Problem 70 - Calculus and its applications

Question

The amount of money, \(A(t)\), in John's savings account after \(t\) years is modeled by the differential equation \(d A / d t=0.0325 A\) a) What is the continuous growth rate? b) Find the particular solution, \(A(t),\) if John's account is worth \(\$ 2582.58\) after 1 yr. c) Find the amount that John deposited initially.

Step-by-Step Solution

Answer

a) Continuous growth rate is 0.0325. b) Particular solution is \(A(t) = 2500e^{0.0325t}\). c) Initial deposit was \$2500.

Step 1: Identifying the Continuous Growth Rate

The differential equation given is \(\frac{dA}{dt} = 0.0325A\). This equation is in the form \(\frac{dA}{dt} = kA\), where \(k\) is the continuous growth rate. Therefore, the continuous growth rate is \(k = 0.0325\).

Step 2: Finding the General Solution of the Differential Equation

The general solution to the differential equation \(\frac{dA}{dt} = 0.0325A\) can be solved by using the method of separation of variables. Integrating both sides, we get \(\int \frac{1}{A} \, dA = \int 0.0325 \, dt\). Solving these integrals, we obtain \(\ln |A| = 0.0325t + C\). By exponentiating both sides, we get \(A(t) = Ce^{0.0325t}\), where \(C\) is a constant determined by initial conditions.

Step 3: Applying Initial Condition to Find Particular Solution

We use the initial condition \(A(1) = 2582.58\) to find \(C\). Substituting into the equation, we get \(2582.58 = Ce^{0.0325 \times 1}\). Solving for \(C\), we have \(C = \frac{2582.58}{e^{0.0325}}\). Calculating \(e^{0.0325} \approx 1.0330\), so \(C \approx \frac{2582.58}{1.0330} \approx 2500\). Therefore, the particular solution is \(A(t) = 2500e^{0.0325t}\).

Step 4: Finding the Initial Amount Deposited

The initial amount deposited, \(A(0)\), is calculated by evaluating the particular solution at \(t = 0\). Substituting \(t = 0\) into \(A(t) = 2500e^{0.0325t}\), we get \(A(0) = 2500e^{0} = 2500\). Thus, the initial deposit was \(\$2500\).

Key Concepts

Exponential Growth

Exponential Growth occurs when the rate of change of a quantity is directly proportional to the current quantity itself. This means that as the quantity increases, the rate at which it grows also increases. In the context of the differential equation \( \frac{dA}{dt} = 0.0325A \), the rate of growth is 3.25% of the amount in the account at that moment.
This is a hallmark of exponential growth and can be visualized as a curve that rises increasingly steeper over time.

• It is characterized by a constant percentage increase, in this case, 3.25% annually.
• The form \( \frac{dA}{dt} = kA \) is typical for exponential growth, where \( k \) is the growth rate constant.

The idea of exponential growth is crucial in understanding how investments can increase over time due to the power of compound interest.

Initial Conditions

Initial conditions provide a specific point that helps determine the particular solution of a differential equation. They are like a starting guideline which allows integration of the differential equation to be more accurate.
For this exercise, the initial condition is given as \( A(1) = 2582.58 \), meaning that after 1 year, the amount in the account should be $2582.58.

• By using this condition, we substitute it into the equation to find the constant \( C \).
• This step bridges the gap between a general solution and a solution that accurately reflects the real-world scenario or the problem statement.

Knowing the initial condition enables us to fully tailor the mathematical model to fit real-life data, ensuring that predictions and outcomes align with actual values.

Separation of Variables

Separation of Variables is a powerful technique used to solve differential equations. It involves rearranging the equation to separate the variables on opposite sides of the equation, resulting in one side containing only the dependent variable and the other side only the independent variable.
For the equation \( \frac{dA}{dt} = 0.0325A \), we achieve this by dividing both sides by \( A \) to isolate the differential terms:
\( \int \frac{1}{A} \, dA = \int 0.0325 \, dt \).

• This step allows us to integrate each side independently.
• The result of these integrations leads to the function \( \ln |A| = 0.0325t + C \).

This methodology transforms a complex differential equation into a simpler form, making it easier to solve and understand. It exemplifies how intricate problems can be systematically reduced to more manageable parts through logical manipulation.