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If \(\$ 1\) is invested over a 10-year period, then the balance \(A\), where \(t\) represents the time in years, is given by \(A=1+0.06[t]\) or \(A=[1+(0.055 / 365)]^{[365 r]}\) depending on whether the interest is simple interest at \(6 \%\) or compound interest at \(5 \frac{1}{2} \%\) compounded daily. Use a graphing utility to graph each function in the same viewing window. Which grows at a greater rate?

Short Answer

Expert verified
The graph which ends higher at the end of the 10 year period represents the model that grows at a greater rate. The graph of these functions will definitively answer whether the simple or compound interest represent the higher growth over 10 years.

Step by step solution

01

Setting up the simple interest function

The function to represent the simple interest over time (in years) is \(A=1+0.06t\).
02

Setting up the compound interest function

The function to represent the compound interest over time (in years) is \(A=[1+(0.055 / 365)]^{[365 t]}\). This function takes into account that the interest is calculated daily.
03

Graphing the functions

To visualize these functions, use a graphing tool. Simply enter the functions as they are derived and plot them over a 10 year period. In most graphing utilities, this can be done by typing in the function and setting the x-axis (which represents time in this case) to range from 0 to 10.
04

Analyzing the graphs

Once the functions have been graphed, observe which line is higher at the end of the 10 year period (at \(t=10\)). The line that ends higher indicates the model which results in more growth over the 10 year period.

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Key Concepts

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

Simple Interest Formula
Simple interest is a financial concept where the interest charge is calculated only on the principal amount, or the initial sum of money. The formula used for calculating simple interest is given by:

\( I = P \times r \times t \),

where \( I \) represents the interest earned, \( P \) is the principal amount (the initial sum of money), \( r \) is the annual interest rate in decimal, and \( t \) is the time in years. For instance, if you invest \( 1 \) dollar at a simple interest rate of 6%, the function that represents your balance over time would be \( A=1+0.06t \). Here, each year, the interest is calculated as 6% of the initial \( 1 \) dollar.
Compound Interest Formula
Unlike simple interest that is only calculated on the principal, compound interest is calculated on the principal amount and the accumulated interest of prior periods. The compound interest formula can look daunting at first, but it simply reflects the process of interest accumulating over time. It is given by:

\( A = P(1 + \frac{r}{n})^{nt} \),

where \( A \) is the amount of money obtained after \( t \) years including interest, \( P \) is the principal amount, \( r \) is the annual interest rate (in decimal form), \( n \) is the number of times that interest is compounded per year, and \( t \) is the time the money is invested or borrowed for, in years. In the given problem, the annual rate is 5.5% or 0.055, and since the interest is compounded daily, \( n = 365 \). Therefore, the function for compound interest becomes \( A=[1+(0.055 / 365)]^{[365t]} \).
Graphing Financial Functions
Graphing financial functions like those representing simple and compound interest is an effective way to visualize the growth of investments over time. To do this, you can enter the simple and compound interest functions into a graphing calculator or software, setting the horizontal axis to represent time (in years) and the vertical axis to represent the balance.

For our simple interest function \( A=1+0.06t \), we'll see a straight line that steadily slopes upwards. On the other hand, the compound interest graph \( A=[1+(0.055 / 365)]^{[365t]} \) will curve upwards, increasing faster as time goes on. This is because the interest is being calculated on a growing balance due to compounding. It's important to have both functions in the same viewing window to properly compare their growth over the same period.
Analyzing Interest Rates
Analyzing interest rates is crucial in understanding how fast your money can grow through different investment options. A higher interest rate does not automatically mean greater returns, especially when comparing simple and compound interests.

Simple interest tends to result in slower growth since it's always calculated on the initial principal. Compound interest, especially when compounded frequently (like daily), can lead to exponential growth. By graphing and comparing, you can analyze which investment grows at a greater rate. Normally, compound interest will outpace simple interest over time, as the effect of compounding becomes increasingly significant. In the exercise, by observing which line (simple or compound interest) is higher at the end of 10 years, you conclude which investment would yield more. This analytical skill can be beneficial for making informed financial decisions in the real world.

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Most popular questions from this chapter

Comparing Models A cup of water at an initial temperature of \(78^{\circ} \mathrm{C}\) is placed in a room at a constant temperature of \(21^{\circ} \mathrm{C} .\) The temperature of the water is measured every 5 minutes during a half-hour period. The results are recorded as ordered pairs of the form \((t, T),\) where \(t\) is the time (in minutes) and \(T\) is the temperature (in degrees Celsius). $$ \begin{array}{l}{\left(0,78.0^{\circ}\right),\left(5,66.0^{\circ}\right),\left(10,57.5^{\circ}\right),\left(15,51.2^{\circ}\right)} \\\ {\left(20,46.3^{\circ}\right),\left(25,42.4^{\circ}\right),\left(30,39.6^{\circ}\right)}\end{array} $$ (a) The graph of the model for the data should be asymptotic with the graph of the temperature of the room. Subtract the room temperature from each of the temperatures in the ordered pairs. Use a graphing utility to plot the data points \((t, T)\) and \((t, T-21)\) . (b) An exponential model for the data \((t, T-21)\) is given by \(T-21=54.4(0.964)^{t} .\) Solve for \(T\) and graph the model. Compare the result with the plot of the original data. (c) Take the natural logarithms of the revised temperatures. Use the graphing utility to plot the points \((t, \ln (T-21))\) and observe that the points appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. This rasulting line has the form \(\ln (T-21)=a t+b\) . Solve for \(T,\) and verify that the result is equivalent to the model in part (b). (d) Fit a rational model to the data. Take the reciprocals of the \(y\) -coordinates of the revised data points to generate the points $$ \left(t, \frac{1}{T-21}\right) $$ Use the graphing utility to graph these points and observe that they appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. The resulting line has the form $$ \frac{1}{T-21}=a t+b $$ Solve for \(T,\) and use the graphing utility to graph the rational function and the original data points. (e) Why did taking the logarithms of the temperatures lead to a linear scatter plot? Why did taking the reciprocals of the temperatures lead to a linear scatter plot?

Using Properties of Logarithms In Exercises \(15-20\) , use the properties of logarithms to rewrite and simplify the logarithmic expression. $$\ln \left(5 e^{6}\right)$$

Using Properties of Logarithms In Exercises \(21-36,\) find the exact value of the logarithmic expression without using a calculator. (If this is not possible, then state the reason.) $$\log _{4} 16^{2}$$

Graphical Analysis Use a graphing utility to graph each pair of functions in the same viewing window. Describe any similarities and differences in the graphs. (a) \(y_{1}=2^{x}, y_{2}=x^{2}\) (b) \(y_{1}=3^{x}, y_{2}=x^{3}\)

Condensing a Logarithmic Expression In Exercises \(67-82\) , condense the expression to the logarithm of a single quantity. $$2 \ln 8+5 \ln (z-4)$$

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