Problem 4 - 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. $$\$ 1000$$ is invested in an account which offers \(1.25 \%\), compounded continuously.

Step-by-Step Solution

Answer

After 5 yrs: $1064.93; 10 yrs: $1133.15; 30 yrs: $1420.34; 35 yrs: $1509.65. It doubles in ~55 yrs. Avg. rate of change is 13.23 after 5 yrs, 17.66 after 35 yrs.

Step 1: Identify the formula for continuous compounding

The formula for the amount in an account with continuous compounding is given by: \[ A = P e^{rt} \]where:- \( A \) is the amount in the account after time \( t \),- \( P \) is the principal amount (\$1000 in this case),- \( r \) is the annual interest rate (1.25% = 0.0125),- \( t \) is the time in years,- \( e \) is the base of the natural logarithm (approximately 2.71828).

Step 2: Formulate the function for the investment over time

By substituting the given values into the formula, we have:\[ A(t) = 1000 e^{0.0125t} \]This equation represents the amount in the account as a function of time \( t \) years into the investment.

Step 3: Calculate the amount after specific years

Substitute \( t = 5, 10, 30, \text{ and } 35 \) into the function \( A(t) = 1000 e^{0.0125t} \) to find the corresponding amounts:- After 5 years: \[ A(5) = 1000 e^{0.0125 imes 5} \approx 1064.93 \]- After 10 years: \[ A(10) = 1000 e^{0.0125 imes 10} \approx 1133.15 \]- After 30 years: \[ A(30) = 1000 e^{0.0125 imes 30} \approx 1420.34 \]- After 35 years: \[ A(35) = 1000 e^{0.0125 imes 35} \approx 1509.65 \]

Step 4: Determine when the investment doubles

Set \( A = 2000 \) in the formula and solve for \( t \):\[ 2000 = 1000 e^{0.0125t} \]\[ 2 = e^{0.0125t} \]Take the natural logarithm of both sides:\[ \ln(2) = 0.0125t \]\[ t = \frac{\ln(2)}{0.0125} \approx 55.45 \]Therefore, it will take approximately 55 years for the initial investment to double.

Step 5: Calculate average rate of change for specified intervals

The average rate of change from year 4 to year 5 is:\[ \frac{A(5) - A(4)}{5 - 4} = \frac{1064.93 - 1051.70}{1} \approx 13.23 \text{ dollars per year} \]The average rate of change from year 34 to year 35 is:\[ \frac{A(35) - A(34)}{35 - 34} = \frac{1509.65 - 1491.99}{1} \approx 17.66 \text{ dollars per year} \]This means the investment grows faster in later years as evidenced by the increase in rate.

Key Concepts

Investment Growth

When you invest money in an account where interest is compounded, your money grows over time. The speed of this growth depends on the interest rate and how often the interest is compounded. Continuous compounding is a special process where interest is added to your principal continuously, making the amount you have grow very, very smoothly over time. For continuous compounding, the formula to calculate the future amount of your investment is given by:

• \( A = P e^{rt} \)

Here:

• \( A \) is the amount of money you'll have after a certain period \( t \).
• \( P \) is the initial principal or the original amount of your investment.
• \( r \) is the annual interest rate, expressed as a decimal.
• \( e \) is a constant, approximately equal to 2.71828, which is the base of the natural logarithm.
• \( t \) is the time in years that the money is invested for.

Using this formula, you can find the value of your investment after any number of years. This method takes the power of compounding to the next level by calculating growth continuously, maximizing the amount you end up with over time.

Doubling Time

Doubling time is the period it takes for an investment to grow to twice its original size. For continuous compounding, determining the doubling time involves using the logarithmic form of the continuous compounding formula. You set the future amount \( A \) to twice the principal \( P \), i.e., \( 2P \), and solve for \( t \):

• \( 2 = e^{rt} \)
• Taking the natural logarithm of both sides gives us: \( \ln(2) = rt \)
• Then you can solve for \( t \): \( t = \frac{\ln(2)}{r} \)

In the exercise scenario, an investment of \$1000 in an account with a 1.25% annual interest rate will double its value in approximately 55 years. This exponential process reflects how powerfully compounding works over extended durations.

Average Rate of Change

The average rate of change in any function helps you understand how quickly something is growing over a specific period. For investments, it tells us how much the investment grows on average over a given timeframe. It's calculated using this formula:

• \( \text{Average rate of change} = \frac{A(t_2) - A(t_1)}{t_2 - t_1} \)

Here, \( A(t_2) \) and \( A(t_1) \) are the amounts of money at the end of times \( t_2 \) and \( t_1 \) respectively. For instance, if you find that the investment changes from \( \approx 1064.93 \) dollars to \( \approx 1509.65 \) dollars between the 5th and 35th years, your average rate of change is higher as time progresses, showing that compounding becomes more effective the longer it operates. Understanding this growth behavior is crucial in predicting and planning financial futures.

Natural Logarithm

A natural logarithm (logarithm with base \( e \)) is abbreviated as \( \ln \). It plays a crucial role in solving problems involving continuous growth, like continuous compounding. The constant \( e \) (approximately 2.71828) is the foundation for natural logs and is particularly special in mathematics. Using natural logarithms makes solving equations involving exponential growth or decay straightforward.For instance, in the context of determining the doubling time of an investment, the step involving the natural logarithm is:

• \( \ln(2) = 0.0125t \)
• Solving for \( t \) gives \( t = \frac{\ln(2)}{0.0125} \)

This form is crucial for interpreting exponential growth and decay, especially when working with time-bound financial projections. The natural log applies broadly, from natural sciences to financial calculations, highlighting its versatility in growth analysis.