Problem 16 - Calculus and its applications

Question

Find the accumulated present value of each continuous income stream at rate \(R(t),\) for the given time \(T\) and interest rate \(k\) compounded continuously. $$R(t)=\$ 520,000, \quad T=25 \mathrm{yr}, \quad k=6 \%$$

Step-by-Step Solution

Answer

The accumulated present value is approximately $6,722,667.

Step 1: Understand the Accumulated Present Value Formula

The accumulated present value (APV) of a continuous income stream with rate \( R(t) \) over time \( T \) and interest rate \( k \) is given by the integral \( \int_0^T R(t) e^{-kt} \, dt \). In our case, \( R(t) = 520,000 \), \( T = 25 \) years, and \( k = 0.06 \) (since 6% equals 0.06 in decimal form).

Step 2: Set Up the Integral

Substitute the values of \( R(t) \), \( T \), and \( k \) into the formula to get \(\int_0^{25} 520,000 \cdot e^{-0.06t} \, dt \).

Step 3: Evaluate the Integral

The integral \( \int 520,000 e^{-0.06t} \, dt \) can be solved using the formula \( \int e^{at} \, dt = \frac{1}{a} e^{at} + C \). Thus,\[ \int 520,000 e^{-0.06t} \, dt = 520,000 \cdot \left( -\frac{1}{0.06} e^{-0.06t} \right) + C. \] After simplifying, we have \( -\frac{520,000}{0.06} e^{-0.06t} \).

Step 4: Calculate the Definite Integral

Evaluate the definite integral from \( t = 0 \) to \( t = 25 \):\[ \left[ -\frac{520,000}{0.06} e^{-0.06t} \right]_0^{25} = -\frac{520,000}{0.06} \left( e^{-0.06 \cdot 25} - e^{0} \right) = -\frac{520,000}{0.06} \left( e^{-1.5} - 1 \right). \]

Step 5: Compute the Numerical Solution

Calculate \( e^{-1.5} \) to find its approximate value. It is approximately 0.2231. Substitute this into the previous expression:\[ -\frac{520,000}{0.06} (0.2231 - 1) = \frac{520,000}{0.06} (1 - 0.2231) = \frac{520,000}{0.06} \cdot 0.7769. \]Compute the result, which gives approximately 6,722,667.

Key Concepts

Accumulated Present Value

Accumulated Present Value, often abbreviated as APV, is a fundamental concept in finance and mathematics. It allows us to determine how much a continuous stream of income is worth right now, rather than in the future. This is important because it accounts for the time value of money, a key principle in finance. The time value of money suggests that a dollar today is worth more than a dollar in the future, due to its potential earning capacity.
This is calculated using the integral formula:

• APV = \( \int_0^T R(t) e^{-kt} \, dt \)

In this formula:

• \( R(t) \) represents the income rate at time \( t \).
• \( T \) is the total time period of the income.
• \( k \) is the interest rate expressed in decimal.
• The term \( e^{-kt} \) is derived from the concept of continuous compounding, reducing future values to present day values.

Understanding this concept helps in making informed decisions today by recognizing the equivalency of future earnings.

Integration

Integration is a mathematical process used to find areas, among other things, under curves. It is essentially the reverse of differentiation. In finance, integration is crucial in calculating the accumulated value of streams of cash flows.

• To solve the problem of accumulated present value, we employ the definite integral, which allows us to sum up these continuous income streams over a certain period.

When we integrate a function like \( 520,000 e^{-0.06t} \), we accumulate its values from \( t = 0 \) to \( t = 25 \).
There are basic integration rules we use:

• \( \int e^{at} \, dt = \frac{1}{a} e^{at} + C \) for constant \( a \).

By mastering integration, you can calculate not just present values, but also areas, volumes, and many other interesting concepts in mathematics.

Compound Interest

Compound interest is a key principle in finance where interest is calculated not only on the initial principal but also on the accumulated interest of previous periods. This snowballing effect is why loans and investments can grow exponentially. In the context of the exercise, we consider interest compounding continuously.
For continuous compounding, we use the formula:

• \( A = Pe^{rt} \)

Where:

• \( A \) is the amount of money accumulated after \( n \) years, including interest.
• \( P \) is the principal amount (initial value).
• \( e \) is the base of the natural logarithm.
• \( r \) represents the annual interest rate (in decimal).
• \( t \) is the time the money is invested.

Compound interest can significantly affect the value of investments, making it a powerful tool for wealth growth.

Definite Integral

The definite integral is a type of integral with set upper and lower limits. It calculates the accumulated area under a curve between two points on the x-axis. In financial terms, it is used to sum up continuous income or costs over a certain period.
For our exercise, the definite integral helps in calculating the accumulated present value. This is performed by evaluating the integral from an initial point (\( t = 0 \)) to a final point (\( t = 25 \)).
We can solve this by:

• Finding the antiderivative of the function.
• Evaluating it at the upper and lower limits, \( T \) and \( 0 \) respectively.

The outcome provides the total accumulated value at the present time. Understanding definite integrals is crucial for applying calculus to real-life financial problems such as mortgage calculations and investment analysis.