Problem 9 - Applied Calculus

Question

A bank account that earns \(10 \%\) interest compounded continuously has an initial balance of zero. Money is deposited into the account at a continuous rate of \(\$ 1000\) per year. (a) Write a differential equation that describes the rate of change of the balance \(B=f(t)\). (b) Solve the differential equation.

Step-by-Step Solution

Answer

The balance differential equation is \( \frac{dB}{dt} = 0.1B + 1000 \). Solving gives \( B(t) = 10000e^{0.1t} - 10000 \).

Step 1: Understanding the Problem

The problem involves modeling a bank account with continuous compounding interest and continuous deposits. We'll describe the rate of change of the balance using a differential equation.

Step 2: Formulate the Differential Equation

Let \( B(t) \) be the balance at time \( t \), with interest rate \( r = 0.1 \). Money is deposited at a rate of \( 1000 \) per year. The differential equation describing the rate of change in balance is given by: \[ \frac{dB}{dt} = rB(t) + 1000 = 0.1B(t) + 1000. \]

Step 3: Find the General Solution

To solve \( \frac{dB}{dt} = 0.1B + 1000 \), first solve the homogeneous equation \( \frac{dB}{dt} = 0.1B \), which gives \( B_h(t) = Ce^{0.1t} \). For the particular solution, assume \( B_p(t) = K \), so \( 0 = 0.1K + 1000 \). Solving yields \( K = -10000 \). Thus, the general solution is: \[ B(t) = Ce^{0.1t} - 10000. \]

Step 4: Apply Initial Conditions

Since the initial balance is zero, set \( B(0) = 0 \). This provides \( 0 = C - 10000 \), resulting in \( C = 10000 \). Thus the specific solution becomes: \( B(t) = 10000e^{0.1t} - 10000. \)

Key Concepts

Continuous Compounding

Continuous compounding is a fascinating concept in finance. It represents how often interest is calculated on a principal sum. In this case, the interest is compounded continuously, meaning that it is calculated and added back to the balance at every possible moment.
When interest compounds continuously, the process is mathematically described using exponential growth.
The formula for continuous compounding is:

• \( A = Pe^{rt} \)

where:

• \( A \) is the amount of money accumulated after time \( t \).
• \( P \) is the principal amount (initial balance).
• \( r \) is the interest rate (as a decimal).
• \( t \) is the time the money is invested for.

Interest Rate

The interest rate in this problem is given as 10%, which is equivalent to 0.1 when expressed as a decimal. This rate is one of the critical factors determining how quickly the balance in the account grows.
In a differential equation modeling continuous compounding, the interest rate \( r \) influences the exponential growth factor.
It is crucial to convert percentage figures into decimals when using them in mathematical equations. Thus, a 10% interest rate becomes:

• \( r = 0.1 \)

This decimal representation allows the differential equation to accurately depict how the balance changes over time due to interest.

Initial Balance

In the exercise, the initial balance of the bank account is stated to be zero. Despite starting with zero, the account can grow due to continuous deposits and interest compounding.
The initial balance is a vital initial condition in solving differential equations. It allows us to find the constant in the general solution.
When the initial balance, \( B(0) \), is used, it helps tailor the general solution to fit the specific scenario. For example, initially:

• \( B(0) = 0 \)

Applying this to the solution gives a complete solution that reflects the growth from a zero starting point.

General Solution

A general solution of a differential equation incorporates an arbitrary constant that accounts for various initial conditions. The given general solution for the account balance is:

• \[ B(t) = Ce^{0.1t} - 10000 \]

Where \( C \) is an arbitrary constant determined by the initial condition.
In our specific problem, substituting the initial balance provides:

• \( B(0) = 0 \) yields \( C = 10000 \)

Thus, the specific solution becomes:

• \[ B(t) = 10000e^{0.1t} - 10000 \]

This solution demonstrates how continuous deposits and compounding grow the balance over time.