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Chapter 15: Problem 6

Suppose you invest \(\$ 10,000\) in an account with a nominal annual interest rate of \(5 \%\). How much money will you have 10 years later if the interest is compounded (a) quarterly? (b) daily? (c) continuously?

Short Answer

Expert verified
After 10 years, \$ 10,000 becomes (a) approximately \$ 16,386.16 when compounded quarterly; (b) \$ 16,470.09 when compounded daily; (c) \$ 16,486.74 when compounded continuously.

Step by step solution

01

Understanding the Compound Interest Formula

The formula for compound interest is \(A = P (1 + r/n) ^ {nt}\), where:\n\n\(A\) = the amount of money accumulated after n years, including interest.\n\(P\) = principle amount (the initial money you start with).\n\(r\) = annual interest rate (in decimal).\n\(n\) = number of times that interest applied per time period.\n\(t\) = time the money is invested for in years.\nIn this case, Principle Amount \(P = \$ 10,000\), nominal annual interest rate \(r = 5% = 0.05\) and time \(t = 10\) years.
02

Calculating Quarterly Compound Interest

For quarterly compounding, interest is applied four times a year. So, \(n = 4\). Substituting \(P = \$10,000\), \(r = 0.05\), \(n = 4\) and \(t = 10\) into the compound interest formula, we get the accumulated amount \(A = \$10,000 * (1 + 0.05/4) ^ {(4*10)}\).
03

Computing Daily Compound Interest

For daily compounding, interest is applied 365 times a year. So, \(n = 365\). Substituting \(P = \$10,000\), \(r = 0.05\), \(n = 365\) and \(t = 10\) into the compound interest formula, we get the accumulated amount \(A = \$10,000 * (1 + 0.05/365) ^ {(365*10)}\).
04

Figuring Out Continuously Compounded Interest

The formula for continuous compound interest is \(A = Pe^{rt}\), where \(e\) is the base of the natural logarithm (\(~2.7183\)). Substituting \(P = \$10,000\), \(r = 0.05\) and \(t = 10\) into this formula, we get the accumulated amount \(A = \$10,000 * e^{(0.05*10)}\).
05

Perform the Calculations

Now, compute the amounts on a calculator to get the final answers.

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Key Concepts

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

Annual Interest Rate
The annual interest rate is pivotal as it determines how much interest you earn on an investment or pay on a loan over the span of a year. It is expressed as a percentage of the principal amount, which is the initial sum borrowed or invested. In our exercise, the annual interest rate is given as 5%. To apply this rate in calculations, it's converted into a decimal by dividing by 100, resulting in 0.05. This rate is then used in the compound interest formula to compute the growth of the investment over time.
When comparing investment options or loans, the annual interest rate serves as a crucial benchmark, although it doesn't account for the frequency of compounding, which can significantly affect the total interest accumulated or paid.
Compounding Frequency
Compounding frequency refers to the number of times interest is applied to the principal in a year. Common frequencies include annually, semi-annually, quarterly, monthly, daily, or continuously. Higher compounding frequencies can lead to more substantial investment growth since interest is calculated on an ever-increasing balance. In our exercise, the scenarios of quarterly and daily compounding demonstrate this concept.
  • Quarterly compounding means interest is added four times a year (n=4).
  • Daily compounding implies interest is calculated each day of the year (n=365).
The compound interest formula incorporates n to show the effect of these different frequencies. As n increases, so does the amount of interest accrued, illustrating the power of compounding on investments.
Exponential Growth
Exponential growth in finance refers to the increase in investment value at a rate that becomes ever more rapid in proportion to the growing total number. It is distinctly different from linear growth, where a fixed amount is added over time. In the context of compound interest, the investment grows exponentially because the interest earned in each period is added to the principal, which then earns additional interest.
The compound interest formula represents this exponential growth through the exponent nt, where the principal amount is multiplied by the growth factor (1+r/n) raised to the power of the total number of compounding periods. This exponential factor is why compounding frequency has a significant impact on your investment's growth over time.
Continuously Compounded Interest
Continuously compounded interest is the mathematical limit of the compound interest formula where the number of compounding periods per year approaches infinity. It shows the most aggressive interest accumulation on an investment. The formula for continuous compounding, represented by Pe^(rt), where e is Euler's number (~2.7183), is derived from the general compound interest formula. Unlike regular compounding, which occurs at specific intervals, continuous compounding accrues at every moment.
In our example, continuous compounding for a 10-year investment at an annual interest rate of 5% is computed using this unique formula. The result is slightly higher than with daily or quarterly compounding, highlighting how continuous growth can incrementally increase an investment's return.

