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Chapter 5: Problem 57

Compound Interest In Exercises 57-60, complete the table to determine the balance \(A\) for $$\$ 12,000$$ invested at rate \(r\) for \(t\) years, compounded continuously. \(\begin{array}{|l|l|l|l|l|l|} \hline t & 10 & 20 & 30 & 40 & 50 \\ \hline A & & & & & \\ \hline \end{array}\) $$ r=4 \% $$

Short Answer

Expert verified
The balances after 10, 20, 30, 40 and 50 years are approximately \$16,537.15, \$22,716.89, \$31,169.90, \$42,782.30 and \$58,686.13 respectively.

Step by step solution

01

Interpret the Exercise

Determine that you want to find the final amount \( A \) after an initial investment of \$12,000 at an interest rate of 4% compounded continuously for the durations of 10, 20, 30, 40, and 50 years.
02

Convert Percentage to Decimal

Convert 4% into decimal form for use in the continuous compound interest formula. This is done by dividing 4 by 100, giving a result of 0.04.
03

Perform Calculations

Use the continuous compound interest formula to calculate the final amount for each time interval. For 10 years, it's \( 12000 * e^{0.04 * 10} \). For 20 years, it's \( 12000 * e^{0.04 * 20} \) and so on for 30, 40, and 50 years.
04

Round the Results

Round the final answers to create a readable table of results.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuous Compounding
Continuous compounding is a method that lets an investment grow exponentially by calculating interest at every possible moment. Unlike regular compounding, which might occur annually, quarterly, or daily, continuous compounding assumes that accruing interest happens perpetually without pause. This means the frequency of compounding is infinite, theoretically providing an investor with more profits.

To compute continuously compounded interest, we use the formula
\[ A = P \cdot e^{rt} \]
where:
  • \( A \) is the final amount
  • \( P \) is the initial principal balance (\$12,000 in our example)
  • \( r \) is the annual interest rate (expressed as a decimal, so 4\% becomes 0.04)
  • \( t \) is the time the money is invested for, in years
The constant \( e \) is approximately 2.71828. This formula helps us understand how powerful continuous compounding can be for investment growth over extended periods of time.
Investment Growth
Investment growth with continuous compounding is fairly straightforward once we understand the formula. Using our example, imagine placing \$12,000 into an account with a 4\% interest rate compounded continuously. Over time, this investment will grow significantly due to the constant calculation of interest.

As time progresses, the amount of money accrued from the interest becomes substantial. For instance, at 10 years, the investment will grow to
\[ 12000 \cdot e^{0.4} \]
At 20 years, it will be
\[ 12000 \cdot e^{0.8} \]
Observe how the time factor \( t \) influences the result. In essence, the longer the investment period, the greater the growth due to the exponential nature of the formula.

Continuous compounding assures compound interest is computed at every possible time segment, furiously growing your investment like a snowball gaining mass as it rolls downhill.
Interest Rate Calculation
Calculating the interest rate in the context of continuous compounding is crucial for accurate financial projections. Interest rates are often presented as percentages. For calculation purposes, these percentages need to be converted into decimals.

In our exercise, a 4\% interest rate is transformed into a decimal by dividing by 100, resulting in 0.04. This step is fundamental since it standardizes the rate for use in continuous formulas.

Properly converting and using the interest rate allows us to plug this into the formula
\[ A = P \cdot e^{rt} \]
ensuring we obtain the accurate final balance of an investment. Even slight errors in interest rate calculations can lead to significant miscalculations in the amount of returned investment over long periods. The transformation from percentage to decimal should be precise and methodical to ensure success in quantitative evaluations.

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Most popular questions from this chapter

\(e^{-0.5}=0.6065 \ldots\)

\(e^{-4.1}=0.0165 \ldots\)

Carbon 14 dating assumes that the carbon dioxide on Earth today has the same radioactive content as it did centuries ago. If this is true, the amount of \({ }^{14} \mathrm{C}\) absorbed by a tree that grew several centuries ago should be the same as the amount of \({ }^{14} \mathrm{C}\) absorbed by a tree growing today. A piece of ancient charcoal contains only \(15 \%\) as much radioactive carbon as a piece of modern charcoal. How long ago was the tree burned to make the ancient charcoal if the half-life of \({ }^{14} \mathrm{C}\) is 5715 years?

A $$\$ 120,000$$ home mortgage for 35 years at \(7 \frac{1}{2} \%\) has a monthly payment of $$\$ 809.39$$. Part of the monthly payment is paid toward the interest charge on the unpaid balance, and the remainder of the payment is used to reduce the principal. The amount that is paid toward the interest is $$ u=M-\left(M-\frac{P r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t} $$ and the amount that is paid toward the reduction of the principal is $$ v=\left(M-\frac{P r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t} $$ In these formulas, \(P\) is the size of the mortgage, \(r\) is the interest rate, \(M\) is the monthly payment, and \(t\) is the time in years. (a) Use a graphing utility to graph each function in the same viewing window. (The viewing window should show all 35 years of mortgage payments.) (b) In the early years of the mortgage, is the larger part of the monthly payment paid toward the interest or the principal? Approximate the time when the monthly payment is evenly divided between interest and principal reduction. (c) Repeat parts (a) and (b) for a repayment period of 20 years \((M=\$ 966.71)\). What can you conclude?

In Exercises 73-78, (a) plot the points, (b) find the distance between the points, (c) find the midpoint of the line segment joining the points, and (d) find the slope of the line passing through the points. $$ (4,-3),(-6,1) $$
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