Problem 35 - Thomas Calculus in SI Units

Question

Suppose the amount of oil pumped from one of the canyon wells in southern California decreases at the continuous rate of \(10 \%\) per year. When will the well's output fall to one-fifth of its present value?

Step-by-Step Solution

Answer

The well's output will fall to one-fifth of its present value in about 16.09 years.

Step 1: Understand the problem statement

We need to determine when the output from the well, which decreases at a continuous rate of 10% per year, reaches one-fifth of its current value. Essentially, we are solving for time (t) in years.

Step 2: Set up the exponential decay formula

Since the amount of oil decreases continuously, we use the exponential decay formula: \( A(t) = A_0 e^{-rt} \). Here, \( A_0 \) is the initial amount, \( r \) is the rate of decay (0.10), and \( t \) is the time in years.

Step 3: Establish the condition for one-fifth of the present value

We want the output to become \( \frac{1}{5} \) of its current value. Substitute into the decay formula: \( \frac{1}{5}A_0 = A_0 e^{-0.10t} \).

Step 4: Simplify the equation

Divide both sides by \( A_0 \) (assuming \( A_0 eq 0 \)) to get \( \frac{1}{5} = e^{-0.10t} \).

Step 5: Solve for \( t \) using natural logarithms

Take the natural logarithm of both sides to solve for \( t \):\[ \ln\left(\frac{1}{5}\right) = -0.10t \]Then, isolate \( t \):\[ t = \frac{\ln\left(\frac{1}{5}\right)}{-0.10} \]

Step 6: Calculate the value of \( t \)

Compute the value of \( t \):\[\ln\left(\frac{1}{5}\right) \approx -1.60944 \]\[t = \frac{-1.60944}{-0.10} \approx 16.0944\]Therefore, it takes approximately 16.09 years.

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. These functions are at the heart of various natural phenomena and many equations in science, particularly those depicting growth and decay processes.

A key characteristic of exponential functions is their continuous and consistent rate of change. In simpler terms, these functions either grow or decay at a constant percentage rate over time. The general formula for an exponential function is given by:

• \( A(t) = A_0 e^{rt} \) for growth
• \( A(t) = A_0 e^{-rt} \) for decay

Here, \( A(t) \) is the value of the function at time \( t \), \( A_0 \) is the initial value, \( e \) is the base of the natural logarithm (approximately 2.71828), and \( r \) is the rate of growth or decay expressed as a decimal. This structure allows us to model scenarios such as population growth, radioactive decay, and in our case, the decrease in oil output from a well.

Understanding how these functions work is crucial, especially when predicting future values given a continuous rate of change. Take careful note of the signs in the exponent to determine if the scenario depicts growth (positive) or decay (negative).

Natural Logarithms

Natural logarithms are a specific type of logarithm where the base is the mathematical constant \( e \) (approximately 2.71828). They are incredibly useful in equations where processes are measured continuously, just like exponential functions.

The natural logarithm of a number is written as \( \ln(x) \), and it provides the exponent to which \( e \) must be raised to obtain that number. For example, if \( e^y = x \), then \( \ln(x) = y \). These logarithms simplify many complex exponential equations, especially when isolating variables like time or rates.

• When solving exponential decay problems, natural logarithms help in rearranging the formula to solve for time \( t \).
• The formula \( \ln(x) \) is instrumental in transitioning from exponential form to linear form.

To solve for \( t \) in exponential decay, take the natural logarithm of both sides of an equation. This process turns multiplicative relationships into additive ones, which are much easier to solve. In the oil well problem, natural logarithms help us find the exact point in time when the well's output reduces to one-fifth of its original value.

Logarithmic Equations

Logarithmic equations involve variables contained within logarithm functions. They are crucial in situations where the behavior of an exponential function needs to be studied in reverse. Practically speaking, they enable us to calculate unknown exponents.

To solve a logarithmic equation, you essentially reverse the process of taking a power of a number. Here’s a simple way to understand it:

• Convert the logarithmic equation into an exponential form to isolate the variable. For example, if given \( \log_b(y) = x \), then the equivalent statement is \( b^x = y \).
• In our context involving continuous decay, natural logarithms (where base \( b \) is \( e \)) become instrumental since they simplify the mathematics involved.

These tools allow one to convert an equation like \( \ln \left( \frac{1}{5} \right) = -0.10t \) into its exponential counterpart to extract the value of \( t \). Calculating logarithms accurately is key. It's how you find out exactly when a particular event, like the oil well's production drop, will occur. Understanding logarithmic equations helps reveal the time variable hidden in exponential decay functions.