Problem 40 - Calculus: Graphical, Numerical, Algebraic

Question

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

Step-by-Step Solution

Answer

The well's output will fall to one-fifth of its present level in approximately 16 years.

Step 1: Identify the given values

Given the present output as 'P', the rate of decrease as \(r=0.10\) per year. The output level to which it should fall is \(\frac{1}{5}\) th of its current output. Let's label this as 'A'. To apply the decay formula \(A = Pe^{rt}\), we need to adjust the function for decay, so r will be negative, giving us the function \(A = Pe^{-rt}\).

Step 2: Adjust the formula and apply the given values

Firstly, replace 'A' with \(\frac{1}{5}P\), and 'P' with 'P' in the formula, producing the equation \(\frac{1}{5}P = Pe^{-0.10t}\). 'P' on both sides of the formula cancels out, simplifying this to \(\frac{1}{5} = e^{-0.10t}\).

Step 3: Isolate the variable t

To isolate 't', first apply natural log on both sides which gives \(-ln(5) = -0.10t\). Then, to completely isolate 't' divide both sides by \(-0.10\) leading to \(t = \frac{ln(5)}{0.10}\).

Step 4: Solve for t

Finally, we can calculate the time as \(t = \frac{ln(5)}{0.10} = 16.1\) years approximately. So, it will take around 16 years for the well's output to fall to one-fifth of its present level.

Key Concepts

Calculus Applications

In calculus, one of the primary applications is to model real-world phenomena with mathematical equations that describe change over time. In this particular problem, we are tasked with modeling the depletion of oil, where the rate at which oil is being pumped decreases exponentially. Calculus provides a framework for exploring these situations with precision.
Here, exponential decay is used to represent the decreasing output of an oil well. The calculus application comes into play by understanding how this decay works: the rate of change is proportional to the quantity present. This fundamental understanding helps us predict the time it takes for the well's output to decrease to a certain level - in this case, to one-fifth of its original level. By setting up the exponential decay function and solving it, calculus allows us to make precise predictions about the situation.

Natural Logarithms

Natural logarithms (often denoted with 'ln') are logarithms with the base 'e', where 'e' is approximately 2.71828. They are called "natural" because they arise naturally in the process of solving for time in continuous growth or decay models, such as our oil depletion example.
In our problem, we encounter a step where natural logs are necessary. After setting up the equation \(\frac{1}{5} = e^{-0.10t}\),
we need to isolate 't'. By taking the natural logarithm of both sides, we use the property that \(ln(e^x) = x\) to simplify the equation. This yields \(-ln(5) = -0.10t\).
Natural logarithms thus provide the tool to unravel 't' from the exponent in an exponential function, allowing us to solve the problem efficiently.

Exponential Functions

Exponential functions model processes where a quantity changes at a rate proportional to its current value. The general form of an exponential function is \(A = Pe^{rt}\),
where 'A' is the amount after time 't', 'P' is the initial amount, 'r' is the rate of change, and 'e' is the base of the natural logarithm.
In our scenario of oil depletion, the rate of change is negative due to the continuous decrease in output, leading us to use the formula \(A = Pe^{-rt}\).
This formula captures exponential decay, a common real-world application where things decrease steadily over time. By substituting the given values into the function, we establish the equation needed to determine the necessary time 't' until the oil output falls to one-fifth of its present level. Exponential functions thus offer a powerful tool for analyzing real-world situations where change is not linear.