Problem 6 - 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 constant rate of 1000 dollars per year. (a) Write a differential equation that describes the rate of change of the balance \(B=f(t)\). (b) Solve the differential equation to find the balance as a function of time.

Step-by-Step Solution

Answer

The differential equation is \( \frac{dB}{dt} = 0.1B + 1000 \) and the balance is \( B(t) = 10000(e^{0.1t} - 1) \).

Step 1: Identify the type of differential equation

The differential equation will involve the rate of change of the balance in the account. Since the interest is compounded continuously, we use the formula for continuously compounding interest, and additionally, a constant deposit rate is involved.

Step 2: Formulate the differential equation

Let the balance of the account at time \( t \) be \( B(t) \). The rate of change of \( B(t) \) is given by the sum of the interest earned continuously, which is \( 0.1B(t) \), and the constant deposit rate, 1000. Thus, the differential equation becomes: \[ \frac{dB}{dt} = 0.1B + 1000. \]

Step 3: Solve the differential equation

This is a first-order linear differential equation of the form \( \frac{dB}{dt} - 0.1B = 1000 \). Use the integrating factor method to solve it. The integrating factor is \( e^{-0.1t} \). Multiply through by the integrating factor: \[ e^{-0.1t}\frac{dB}{dt} - 0.1e^{-0.1t}B = 1000e^{-0.1t}. \]

Step 4: Simplify and integrate

The left side of the equation is the derivative of \( e^{-0.1t}B \). Integrate both sides with respect to \( t \): \[ \int \frac{d}{dt}(e^{-0.1t}B) \, dt = \int 1000e^{-0.1t} \, dt. \] Thus, \[ e^{-0.1t}B = -10000e^{-0.1t} + C. \]

Step 5: Solve for B(t)

Isolate \( B(t) \) by multiplying through by \( e^{0.1t} \): \[ B(t) = -10000 + Ce^{0.1t}. \] Since the initial balance is zero \( B(0) = 0 \), substitute \( t = 0 \) into the equation: \[ 0 = -10000 + C. \] Therefore, \( C = 10000 \).

Step 6: Final expression for B(t)

Substitute \( C = 10000 \) back into the expression for \( B(t) \): \[ B(t) = -10000 + 10000e^{0.1t}. \] Simplify to: \[ B(t) = 10000(e^{0.1t} - 1). \]

Key Concepts

Continuous Compounding Interest

Continuous compounding interest is a fascinating concept in finance. It assumes that interest is calculated and added to the principal balance at every moment, continuously.
This method results in a little more interest accrued than with traditional compounding methods.

In continuous compounding, the growth of your investment is exponential. The mathematical model for continuous compounding is inspired by natural exponential growth and can be expressed by the formula:

• \( A = Pe^{rt} \)

where:

• \( A \) is the final amount
• \( P \) is the principal balance
• \( r \) is the annual interest rate
• \( t \) is the time in years.

By using this formula, you can see how investments grow over time when interest is added continuously.

First-order Linear Differential Equation

A first-order linear differential equation is a type of equation that involves the derivatives of a function with respect to one variable, like time. In our context, it represents how the balance changes in a bank account.
The general form is:

• \( \frac{dy}{dt} + Py = Q \)

where:

• \( y \) is the unknown function
• \( P \) and \( Q \) are functions of \( t \)

These equations model many real-world situations, often pertaining to rates and growth processes.

In the exercise, the equation for the bank balance was:

• \( \frac{dB}{dt} = 0.1B + 1000 \)

This equation incorporates continuous compounding interest and constant deposits.

Integrating Factor Method

The integrating factor method is a clever technique used to solve first-order linear differential equations. It's invaluable when the differential equation isn't readily separable.
The core idea is to multiply the entire differential equation by an integrating factor, which greatly simplifies the equation.

For our equation \( \frac{dB}{dt} - 0.1B = 1000 \), the integrating factor is \( e^{-0.1t} \).

• When the equation is multiplied by this factor, the left-hand side becomes the derivative of a product.

This clever manipulation allows the equation to be integrated straightforwardly.

After integration, constants are adjusted based on initial conditions, such as the initial balance of zero, to find the specific solution.

Rate of Change

The rate of change is a fundamental concept in calculus and describes how a quantity changes over time. In a differential equation, it is represented by a derivative, like \( \frac{dB}{dt} \) in our example.
This represents how quickly or slowly the bank balance \( B \) changes with respect to time \( t \).

In the given problem, the rate of the balance change is affected by two factors:

• Interest accruing continuously due to the 10% rate
• The constant deposit of 1000 dollars per year.

Together, these create a dynamic interplay that causes the balance to grow exponentially over time, showcasing the beauty and power of differential equations in modeling real-world financial scenarios.