Problem 56 - Applied Calculus

Question

a. Solve without using a graphing calculator. b. Verify your answer to part (a) using a graphing calculator. Present Value of a Continuous Stream of Income An oil well generates a continuous stream of income of \(60 t\) thousand dollars per year, where \(t\) is the number of years that the rig has been in operation. Find the present value of this stream of income over the first 20 years at a continuous interest rate of \(5 \% .\)

Step-by-Step Solution

Answer

The present value is approximately 6341.81 thousand dollars.

Step 1: Understanding the Problem

We need to find the present value of a continuous stream of income for an oil well. The income is generated continuously as \(60t\) thousand dollars per year. Our task is to calculate this over the first 20 years with a 5% continuous interest rate.

Step 2: Setting Up the Integral

The present value \(PV\) of a continuous income stream \( R(t) \) over the interval \([a, b]\) is given by the formula: \[ PV = \int_{a}^{b} R(t)e^{-rt} \, dt \] where \(r\) is the interest rate. In this case, \(R(t) = 60t\) and \(r = 0.05\). We're integrating from \(t = 0\) to \(t = 20\).

Step 3: Calculating the Integral

We plug the values into the formula: \[ PV = \int_{0}^{20} 60t \cdot e^{-0.05t} \, dt \]This is an integration by parts problem, where \(u = 60t\) and \(dv = e^{-0.05t} \, dt\).

Step 4: Applying Integration by Parts

For integration by parts, \( dv = e^{-0.05t} \, dt \), then \( v = \frac{-1}{0.05} e^{-0.05t} = -20 e^{-0.05t} \).Let \( u = 60t \), then \( du = 60 \, dt \).Using integration by parts: \[ \int u \, dv = uv - \int v \, du \]\[ \int 60t e^{-0.05t} \, dt = -20 \cdot 60t e^{-0.05t} + \int 1200 e^{-0.05t} \, dt \]\[ = -1200t e^{-0.05t} - 24000 e^{-0.05t} \] evaluated from 0 to 20.

Step 5: Evaluating the Integral from 0 to 20

Plug the limits back into the antiderivative:\[ -1200(20)e^{-0.05 \times 20} - 24000e^{-0.05 \times 20} + 1200(0)e^{0} + 24000e^{0} \]\[ = -24000 \times e^{-1} - 24000 \times e^{-1} + 24000 \]Use \( e^{-1} \approx 0.367879 \):\[ = -24000 \times 0.367879 - 24000 \times 0.367879 + 24000 \]\[ \approx -24000 \times 0.735758 + 24000 \]\[ \approx -17658.19 + 24000 \]\[ = 6341.81 \] thousand dollars.

Step 6: Verifying with a Graphing Calculator

Using a graphing calculator, input the function \( 60t e^{-0.05t} \) and integrate from 0 to 20. The graphing calculator should confirm the manual calculation of the present value as approximately 6341.81 thousand dollars.

Key Concepts

Continuous Stream of Income

A continuous stream of income is a financial concept where income is generated consistently over time. In the context of this exercise, the oil well generates a continuous income of \(60t\) thousand dollars per year. Here, \(t\) represents each year since the well started operating.

This continuous generation of income is crucial in fields like finance and economics because it allows for a steady cash flow that can be modeled using calculus. Unlike discrete streams, continuous streams do not have obvious breaks or intervals, making them more predictable and easier to integrate over a given period.

• No gaps or interruptions in the cash flow.
• Represented mathematically using functions.
• Commonly modelled with integral calculus to adjust for interest rates.

Understanding continuous income helps us in calculating the present value, which requires accounting for time and interest continuously.

Integration by Parts

Integration by parts is a technique used in integral calculus to integrate products of functions. Rather than integrating directly, we split the product into two parts, chose one to differentiate and the other to integrate. The formula for integration by parts is derived from the product rule for differentiation: \[\int u \, dv = uv - \int v \, du\]In our example, \( u = 60t \) and \( dv = e^{-0.05t} \, dt \). This choice is strategic: after computing the derivatives and anti-derivatives, the integration becomes simpler.

The steps for applying integration by parts include:

• Select \(u\) and \(dv\) such that \(du\) and \(v\) are easy to find.
• Find \(v\) by integrating \(dv\).
• Perform the integration using the formula, simplifying where necessary.
• Evaluate the result over the given limits.

This technique is particularly useful when dealing with exponential functions coupled with polynomials, as in our scenario with \(60t \cdot e^{-0.05t}\).

Continuous Interest Rate

The continuous interest rate is a concept where interest is compounded constantly rather than at discrete intervals like annually or quarterly. It's often represented by the constant \(r\) in mathematical formulas.

In the exercise, a continuous interest rate of 5% is used. This means the cash flow generated by the oil well is discounted continuously over the time period of 20 years. Mathematically, this is expressed using the exponential decay factor \( e^{-rt} \), where \(r = 0.05\).

• Provides a smoother and more accurate calculation for growth or decay.
• Used in present value calculations for continuous cash flows.
• Integral to many financial and economic models.

Continuous compounding forms the basis for determining present value with continuous income streams, offering a more precise method compared to simple or periodic compounding.

Integral Calculus

Integral calculus is a branch of mathematics concerned with integration, which is the process of finding integrals. In simpler terms, it allows us to calculate things like areas, volumes, and other quantities that are cumulative in nature.

In this real-world problem, integral calculus is used to determine the present value of the continuous income stream generated by the oil well. By setting up the integral of \(60t \cdot e^{-0.05t}\), we're able to find the accumulated value of income over time, accounting for interest.

• Calculates accumulations, such as total income over time.
• Helps in switching from a rate of change (derivative) to total accumulation (integral).
• Provides tools for evaluating areas under curves, which in finance translates to calculating present value.

The application of integral calculus in finance is crucial for evaluating continuous cash flows and understanding the full value of such income over a defined period.