Problem 20 - Calculus Early Transcendentals

Question

Devise the exponential growth function that fits the given data; then answer the accompanying questions. Be sure to identify the reference point\((t=0)\) and units of time. 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

Question: Using the derived exponential growth formula, calculate the number of years after 2010 when the oil consumption reaches 10 million barrels. Answer: To find the number of years after 2010 when the oil consumption reaches 10 million barrels, we can use the function we found in step 4, and plug in the value for the oil consumption: \(80(e^{0.015t} - 1) = 10\) Solving for t, we get: \(t = \frac{\ln(1.125)}{0.015}\) So the number of years after 2010 when the oil consumption reaches 10 million barrels is approximately \(\frac{\ln(1.125)}{0.015}\) years.

Step 1: Identify the given information and important values

The data given in the problem is: 1. The initial rate of consumption (at t=0) is 1.2 million barrels/yr. 2. The rate at which oil is consumed increases at 1.5% per year. Let's denote the initial consumption as \(C_{0}\), the rate at which the consumption increases as \(r\), and let \(C(t)\) be the oil consumption rate at any time t. We have the values given as: \(C_{0} = 1.2\) million barrels/yr \(r = 0.015\) (since 1.5% means 1.5/100)

Step 2: Apply the exponential growth formula

Now, we'll apply the exponential growth formula to our given values: \(C(t) = C_{0}e^{rt}\) In this case, \(C(t) = 1.2e^{0.015t}\) (in million barrels per year)

Step 3: Solve part (a) - Oil consumption in 2010

To find the oil consumption in the year 2010, we'll need to find the oil consumed between \(t=0\) and \(t=1\). First, let's find the oil consumption rate at t = 1: \(C(1) = 1.2e^{0.015(1)}\) Now, integrate this function from \(t=0\) to \(t=1\): \(\int_{0}^{1} C(t) dt = \int_{0}^{1} 1.2e^{0.015t} dt\) To solve this integral, let's do the following substitution: Let \(u = 0.015t\) then \(du = 0.015 dt\) and \(dt = \frac{1}{0.015} du\) After the substitution, the integral becomes: \(\int_{0}^{1} 1.2e^{0.015t} dt = 1.2(1/0.015)\int_{0}^{0.015} e^{u} du\) Now we can integrate: \(1.2(1/0.015)(e^u|_{0}^{0.015})\) Evaluate the integral: \(1.2(1/0.015)(e^{0.015} - e^0)\) Simplify: \(oil\_consumed = 80(e^{0.015} - 1)\) million barrels

Step 4: Answer part (b) - Function for oil consumed

To find the total oil consumed between \(t=0\) and any future time \(t\), we need to integrate the function \(C(t)\) over the time range: \(Oil\_consumed(t) = \int_{0}^{t} C(s) ds\) We already have the expression for \(C(t) = 1.2e^{0.015t}\), so we'll substitute it into the integral: \(Oil\_consumed(t) = \int_{0}^{t} 1.2e^{0.015s} ds\) Now, perform the same substitution as we did before, with \(u = 0.015s\): \(Oil\_consumed(t) = 1.2(1/0.015)\int_{0}^{0.015t} e^{u} du\) Which yields: \(Oil\_consumed(t) = 80(e^{0.015t} - 1)\) million barrels

Step 5: Solve part (c) - Find when oil consumption reaches 10 million barrels

We are now given that the oil consumption reaches 10 million barrels and need to find the time t when this occurs. we can use the function we found in step 4: \(80(e^{0.015t} - 1) = 10\) Now, we need to solve for t: \(e^{0.015t} - 1 = \frac{10}{80}\) \(e^{0.015t} = 1.125\) \(0.015t = \ln(1.125)\) \(t = \frac{\ln(1.125)}{0.015}\) So that the oil consumption is 10 million barrels, the number of years after 2010 is: \(years = \frac{\ln(1.125)}{0.015}\) years

Key Concepts

Oil Consumption

Understanding oil consumption is crucial as it reflects the rate at which resources are used and helps in planning for future energy needs. In this exercise, the consumption rate increases over time, making it an example of exponential growth in real-world applications. Initially, in 2010, the country began with an oil consumption rate of 1.2 million barrels per year. Over time, this rate increases by 1.5% annually. Such a scenario is common when resources are being depleted at a growing rate due to increased demand.

• Initial rate: 1.2 million barrels/year
• Growth rate: 1.5% per year

Tracking this consumption helps in understanding how quickly resources diminish and planning for alternative solutions or conservation methods.

Integration

Integration is a fundamental mathematical process used to find the area under a curve, which in this context represents the total oil consumed over a period of time. By integrating the function that models the consumption rate, we can determine the total quantity of oil used from a starting point until a designated time.

• Integration helps accumulate the rate from a point in time, giving the cumulative oil consumed.
• It involves calculating the definite integral of the consumption rate function over the desired time interval.

The integral in this problem, \[\int_{0}^{t} 1.2e^{0.015s} ds\]presents the accumulated consumption in terms of years, showing how integration converts a rate into a total amount used over that span.

Exponential Growth Formula

The exponential growth formula is pivotal in modeling situations where a quantity increases proportionally to its current value, such as oil consumption growth. The basic formula is:\[C(t) = C_{0} e^{rt}\]where:

• \(C_{0}\) is the initial amount.
• \(r\) is the growth rate expressed as a decimal.
• \(t\) is the time period.
• \(e\) is the base of natural logarithms.

This formula enables predicting future consumption rates and is easily adapted to reflect different growth rates or initial values, making it a critical tool for planning in resource management and economics.

Exponential Equation Solving

Solving exponential equations involves finding the unknown variable, in this case time, which fits the equation given a specific condition, such as a set total of 10 million barrels. The process is:

• Set up the equation based on the condition: \(80(e^{0.015t} - 1) = 10\)
• Solve for \(t\) by isolating the exponential term.
• Use logarithms to deal with exponents, leading to: \(t = \frac{\ln(1.125)}{0.015}\)

This method allows solving real-world problems such as predicting when a resource reaches a certain level or threshold. Understanding this process is essential for students in developing skills applicable to various scientific and economic fields.