Problem 9 - Calculus and its applications

Question

Find the accumulated future value of each continuous income stream at rate \(\mathrm{R}(t),\) for the given time \(\mathrm{T}\) and interest rate \(k\) compounded continuously. $$ R(t)=\$ 50,000, \quad T=22 \mathrm{yr}, \quad k=5 \% $$

Step-by-Step Solution

Answer

The accumulated future value is approximately $2,007,606.40.

Step 1: Understanding the Formula

To find the accumulated future value of a continuous income stream, we use the formula \( A = \int_0^T R(t) \cdot e^{k(T-t)} \, dt \). Here, \( R(t) \) is the continuous income rate, \( T \) is the total time in years, and \( k \) is the continuously compounded interest rate.

Step 2: Setting Up the Equation

Given \( R(t) = 50,000 \) dollars per year, \( T = 22 \) years, and \( k = 5\% = 0.05 \), we substitute into the formula: \( A = \int_0^{22} 50,000 \cdot e^{0.05(22-t)} \, dt \).

Step 3: Integration Setup

We need to integrate the function \( 50,000 \cdot e^{1.1 - 0.05t} \) over \( t \) from 0 to 22. Simplifying, the equation becomes \( A = 50,000 \cdot \int_0^{22} e^{1.1} \cdot e^{-0.05t} \, dt \).

Step 4: Calculating the Integral

Factor out constants: \( A = 50,000 \cdot e^{1.1} \cdot \int_0^{22} e^{-0.05t} \, dt \). The integral becomes \( \int e^{-0.05t} \, dt \), which is \( -\frac{1}{0.05} e^{-0.05t} + C \).

Step 5: Evaluating the Definite Integral

Evaluate the integral from 0 to 22: \( \left[-20 e^{-0.05t}\right]_0^{22} = -20(e^{-1.1} - e^{{0}}) = -20(e^{-1.1} - 1) \).

Step 6: Calculating the Accumulated Value

Substitute back: \( A = 50,000 \cdot e^{1.1} \cdot 20(1 - e^{-1.1}) \). Calculate using known values for \( e^{1.1} \approx 3.004 \) and \( e^{-1.1} \approx 0.333 \).

Step 7: Final Computation

Plug in the values to calculate \( A = 50,000 \cdot 3.004 \cdot 20 \cdot (1 - 0.333) \). We get \( A \approx 50,000 \cdot 3.004 \cdot 20 \cdot 0.667 \). Calculate to find the accumulated value.

Key Concepts

Accumulated Future Value

The accumulated future value refers to the total value of a stream of continuous income at a future date, considering compound interest over time. It's important for determining how investments grow with continuous cash inflows.
To calculate this, we need to know:

• The income stream rate, \( R(t) \), which is the consistent income earned over time.
• The duration \( T \), reflecting the period the money will earn interest.
• The interest rate \( k \), applied continuously.

The formula used is:\[A = \int_0^T R(t) \cdot e^{k(T-t)} \, dt\]This formula lets you accumulate contributions made to an account over time and see how they grow when interest compounds continuously. In the example given, \( R(t) = 50,000 \), \( T = 22 \) years, and interest rate \( k = 5\% \). Calculating this can give a substantial sum at the end of the investment period.

Continuous Compounding

Continuous compounding involves calculating interest in a manner where it's compounded an infinite number of times per year. This maximizes earnings compared to typical compounding methods. For instance, instead of yearly or semi-annually, it assumes the rate is applied at every possible moment, providing a better estimate of future value.
The formula involved with continuous compounding includes the exponential function, \( e \), and is generally represented as:\[ A = P \cdot e^{rt} \]where:

• \( P \) is the principal amount.
• \( r \) is the annual interest rate.
• \( t \) is the time in years.

The distinction of continuous compounding versus regular methods is its ability to reflect growth most accurately, important in maximizing applications like investments or savings consistently growing over time.

Definite Integral

A definite integral is a mathematical concept used to calculate the accumulated quantity over a specified interval. In the context of financial mathematics, it helps compute accumulated income over time considering interest.
Fundamentally, integrating a function involves summing up small slices or bits of data, from the start to the end of a given range. Here, the function involves a capital influx over time with continuously compounding interest:
\[A = \int_0^T R(t) \cdot e^{k(T-t)} \, dt\]With this expression:

• \( R(t) \cdot e^{k(T-t)} \) represents the growing income over time.
• Integration bounds \( 0 \) to \( T \) reflect the time period for computation.

Definite integrals are key in accurately calculating accumulated values in real-world finance, offering a clear view of potential financial growth outcomes.

Exponential Functions

Exponential functions are crucial in expressing rapidly changing growth or decay processes seen in continuously compounding interest. The function \( e^x \) features a constant base \( e \), approximately equal to 2.718, symbolizing natural growth rates.
These functions are characterized by a constant ratio of change, meaning each unit of the independent variable shifts the output by a fixed factor.

• Exponential growth in finance utilizes \( e^{kt} \) to describe how investments increase continually over time.
• Unlike linear functions, exponential forms are non-linear, reflecting faster growth.

The beauty of exponential functions in finance lies in their ability to model accumulated interest and growth. This makes them indispensable to financial calculations, allowing predictions and evaluations of long-term financial behavior effectively.