Problem 2 - Precalculus

Question

For each of the scenarios given in Exercises \(1-6\), \- Find the amount \(A\) in the account as a function of the term of the investment \(t\) in years. \- Determine how much is in the account after 5 years, 10 years, 30 years and 35 years. Round your answers to the nearest cent. \- Determine how long will it take for the initial investment to double. Round your answer to the nearest year. \- Find and interpret the average rate of change of the amount in the account from the end of the fourth year to the end of the fifth year, and from the end of the thirty-fourth year to the end of the thirty-fifth year. Round your answer to two decimal places. $$\$ 500$$ is invested in an account which offers \(0.75 \%\), compounded continuously.

Step-by-Step Solution

Answer

Amount after 5, 10, 30, 35 years: \( \$519.00, \$538.40, \$621.53, \$645.16 \). Doubling time: 92 years. Average rate: 4-5 years \(3.87\), 34-35 years \(3.61\).

Step 1: Define the Compound Interest Formula

For continuous compounding, the formula for the amount in the account after time \( t \) years is given by \( A(t) = P e^{rt} \), where \( P \) is the principal amount, \( r \) is the annual interest rate, and \( e \) is Euler's number (approximately 2.71828).

Step 2: Apply the Given Values

In this problem, \( P = 500 \) and \( r = 0.0075 \). Using these values, the formula becomes \( A(t) = 500 e^{0.0075t} \). This represents the amount as a function of time \( t \).

Step 3: Determine Amount After Specific Years

Calculate \( A(t) \) for \( t = 5, 10, 30, \) and \( 35 \):\[\begin{align*}A(5) & = 500 e^{0.0075 \times 5} \ A(10) & = 500 e^{0.0075 \times 10} \ A(30) & = 500 e^{0.0075 \times 30} \ A(35) & = 500 e^{0.0075 \times 35} \end{align*}\]

Step 4: Calculate Doubling Time

To find when the investment doubles, set \( A(t) = 2P = 1000 \). Solve for \( t \) in \( 1000 = 500 e^{0.0075t} \):\[2 = e^{0.0075t} \]Taking natural logarithms of both sides gives:\[ t = \frac{\ln 2}{0.0075} \]

Step 5: Compute Average Rate of Change

The average rate of change from year 4 to year 5 is:\[ \frac{A(5) - A(4)}{5 - 4}\]And from year 34 to year 35:\[ \frac{A(35) - A(34)}{35 - 34} \]

Step 6: Numerical Calculations

Compute the numerical values. For \( t = 5 \), \( A(5) \approx 519.00 \). For \( t = 10 \), \( A(10) \approx 538.40 \). For \( t = 30 \), \( A(30) \approx 621.53 \). For \( t = 35 \), \( A(35) \approx 645.16 \). Doubling time \( t \approx 92 \) years. Average rate from year 4 to 5: \( \approx 3.87 \). Average rate from year 34 to 35: \( \approx 3.61 \).

Step 7: Interpretation

The amount in the account grows slightly each year due to compounding interest, reaching higher values over longer periods. Doubling time indicates the period required for the principal to double at the current interest rate. The average rate of change shows the growth over short time intervals, reflecting how growth rates might start higher and slightly decrease as time progresses.

Key Concepts

Continuous Compounding

Continuous compounding refers to the mathematical limit that compound interest can reach if it is calculated and set to add to the principal an infinite number of times over a fixed period. Unlike daily, monthly, or annual compounding, continuous compounding happens at every possible instant. This concept uses the formula:

• For the amount in the account, \( A(t) = P e^{rt} \), where:
• \( P \) is the principal investment amount,
• \( r \) is the annual interest rate,
• \( e \approx 2.71828 \) (Euler's number), and
• \( t \) is the time in years.

In our exercise, with a principal of \$500 and a continuous rate of \( 0.75\% \), we set up the formula as \( A(t) = 500 e^{0.0075t} \). As time progresses, the compounding effect results in exponential growth, which we will explore next.

Exponential Growth

Exponential growth describes the process where the value increases by a consistent percentage over equal increments of time. In finance, this concept is illustrated by the growth of investments with compound interest, where the value of the account increases more substantially over time. Using the continuous compounding formula, the exponential growth manifests as:

• The exponential function \( e^{rt} \) causes the investment to grow faster as time goes on.
• This "acceleration" of growth results from the interest earning interest over continuous periods.

For example, when \( t = 5 \), \( A(5) \approx \\(519.00 \); when \( t = 10 \), \( A(10) \approx \\)538.40 \). Notice how the amount for each successive period grows more than the previous period due to exponential growth.

Doubling Time

Doubling time is a key financial concept that helps investors understand how quickly an initial investment will double at a specific interest rate with continuous compounding. To find the doubling time, set the final amount \( A(t) \) to twice the initial investment \( 2P \) and solve:

• \( A(t) = 2P \)
• \( 2 = e^{0.0075t} \)
• Taking natural logarithms results in \( t = \frac{\ln 2}{0.0075} \)

For our exercise, the doubling time calculates to approximately 92 years. This means it will take about 92 years for the \\(500 investment to grow to \\)1000, given the current rate of interest compounded continuously.

Average Rate of Change

The average rate of change provides insight into how the investment's value changes over a specific time period, giving a glimpse of short-term growth trends. It is calculated using the formula \( \frac{A(b) - A(a)}{b - a} \), where \( a \) and \( b \) are time periods:

• From year 4 to 5: \( \frac{A(5) - A(4)}{1} \approx 3.87 \)
• From year 34 to 35: \( \frac{A(35) - A(34)}{1} \approx 3.61 \)

The decreased average rate from year 34 to 35 compared to year 4 to 5 indicates a slower growth rate as time progresses, which is a common pattern in continuously compounding interest scenarios. It reflects how the power of interest diminishes relative to the accumulated amount over time.