Problem 76 - Precalculus Mathematics for Calculus

Question

Compound Interest A woman invests \(\$ 6500\) in an account that pays \(6 \%\) interest per year, compounded continuously. (a) What is the amount after 2 years? (b) How long will it take for the amount to be \(\$ 8000 ?\)

Step-by-Step Solution

Answer

(a) \(\approx \$ 7338.75\), (b) \(\approx 3.45 \) years.

Step 1: Understand Compound Interest Formula

For continuous compounding, the amount \( A \) after time \( t \) can be calculated from the initial principal \( P \), the interest rate \( r \), and time \( t \) in years using the formula:\[ A = P \times e^{rt} \] where \( e \approx 2.71828 \).

Step 2: Calculate Amount After 2 Years

Using the formula \( A = P \times e^{rt} \), substitute \( P = 6500 \), \( r = 0.06 \), and \( t = 2 \):\[ A = 6500 \times e^{0.06 \times 2} \]Using a calculator, \( e^{0.12} \approx 1.1275 \), so:\[ A = 6500 \times 1.1275 \approx 7338.75 \] Thus, the amount after 2 years is approximately \( \$ 7338.75 \).

Step 3: Set up the Equation for Part (b)

To find when the account will grow to \( \$ 8000 \), use the same formula but solve for \( t \): \[ 8000 = 6500 \times e^{0.06t} \]

Step 4: Solve for Time \( t \)

Divide both sides by \( 6500 \):\[ \frac{8000}{6500} = e^{0.06t} \]Simplify the fraction, \( \frac{8000}{6500} \approx 1.23077 \).Take the natural logarithm of both sides:\[ \ln(1.23077) = 0.06t \]Using a calculator, \( \ln(1.23077) \approx 0.2072 \).Then solve for \( t \):\[ t = \frac{0.2072}{0.06} \approx 3.453 \]So, it will take approximately 3.45 years for the amount to grow to \( \$ 8000 \).

Key Concepts

Continuous Compounding

Continuous compounding is a powerful concept in finance. It refers to the idea that interest is calculated and added to the principal balance at every possible moment.
This means that interest is continually being applied not only to the initial principal but also on any interest already earned, making your money grow faster.
The formula for continuous compounding is given by \[ A = P \times e^{rt} \] where:

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

This mechanism allows for exponential growth of investments, meaning your returns can be significantly higher compared to typical compound interest calculated at discrete intervals like annually or quarterly.

Exponential Growth

Exponential growth occurs when the rate of growth is directly proportional to the current size or amount.
In the context of compound interest, this is represented by the exponent in the formula \[ e^{rt} \].
This implies that over time, the value of the investment doesn't just increase by a fixed amount but rather grows larger at an accelerating rate.

• As the principal amounts grow due to accrued interest, the rate becomes more potent, further increasing the investment amount.
• This creates a snowball effect, leading to substantial increases over time.

Understanding this concept is important when planning long-term financial goals as it shows how small changes in rate or time can have a large impact on future value.

Natural Logarithm

The natural logarithm is a logarithm to the base \( e \), and it is denoted by \( \ln \).
It is especially useful when dealing with continuous compounding because \( e \) is the base of the exponential function used in the compounding formula.
In the context of compound interest, solving for time \( t \) involves using the natural logarithm. For example, if you need to determine how long it will take for an investment to reach a certain amount, you rearrange and solve \[ \ln(\text{final amount}/\text{initial amount}) = rt \]
Here's how the natural logarithm function helps:

• It allows us to "undo" the exponentiation in the formula \( e^{rt} \) when solving for \( t \).
• It helps in determining growth periods and is vital for interpreting exponential obligations in simpler arithmetic terms.

Grasping the concept of natural logarithms enhances your ability to solve and understand various problems involving continuous compounding.