). Evaluating this expression from 32 to 157 provides the total change, representing the oil produced:

• \( 200 (\ln(157) - \ln(32)) \)

Finally, combining this with the evaluation of other parts of the integral gives the total expected oil yield for the first five years, approximately 313.98 thousand barrels. This shows the power of calculus in making complex real-world predictions.