Problem 31 - Calculus

Question

(a) Let \(B\) be the balance at time \(t\) of a bank account that earns interest at a rate of \(r \%,\) compounded continuously. What is the differential equation describing the rate at which the balance changes? What is the constant of proportionality, in terms of \(r ?\) (b) Find the equilibrium solution to the differential equation. Determine whether the equilibrium is stable or unstable and explain what this means about the bank account. (c) What is the solution to this differential equation? (d) Sketch the graph of \(B\) as function of \(t\) for an account that starts with \(\$ 1000\) and earns interest at the following rates: (i) \(4 \%\) (ii) \(10 \%\) (iii) \(15 \%\)

Step-by-Step Solution

Answer

(a) \( \frac{dB}{dt} = \frac{r}{100}B \), constant is \( r/100 \). (b) Equilibrium: \( B = 0 \), stable. (c) Solution: \( B = Ae^{\frac{r}{100}t} \). (d) Graphs are exponential growth curves starting at \$1000.

Step 1: Understanding the Differential Equation

The balance changes due to continuously compounded interest. The rate is proportional to the current balance, so we express this as a differential equation: \( \frac{dB}{dt} = rB \). Here, \( r \) should be expressed as a fraction (i.e., \( r \% = \frac{r}{100} \)).

Step 2: Constant of Proportionality

The constant of proportionality in the differential equation \( \frac{dB}{dt} = rB \) is the rate \( r \) expressed as a decimal format (\( r/100 \)). Therefore, the equation becomes: \( \frac{dB}{dt} = \frac{r}{100}B \).

Step 3: Equilibrium Solution Analysis

The equilibrium solution occurs where \( \frac{dB}{dt} = 0 \). Since \( \frac{dB}{dt} = \frac{r}{100}B \), the equilibrium occurs at \( B = 0 \). This is a stable equilibrium as any nonzero initial balance will grow over time, moving away from zero.

Step 4: Solution to the Differential Equation

To solve \( \frac{dB}{dt} = \frac{r}{100}B \), we separate variables: \( \frac{1}{B}dB = \frac{r}{100}dt \). Integrating both sides gives: \( \ln|B| = \frac{r}{100}t + C \). Exponentiating both sides: \( B = e^{C}e^{\frac{r}{100}t} \). Let \( C = \ln(A) \), so \( B = Ae^{\frac{r}{100}t} \), where \( A \) is the initial balance.

Step 5: Graph Sketching

For an initial balance of \( \$1000 \), calculate \( B(t) = 1000e^{\frac{r}{100}t} \): 1. For \( r = 4\% \): \( B(t) = 1000e^{0.04t} \).2. For \( r = 10\% \): \( B(t) = 1000e^{0.1t} \).3. For \( r = 15\% \): \( B(t) = 1000e^{0.15t} \).These are exponential growth functions starting at 1000, with steeper curves for higher \( r \) values.

Key Concepts

Exponential Growth

Exponential growth is a fundamental concept in understanding how quantities increase rapidly over time. In the context of a differential equation, exponential growth refers to a process where the rate of change of a variable is proportional to its current value.
If you imagine the balance in a bank account, this is exactly what happens. As the amount increases, the rate of increase accelerates, leading to fast growth.

Here are some key points about exponential growth:

• The form of the solution is typically expressed as \( B(t) = Ae^{kt} \), where \( A \) is the initial amount and \( k \) is the growth rate.
• This type of growth is non-linear, which means each day, week, or year the balance grows more rapidly than the previous time period.
• Exponential growth can be visualized as a curve that gets steeper over time, indicating rapid acceleration.

Understanding exponential growth helps us predict how quickly phenomena such as finances or biological populations can increase if left unchecked.

Continuously Compounded Interest

Continuously compounded interest is a concept that describes how interest is not just applied periodically (like annually or monthly) but continuously. This means the balance is always growing, minute by minute, second by second.
This can be described using a differential equation \( \frac{dB}{dt} = \frac{r}{100}B \), where the change in balance \( B \) over time \( t \) is proportional to the current balance.

Key features of continuously compounded interest include:

• The formula for calculating this interest is \( B(t) = Ae^{\frac{r}{100}t} \), where \( A \) is the initial principal, \( r \) is the rate as a percentage, and \( e \) is the base of the natural logarithm (~2.718).
• This approach maximizes the interest accrued, because every tiny increase in balance immediately contributes to future interest calculations.
• It's often used in financial models where precision is key, such as banking and investment.

Understanding continuously compounded interest is crucial for making informed financial decisions, particularly for long-term savings and investments.

Equilibrium Solutions

An equilibrium solution in differential equations refers to a steady state where the rate of change is zero. For a bank account with continuously compounded interest, we consider where \( \frac{dB}{dt} = 0 \).
In our example, the equilibrium occurs when the balance \( B = 0 \). This means, theoretically, there is no change when there is no balance, which seems trivial but provides insights into system stability.

Important aspects of equilibrium solutions are:

• An equilibrium can be classified into stable or unstable. In our example, since any positive balance will grow, this points to a stable equilibrium away from zero.
• Stability indicates whether small changes will return to equilibrium or move away from it.
• This concept is important for analyzing systems that self-correct or perpetuate growth, such as biological systems or economies.

By understanding equilibrium solutions, one can predict how systems respond to changes and determine conditions for long-term sustainability.