Problem 7 - Applied Calculus

Question

Money in a bank account grows continuously at an annual rate of \(r\) (when the interest rate is \(5 \%, r=0.05\), and so on). Suppose \(\$ 1000\) is put into the account in 2000 . (a) Write a differential equation satisfied by \(M\), the amount of money in the account at time \(t\), measured in years since 2000 . (b) Solve the differential equation. (c) Sketch the solution until the year 2030 for interest rates of \(5 \%\) and \(10 \%\).

Step-by-Step Solution

Answer

(a) \( \frac{dM}{dt} = rM \), (b) \( M(t) = 1000 e^{rt} \), (c) Plot from 2000 to 2030 for 5% and 10% rates.

Step 1: Formulate the Differential Equation

Since the money in the bank account grows continuously, we can describe its growth using the differential equation \( \frac{dM}{dt} = rM \), where \( M \) is the amount of money at time \( t \), and \( r \) is the continuous growth rate. If the interest rate is specified as 5\%, \( r = 0.05 \).

Step 2: Solve the Differential Equation

The differential equation \( \frac{dM}{dt} = rM \) is separable. We can rewrite it as \( \frac{1}{M} \frac{dM}{dt} = r \). Integrating both sides with respect to \( t \), we get \( \int \frac{1}{M} dM = \int r \, dt \), which results in \( \ln|M| = rt + C \), where \( C \) is an integration constant. Solving for \( M \), we get \( M(t) = e^{C}e^{rt} \). We simplify this further as \( M(t) = M_0 e^{rt} \), where \( M_0 = e^C \) is the initial amount of money, given to be \( 1000 \). Thus, \( M(t) = 1000 e^{rt} \).

Step 3: Sketch the Solution for Different Interest Rates

Given the function \( M(t) = 1000 e^{rt} \), we will sketch it from 2000 (\( t = 0 \)) to 2030 (\( t = 30 \)) for two cases: when \( r = 0.05 \) and when \( r = 0.10 \). For \( r = 0.05 \), calculate \( M(30) = 1000 e^{0.05 \times 30} \). Similarly, for \( r = 0.10 \), calculate \( M(30) = 1000 e^{0.10 \times 30} \). These calculations show the growth in the account for both interest rates by 2030.

Key Concepts

Differential Equation

A differential equation is essentially an equation that involves functions and their derivatives. In simpler terms, it describes how a certain quantity changes. In the context of our exercise, the quantity is the money in the bank account, denoted as \(M\), which changes or grows over time.

The growth can be captured by the differential equation \( \frac{dM}{dt} = rM \). Here, \( \frac{dM}{dt} \) represents the rate of change of money with respect to time \(t\). The variable \(r\) stands for the continuous annual growth rate, such as \(0.05\) for a 5% interest rate.

This form of differential equation is characteristic of continuous compound interest scenarios, forming the foundation to predict how investments increase over time. Understanding it involves knowing that the change in money is directly proportional to the current amount due to the constant rate \(r\). This proportionality is what makes the equation exponential in nature, leading us to our next concept: separable differential equations.

Separable Differential Equation

Separable differential equations are a specific type of differential equations where we can separate the variables involved, allowing each side of the equation to focus on different variables. This makes them much easier to solve by using integration.

In our exercise, we start with \( \frac{dM}{dt} = rM \). To separate the differentials, we rewrite it as \( \frac{1}{M} \frac{dM}{dt} = r \). This way, all the terms involving \(M\) are on one side and all terms involving \(t\) are on the other.

The next step is integration. Integrating the left side with respect to \(M\) provides \( \int \frac{1}{M} \, dM = \ln|M| \). Integrating the right side with respect to \(t\) results in \( \int r \, dt = rt + C \), \(C\) being the constant of integration.

These steps bring us to \( \ln|M| = rt + C \). Simplifying, we express the solution as \( M(t) = M_0 e^{rt} \), where \(M_0 = e^C\) symbolizes the initial amount. Solving separable differential equations like this one is straightforward, making them a vital tool in dealing with exponential growth problems.

Exponential Growth

Exponential growth happens when a quantity increases at a rate proportional to its current value. This type of growth is named exponential because the solution to the growth differential equation, \( \frac{dM}{dt} = rM \), leads to an exponential function.

In the context of our exercise, the function \( M(t) = M_0 e^{rt} \) represents the amount of money in the account at any given time \(t\). "Exponential" refers to the function \( e^{rt} \). This formula essentially says that the money grows exponentially over time, providing a way to calculate the future value of an investment based on its current value.

In practical terms, continuous compound interest utilizes exponential growth, as shown when calculating the money amount in future years, like 2030. This growth can be quickly computed using \(e^{rt}\), making understanding exponential functions critical for financial planning and predicting investment returns. For instance, over 30 years, with an interest rate of 5% or 10%, the initial deposit greatly increases, illustrating the significant impact of compounding.