Problem 6 - 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. $$\$ 5000\( is invested in an account which offers \)2.125 \%$, compounded continuously.

Step-by-Step Solution

Answer

After 5 years: \$5623.64, 10 years: \$6327.90, 30 years: \$9478.70, 35 years: \$10638.21. It takes 33 years to double. Avg. rate change: Year 4-5: \$117.44, Year 34-35: \$164.95.

Step 1: Formula for Continuous Compounding

The formula for the amount in an account with continuous compounding is given by \( A(t) = P e^{rt} \), where \( P \) is the initial principal, \( r \) is the interest rate as a decimal, and \( t \) is the time in years. Here, \( P = 5000 \) and \( r = 0.02125 \). Thus, the function is \( A(t) = 5000 e^{0.02125t} \).

Step 2: Calculate Amounts for Given Years

Calculate \( A(t) \) for the specified years: 5, 10, 30, and 35.- For 5 years: \( A(5) = 5000 e^{0.02125 \times 5} \approx 5000 e^{0.10625} \approx 5623.64 \).- For 10 years: \( A(10) = 5000 e^{0.02125 \times 10} \approx 5000 e^{0.2125} \approx 6327.90 \).- For 30 years: \( A(30) = 5000 e^{0.02125 \times 30} \approx 5000 e^{0.6375} \approx 9478.70 \).- For 35 years: \( A(35) = 5000 e^{0.02125 \times 35} \approx 5000 e^{0.74375} \approx 10638.21 \).

Step 3: Determine Time to Double the Investment

To find how long it takes for the investment to double, solve \( 2P = P e^{rt} \) for \( t \). Simplify to \( 2 = e^{0.02125t} \). Take the natural logarithm of both sides: \( \ln(2) = 0.02125t \). Therefore, \( t = \frac{\ln(2)}{0.02125} \approx 32.64 \) years. Rounded to the nearest year, it will take 33 years to double the investment.

Step 4: Average Rate of Change from Year 4 to Year 5

Calculate the average rate of change from years 4 to 5: \( \frac{A(5) - A(4)}{5 - 4} \).- First, find \( A(4) \): \( A(4) = 5000 e^{0.02125 \times 4} \approx 5000 e^{0.085} \approx 5506.20 \).- Then, \( A(5) = 5623.64 \).- The average rate of change is \( \frac{5623.64 - 5506.20}{1} = 117.44 \). So, the account increases by an average of \$117.44 from year 4 to year 5.

Step 5: Average Rate of Change from Year 34 to Year 35

Calculate the average rate of change from years 34 to 35: \( \frac{A(35) - A(34)}{35 - 34} \).- Find \( A(34) \): \( A(34) = 5000 e^{0.02125 \times 34} \approx 5000 e^{0.7225} \approx 10473.26 \).- \( A(35) = 10638.21 \).- The average rate of change is \( \frac{10638.21 - 10473.26}{1} = 164.95 \). Thus, the account increases by an average of \$164.95 from year 34 to year 35.

Key Concepts

Exponential Growth

Exponential growth is a powerful concept in mathematics and finance that describes how quantities can increase dramatically over time. In the context of continuous compounding, exponential growth helps us understand how an investment can grow over time when interest is continuously added to the principal.

The idea is that the interest earned not only applies to the initial amount invested, or principal, but also to the interest that has already been accumulated. This results in the total value growing at an increasing rate.

• Imagine planting a tree—it starts small, but over time, its growth accelerates as it sheds seeds that grow into new trees. This is similar to how continuously compounding interest works.
• The mathematical representation of exponential growth using continuous compounding is given by the formula: \( A(t) = Pe^{rt} \).

This formula describes how the amount \( A(t) \) in an account changes over time \( t \), where \( P \) is the principal and \( r \) is the interest rate. As \( t \) increases, \( A(t) \) grows exponentially. Continuous compounding effectively allows the investment to "multiply" over time, mimicking nature's way of expanding populations.

Interest Rate Formula

The interest rate formula is a critical component in calculating how much an investment will grow over time, especially when compounded continuously. When given an initial investment, known as the principal \( P \), and a continuously compounded interest rate \( r \), we can use the following equation:

The formula for continuous compounding is: \[ A(t) = Pe^{rt} \]

- **\( P \)**: Principal or initial amount invested.- **\( r \)**: Annual interest rate expressed as a decimal (e.g., 2.125% becomes 0.02125).- **\( e \)**: Euler's number, approximately equal to 2.71828, a constant that serves as the base of the natural logarithms.- **\( t \)**: Time the money is invested for, in years.

This formula allows us to calculate the future value of an investment given the continuous nature of compounding. It showcases the power of exponential functions in determining wealth accumulation and can effectively forecast the future worth of today's investments.

Understanding this formula enables investors to better plan for their financial futures by providing a clear view of how interest rates affect investments over time.

Doubling Time

Doubling time in finance refers to the period required for an investment to grow to twice its size, based on a fixed annual interest rate and continuous compounding. This concept is vital for investors who want to understand when their investments will significantly increase without needing regular deposits.

To determine the doubling time, we use a rearranged form of the continuous compounding formula:\[ 2P = Pe^{rt} \]Simplifying gives us:\[ 2 = e^{rt} \]
By taking the natural logarithm of both sides, we obtain:\[ \ln(2) = rt \]Solving for \( t \):\[ t = \frac{\ln(2)}{r} \]
This tells us the time it will take for an investment to double, given a specific interest rate \( r \). For example, if \( r = 0.02125 \), then:\[ t = \frac{\ln(2)}{0.02125} \approx 32.64 \] years
This method shows that, under a continuous compounding interest rate, it will take approximately 33 years for a $5000 investment to double. The notion of doubling time provides a clear and tangible metric to consider long-term financial growth outcomes.

Average Rate of Change

The average rate of change is a valuable concept for understanding how a quantity, such as an investment, changes over a specific time period. It provides insights into the performance of an investment over short durations within a longer timeframe.

Mathematically, the average rate of change of an investment from time \( t_1 \) to \( t_2 \) is defined as:\[ \frac{A(t_2) - A(t_1)}{t_2 - t_1} \]

This formula gives the average amount by which the investment increased per year during the specified time interval. For example:

• From year 4 to year 5, using previously calculated figures, the average rate of change was \\(117.44.
• From year 34 to year 35, the average rate of change was \\)164.95.

By understanding these figures, investors can evaluate how much growth their investment experienced between specific time periods and identify trends over different phases of their investment journey. Recognizing shifts in the average rate of change can indicate potential adjustments in economic conditions influencing the investment.