Problem 39 - Calculus and its applications

Question

In \(2012,34.1 \%\) of all glass containers were recycled. A beverage company used \(400,000 \mathrm{lb}\) of glass containers per year. After recycling, the amount of glass, in pounds, still in use after \(t\) years is given by \(N(t)=400,000(0.341)^{t}\). a) Find \(N^{\prime}(t)\) b) Interpret the meaning of \(N^{\prime}(t)\).

Step-by-Step Solution

Answer

a) \(N^{\prime}(t) \approx -430,000 \cdot (0.341)^t\); b) It shows the yearly decrease rate of recycled glass in use.

Step 1: Understand the function

The function given is \(N(t) = 400,000(0.341)^t\), which represents the amount of glass still in use after \(t\) years. The base \(0.341\) indicates the percentage of glass that continues to be recycled yearly.

Step 2: Differentiate the function

To find \(N^{\prime}(t)\), we need to differentiate \(N(t) = 400,000(0.341)^t\) with respect to \(t\). Using the formula for differentiation of an exponential function \((a^t)' = a^t \ln(a)\), we have:\[ N^{\prime}(t) = 400,000 \cdot \left( (0.341)^t \ln(0.341) \right) \].Thus, the derivative is:\[ N^{\prime}(t) = 400,000 \cdot (0.341)^t \cdot \ln(0.341) \].

Step 3: Calculate \(\ln(0.341)\)

First, calculate the natural logarithm of \(0.341\). Using a calculator,\[ \ln(0.341) \approx -1.075 \].

Step 4: Express the derivative

Substitute \(\ln(0.341)\) into the derivative:\[ N^{\prime}(t) \approx 400,000 \cdot (0.341)^t \cdot (-1.075) \].So, the derivative is:\[ N^{\prime}(t) \approx -430,000 \cdot (0.341)^t \].

Step 5: Interpret \(N^{\prime}(t)\)

The derivative \(N^{\prime}(t)\) represents the rate of change of the amount of glass still in use with respect to time. Since it is negative, it means that the amount of glass decreases each year as more is recycled. The magnitude indicates how quickly it decreases.

Key Concepts

Exponential functions

Exponential functions are a type of mathematical expression where a constant base is raised to a variable exponent. These functions are crucial in modeling growth or decay processes, such as populations, interest rates, and radioactive decay. In the problem, we see an example of an exponential decay function, which represents the recycling of glass containers. Here, the function \( N(t) = 400,000(0.341)^t \) illustrates how much glass remains in use after \( t \) years.

The base, 0.341, is particularly important as it represents the fraction of glass continuously recycled each year. This fraction is less than 1, indicating that it's a decay process, where the total amount decreases over time. Exponential functions like this are defined by the following key features:

• The rate of change is proportional to the current quantity.
• The graph of the function is a curve, typically either increasing or decreasing exponentially.

Understanding how these functions work helps in predicting future behaviors, making them invaluable in scientific and business applications.

Rate of change

The rate of change is a concept that measures how a quantity changes over time. In calculus, the rate of change of a function is represented by its derivative. For the glass container problem, the derivative \( N^{\prime}(t) \) gives us the rate at which the amount of glass still in use decreases over the years.

Calculating the derivative of an exponential function like \( N(t) \) involves using the natural logarithm of the base as part of the differentiation process. For the given problem, the derivative is \[ N^{\prime}(t) \approx -430,000 \cdot (0.341)^t \]. This expression tells us several things:

• The negative sign indicates a decrease, confirming that the amount of glass is reducing over time.
• The magnitude of the derivative tells us how quickly the amount of glass is declining annually.
• If \( t \) increases, the amount of glass decreases faster initially, but the speed of decline slows down as \( t \) becomes larger.

Understanding rates of change is crucial in various fields, as it helps to describe how and why changes are occurring in a process.

Natural logarithm

The natural logarithm is a specific logarithm with the base \( e \), an irrational constant approximately equal to 2.71828. It’s denoted as \( \ln(x) \) and is an essential function in calculus, particularly for problems involving exponential growth or decay.

In the problem, the natural logarithm is used during differentiation. When you differentiate \( (a^t) \), the result involves \( \ln(a) \). For instance, \( \ln(0.341) \) plays a role in finding \( N^{\prime}(t) \). The approximation \( \ln(0.341) \approx -1.075 \) helps simplify the derivative expression.

Key insights about natural logarithms include:

• They convert multiplicative processes into additive ones, simplifying calculations.
• They have many applications, such as solving for time in compound interest or decay equations.
• In differentiation, they help find the actual rate at which exponential functions change.

Understanding \( \ln(x) \) is vital as it directly corresponds to how quickly an exponential function is growing or decaying, linking back to real-world contexts like the recycling process in this problem.