Problem 3 - 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 monthly.

Step-by-Step Solution

Answer

After 5 years: $1064.05, after 10 years: $1132.70, after 30 years: $1443.78, after 35 years: $1535.05. It takes about 55 years to double. Average rate: $12.74 from year 4 to 5, $13.43 from year 34 to 35.

Step 1: Identify the Compound Interest Formula

The general formula for compound interest 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 (initial investment).- \(r\) is the annual interest rate (decimal).- \(n\) is the number of times that interest is compounded per year.- \(t\) is the time in years. In this problem, \(P = 1000\), \(r = 0.0125\), and \(n = 12\) because interest is compounded monthly.

Step 2: Write the Function

Using the compound interest formula from Step 1, we identify the function for the amount \(A\) in the account as:\[ A(t) = 1000 \left(1 + \frac{0.0125}{12}\right)^{12t} \]This function models the amount in the account in terms of the years \(t\).

Step 3: Calculate Amount for Each Year

To find the amount after 5, 10, 30, and 35 years, substitute \(t\) with these values in \(A(t)\):- After 5 years: \[ A(5) = 1000 \left(1 + \frac{0.0125}{12}\right)^{60} \approx 1064.05 \]- After 10 years: \[ A(10) = 1000 \left(1 + \frac{0.0125}{12}\right)^{120} \approx 1132.70 \]- After 30 years: \[ A(30) = 1000 \left(1 + \frac{0.0125}{12}\right)^{360} \approx 1443.78 \]- After 35 years: \[ A(35) = 1000 \left(1 + \frac{0.0125}{12}\right)^{420} \approx 1535.05 \]

Step 4: Determine Investment Double Time

To find when the investment doubles, solve for \(t\) in \(A(t) = 2000\):\[ 2000 = 1000 \left(1 + \frac{0.0125}{12}\right)^{12t} \]Divide both sides by 1000:\[ 2 = \left(1 + \frac{0.0125}{12}\right)^{12t} \]Take the natural logarithm of both sides:\[ \ln{2} = 12t \cdot \ln\left(1 + \frac{0.0125}{12}\right) \]Solve for \(t\):\[ t \approx \frac{\ln{2}}{12 \cdot \ln\left(1 + \frac{0.0125}{12}\right)} \approx 55 \text{ years} \]

Step 5: Calculate Average Rate of Change

The average rate of change is calculated as:\[ \text{Average rate of change} = \frac{A(t_2) - A(t_1)}{t_2 - t_1} \]- From year 4 to 5: \[ A(4) = 1000 \left(1 + \frac{0.0125}{12}\right)^{48} \approx 1051.31 \] \[ \text{Average rate} = \frac{A(5) - A(4)}{5-4} = \frac{1064.05 - 1051.31}{1} \approx 12.74 \]- From year 34 to 35: \[ A(34) = 1000 \left(rac{0.0125}{12}\right)^{408} \approx 1521.62 \] \[ \text{Average rate} = \frac{A(35) - A(34)}{35-34} = \frac{1535.05 - 1521.62}{1} \approx 13.43 \]

Key Concepts

Investment Growth

Investment growth is a process experienced when the value of an initial investment increases over time. This growth is typically influenced by factors such as interest rates, the compounding frequency, and the length of time the money is invested. In the case of this exercise, we are working with compound interest, which means the interest earned is added to the principal. Then, in subsequent periods, you earn interest on the new principal, including any interest paid. This creates a snowball effect where the investment grows progressively over time.

In our exercise, a sum of $1000 is invested with a 1.25% annual interest rate, compounded monthly. With compound interest, the amount of interest earned is reinvested each month, resulting in exponential growth over time.

• For example, after 5 years, this initial $1000 will increase to approximately $1064.05.
• After 10 years, the amount rises to about $1132.70.
• Continuing this pattern, after 30 years, it reaches $1443.78.
• Finally, after 35 years, the total is approximately $1535.05.

The pattern noticed here derives from the compound interest formula, where the investment grows at an increasing rate over time, surpassing what would be anticipated with simple interest.

Average Rate of Change

The average rate of change is an important tool in understanding how quickly a quantity is changing over a specified period. It gives us a sense of how much an entity, in this case, the investment amount, has increased or decreased per unit of time. In terms of our investment problem, it helps observe how much the investment's value shifts between two time points.

To calculate this, you subtract the initial value from the final value and then divide by the period over which the change happened.

• From the end of year 4 to the end of year 5, the investment grew from $1051.31 to $1064.05, leading to an average rate of change of about $12.74 per year.
• Similarly, from year 34 to year 35, the investment increased from $1521.62 to $1535.05, which results in an average rate of change of $13.43 per year.

These calculations show how the investment's growth speed varies slightly as time goes on, with a tendency to increase slowly over the decades due to the compounding effects and higher base amounts generating interest.

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. They are crucial in modeling processes where quantities grow or decay by a fixed percentage per unit of time, such as in compound interest situations. In our exercise, the exponential function describes how the investment value, relying on compound interest, increases over time.

The general form of an exponential growth function used here is \[ A(t) = P \left(1 + \frac{r}{n}\right)^{nt} \] where:

• \(P\) is the initial amount.
• \(r\) is the annual interest rate.
• \(n\) is the number of compounding periods per year.
• \(t\) is the time in years.

In this exercise, by plugging in \(P = 1000\), \(r = 0.0125\), and \(n = 12\), the function becomes \[ A(t) = 1000 \left(1 + \frac{0.0125}{12}\right)^{12t} \].

Through this function, you can determine how the investment evolves at any time \(t\). It highlights the nature of exponential growth, where the larger the quantity grows, the faster the increased compounding will accelerate its growth.