Question

Suppose that \(r(t)=r_{0} e^{-k t},\) with \(k>0,\) is the rate at which a nation extracts oil, where \(r_{0}=10^{7}\) barrels \(/\) yr is the current rate of extraction. Suppose also that the estimate of the total oil reserve is \(2 \times 10^{9}\) barrels. a. Find \(Q(t),\) the total amount of oil extracted by the nation after \(t\) years. b. Evaluate \(\lim _{t \rightarrow \infty} Q(t)\) and explain the meaning of this limit. c. Find the minimum decay constant \(k\) for which the total oil reserves will last forever. d. Suppose \(r_{0}=2 \times 10^{7}\) barrels/yr and the decay constant \(k\) is the minimum value found in part (c). How long will the total oil reserves last?

Short answer

a. The total amount of oil extracted after t years is given by the function \(Q(t) = \frac{r_0}{k}(1 - e^{-kt})\). b. The limit of \(Q(t)\) as \(t\) approaches infinity is \(\frac{r_0}{k}\), which represents the maximum amount of oil that can be extracted. c. The minimum decay constant (\(k\)) for which the oil reserves will last forever is \(\frac{1}{200}\). d. The duration for the oil reserves to last with a doubled initial extraction rate (\(r_0\)) and the minimum decay constant (\(k\)) is approximately 138.63 years.

Step-by-step solution

Integrate the rate function

To find the total oil extracted after t years, we need to integrate the oil extraction rate function with respect to time: \(Q(t) = \int r(t) \ dt = \int r_0 e^{-kt} \ dt \)

Find the antiderivative of the exponential function

By taking the antiderivative with respect to time, we get: \(Q(t) = -\frac{r_0}{k}e^{-kt} + C\)

Calculate the constant of integration

At the beginning (\(t=0\)), the total amount of oil extracted is 0, so: \(Q(0) = -\frac{r_0}{k}e^{0} + C = 0 \Rightarrow C = \frac{r_0}{k}\) Now, we can rewrite the function for the total oil extracted as: \(Q(t) = -\frac{r_0}{k}e^{-kt} + \frac{r_0}{k} = \frac{r_0}{k}(1 - e^{-kt})\)

Evaluate the limit of Q(t)

To find the value of the limit as \(t \rightarrow \infty\), use L'Hôpital's rule: \(\lim_{t \rightarrow \infty} Q(t) = \lim_{t\rightarrow \infty} \frac{r_0}{k}(1 - e^{-kt}) = \frac{r_0}{k}\) This means that the total oil extracted asymptotically approaches \(\frac{r_0}{k}\) as time goes to infinity. This also represents the maximum amount of oil that can be extracted.

Determine the minimum decay constant

To find the value of k that will make the oil reserves last forever, we set the total oil extraction limit equal to the total oil reserve. \(\frac{r_0}{k} = 2\times 10^9\) Plugging in \(r_0 = 10^7\) barrels/yr, we can solve for k: \(k = \frac{10^7}{2\times 10^9} = \frac{1}{200}\) Hence, the minimum decay constant, \(k = \frac{1}{200}\).

Calculate the time the reserves will last

Now, let's find how long the oil reserves will last if \(r_0 = 2 \times 10^7\) barrels/yr and \(k = \frac{1}{200}\). We know: \(\frac{r_0}{k} = 2\times 10^9\) Replacing \(r_0\) with twice its original value and \(k\) with the minimum value obtained in the previous step, we have: \(\frac{2\times 10^7}{\frac{1}{200}} = 2\times 10^9\) Solve for t: \(2\times 10^9 = \frac{2\times 10^7}{\frac{1}{200}}(1 - e^{-\frac{1}{200}t})\) Simplify and isolate the exponential term: \(e^{-\frac{1}{200}t} = 1 - \frac{2\times 10^9}{4\times 10^9} = \frac{1}{2}\) Take the natural logarithm of both sides: \(-\frac{1}{200}t = \ln\left(\frac{1}{2}\right)\) Solve for t: \(t = (-200) \ln\left(\frac{1}{2}\right) \approx 138.63\) years Thus, the oil reserves will last approximately 138.63 years with \(r_0 = 2 \times 10^7\) barrels/yr and \(k = \frac{1}{200}\).

