Question

Imagine that today you deposit \(\$ B\) in a savings account that earns interest at a rate of \(p \%\) per year compounded continuously. The goal is to draw an income of \(\$ I\) per year from the account forever. The amount of money that must be deposited is \(B=I \int_{0}^{\infty} e^{-n t} d t,\) where \(r=p / 100 .\) Suppose you find an account that earns \(12 \%\) interest annully and you wish to have an income from the account of \(\$ 5000\) per year. How much must you deposit today?

Short answer

Answer: Approximately $41,666.67 must be deposited today.

Step-by-step solution

Define the variables and find n

According to the problem, we have the annual interest rate given as \(p = 12\%\). We need to find the proportion, \(n\). Using the formula \(n = \frac{p}{100}\), we can find the proportion. \(n = \frac{12}{100} = 0.12\)

Evaluate the integral

Now, we need to evaluate the given integral, \(\int_{0}^{\infty} e^{-nt} dt\): \(\int_{0}^{\infty} e^{-0.12t} dt\) To evaluate this integral, we can use the exponential integral rule, which states that: \(\int e^{at} dt = \frac{1}{a} e^{at} + C\) Here, we have \(a = -0.12\). Now let's evaluate the integral: \(= -\frac{1}{0.12} e^{-0.12t} \Big|_0^\infty\) Now we need to apply the limits: \(= -\frac{1}{0.12} \left[ \lim_{t \to \infty} e^{-0.12t} - e^0 \right]\) Since \(e^{-0.12t}\) goes to 0 as \(t\) approaches infinity, we get: \(= -\frac{1}{0.12} \left[ 0 - 1 \right]\) \(= \frac{1}{0.12}\)

Calculate the initial deposit amount

To find the initial deposit amount that must be deposited, we use the formula \(B = I \int_{0}^{\infty} e^{-nt} dt\). We found the value of the integral and now we just need to multiply it by the desired income, \(I = 5000\). \(B = 5000 \cdot \frac{1}{0.12}\) \(B = \frac{5000}{0.12}\) \(B = 41666.67\) So, you must deposit approximately \(\$41666.67\) today to get a \(\$5,000\) income per year from the account forever.

Key concepts

Exponential Integration

Exponential integration is a fundamental concept in calculus, especially when dealing with continuously compounded interest. It involves the integration of functions in the form \( e^{ax} \). These functions are common in mathematical models related to growth and decay processes.

When evaluating the integral \( \int e^{at} \, dt \), the integral is \( \frac{1}{a} e^{at} + C \), where \( C \) is the constant of integration. For definite integrals, such as in continuous compound interest problems, you apply limits, which eliminate the constant \( C \).

• The term \( a \) represents the rate at which the function is growing or decaying. A positive \( a \) indicates growth, while a negative \( a \) indicates decay.
• Exponential decay, such as \( e^{-nt} \), is crucial in financial calculations where interest is compounded continuously over time.

In our example, we deal with exponential decay because we want to maintain an income forever, meaning we are using an integral from \( t = 0 \) to \( t = \infty \), which characterizes a perpetuity.

Calculus Applications

Calculus plays a pivotal role in solving real-world financial problems, like calculating the principal deposit needed to generate a perpetual income through continuous compounding. In these scenarios, differentiation and integration are employed to model how financial values change over time.
Finance professionals use exponential integrals to compute the present value of future income streams, accounting for continuous compounding of interest. Thanks to calculus, they can assess how changing interest rates or desired income affects the required principal deposit.

• The formula \( B = I \int_{0}^{\infty} e^{-nt} \, dt \) arises from the process of finding the present value of a continuous annuity.
• Differentiation helps in understanding how small changes in variables, like interest rates, affect the outcomes.
• Integration, especially of exponential functions, is essential when dealing with infinite time horizons, as with continuous income streams.

By understanding these concepts, you can confidently approach various financial calculations that involve investment over time, ensuring you're prepared for diverse economic situations.

Infinite Series

Infinite series are sequences of numbers or functions that continue indefinitely. They are representations of sums that can extend to infinity, such as the integral we evaluated. While they might seem abstract, infinite series have practical applications, including in calculus when calculating continuously compounded interest.
In our context, the evaluation \( \int_{0}^{\infty} e^{-nt} \, dt \) is essentially finding the sum of an infinite series of exponential decay terms. Such expressions indicate how functions shrink to a negligible size over an infinite period.

• Infinite series help in understanding the behavior of investments or functions over long (infinite) horizons.
• They simplify seemingly complex calculations by dealing with convergence, which tells us how an infinite series can still sum to a finite value.
• Convergence ensures that, although we are summing an infinite number of terms, the result remains practical and applicable.

By mastering the concept of infinite series, you learn to see the logical framework that underpins the financial calculations we perform, making you adept at interpreting and applying them to real-world conditions.