Problem 13 - Applied Calculus

Question

A bank account earns \(5 \%\) annual interest, compounded continuously. Money is deposited in a continuous cash flow at a rate of \(\$ 1200\) per year into the account. (a) Write a differential equation that describes the rate at which the balance \(B=f(t)\) is changing. (b) Solve the differential equation given an initial balance \(B_{0}=0\). (c) Find the balance after 5 years.

Step-by-Step Solution

Answer

The balance after 5 years is approximately \(\$6818.79\).

Step 1: Understanding the Problem

We are modeling a bank account's changing balance with both continuous interest and continuous deposits. We need to express this as a differential equation and solve it with the given initial condition.

Step 2: Formulating the Differential Equation

To describe the rate of change of the balance, note that continuously compounded interest adds a term \(0.05B\) to the growth rate. Additionally, a continuous deposit rate of \(1200\) adds another term. Thus, the differential equation for the balance \(B\) is \(\frac{dB}{dt} = 0.05B + 1200\).

Step 3: Solving the Differential Equation

The differential equation \(\frac{dB}{dt} = 0.05B + 1200\) is a first-order linear ordinary differential equation. Its general solution has the form \(B(t) = Ce^{0.05t} + k\), where \(k\) is a particular solution. We determine \(k\) by considering the steady-state solution and integrating the deposit: \(k = 1200/0.05 = 24000\).

Step 4: Applying Initial Conditions

Given the initial balance \(B(0)=0\), we can substitute into the general solution \(0 = Ce^{0.05 \cdot 0} + 24000\) to find \(C = -24000\). Thus, the solution for \(B(t)\) is \(B(t) = -24000e^{0.05t} + 24000\).

Step 5: Finding the Balance After 5 Years

Substitute \(t=5\) into our equation for \(B(t)\): \(B(5) = -24000e^{0.05 \cdot 5} + 24000\). Calculate \(B(5)\) to find the balance after 5 years: \(B(5) = -24000e^{0.25} + 24000 \approx 24000(1-e^{0.25}) \approx 24000(1 - 1.284) \approx 6818.79\).

Step 6: Final Result Interpretation

Our calculated balance after 5 years includes only continuous deposits and interest. The process confirms that, even with a zero initial balance, regular contributions and interest grow the balance significantly over time.

Key Concepts

Continuously Compounded Interest

Continuously compounded interest is a method of calculating interest where the interest amount grows exponentially because it is calculated and added to the account balance at every instant. Most commonly, this concept is illustrated with formulas involving the exponential function.
To understand this better, imagine your balance earning a small amount of interest not just daily or monthly, but continuously, every microsecond of every day. Mathematically, if you have a principal amount, say $100, and an annual interest rate, say 5%, the continuously compounded value would be calculated using the formula:

• \[ A = Pe^{rt} \]

- \( A \) represents the future value of the investment/loan, including interest.- \( P \) is the principal investment amount (initial deposit or loan amount).- \( r \) is the annual interest rate (decimal).- \( t \) is the time, in years, the money is invested or borrowed for.- \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
In the context of our exercise, the continuously compounded interest adds a term \(0.05B\) to the rate of change of the account balance \(B\), leading to an exponential growth in the balance over time.

Continuous Cash Flow

Continuous cash flow refers to a scenario where money is being added (or withdrawn) from an account not in lump sums or regular intervals, but constantly, over time. This steady flow is typically measured as a rate, such as dollars per year.
In our exercise, the account experiences a continuous deposit of $1200 per year. This continuous addition contributes to increasing the account balance constantly rather than in discrete intervals as in regular deposits.

• Think of it like a faucet leaking water into a bucket – the water level (or cash balance) increases gradually over continuous time.
• Mathematically, in our differential equation, the continuous cash flow is represented by the constant addition term \(+1200\) added directly into the account balance's rate of change.

This concept contrasts with discrete cash flows, where contributions occur at specific times. Understanding continuous cash flow is essential for solving the differential equation representing the change in an account balance.

Ordinary Differential Equation

An ordinary differential equation (ODE) is an equation involving a function and its derivatives. It expresses a relationship between varying quantities and how they change over time.
In our situation, the differential equation that describes the balance of the account is:

• \[ \frac{dB}{dt} = 0.05B + 1200 \]

This equation is a first-order linear ODE, where:- \( \frac{dB}{dt} \) is the rate of change of the balance.- The term \(0.05B\) arises from the continuously compounded interest.- The term \(1200\) represents the continuous cash flow deposit.
To solve the ODE, you calculate the general solution considering both the interest and deposit term effects over time. The solution of this ODE helps understand how the balance changes over time given both the continuous deposit and interest.

Initial Condition

An initial condition in the context of differential equations specifies the value of the function at a particular time, often at the beginning (\(t=0\)). It helps in finding a particular solution rather than just a family of solutions.
For example, knowing that initially, \(B(0)=0\) allows us to uniquely determine the integration constant when solving our ordinary differential equation.

• It ensures that our solution precisely reflects the specifics of the problem setting we are working with.
• With the initial condition \(B(0) = 0\), the particular solution to the ODE becomes \[ B(t) = -24000e^{0.05t} + 24000 \]
• This means after setting \(t=0\), the balance is zero, aligning with the initial state before interest and deposits start accruing.

Initial conditions are essential in modeling real-world problems accurately, ensuring that the solution behaves correctly at the starting point.