Problem 12 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Money in a certain trust-fund account is earning \(5 \%\) interest per year compounded continuously. Suppose money is being withdrawn from the account at a constant rate of \(\$ 2000\) per year. For the sake of our model, assume that money is being withdrawn continuously. The account begins with \(\$ 30,000\). Let \(M=M(t)\) be the amount of money in the account at time \(t\), where \(t\) is in years. Write a differential equation modeling the situation. What is the initial condition?

Step-by-Step Solution

Answer

The differential equation modeling the situation is \(dM/dt = 0.05M - 2000\), and the initial condition is \(M(0) = 30000\).

Step 1: Understand the Interest Compounding Factor

Firstly, knowing that the account earns \(5%\) interest per year continuously, we can write it as \(0.05M\), where \(M\) is the amount of money in the account at time \(t\). The \(0.05\) comes from the rate of interest.

Step 2: Understand the Withdrawals

Secondly, the money is being withdrawn from the account at a constant rate of \(\$2000\) per year. This can be written as \(-2000\), because the money is being taken out of the account.

Step 3: Form the Differential Equation

Next, we put together our expressions for the interest and withdrawals to form the desired differential equation. The rate of change of money in the account, \(dM/dt\), is the sum of the interest gained and the money withdrawn. So, \(dM/dt = 0.05M - 2000\). This is the differential equation that models the situation.

Step 4: Determine the Initial Condition

The initial condition is simply the amount of money in the account at time \(t=0\), which is given as \$30000. So \(M(0) = 30000\).

Key Concepts

Continuous Compound Interest

Continuous compound interest is a financial concept where interest is calculated endlessly throughout the time, with the interest amount being instantly reinvested or compounded. Unlike simple interest that is calculated on the initial principal only, or standard compound interest that is calculated on set periods like annually or monthly, continuous compounding computes interest at every possible instant and adds it to the principal.

Mathematically, this type of interest growth is described by an exponential function. The formula that represents continuous compounding is given by the expression \( P = Pe^{rt} \) where\( P \) is the principal amount, \( e \) is the base of the natural logarithms, \( r \) is the interest rate, and \( t \) is time. In the context of our problem, the interest rate is 5%, or 0.05 in decimal form. As money is being withdrawn continuously, the change in the account's balance is affected by both the continuous compounding and the withdrawals.

Exponential Decay Model

The exponential decay model is a mathematical representation of processes where the quantity decreases at a rate proportional to its current value. This model is fundamentally important in various scientific fields, including physics, biology, and economics. In finance, it's often used to model the depreciation of assets or diminishing account balances.

For our trust-fund account scenario, the continuous withdrawals of $2000 per year introduce an exponential decay in the amount of money. If we had no withdrawals, the account would grow exponentially due to continuous compound interest. With the withdrawals, the exponential growth rate gets reduced. The differential equation resulting from this situation has the general form \( \frac{dM}{dt} = rM - W \) where \(\frac{dM}{dt}\) is the rate of change of the money in the account, \( M \) is the amount of money at time \( t \), \( r \) is the continuous interest rate, and \( W \) is the rate of withdrawal. In our case, \( r = 0.05 \) and \( W = 2000 \) representing continuous compound interest and exponential decay respectively.

Initial Value Problem

An initial value problem in the context of differential equations consists of finding a function that satisfies a given differential equation and that also meets specific initial conditions. These problems are pivotal because they allow us to determine a unique solution that models a physical situation accurately at a given starting point.

In our trust-fund account example, we are looking for the function \( M(t) \) that indicates the amount of money at any time \( t \) based on a differential equation that considers both interest and withdrawals. The initial value given, \( M(0) = 30000 \), provides the starting amount in the account when \( t=0 \). This information is crucial as it ensures the solution to our differential equation accurately reflects the real-world situation from the moment we start tracking the account balance. Together, the differential equation and the initial condition allow us to predict the future balance of the account over time.