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Most popular questions from this chapter

Solve the following differential equations. Use substitution to convert them to the form \(\frac{d y}{d t}=k y\) (a) \(\frac{d y}{d t}=3 y-6\) (b) \(\frac{d y}{d t}=y+1\) (c) \(\frac{d y}{d t}=4-2 y\)

When a population has unlimited resources and is free from disease and strife, the rate at which the population grows is proportional to the population. Assume that both the bee and the mosquito populations described below behave according to this model. In both scenarios you are given enough information to find the proportionality constant \(k\). In one case, the information allows you to find \(k\) solely using the differential equation, without requiring that you solve it. In the other scenario, you must actually solve the differential equation in order to find \(k\). (a) Let \(M=M(t)\) be the mosquito population at time \(t, t\) in weeks. At \(t=0\), there are 1000 mosquitoes. Suppose that when there are 5000 mosquitoes, the population is growing at a rate of 250 mosquitoes per week. Write a differential equation reflecting the situation. Include a value for \(k\), the proportionality constant. (b) Let \(B=B(t)\) be the bee population at time \(t, t\) in weeks. At \(t=0\), there are 600 bees. When \(t=10\), there are 800 bees. Write a differential equation reflecting the situation. Include a value for \(k\), the proportionality constant.

Newton's law of cooling in its more general form tells us that the rate at which the temperature between an object and its environment changes is proportional to the difference in temperatures. In other words, if \(D(t)\) is the temperature difference, then \(\frac{d D}{d t}=k D\) (a) Solve the differential equation \(\frac{d D}{d t}=k D\) for \(D(t)\). (b) Suppose a hot object is placed in a room whose temperature is kept constant at \(R\) degrees. Let \(T(t)\) be the temperature of the object. Newton's law says that the hot object will cool at a rate proportional to the difference in temperature between the object and its environment. Write a differential equation reflecting this statement and involving \(T\). Explain why this differential equation is equivalent to the previous one. (c) What is the sign of the constant of proportionality in the equation you wrote in part (b)? Explain. (d) Suppose that instead of a hot object we now consider a cold object. Suppose that we are interested in the temperature of a cold cup of lemonade as it warms up to room temperature. Let \(L(t)\) represent the temperature of the lemonade at time \(t\) and assume that it sits in a room that is kept at 65 degrees. At time \(t=0\), the lemonade is at 40 degrees. 15 minutes later it has warmed to 50 degrees. i. Sketch a graph of \(L(t)\) using your intuition and the information given. ii. Is \(L(t)\) increasing at an increasing rate, or a decreasing rate?

Evaluate the following limits. Keep in mind that limit calculations can be subtle-don't ad lib, but instead keep the limits we looked at firmly in your mind and use substitution in order to make the transfer to the problems here. You can determine whether or not your answer is in the ballpark by using your calculator. (a) \(\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}\) (b) \(\lim _{s \rightarrow \infty}\left(1+\frac{1}{s}\right)^{3 s}\) (c) \(\lim _{r \rightarrow \infty}\left(1+\frac{0.3}{r}\right)^{r}\) (d) \(\lim _{w \rightarrow \infty}\left(1+2 w^{-1}\right)^{w}\) (e) \(\lim _{w \rightarrow \infty}\left(1+(2 w)^{-1}\right)^{w}\)

Suppose that a person invests \(\$ 10,000\) in a venture that pays interest at a nominal rate of \(8 \%\) per year compounded quarterly for the first 5 years and \(3 \%\) per year compounded quarterly for the next 5 years. (a) How much does the \(\$ 10,000\) grow to after 10 years? (b) Suppose there were another investment option that paid interest quarterly at a constant interest rate \(r\). What would \(r\) have to be for the two plans to be equivalent, ignoring taxes? (c) If an investment scheme paid \(3 \%\) interest compounded quarterly for the first 5 years and \(8 \%\) interest compounded quarterly for the next 5 years, would it be better than, worse than, or equivalent to the first scheme?
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