Problem 34
Question
Compound Interest If \(\$ 8000\) is invested in an account for which interest is compounded continuously, find the amount of the investment at the end of 12 years for the following interest rates. \(\begin{array}{llll}{\text { (a) } 2 \%} & {\text { (b) } 3 \%} & {\text { (c) } 4.5 \%} & {\text { (d) } 7 \%}\end{array}\)
Step-by-Step Solution
Verified Answer
At the end of 12 years, the investment will be approximately $10,170 at 2%, $11,466 at 3%, $13,724 at 4.5%, and $18,539 at 7% interest rates.
1Step 1: Understanding Continuous Compounding Formula
The formula for continuous compounding interest is given by \( A = Pe^{rt} \) where \( A \) is the amount after time \( t \), \( P \) is the principal amount, \( r \) is the rate of interest per year, and \( t \) is the time in years. \( e \) is the base of the natural logarithm.
2Step 2: Setting Up the Equation for 2% Interest Rate
For 2% interest rate converted to decimal, \( r = 0.02 \), the principal amount \( P = 8000 \), and \( t = 12 \). Substitute these values into the formula: \[ A = 8000e^{0.02 \times 12} \]
3Step 3: Calculating for 2% Interest Rate
Compute \( e^{0.02 \times 12} = e^{0.24} \). Using a calculator, \( e^{0.24} \approx 1.27125 \). Then multiply by the principal: \[ A = 8000 \times 1.27125 \approx 10170 \] So, the amount will be approximately \( 10170 \).
4Step 4: Setting Up and Calculating for 3% Interest Rate
For 3% interest rate, \( r = 0.03 \). Substitute it into the formula:\[ A = 8000e^{0.03 \times 12} \]Compute \( e^{0.36} \approx 1.43333 \). Then: \[ A = 8000 \times 1.43333 \approx 11466 \]
5Step 5: Setting Up and Calculating for 4.5% Interest Rate
For 4.5% interest rate, \( r = 0.045 \). Substitute it:\[ A = 8000e^{0.045 \times 12} \]Compute \( e^{0.54} \approx 1.71553 \). Then:\[ A = 8000 \times 1.71553 \approx 13724 \]
6Step 6: Setting Up and Calculating for 7% Interest Rate
For 7% interest rate, \( r = 0.07 \). Substitute:\[ A = 8000e^{0.07 \times 12} \]Compute \( e^{0.84} \approx 2.31737 \). Then:\[ A = 8000 \times 2.31737 \approx 18539 \]
Key Concepts
Continuous CompoundingExponential GrowthFinancial Mathematics
Continuous Compounding
When we talk about continuous compounding, we're referring to a situation where interest is calculated and added to the principal balance of an investment constantly, at every infinitesimal moment. This is a bit different from regular compounding, which might occur yearly, quarterly, monthly, or daily. The continuous compounding formula is expressed as \( A = Pe^{rt} \). Here;
- \( A \) is the final amount in your account after time \( t \).
- \( P \) is the initial principal balance.
- \( r \) is the annual interest rate expressed as a decimal.
- \( t \) is the time the money is invested for, in years.
- \( e \) is Euler's number, approximately 2.718, representing the base of the natural logarithm.
Exponential Growth
Exponential growth is a pattern of data that increases faster and faster over time. This is incredibly important in the context of continuously compounding interest. The formula \( A = Pe^{rt} \) not only models continuous compounding but also serves as a classic example of exponential growth.In this framework:
- The interest you earn begins small at first.
- As each instant passes, the amount of interest compounds on itself, creating a snowball effect.
Financial Mathematics
Financial mathematics is essential in understanding how money grows over time and making smart investment choices. It uses models and formulas to help us predict and calculate the behavior of different financial instruments, like loans, bonds, and investments.
Especially relevant is the concept of:
- Compound Interest: Unlike simple interest, compound interest calculates not just on the initial principal but also accumulates on the interest itself.
- Continuous Compounding: This is a specific type of compound interest where calculations and reinvestments happen at an infinite rate, leading to exponential growth.
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