Chapter 5: Problem 9
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. $$ R(t)=\$ 50,000, \quad T=22 \mathrm{yr}, \quad k=5 \% $$
Short Answer
Expert verified
The accumulated future value is approximately $2,007,606.40.
Step by step solution
01
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.
02
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: \( A = \int_0^{22} 50,000 \cdot e^{0.05(22-t)} \, dt \).
03
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 \( A = 50,000 \cdot \int_0^{22} e^{1.1} \cdot e^{-0.05t} \, dt \).
04
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 \( -\frac{1}{0.05} e^{-0.05t} + C \).
05
Evaluating the Definite Integral
Evaluate the integral from 0 to 22: \( \left[-20 e^{-0.05t}\right]_0^{22} = -20(e^{-1.1} - e^{{0}}) = -20(e^{-1.1} - 1) \).
06
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 \( e^{-1.1} \approx 0.333 \).
07
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
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:
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.
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:
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.
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.
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.