Problem 10 - Applied Calculus

Question

A bank account earns \(7 \%\) annual interest compounded continuously. You deposit \(\$ 10,000\) in the account, and withdraw money continuously from the account at a rate of \(\$ 1000\) per year. (a) Write a differential equation for the balance, \(B\), in the account after \(t\) years. (b) What is the equilibrium solution to the differential equation? (This is the amount that must be deposited now for the balance to stay the same over the years.) (c) Find the solution to the differential equation. (d) How much is in the account after 5 years? (e) Graph the solution. What happens to the balance in the long run?

Step-by-Step Solution

Answer

The equilibrium solution is \(14285.71\). After 5 years, the balance is approximately \$10,967.18. Long-term, the balance stabilizes at \$14,285.71.

Step 1: Identify Known Values

We know the interest rate is \(r = 7\% = 0.07\) and the initial deposit is \\(10,000. The withdrawal rate is \\)1,000 per year.

Step 2: Formulate the Differential Equation

The differential equation for continuous compounding and constant withdrawal is \( \frac{dB}{dt} = rB - W \), where \( r \) is the interest rate and \( W \) is the withdrawal rate. Substituting the given values, we have \( \frac{dB}{dt} = 0.07B - 1000 \).

Step 3: Find Equilibrium Solution

The equilibrium solution is when the balance does not change over time, i.e., \( \frac{dB}{dt} = 0 \). Solving \( 0.07B - 1000 = 0 \) gives \( B = \frac{1000}{0.07} = 14285.71 \).

Step 4: Solve the Differential Equation

To solve \( \frac{dB}{dt} = 0.07B - 1000 \), use integrating factors or separation of variables. The general solution is \( B(t) = Ce^{0.07t} + 14285.71 \), where \( C \) is a constant determined by initial conditions.

Step 5: Apply Initial Condition

The initial condition is \( B(0) = 10000 \). Substituting into the equation gives \( 10000 = C + 14285.71 \), so \( C = 10000 - 14285.71 = -4285.71 \). Thus, \( B(t) = -4285.71e^{0.07t} + 14285.71 \).

Step 6: Determine Balance After 5 Years

Substitute \( t = 5 \) into the equation: \( B(5) = -4285.71e^{0.35} + 14285.71 \). Calculating this gives \( B(5) \approx 10,967.18 \).

Step 7: Analyze Long-term Behavior and Graph

As \( t \to \infty \), the term \( -4285.71e^{0.07t} \to 0 \) and \( B(t) \) approaches \( 14285.71 \). In the long run, the balance stabilizes at the equilibrium solution. Graph \( B(t) = -4285.71e^{0.07t} + 14285.71 \) to visualize this stabilization.

Key Concepts

Continuous Compounding

Continuous compounding is an essential concept in finance and differential equations. When we talk about compounding continuously, we imply that interest is added to the principal balance at every moment, theoretically infinitely. This differs from traditional compounding methods where interest might be added annually, quarterly, or monthly.
This concept is modeled by exponential growth, and its effect on the bank balance overtime is governed by the differential equation typically expressed as \( \frac{dB}{dt} = rB \), where \( r \) is the interest rate and \( B \) is the balance. In scenarios where continuous withdrawals are also occurring, you would include the withdrawal rate in the equation.
This can reflect realistic circumstances, such as the given exercise of a bank account with constant withdrawals, leading us to the modified differential equation \( \frac{dB}{dt} = rB - W \). This allows us to analyze how both interest applied over every instant affects the account balance amid consistent outflows.

Equilibrium Solution

The equilibrium solution is a fundamental concept in differential equations, representing a state where the system's output does not change. This occurs when the rate of change is zero. In terms of a bank account, it reflects a balance level where the interest earned exactly offsets the amount withdrawn.
For equilibrium, we set \( \frac{dB}{dt} = 0 \) in the differential equation. For the example of continuous compounding with withdrawals, solving \( 0.07B - 1000 = 0 \) yields the equilibrium balance \( B = 14285.71 \).
This tells you that if you start with this balance, the withdrawals and interest will cancel each other out, keeping the balance stable. It's quite useful for financial planning, allowing one to determine how much needs to be in the account so that withdrawals do not reduce the overall balance over time.

Initial Conditions

Initial conditions are critical when solving differential equations, as they allow you to find a particular solution that fits a specific scenario. The initial condition in our problem was the initial deposit of \( \$10,000 \).
After deriving a general solution from the differential equation, initial conditions help assimilate points of the problem unique to the specific context. For our exercise, starting with the equilibrium solution \( B(t) = Ce^{0.07t} + 14285.71 \), where \( C \) is a constant dependent on initial conditions, the initial condition \( B(0) = 10000 \) allowed us to solve for \( C \).
This involved substituting \( t = 0 \) and solving, resulting in a specific function for \( B(t) \) that accurately predicts how the balance behaves over time given the initial state.

Long-term Behavior

Understanding the long-term behavior of an account with continuous compounding and withdrawals is important for financial planning. As time progresses, the influence of initial conditions diminishes, and certain terms in the differential equation may become negligible.
In our exercise, the function \( B(t) = -4285.71e^{0.07t} + 14285.71 \) describes the balance over time. With \( e^{0.07t} \) approaching zero as \( t \to \infty \), the solution converges to the equilibrium solution \( B = 14285.71 \).
This implies that regardless of the initial deposit, the eventual balance will stabilize, assuming that the withdrawal rate and interest rate remain constant. This has practical significance, suggesting how much balance one can expect to maintain in the long run.

Interest Rate

The interest rate is a pivotal factor affecting how quickly an account grows when compounded continuously. In differential equations, the interest rate, denoted as \( r \), is a coefficient in the equation \( \frac{dB}{dt} = rB - W \).
In financial computations, such as our example, the interest rate was \( 7\% \). In a continuously compounded scenario, even a small change in \( r \) can have significant effects on the long-term growth of an account due to its exponential nature.
Anyone analyzing investment or savings scenarios must grasp how changes in interest rate alter the time it takes for balances to reach certain values, as well as their role in determining equilibrium solutions and their effect on long-term behaviors. Understanding this helps in making informed decisions regarding deposits, withdrawals, and overall financial planning.