Problem 69 - Calculus and its applications

Question

The amount of money, \(A(t)\), in Ina's savings account after \(t\) years is modeled by the differential equation \(d A / d t=0.0418 A\) a) What is the continuous growth rate? b) Find the particular solution, \(A(t),\) if Ina's account is worth \(\$ 3479.02\) after 2 yr. c) Find the amount that Ina deposited initially.

Step-by-Step Solution

Answer

a) 4.18% b) \( A(t) = 3200e^{0.0418t} \) c) Initial deposit is approximately \( 3200 \).

Step 1: Recognize the continuous growth rate

The given differential equation is \( \frac{dA}{dt} = 0.0418A \). Here, the coefficient of \( A \) (i.e., \( 0.0418 \)) represents the continuous growth rate. Therefore, the continuous growth rate is \( 4.18\% \).

Step 2: Formulate the general solution

The equation \( \frac{dA}{dt} = 0.0418A \) is a first-order linear differential equation. Its general solution can be formulated as: \( A(t) = A_0 e^{0.0418t} \), where \( A_0 \) is the initial amount deposited.

Step 3: Use initial conditions to find the particular solution

We're given that after 2 years, the account balance is \( 3479.02 \). Incorporating this into the general solution gives: \( 3479.02 = A_0 e^{0.0418 \times 2} \). We now solve for \( A_0 \).

Step 4: Solve for the initial deposit \( A_0 \)

First, calculate \( e^{0.0836} \). Using a calculator, \( e^{0.0836} \approx 1.0872 \). Now substitute back: \( 3479.02 = A_0 \times 1.0872 \). Solving for \( A_0 \) gives: \[ A_0 = \frac{3479.02}{1.0872} \approx 3200 \].

Step 5: Confirm the particular solution

The specific solution to the differential equation given the initial condition is \( A(t) = 3200e^{0.0418t} \). This confirms that the initial deposit \( A_0 \) is approximately \( 3200 \).

Key Concepts

Continuous Growth Rate

In the world of differential equations, understanding the continuous growth rate is fundamental. The continuous growth rate explains how a quantity increases in a way that is proportional to its current size. In the given differential equation \( \frac{dA}{dt} = 0.0418A \), this concept is portrayed clearly. The coefficient next to \( A \) is the growth factor, representing the rate of change over time. Here it is \( 0.0418 \), which when expressed as a percentage, becomes exactly \( 4.18\% \).
This growth rate tells us that Ina's savings grow continuously by \( 4.18\% \) every year, irrespective of how much money is currently in the account.

• This continuous rate is crucial for long-term forecasts.
• It reflects how interest compounds.
• It ensures understanding of natural growth in financial contexts.

Recognizing this rate from a differential equation is a key skill in analyzing exponential growth scenarios.

Exponential Growth Model

Exponential growth models capture how quantities multiply over time, and they are often represented by the equation \( A(t) = A_0 e^{kt} \). Here, \( A_0 \) is the initial amount, \( e \) is the base of the natural logarithm, and \( k \) represents the growth rate.
In the problem at hand, the equation \( A(t) = A_0 e^{0.0418t} \) models Ina's account balance over time. This setup means that her savings grow exponentially at a continuous rate of \( 4.18\% \) per year.

• Exponential growth differs from simple interest as it compounds continuously.
• Each year's interest contributes to the next year's growth.

This creates a snowball effect, where money doesn't just sit and grow linearly; it accelerates as the account balance increases. Understanding this model is essential in forecasting account balances, population dynamics, and many phenomena in nature and finance.

Initial Value Problem

An initial value problem requires solving a differential equation with a specific initial condition to find a unique solution. In the context of Ina's savings, we begin with the general solution derived from the differential equation: \( A(t) = A_0 e^{0.0418t} \).
However, to find the particular solution that matches the given scenario, we must apply the known initial value condition \( A(2) = 3479.02 \). This requires determining the initial deposit \( A_0 \).
Following the steps:

• Substitute \( A(t) = 3479.02 \) at \( t = 2 \) into the general equation.
• Solve \( 3479.02 = A_0 e^{0.0836} \) for \( A_0 \).

These steps demonstrate how the initial value problem allows us to pinpoint exactly how much money Ina deposited initially, which is approximately \( 3200 \).
This process is a keystone in many mathematical applications, letting us harness specific conditions to predict various outcomes accurately.