Problem 8 - 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 \(\$ 2000\) is put into the account in 2010 . (a) Write a differential equation satisfied by \(M,\) the amount of money in the account at time \(t,\) measured in years since 2010. (b) Solve the differential equation. (c) Sketch the solution until the year 2040 for interest rates of \(5 \%\) and \(10 \%\).

Step-by-Step Solution

Answer

The money grows according to the equation \( M(t) = 2000 e^{rt} \), with different growth rates for each interest rate.

Step 1: Set up the Differential Equation

The problem says that the money grows continuously at an annual rate of \( r \). For continuous growth, the differential equation can be expressed as \( \frac{dM}{dt} = rM \), where \( M(t) \) is the amount of money at time \( t \) and \( r \) is the growth rate.

Step 2: Solve the Differential Equation

The differential equation \( \frac{dM}{dt} = rM \) is separable. It can be rewritten as \( \frac{1}{M} \frac{dM}{dt} = r \). Integrating both sides with respect to \( t \) gives: \( \int \frac{1}{M} dM = \int r \, dt \), resulting in \( \ln(M) = rt + C \), where \( C \) is the integration constant.

Step 3: Express Solution in Terms of M

To isolate \( M \), exponentiate both sides of the equation \( \ln(M) = rt + C \). This results in \( M(t) = e^{rt + C} = e^C \cdot e^{rt} \). Let \( M_0 = e^C \), which is the initial amount of money. Therefore, \( M(t) = M_0 e^{rt} \).

Step 4: Determine Initial Condition

According to the problem, the initial amount of money in 2010 is \( \$2000 \), so \( M_0 = 2000 \). Thus, the solution becomes \( M(t) = 2000 e^{rt} \).

Step 5: Solve for Specific Interest Rates

For a \( 5\% \) interest rate, \( r = 0.05 \), so the amount is \( M(t) = 2000 e^{0.05t} \). For a \( 10\% \) interest rate, \( r = 0.10 \), so the amount is \( M(t) = 2000 e^{0.10t} \).

Step 6: Sketch the Solution

Now, sketch the functions \( M(t) = 2000 e^{0.05t} \) and \( M(t) = 2000 e^{0.10t} \) from 2010 (\( t = 0 \)) to 2040 (\( t = 30 \)). These are exponential growth graphs, with the \( 10\% \) interest rate showing a steeper curve.

Key Concepts

Exponential Growth

Exponential growth is a process where the quantity increases at a rate proportional to its current value. It is a common occurrence in nature, economics, and population dynamics. In the context of money placed in a bank account, exponential growth is indicative of how wealth compounds over time. When the differential equation \(\frac{dM}{dt} = rM\) is applied, it mathematically represents how the rate of change of money \(M\) is proportional to the current amount present, with \(r\) being the constant growth rate.
This relationship means that the more money you have, the faster it grows. The process portrays a J-shaped curve when graphed over time, with the growth becoming more rapid as time passes.

• Start with understanding the base principle of proportionality in growth.
• Use the differential equation to see how this growth is modeled mathematically.
• Recognize that exponential growth is powerful but ideally happens only under conditions of constant rates without external influences.

Continuous Compounding

Continuous compounding refers to the mathematical limit that compound interest can reach when it is calculated and reinvested into an account constantly, in an ongoing manner. Instead of compounding bi-annually or monthly, continuous compounding assumes that the compounding happens all the time without interruption.
In the formula \(M(t) = M_0 e^{rt}\), \(e\) (approximately 2.718) is the base of the natural logarithm, and it arises in scenarios approximating continuous growth.

• Understand that with continuous compounding, interest accumulates instantly, adding to the principal and further earning interest.
• This form of compounding illustrates the potency of exponential formulas in financial growth situations.

When you think about continuous compounding, imagine the bank recalculating interest after every millisecond—resulting in maximum growth potential.

Initial Value Problem

An initial value problem in differential equations specifies the value of the function at a certain point. It serves as an anchor to solve the differential equation uniquely. In our scenario, the initial value problem involves starting with an initial amount \(M_0 = 2000\) in a bank account from the year 2010.
The initial value lets us solve for constants in the solution of the differential equation. After setting up the model of growth \(M(t) = M_0 e^{rt}\), you plug the initial value into the equation, ensuring the model accurately describes the system from the start.

• Initial value acts as the baseline of your financial prediction.
• It is crucial for solving equations that predict future behavior based on initial conditions.

Initial conditions are indispensable in precisely decoding how systems evolve over time.

Interest Rates

Interest rates are crucial in determining the rate of growth for an investment or principal sum over time. Different rates drastically alter how fast or slow the money grows under the exponential growth model.
In the exercise, we saw how a change from a 5% interest rate to a 10% rate results in a much steeper curve on an exponential graph. This change highlights the importance of even small percentage differences when looking at long-term growth.

• Interest rates determine the \( r \) in the exponential formula \(M(t) = M_0 e^{rt}\).
• Higher interest rates, although often risky, lead to faster growth of capital.
• Always consider the compounding frequency in tandem with interest rates to get a complete picture of growth potential.

Understanding how interest rates influence growth will help in making informed financial decisions and planning.