Problem 66 - Applied Calculus

Question

An employer offers to pay workers at the rate of \(30,000 e^{0.04 t}\) dollars per year, while the union demands payment at the rate of \(30,000 e^{0.08 t}\) dollars per year, where \(t=0\) corresponds to the beginning of the contract. Find the accumulated difference in pay between these two rates over the 10 -year life of the contract.

Step-by-Step Solution

Answer

The accumulated difference in pay over 10 years is $93,424.25.

Step 1: Set Up the Integral for Differences in Rates

To find the accumulated difference in pay over 10 years, we need to calculate the difference between the two rate formulas and integrate that difference from 0 to 10. The difference between the union's demanded rate and the employer's offered rate is given by: \[ 30,000 e^{0.08t} - 30,000 e^{0.04t} \].

Step 2: Simplify the Expression

Factor out the common factor of 30,000 from the expression: \[ 30,000 \left( e^{0.08t} - e^{0.04t} \right). \] This makes it easier to integrate.

Step 3: Integrate the Expression

Integrate the simplified difference in pay expression over the time interval from 0 to 10:\[ \int_{0}^{10} 30,000 \left( e^{0.08t} - e^{0.04t} \right) \ dt. \]This integral can be split into two separate integrals:\[ 30,000 \int_{0}^{10} e^{0.08t} \ dt - 30,000 \int_{0}^{10} e^{0.04t} \ dt. \]

Step 4: Integrate Each Term Separately

The integral of \( e^{kt} \) is \( \frac{1}{k} e^{kt} \). Apply this rule to each term:1. \( 30,000 \int_{0}^{10} e^{0.08t} \ dt = 30,000 \left[ \frac{1}{0.08} e^{0.08t} \right]_{0}^{10} = 30,000 \left( \frac{1}{0.08} \right) \left( e^{0.8} - 1 \right). \)2. \( 30,000 \int_{0}^{10} e^{0.04t} \ dt = 30,000 \left[ \frac{1}{0.04} e^{0.04t} \right]_{0}^{10} = 30,000 \left( \frac{1}{0.04} \right) \left( e^{0.4} - 1 \right). \)

Step 5: Evaluate the Definite Integrals

Calculate the values from the integrated expressions:1. For \( e^{0.08t} \): - \( e^{0.8} \approx 2.22554 \) - Result: \( 30,000 \times 12.5 \times (2.22554 - 1) \approx 462,081.75 \)2. For \( e^{0.04t} \): - \( e^{0.4} \approx 1.49182 \) - Result: \( 30,000 \times 25 \times (1.49182 - 1) \approx 368,657.5 \)

Step 6: Calculate the Total Accumulated Difference

Subtract the total accumulated values of the employer's offer from the union's demand to find the difference:\[ 462,081.75 - 368,657.5 = 93,424.25. \]Therefore, the accumulated difference in pay over 10 years is $93,424.25.

Key Concepts

Exponential Functions

Exponential functions are a fundamental concept in calculus and many other areas of mathematics. They are characterized by the equation \( f(t) = a \cdot e^{kt} \), where \(e\) is the base of the natural logarithm, approximately equal to 2.71828. In this function, \(a\) is a constant, \(k\) is the rate of growth or decay, and \(t\) represents time.

These functions model continuous growth or decay processes, making them applicable in various real-world scenarios such as population growth, radioactive decay, and, as in our exercise, salary increases over time.

• Positive \(k\) indicates exponential growth, as seen in the function \(30,000 e^{0.08 t}\), representing the union's demanded payment rate over time.
• Negative \(k\) denotes exponential decay, though not specifically part of this exercise.

Considering exponential growth in this context showcases how quickly certain factors such as salary can increase over a given time, due to the compound nature of the exponential function.

Definite Integral

The definite integral is a core concept in calculus that represents the area under a curve within a certain interval. It is calculated for a function \(f(t)\) over a specified range \([a, b]\) to determine total accumulation during that interval. It is expressed as \( \int_{a}^{b} f(t) \, dt \).

In our example of salary rates, the definite integral is used to calculate the total payment over ten years for each salary rate provided. By finding the difference between the two integrated rates, we determine the accumulated difference in pay due to the varying exponential rates.

• This technique effectively sums up continuous data, like varying salary over time, helping us understand net changes within that period.
• A definite integral always computes an exact value, unlike indefinite integrals that incorporate a constant of integration.

Fortunately, the integral of an exponential function \(e^{kt}\) is particularly manageable, as shown in the solution. By integrating \(e^{0.08t}\) and \(e^{0.04t}\), we leveraged this simplicity to find the exact change in salary.

Accumulated Difference Calculation

Accumulated difference calculation involves determining the total difference between two varying values over a specific time period. In our case, this refers to the difference in pay rates provided by two distinct exponential functions, integrating each over the interval from \(t=0\) to \(t=10\).

This method uses the formula \( \int_{0}^{10} (f_1(t) - f_2(t)) \, dt \), where \(f_1(t)\) and \(f_2(t)\) are two different functions representing the salary rates. By finding this integral, we get the total difference in accumulated salaries over ten years.

• The goal is to highlight the change over years by integrating the rate of difference between the proposed salaries.
• Simplifying the function through factoring makes the integration process more efficient, as shown by factoring out the common multiplier of 30,000.

Understanding this accumulated difference is crucial for making informed decisions based on the financial implications of continuous salary proposals. In simple terms, it tells us how much more one rate would yield compared to another over the contract's lifespan.