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Savings Account Let \(A(t)\) be the balance in a savings account after \(t\) years, and suppose that \(A(t)\) satisfies the differential equation $$ A^{\prime}(t)=.045 A(t), \quad A(0)=3000 $$ (a) How much money was originally deposited in the account? (b) What interest rate is being earned? (c) Find the formula for \(A(t)\) (d) What is the balance after 5 years? (e) Use part (d) and the differential equation to determine how fast the balance is growing after 5 years. (f) How large will the balance be when it is growing at the rate of \(\$ 270\) per year?

Short Answer

Expert verified
The original deposit was 3000 dollars. The interest rate is 4.5%. The formula for the balance is \( A(t) = 3000 e^{0.045t} \). The balance after 5 years is approximately 3756.97 dollars. The growth rate after 5 years is about 169.06 dollars per year. The balance is 6000 dollars when growing at 270 dollars per year.

Step by step solution

01

Original Deposit

To determine the initial deposit, evaluate the given condition for the balance at time zero. The given initial condition is: \( A(0) = 3000 \).This means \(3000\) dollars were originally deposited in the account.
02

Interest Rate Earned

The differential equation is: \( A'(t) = 0.045 A(t) \).This indicates that the interest rate is 4.5% per year since \( r = 0.045 \) can be interpreted as the interest rate.
03

Finding the Formula for A(t)

The differential equation \( A'(t) = 0.045 A(t) \) is a first-order linear differential equation. To solve it, notice it is of the form: \( \frac{dA}{dt} = kA \), where \( k = 0.045 \). The general solution is: \( A(t) = A(0)e^{kt} \).Using the initial condition \( A(0) = 3000 \): \( A(t) = 3000 e^{0.045t} \).
04

Balance After 5 Years

To find the balance after 5 years, substitute \( t = 5 \) into the formula for \( A(t) \): \( A(5) = 3000 e^{0.045 \times 5} \).Calculate the exponent: \( A(5) = 3000 e^{0.225} \).Using a calculator to find the value: \( A(5) \approx 3000 \times 1.252322 \approx 3756.97 \).So, the balance after 5 years is approximately \( 3756.97 \) dollars.
05

Growth Rate After 5 Years

To determine how fast the balance is growing after 5 years, use the derivative from the differential equation: \( A'(t) = 0.045 A(t) \).Substitute \( t = 5 \) and the previously found balance \( A(5) \): \( A'(5) = 0.045 \times 3756.97 \).Calculate the rate of growth: \( A'(5) \approx 169.06 \).So, the balance is growing at a rate of \( 169.06 \) dollars per year after 5 years.
06

Balance for a Growth Rate of 270 per Year

To find the balance when it is growing at the rate of \( 270 \) dollars per year, use the differential equation: \( A'(t) = 0.045 A(t) \).Set \( A'(t) = 270 \) and solve for \( A(t) \): \( 270 = 0.045 A(t) \).Rearrange to solve for \( A(t) \): \( A(t) = \frac{270}{0.045} \).Calculate the value: \( A(t) = 6000 \).So, the balance is \( 6000 \) dollars when it is growing at the rate of \( 270 \) dollars per year.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

differential equations
Differential equations describe how a quantity changes over time. In our exercise, the differential equation is: \(A'(t) = 0.045A(t)\). This describes how the balance \(A(t)\) in a savings account changes with time. The prime symbol \('A'(t)\) signifies a derivative, meaning the rate of change of \(A(t)\). Essentially, this equation states that the rate of growth of the balance is proportional to the current balance.
Importance: Differential equations like this are essential in modeling situations where growth is exponential, such as population growth, radioactive decay, and financial investments. Understanding how to set up and solve these equations helps in predicting future values of a changing quantity.
initial deposit
The initial deposit refers to the amount of money first deposited into the savings account. In our problem, this is denoted by \(A(0)\). According to the initial condition given, \(A(0) = 3000\), the initial deposit is 3000 dollars.
Why it matters: Knowing the initial deposit is crucial for predicting future values in a savings account. It's the foundation on which the interest is calculated and compounded. Without this piece of information, it would be impossible to accurately determine the growth of the balance over time.
interest rate
The interest rate determines how quickly the balance in the account grows. In our differential equation, \(A'(t) = 0.045A(t)\), the interest rate is represented by the constant 0.045, or 4.5% per year. This rate shows that for every dollar in the account, 0.045 dollars are added each year.
Key points:
  • An interest rate can vary or be fixed over different periods.
  • The higher the interest rate, the faster the account balance grows.
  • Financial institutions use interest rates to incentivize savings and offer returns on deposits.
exponential growth
Exponential growth occurs when the growth rate of a quantity is proportional to its current value. In the context of our exercise, the formula \(A(t) = A(0)e^{0.045t}\) illustrates exponential growth. Here, \(A(0)\) is our initial deposit of 3000 dollars, and \(e^{0.045t}\) shows how the balance grows over time.
Explanation:
  • Exponential functions grow faster than linear functions.
  • Exponential growth is common in many natural and financial systems.
  • It's crucial for understanding long-term behaviors of savings and investments.

Example: After 5 years, the balance becomes approximately 3756.97 dollars, showing significant growth from the initial 3000 dollars due to the compounding effect of exponential growth.

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