Problem 54 - Calculus and its applications

Question

Jennifer deposits \(A_{0}=\$ 1200\) into an account that earns \(4.2 \%\) compounded continuously. a) Write the differential equation that represents \(A(t)\), the value of Jennifer's account after t years. b) Find the particular solution of the differential equation from part (a). c) Find \(A(7)\) and \(A^{\prime}(7)\). d) Find \(A^{\prime}(7) / A(7)\), and explain what this number represents.

Step-by-Step Solution

Answer

\( A'(7)/A(7) = 0.042 \), representing the continuous growth rate of 4.2% at year 7.

Step 1: Setting up the differential equation

Since the account earns interest compounded continuously, we can use the formula related to continuous compounding. The differential equation that represents the rate of change in the amount in Jennifer's account is given by \( \frac{dA}{dt} = rA \), where \( r \) is the continuous compounding rate.

Step 2: Insert the given rate

From the problem, the interest rate is given as \( 4.2\% \), or \( r = 0.042 \). Therefore, the differential equation in the form of \( \frac{dA}{dt} = rA \) becomes \( \frac{dA}{dt} = 0.042A \).

Step 3: Solving the differential equation

This is a separable differential equation. Separate the variables to get \( \frac{1}{A} dA = 0.042 \, dt \). Integrating both sides, we get \( \ln|A| = 0.042t + C \). Exponentiating both sides gives \( A(t) = e^{0.042t + C} \).

Step 4: Find the particular solution

Since \( A_{0} = \$1200 \), substitute both \( t = 0 \) and \( A(0) = 1200 \) into the equation to solve for \( C \). This gives \( 1200 = e^{0 + C} \rightarrow e^C = 1200 \). So, \( A(t) = 1200 \cdot e^{0.042t} \).

Step 5: Calculate A(7)

Substitute \( t = 7 \) into the particular solution: \( A(7) = 1200 \cdot e^{0.042 \times 7} \). Evaluating this gives the value of the account after 7 years.

Step 6: Calculate the derivative A'(t)

The derivative of the function \( A(t) = 1200 \cdot e^{0.042t} \) is \( A'(t) = 1200 \cdot 0.042 \cdot e^{0.042t} \).

Step 7: Calculate A'(7)

Substitute \( t = 7 \) into the derivative \( A'(t) = 50.4 \cdot e^{0.042 \times 7} \). This calculation gives the rate of change of the account balance at \( t = 7 \).

Step 8: Find A'(7)/A(7)

Using the expressions for \( A'(7) \) and \( A(7) \), calculate \( \frac{A'(7)}{A(7)} \). This simplifies to \( 0.042 \), indicating the instantaneous growth rate of the account at 7 years, which represents the interest rate as a continuous percentage.

Key Concepts

Differential Equation

A differential equation is a mathematical equation that involves functions and their derivatives. It expresses a relationship between a function and its rates of change. In the context of continuous compounding, differential equations model how an investment grows over time when compounded continuously.
For Jennifer's account with continuous compounding, the differential equation is given by:

• \( \frac{dA}{dt} = rA \)

where:

• \( \frac{dA}{dt} \) represents the rate of change of the account balance over time
• \( r \) is the continuous interest rate, and
• \( A \) is the balance of the account at time \( t \).

Understanding this equation helps us compute how Jennifer's money grows when compounded continuously, demonstrating essential financial principles through differential equations.

Separable Differential Equation

Separable differential equations are a specific type of differential equations where the variables can be separated on opposite sides of the equation. This simplification helps in solving the equation easily. The equation for Jennifer's continuously compounded interest is separable:

• Start with: \( \frac{dA}{dt} = 0.042A \)
• Separate variables: \( \frac{1}{A} dA = 0.042 \, dt \)

Integrating both sides with respect to their variables allows us to find an antiderivative, which leads to:

• \( \ln|A| = 0.042t + C \)

By exponentiating both sides, we obtain a general solution for \( A(t) \). This step is critical in transition from a differential equation to a function that describes Jennifer's account over time. Understanding separable differential equations is key in solving many real-world problems involving rates of change.

Particular Solution

A particular solution to a differential equation is a specific instance of the general solution, adapted to meet certain initial conditions. For Jennifer's differential equation, we find a particular solution by using her initial deposit as an initial condition. Given:

• \( A_0 = 1200 \) when \( t = 0 \)

Substituting these values into the general solution:

• \( A(t) = e^{0.042t + C} \)
• Set \( A(0) = 1200 \)

It allows us to solve for \( C \), leading to:

• \( e^C = 1200 \)

which simplifies to:

• \( A(t) = 1200 \cdot e^{0.042t} \)

This particular solution reflects the specific growth behavior of Jennifer's investment, accounting for the continuous compounding starting from her initial deposit.

Instantaneous Growth Rate

The instantaneous growth rate in the context of continuous compounding is a measure of how quickly an investment is growing at a specific instant. It's represented by the derivative of the account balance function, \( A(t) \), with respect to time. At \( t = 7 \), the computations yield:

• \( A'(7) \), the rate of change of the account balance
• \( A(7) \), the actual account value after 7 years

The ratio \( \frac{A'(7)}{A(7)} \) simplifies to the interest rate, 0.042 in this scenario, indicating that the ratio represents the account's continuous return rate at any given time.
In practical terms, it tells us that the amount in Jennifer's account is growing at a rate of 4.2% per year, without need to adjust for compounding over discrete intervals. This rate is vital in understanding financial scenarios utilizing continuous growth principles.