Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

If \(\$ 375\) is put in the bank today, what will it be worth at the end of 2 years if interest is \(3.5 \%\) and is compounded as specified? (a) Annually (b) Monthly (c) Daily (d) Continuously

Step-by-Step Solution

Answer

(a) $401.71; (b) $402.03; (c) $402.06; (d) $402.19.

Step 1: Understand Compound Interest Formula

The compound interest formula is given by \( A = P \left(1 + \frac{r}{n}\right)^{nt} \) where \( A \) is the amount of money accumulated after n years, including interest. \( P \) is the principal amount (in this case, $375), \( r \) is the annual interest rate (0.035 as a decimal), \( n \) is the number of times interest is compounded per year, and \( t \) is the time the money is invested for in years.

Step 2: Calculate for Annual Compounding

For part (a), interest is compounded annually. Here, \( n = 1 \). Plug in the values into the formula: \[ A = 375 \left(1 + \frac{0.035}{1}\right)^{1 \times 2} = 375 \times 1.035^2. \] Calculate \( A = 375 \times 1.071225 = 401.709375. \) Thus, \( A \approx \$401.71.\)

Step 3: Calculate for Monthly Compounding

For part (b), interest is compounded monthly, thus \( n = 12. \) Substitute the values into the formula: \[ A = 375 \left(1 + \frac{0.035}{12}\right)^{12 \times 2}. \] Calculate \( A = 375 \times 1.0029167^{24} = 375 \times 1.07212794 \approx 402.03. \)

Step 4: Calculate for Daily Compounding

For part (c), interest is compounded daily, so \( n = 365. \) Use the formula: \[ A = 375 \left(1 + \frac{0.035}{365}\right)^{365 \times 2}. \] Calculate \( A = 375 \times 1.07125572 \approx 402.06. \)

Step 5: Calculate for Continuous Compounding

For part (d), continuous compounding uses the formula \( A = Pe^{rt}. \) Therefore, \[ A = 375 \times e^{0.035 \times 2}. \] Calculate \( A = 375 \times e^{0.07} = 375 \times 1.07250818 \approx 402.19. \)

Key Concepts

Understanding Annual Compounding

When we talk about compound interest, annual compounding is one of the simplest forms to understand. Here, interest is added to the principal once a year. This means, throughout the year, no interest is being added, but when the year ends, the interest is calculated for the entire year and added to the principal.

In the given problem, we had an initial amount of \(375. With an interest rate of 3.5%, the principal grows once at the end of each year. The formula we use for calculating this is:

• \( A = P \left(1 + \frac{r}{1}\right)^{1 \times t} \)

So, after the first year, the amount would grow because 3.5% of \)375 is added to it. After the second year, the total amount again increases by adding 3.5% of the new total. The benefit of annual compounding is easy tracking as rates remain consistent throughout the year.

Exploring Monthly Compounding

Monthly compounding involves adding interest to the principal more frequently than annual compounding, specifically each month. Imagine splitting the annual interest into 12 parts and applying each part at the end of every month.

For the exercise, the formula we apply becomes:

• \( A = P \left(1 + \frac{r}{12}\right)^{12 \times t} \)

Every month adds a little bit of interest, noticeably more often than annual compounding, which can lead to slightly more growth. Over two years, the interest compounds multiple times, allowing it to earn interest on previous interest more frequently. This provides a higher final amount when compared to annual compounding.

Delving into Daily Compounding

Daily compounding takes the concept of compounding to a more frequent level compared to monthly. Here, interest is added daily. Each day's interest calculates based on what was present the day before.

The formula used is:

• \( A = P \left(1 + \frac{r}{365}\right)^{365 \times t} \)

Even though each day's interest is a very small fraction of the total, over time these small additions make a difference. This results in a higher overall return compared to both annual and monthly compounding because the interest earns its own interest continuously throughout the year.

Continuous Compounding Explained

Continuous compounding takes the concept of adding interest to a theoretical extreme. It involves adding interest in such a way that compounding happens continuously at every possible instant. Though physically impossible, it's a powerful mathematical concept.

The process uses a special constant, \( e \), to calculate future value, given by the formula:

• \( A = Pe^{rt} \)

This formula captures the compounding process at each instant effectively, maximizing the interest earned. In the problem, we find this gives us the highest return compared to other compounding methods. Continuous compounding allows for the most efficient growth of capital in a perfectly complex system.