Problem 53 - Calculus and its applications

Question

Debra deposits \(A_{0}=\$ 500\) into an account that earns interest at a rate of \(3.75 \%,\) compounded continuously. a) Write the differential equation that represents \(A(t)\), the value of Debra's account after \(t\) years. b) Find the particular solution of the differential equation from part (a). c) Find \(A(5)\) and \(A^{\prime}(5)\). d) Find \(A^{\prime}(5) / A(5),\) and explain what this number represents.

Step-by-Step Solution

Answer

The account's value after 5 years is approximately $589.64, growing at a rate of 3.75% per year continuously.

Step 1: Write the Differential Equation

To model continuous compound interest, we use the formula for continuous growth: \( \frac{dA}{dt} = rA(t) \). Here, \( r = 3.75\% = 0.0375 \). Thus, the differential equation is \( \frac{dA}{dt} = 0.0375A(t) \).

Step 2: Find the General Solution

The differential equation \( \frac{dA}{dt} = 0.0375A(t) \) is a separable differential equation. Its solution is \( A(t) = A_0e^{0.0375t} \), where \( A_0 \) is the initial amount. Given \( A_0 = 500 \), we have \( A(t) = 500e^{0.0375t} \).

Step 3: Find A(5)

To calculate the account value after 5 years, substitute \( t = 5 \) into General Solution: \( A(5) = 500e^{0.0375 \times 5} \). Calculating this gives \( A(5) \approx 500e^{0.1875} \approx \$589.64 \).

Step 4: Find A'(5)

The derivative \( A'(t) = 0.0375A(t) \) represents the rate of change of the account value. Thus, \( A'(5) = 0.0375 \times A(5) \approx 0.0375 \times 589.64 = 22.11 \).

Step 5: Calculate A'(5) / A(5)

\( A'(5) / A(5) = 0.0375 \). This represents the continuous growth rate of the account, meaning the account grows at 3.75% per year at any given moment.

Key Concepts

Continuous Compound Interest

Continuous compound interest is a concept that helps us understand how money grows when interest is added continuously, rather than at discrete intervals. The formula for continuous compound interest is derived from the concept of exponential growth, where interest is recalculated and added to the principal at every possible moment.

In mathematical terms, the formula is given as:

• \( A(t) = A_0 e^{rt} \),

where:

• \( A(t) \) is the amount of money accumulated after \( t \) years, including interest.
• \( A_0 \) is the initial deposit (or the principal amount).
• \( r \) is the annual interest rate (expressed as a decimal).
• \( t \) is the time the money is invested or borrowed, in years.
• \( e \) is the base of the natural logarithm, approximately equal to 2.71828.

Understanding continuous compound interest is important because it shows how investments grow more quickly with continuous compounding compared to other compounding intervals like monthly or yearly. This idea is crucial in finance and savings strategies.

Exponential Growth

Exponential growth describes a process that increases at a consistent rate over time. This type of growth is characterized by the equation \( y(t) = y_0e^{kt} \), similar to the continuous compound interest formula.

The defining feature of exponential growth is that as time progresses, the rate of growth becomes faster and faster. This is because the growth is proportional to the current size; the more it grows, the faster it grows. Here’s a breakdown:

• \( y(t) \) represents the quantity or value after time \( t \).
• \( y_0 \) is the initial quantity or size.
• \( k \) is the growth constant, similar to the interest rate in continuous compounding.
• \( e \), the base of the natural logarithm, underlies the constant and continuous nature of exponential functions.

Exponential growth appears in various real-world contexts, such as populations growin rapidly, investments increasing with compound interest, and even the spread of diseases like in epidemiology. Recognizing when a situation is exhibiting exponential growth helps in predicting future outcomes and making informed decisions.

Calculus Problem Solving

Calculus is a branch of mathematics that focuses on rates of change and the accumulation of quantities. It provides tools to solve complex problems involving functions and limits, such as those seen in differential and integral calculus.

When dealing with continuous compound interest and exponential growth, calculus becomes especially useful. Let's see how it helps:

• **Differential Equations**: These are equations involving derivatives that describe how a quantity changes over time. In our example, the differential equation \( \frac{dA}{dt} = rA(t) \) models how an investment grows continuously.
• **Separable Differential Equations**: These can be solved by separating the variables, making them easier to integrate. Solving them provides us with the general solution which describes the behavior of a function over time.
• **Finding Derivatives**: Derivatives determine the instantaneous rate of change. In the problem, \( A'(t) = 0.0375A(t) \) was used to find the rate at which the account value increases after 5 years. It shows how the value changes at any point in time.
• **Analyzing Growth with Ratios**: The calculation of \( \frac{A'(5)}{A(5)} \) revealed the continuous growth rate (3.75% in this case). It tells us how much the account is growing compared to its current value.

Calculus is indispensable in problem-solving for financial models, scientific research, and engineering projects, providing a deeper understanding of dynamic systems.