Problem 11 - Applied Calculus

Question

Money in an account earns interest at a continuous rate of 8\% per year, and payments are made continuously out of the account at the rate of $$\$ 5000\( a year. The account initially contains $$\$ 50,000 .\) Write a differential equation for the amount of money in the account, \(B\), in \(t\) years. Solve the differential equation. Does the account ever run out of money? If so, when?

Step-by-Step Solution

Answer

The account runs out of money in about 8.95 years.

Step 1: Understand the Problem

The problem involves a continuously compounding interest at 8% and a continuous withdrawal rate of $5000 per year. We need to write a differential equation for the balance in the account over time and solve it.

Step 2: Formulate the Differential Equation

The differential equation will account for the rate of change in balance due to interest and withdrawals. Given the balance is compounded continuously, the interest part is \(0.08B\). Since money is withdrawn at $5000/year, the equation becomes \( \frac{dB}{dt} = 0.08B - 5000 \).

Step 3: Solve the Differential Equation

To solve \( \frac{dB}{dt} = 0.08B - 5000 \), rewrite it as \( \frac{dB}{dt} - 0.08B = -5000 \). This is a first-order linear differential equation. The integrating factor is \( e^{0.08t} \). Multiply the entire equation by this factor: \( e^{0.08t} \frac{dB}{dt} - 0.08e^{0.08t}B = -5000e^{0.08t} \).

Step 4: Integrate the Equation

The left side becomes \( \frac{d}{dt}(e^{0.08t}B) \). Integrate both sides: \( \int \frac{d}{dt}(e^{0.08t}B) \, dt = \int -5000e^{0.08t} \, dt \). The left side gives \( e^{0.08t}B \) after integration, and the right side integrates to \( -\frac{5000}{0.08}e^{0.08t} + C \).

Step 5: Solve for the Constant

Given \( e^{0.08t}B = -\frac{5000}{0.08}e^{0.08t} + C \), solve for \( B \): \( B(t) = -\frac{5000}{0.08} + Ce^{-0.08t} \). Use the initial condition \( B(0) = 50000 \) to find \( C \). Set \( e^{0.08 \cdot 0}B(0) = -\frac{5000}{0.08} + C \), solving \( 50000 = -62500 + C \), giving \( C = 112500 \).

Step 6: Write the Particular Solution

Substitute \( C \) back into \( B(t) \) to get the particular solution: \( B(t) = -62500 + 112500e^{-0.08t} \).

Step 7: Determine if the Account Runs Out of Money

Set \( B(t) = 0 \) and solve for \( t \): \( 0 = -62500 + 112500e^{-0.08t} \). Solving gives \( e^{-0.08t} = \frac{62500}{112500} \). Take the natural log: \( -0.08t = \ln \left( \frac{625}{1125} \right) \), solving gives \( t \approx 8.95 \) years.

Key Concepts

Continuous Interest

Continuous interest refers to the compounding of interest infinitely many times each year. In practical terms, it means that the interest is being added back to the principal balance at every possible instant of time. This is the purest form of interest compounding.
When the interest is continuous, the formula involves the exponential function, and can be expressed as \(A = P \, e^{rt}\), where \(A\) is the amount of money accumulated after time \(t\), \(P\) is the principal amount, \(e\) is the base of the natural logarithm, approximately equal to 2.71828, and \(r\) is the rate of interest per period.

• The key advantage of continuous compounding is that it maximizes the earned interest over time.
• Continuous interest is a critical concept in differential equations when dealing with financial models.

In the context of the problem, understanding continuous interest enables us to properly form the differential equation \( \frac{dB}{dt} = 0.08B - 5000 \), which governs the change in balance in the account.

Initial Value Problems

An initial value problem is a type of differential equation together with a specification of the value of the unknown function at a given point in time, often called the initial condition.
Initial value problems are crucial because they define the particular solution to a differential equation that meets certain starting conditions. This makes them vital for applications where the history or past states need to be factored into predictions or calculations.

• The general form is \( y(t_0) = y_0 \), where \( t_0 \) is the initial time and \( y_0 \) is the initial value.
• Solving these types of problems involves determining a specific, rather than a general, function.

For example, in our exercise, the initial condition given is \( B(0) = 50000 \). This tells us the account started with \$ 50,000, which helps find the constant \( C \) in the particular solution to the differential equation.

Compound Interest

Compound interest is the addition of interest to the principal sum of a deposit. It can be compounded annually, semi-annually, quarterly, monthly, daily, or even continuously, as seen earlier.
The formula for compound interest, not continuous, is generally expressed as \(A = P(1 + \frac{r}{n})^{nt}\), where:

• \( A \) is the future value of the investment/loan, including interest,
• \( P \) is the principal investment amount,
• \( r \) is the annual interest rate (decimal),
• \( n \) is the number of times interest is compounded per year,
• \( t \) is the number of years.

With continuous compounding, compound interest becomes particularly powerful because the more frequently interest is applied, the greater the returns.
The transition from compound to continuous interest involves taking the limit as \( n \) approaches infinity, where \( A = P \, e^{rt} \) is derived.

Financial Mathematics

Financial mathematics is a field focused on applying mathematical techniques and theories to solve problems in finance. It involves understanding the principles of compound interest, discounts, loans, and investments.
Differential equations are used extensively in financial mathematics to model scenarios involving complex financial instruments and transactions, including options pricing and the calculation of interest.

• Understanding the math behind financial processes helps in better investment planning and risk management.
• Financial mathematics includes continuous and discrete models for predicting future values based on present conditions and trends.
• It bridges the gap between real-world financial problems and mathematical solutions.

In this exercise, financial mathematics is at play to determine when the account balance reaches zero, which requires solving the differential equation for \(B(t)\) and understanding the implications of continuous withdrawals and interest on the future balance.