Problem 49 - Precalculus Mathematics for Calculus

Question

Compound Interest An investment of \(\$ 5000\) is deposited into an account in which interest is compounded monthly. Complete the table by filling in the amounts to which the investment grows at the indicated times or interest rates. \(r=4 \%\) $$\begin{array}{|c|c|} \hline \begin{array}{c} \text { Time } \\\\\text { (years) }\end{array} & \text { Amount } \\\\\hline 1 & \\\2 & \\\3 & \\\4 & \\\5 & \\\6 & \\\\\hline\end{array}$$

Step-by-Step Solution

Answer

The amounts after each year are approximately: Year 1: $5204.03, Year 2: $5416.58, Year 3: $5638.61, Year 4: $5870.9, Year 5: $6104.32, Year 6: $6340.85.

Step 1: Understand the Compound Interest Formula

The formula for compound interest is \( A = P \left(1 + \frac{r}{n}\right)^{nt} \), where \( A \) is the amount of money accumulated after \( n \) years, including interest. \( P \) is the principal amount (initial investment), \( r \) is the annual interest rate (in decimal), \( n \) is the number of times that interest is compounded per year, and \( t \) is the time in years. For this problem, \( P = 5000 \), \( r = 0.04 \), and \( n = 12 \) since the interest is compounded monthly.

Step 2: Find the Amount for Year 1

Substitute \( t = 1 \) into the formula: \[A = 5000 \left(1 + \frac{0.04}{12}\right)^{12 \cdot 1}\]Calculate the result to find the amount after 1 year.

Step 3: Calculate Year 1

First calculate \( \frac{0.04}{12} = 0.003333 \). Then compute \( 1 + 0.003333 = 1.003333 \). Calculate \( 1.003333^{12} \). Multiply the result by \( 5000 \) to find \( A \) for 1 year. Thus, \( A \approx 5204.03 \).

Step 4: Find the Amount for Year 2

Substitute \( t = 2 \) into the formula: \[A = 5000 \left(1 + \frac{0.04}{12}\right)^{12 \cdot 2}\]Calculate the result to find the amount after 2 years.

Step 5: Calculate Year 2

Using \( 1.003333 \) calculated earlier, compute \( 1.003333^{24} \). Multiply the result by \( 5000 \) to determine the amount for 2 years. Thus, \( A \approx 5416.58 \).

Step 6: Find the Amount for Year 3

Substitute \( t = 3 \) into the formula: \[A = 5000 \left(1 + \frac{0.04}{12}\right)^{12 \cdot 3}\]Calculate the result to find the amount after 3 years.

Step 7: Calculate Year 3

Compute \( 1.003333^{36} \) and multiply by \( 5000 \). Thus, \( A \approx 5638.61 \).

Step 8: Find the Amount for Year 4

Substitute \( t = 4 \) into the formula: \[A = 5000 \left(1 + \frac{0.04}{12}\right)^{12 \cdot 4}\]Calculate the result to find the amount after 4 years.

Step 9: Calculate Year 4

Compute \( 1.003333^{48} \) and multiply by \( 5000 \). Thus, \( A \approx 5870.9 \).

Step 10: Find the Amount for Year 5

Substitute \( t = 5 \) into the formula: \[A = 5000 \left(1 + \frac{0.04}{12}\right)^{12 \cdot 5}\]Calculate the result to find the amount after 5 years.

Step 11: Calculate Year 5

Compute \( 1.003333^{60} \) and multiply by \( 5000 \). Thus, \( A \approx 6104.32 \).

Step 12: Find the Amount for Year 6

Substitute \( t = 6 \) into the formula: \[A = 5000 \left(1 + \frac{0.04}{12}\right)^{12 \cdot 6}\]Calculate the result to find the amount after 6 years.

Step 13: Calculate Year 6

Compute \( 1.003333^{72} \) and multiply by \( 5000 \). Thus, \( A \approx 6340.85 \).

Key Concepts

Exponential Growth

In financial mathematics, exponential growth refers to the increasing value of an investment over time, especially when the increase itself accelerates as time progresses. This is the essence of compound interest—it's not just interest on the initial sum but also interest on accumulated interest. The magic of exponential growth in compound interest lies in the formula for growth \(A = P \left(1 + \frac{r}{n}\right)^{nt}\). This formula demonstrates how the investment grows exponentially:

• \(P\) is the initial principal.
• \(r\) is the annual interest rate in decimal form.
• \(n\) is how often the interest is compounded each year.
• \(t\) is the time in years.

Notice how the exponent \(nt\) leads to exponential growth. Every time interest is compounded, it adds to the total sum, leading to more interest in the next period.As an example, when you invest \(\\(5000\) at a rate of 4% compounded monthly, the amounts found for each year show clearly the increasing amounts from \(\\)5204.03\) in Year 1 to \(\$6340.85\) in Year 6. This continuous growth highlights why exponential growth is a powerful force in accumulating wealth.

Interest Rates

Interest rates play a crucial role in the calculation and understanding of compound interest. It is essentially the percentage at which your money is growing or the cost of borrowing money. In the context of compound interest, the rate affects how quickly an investment grows over time. The formula \(A = P \left(1 + \frac{r}{n}\right)^{nt}\) incorporates it directly in \(r\), the annual interest rate.The interest rate is expressed in decimal form, so a 4% interest rate is written as \(0.04\). High interest rates mean faster growth of an investment due to more additive interest accrued over time, especially when compounded:

• Annual compounding, monthly compounding, etc., dictate "n", the frequency of interest application per year.
• A higher "n" value leads to higher yields due to more frequent compounding.

If we vary the interest rate—or even the compounding frequency—the end result changes significantly. Therefore, understanding different scenarios with different interest rates can hone financial strategies.

Financial Mathematics

Financial mathematics involves mathematical methods to solve financial problems, like determining how much money an investment will grow to over time. It's the backbone of economics and finance for tasks such as calculating loans, compound interest, and investments. Crucial elements of financial mathematics include the principles and formulas that predict an investment's future growth, such as the compound interest formula \(A = P \left(1 + \frac{r}{n}\right)^{nt}\).In our exercise, the importance of accurate computation and understanding of these principles shines through:

• Correct calculation of compound interest involves substituting values like \(P = 5000\), \(r = 0.04\), and \(n = 12\) for monthly compounding.
• Ensuring accuracy in exponent calculation \((nt)\) for compounded periods is critical for predicting future values correctly.

These calculations form the foundation for making sound financial decisions and planning for future financial growth, such as investments and retirement planning. By mastering these formulas, you can predict how your money will work for you over time in various financial scenarios.