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Chapter 4: Problem 13

Find the value of $$\$ 1000$$ deposited in a bank at \(10 \%\) interest for 8 years compounded: a. annually. b. quarterly. c. continuously.

Short Answer

Expert verified
a. \(\$ 2143.60\), b. \(\$ 2219.64\), c. \(\$ 2225.54\)

Step by step solution

01

Understanding Compounding Annually

When interest is compounded annually, it is added to the principal at the end of each year. The formula for annual compounding is: \[ A = P (1 + r)^n \] 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 years.
02

Calculate Annually Compounded Amount

For annually compounding, use the given values: - \(P = 1000\)- \(r = 0.1\) (10%) - \(n = 8\)Plug these into the formula:\[ A = 1000 (1 + 0.1)^8 \] Now, calculate:\[ A = 1000 (1.1)^8 \] \[ A = 1000 \times 2.1436 = 2143.60 \] Thus, the amount after 8 years is \(\$ 2143.60\).
03

Understanding Compounding Quarterly

Quarterly compounding means the interest is added four times a year. The formula becomes:\[ A = P \left(1 + \frac{r}{n_t}\right)^{n_t \times n} \] where:- \(n_t\) is the number of compounding periods per year.
04

Calculate Quarterly Compounded Amount

Here, use:- \(P = 1000\)- \(r = 0.1\) - \(n_t = 4\)- \(n = 8\)Plug these into the formula:\[ A = 1000 \left(1 + \frac{0.1}{4}\right)^{4 \times 8} \] Simplify and solve:\[ A = 1000 \left(1.025\right)^{32} \] \[ A \approx 1000 \times 2.21964 = 2219.64 \] The amount after 8 years is \(\$2219.64\).
05

Understanding Continuous Compounding

Continuous compounding uses a different formula, which is:\[ A = P e^{rt} \] where:- \(e\) is the base of the natural logarithm (approximately 2.71828).- \(t\) is the time the money is invested for in years.
06

Calculate Continuously Compounded Amount

Implement the continuous compounding formula with:- \(P = 1000\)- \(r = 0.1\)- \(t = 8\)Plug these into the formula:\[ A = 1000 e^{0.1 \times 8} \] Calculate:\[ A = 1000 e^{0.8} \]Using \(e^{0.8} \approx 2.22554\), we get:\[ A \approx 1000 \times 2.22554 = 2225.54 \] The amount after 8 years is \(\$2225.54\).

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Key Concepts

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

Annual Compounding
Annual compounding is a method where the interest is calculated once a year. It is added to the principal amount at the end of each year. This method is straightforward and makes it easy to predict the growth of an investment over time.

To calculate the total amount after a certain number of years with annual compounding, we use the formula:
  • \[ A = P (1 + r)^n \]
Here:
  • \(A\) is the total amount after \(n\) years, including the interest.
  • \(P\) is the initial amount or principal.
  • \(r\) is the annual interest rate expressed as a decimal.
  • \(n\) is the number of years the money is invested or borrowed for.
Let's consider the example where \(P = 1000\) dollars, \(r = 0.1\), and \(n = 8\). By plugging these values into the formula, we calculate:
  • \[ A = 1000 (1 + 0.1)^8 = 1000 \times 1.1^8 \]
  • \[ A = 1000 \times 2.1436 \approx 2143.60 \]
After 8 years, the investment grows to approximately \$2143.60 with annual compounding.
Quarterly Compounding
Quarterly compounding involves the interest being calculated and added to the principal four times a year, or every quarter. This means the interest is calculated more frequently, leading to potentially greater growth of the investment compared to annual compounding.

For quarterly compounding, the formula used is:
  • \[ A = P \left(1 + \frac{r}{n_t}\right)^{n_t \times n} \]
Where:
  • \(n_t\) is the number of times the interest is compounded per year. For quarterly, \(n_t = 4\).
The formula factors in the more frequent application of interest. Using the example values \(P = 1000\), \(r = 0.1\), \(n_t = 4\), and \(n = 8\):
  • \[ A = 1000 \left(1 + \frac{0.1}{4}\right)^{4 \times 8} \]
  • \[ A = 1000 (1.025)^{32} \approx 1000 \times 2.21964 = 2219.64 \]
Thus, in 8 years, the investment becomes approximately \$2219.64 through quarterly compounding.
Continuous Compounding
Continuous compounding is an advanced way of calculating interest, assuming that interest is constantly being added to the principal. This means that interest is applied at an infinitely small period intervals leading to the formula involving the mathematical constant \(e\) (approximately 2.71828).

The formula for continuous compounding is:
  • \[ A = P e^{rt} \]
Here:
  • \(e\) is the base of the natural logarithm.
  • \(t\) is the time in years.
For an investment with \(P = 1000\) dollars, \(r = 0.1\), and \(t = 8\), the calculation goes as follows:
  • \[ A = 1000 e^{0.1 \times 8} = 1000 e^{0.8} \]
  • \(e^{0.8} \approx 2.22554\)
  • \[ A = 1000 \times 2.22554 \approx 2225.54 \]
Therefore, using continuous compounding, after 8 years, the amount accumulates to approximately \$2225.54.

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Most popular questions from this chapter

The following problems extend and augment the material presented in the text. a. Show that for a demand function of the form \(D(p)=c / p^{n}\), where \(c\) and \(n\) are positive constants, the elasticity is constant. b. What type of demand function has elasticity equal to 1 for every value of \(p\) ?

The fastest times for the marathon ( \(26.2\) miles) for male runners aged 35 to 80 are approximated by the function $$ f(x)=\left\\{\begin{array}{ll} 106.2 e^{0.0063 x} & \text { if } x \leq 58.2 \\ 850.4 e^{0.000614 x^{2}-0.0652 x} & \text { if } x>58.2 \end{array}\right. $$ in minutes, where \(x\) is the age of the runner. a. Graph this function on the window \([35,80]\) by \([0,240] .\) [Hint: On some graphing calculators, enter \(y_{1}=\left(106.2 e^{0.0063 x}\right)(x \leq 58.2)+\) \(\left.\left(850.4 e^{0.000614 x^{2}-0.0652 x}\right)(x>58.2) .\right]\) b. Find \(f(35)\) and \(f^{\prime}(35)\) and interpret these numbers. [Hint: Use NDERIV or \(d y / d x .\) ] c. Find \(f(80)\) and \(f^{\prime}(80)\) and interpret these numbers.

A European oil-producing country estimates that the demand for its oil (in millions of barrels per day) is \(D(p)=3.5 e^{-0.06 p}\), where \(p\) is the price of a barrel of oil. To raise its revenues, should it raise or lower its price from its current level of $$\$ 120$$ per barrel?

The world population (in billions) is predicted to be \(P(t)=6.45 e^{0.0175 t}\), where \(t\) is the number of years after \(2005 .\) Find the instantaneous rate of change of the population in the year 2015 .

For which of the following two items would you expect demand to be elastic and for which inelastic: heating oil, olive oil.
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