Key concepts

Exponential Functions

Exponential functions are fundamental in calculus, particularly when dealing with growth and decay processes. The general form of an exponential function can be written as \(f(t) = r_0 e^{-kt}\), where \(r_0\) is the initial value, and \(k\) is the rate of decay or growth. This type of function is commonly used to model scenarios where a quantity decreases or increases gradually over time. In our problem, \(r(t) = r_0 e^{-kt}\) represents the rate at which oil is extracted over time.
Key points of understanding exponential functions in this context:

• The negative exponent indicates a decay process, meaning the rate of oil extraction decreases as time goes by.
• \(e\) is Euler's number (approximately 2.718), a constant that appears frequently when continuous growth or decay is involved.
• The term \(k\) determines the speed of the decay; a larger \(k\) results in a faster rate of decrease.

Mastery of these functions allows you to predict how quantities like oil reserves change over time.

Integration

Integration is a core concept in calculus, allowing us to find the total accumulation of a quantity given its rate of change. In contexts like this exercise, it helps compute the total amount of oil extracted over a timeframe, given a rate function. To find the function \(Q(t)\), representing total oil extracted, we integrated the rate function \(r(t) = r_0 e^{-kt}\):
\[Q(t) = \int r_0 e^{-kt} \, dt = -\frac{r_0}{k} e^{-kt} + C\]Integration involves finding an antiderivative, with the integral adding a constant of integration, \(C\). By setting \(Q(0) = 0\), we solve for \(C\). This step ensures our model fits initial conditions where no oil is extracted at \(t=0\). Ultimately, we express the total oil extracted as:
\[Q(t) = \frac{r_0}{k} (1 - e^{-kt})\]
This demonstrates how integration connects rate functions (derivatives) to total changes (integrals) in quantities over time.

Limits

The concept of limits in calculus concerns what value a function approaches as the input approaches some point, often infinity. In the oil extraction problem, assessing \(\lim_{t \to \infty} Q(t)\) allows us to calculate the theoretical maximum amount of oil that can be extracted as time proceeds indefinitely.

• For \(Q(t) = \frac{r_0}{k}(1 - e^{-kt})\), as \(t\) heads towards infinity, the exponential term \(e^{-kt}\) approaches zero because \(k > 0\).
• Therefore, \(\lim_{t \to \infty} Q(t) = \frac{r_0}{k}(1 - 0) = \frac{r_0}{k}\).

This limit tells us that eventually, the total amount of oil extracted will level off at \(\frac{r_0}{k}\). This represents the total available reserves of oil that can be extracted given the current conditions. Understanding limits helps in predicting long-term outcomes and behaviors of systems modeled by calculus.

Natural Logarithms

Natural logarithms are the inverse functions of exponential functions and play a crucial role in solving equations involving exponential decay or growth. We use the natural logarithm, denoted \(\ln\), when solving for time or other quantities within exponential expressions. In the exercise, we solve for \(t\) using the equation:\[e^{\frac{-1}{200}t} = \frac{1}{2}\]Taking the natural logarithm of both sides, we find:\[-\frac{1}{200}t = \ln\left(\frac{1}{2}\right)\]By rearranging and solving, \(t = -200 \ln\left(\frac{1}{2}\right)\). The solution tells us that the reserves will last approximately 138.63 years under given conditions. A natural logarithm is useful because it simplifies the calculation of time when working with exponential equations, aiding in determining how long a process like oil extraction will take under specific parameters. This understanding is essential for interpreting scenarios involving growth or decay over time.