Problem 18 - Essential Calculus Early Transcendentals

Question

(a) If \(\$ 1000\) is borrowed at 8\(\%\) interest, find the amounts due at the end of 3 years if the interest is compounded (i) annually, (ii) quarterly, (iii) monthly, (iv) weekly, (v) daily, (vi) hourly, and (vii) continuously. (b) Suppose \(\$ 1000\) is borrowed and the interest is compounded continuously. If \(A(t)\) is the amount due after \(t\) years, where \(0 \leqslant t \leqslant 3,\) graph \(A(t)\) for each of the interest rates \(6 \%, 8 \%,\) and 10\(\%\) on a common screen.

Step-by-Step Solution

Answer

The amounts due after 3 years: (i) $1259.71, (ii) $1268.24, (iii) $1269.73, (iv) $1270.51, (v) $1270.95, (vi) $1271.03, (vii) $1271.25. For continuous compounding, plot graphs for different rates.

Step 1: Understanding Interest Formula

To find the amount due for compound interest compounded at different intervals, we use the formula: \( A = P \left(1 + \frac{r}{n}\right)^{nt} \), where \( A \) is the amount, \( P \) is the principal, \( r \) is the interest rate, \( n \) is the number of times interest is compounded per year, and \( t \) is the time in years.

Step 2: Compounding Annually

For interest compounded annually, \( n = 1 \). Using \( r = 0.08, \) and \( t = 3 \), the formula becomes: \( A = 1000 \left(1 + \frac{0.08}{1}\right)^{3} \). Calculating, \( A = 1000 \times 1.08^3 \approx 1259.71. \)

Step 3: Compounding Quarterly

For interest compounded quarterly, \( n = 4 \). The formula becomes: \( A = 1000 \left(1 + \frac{0.08}{4}\right)^{12} \). Calculating, \( A = 1000 \times 1.02^{12} \approx 1268.24. \)

Step 4: Compounding Monthly

For interest compounded monthly, \( n = 12 \). The formula becomes: \( A = 1000 \left(1 + \frac{0.08}{12}\right)^{36} \). Calculating, \( A = 1000 \times 1.0067^{36} \approx 1269.73. \)

Step 5: Compounding Weekly

For interest compounded weekly, \( n = 52 \). The formula becomes: \( A = 1000 \left(1 + \frac{0.08}{52}\right)^{156} \). Calculating, \( A = 1000 \times 1.001538^{156} \approx 1270.51. \)

Step 6: Compounding Daily

For interest compounded daily, \( n = 365 \). The formula becomes: \( A = 1000 \left(1 + \frac{0.08}{365}\right)^{1095} \). Calculating, \( A = 1000 \times 1.0002192^{1095} \approx 1270.95. \)

Step 7: Compounding Hourly

For interest compounded hourly, \( n = 8760 \). The formula becomes: \( A = 1000 \left(1 + \frac{0.08}{8760}\right)^{26280} \). Calculating, \( A \approx 1271.03. \)

Step 8: Continuous Compounding

For continuously compounded interest, use \( A = Pe^{rt} \). Substitute \( P = 1000, r = 0.08, \text{ and } t = 3 \): \( A = 1000e^{0.08 \times 3} \. \) Calculating, \( A \approx 1271.25. \)

Step 9: Understanding Continuous Compounding for Graph

The formula for continuous compounding is \( A(t) = Pe^{rt} \). For \( P = 1000 \), and \( r = 0.08 \), find \( A(t) \): \( A(t) = 1000e^{0.08t} \).

Step 10: Graphing Continuous Compounding

To graph \( A(t) \) for each of the interest rates: \( A(t) = 1000e^{0.06t} \), \( A(t) = 1000e^{0.08t} \), and \( A(t) = 1000e^{0.10t} \), plot these equations on the common vertical axis of amount and horizontal axis of years, ranging from \( t = 0 \) to \( t = 3 \).

Key Concepts

Interest Rate

An interest rate is the percentage at which you are charged for borrowing money or rewarded for saving it. It's like a fee for using money that belongs to someone else. A higher interest rate typically means higher costs for borrowers, but also higher returns for savers. When dealing with loans or investments, understanding the interest rate allows you to calculate how much you will pay or earn over time. For example, in our exercise, the rate is 8%. This means for every dollar borrowed, you will have to pay 8 cents per year as a cost of borrowing if compounded annually.
Compounding intervals matter. The more frequently interest is compounded, the more interest you'll pay. This is because with each compounding period, the interest is calculated on a slightly larger amount of money as interest is added back to the principal. Compounding can occur annually, quarterly, monthly, or even more frequently.
Understanding how different compounding periods affect the amount can help you make informed financial decisions, ensuring you choose the best loan or investment option for your financial goals.

Continuous Compounding

Continuous compounding is a concept where interest is calculated and added to the principal continuously. This means that interest is being added in such tiny fractions that you are effectively earning or paying interest at every moment.
To calculate continuous compounding, the formula used is:

• \[ A = Pe^{rt} \]

Where:

• \(A\) is the total amount after time \(t\)
• \(P\) is the principal amount
• \(r\) is the annual interest rate (in decimal form)
• \(t\) is the time in years
• \(e\) is the base of the natural logarithm, approximately equal to 2.71828

Continuous compounding often results in slightly more interest compared to even the most frequent compounding intervals like daily or hourly.
This method is especially useful in financial mathematics, as it simplifies many calculations due to its elegant formula. In our example, using a continuous interest rate of 8%, the amount after 3 years becomes \(1271.25\) for a \(\$1000\) principal.

Graphing Functions

Graphing functions is a powerful way to visually understand how different variables interact. In the context of compound interest, it allows us to see how the amount owed grows over time at different interest rates. The general form of a compound interest function is

• \[ A(t) = Pe^{rt} \]

which describes a curve that rises more steeply as either the rate \(r\) increases, or as more time \(t\) passes. By graphing these functions, students can compare how fast increasing interest rates or longer time periods affect the accumulated amount.
In our exercise, we were asked to plot the amount \(A(t)\) for principal \(P = 1000\) and rates of 6%, 8%, and 10% over a period of 3 years. As you'd expect, the graph shows that higher interest rates lead to higher amounts owed at any time point.
Creating and analyzing these graphs is a practical skill in any discipline that involves growth, decay, or change over time, making it invaluable beyond just calculating interest.