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Chapter 3: Problem 94

A principal \(P\), invested at \(5 \frac{1}{2} \%\) and compounded continuously, increases to an amount \(K\) times the original principal after \(t\) years, where \(t\) is given by \(t=(\ln K) / 0.055\). (a) Complete the table and interpret your results. $$ \begin{array}{|l|l|l|l|l|l|l|l|} \hline K & 1 & 2 & 4 & 6 & 8 & 10 & 12 \\ \hline t & & & & & & & \\ \hline \end{array} $$ (b) Sketch a graph of the function.

Short Answer

Expert verified
Using the given equation for \(t\), \(K=1\) corresponds to \(t=0\), \(K=2\) corresponds to \(t\approx12.71\), and so on. This shows that more time is needed for larger growth factors of the principal. The graph will be a curve reflecting the natural logarithm function, increasing as \(K\) increases, and only defined for \(K>0\).

Step by step solution

01

Calculate the value of \(t\) for each \(K\)

To complete the table begin by substituting each value of \(K\) into the equation \(t=(\ln K) / 0.055\) to get the corresponding value of \(t\). For instance, for \(K=1\), the calculation will be \(t = (\ln 1) / 0.055 = 0\). Repeat this process for all the other values of \(K\) in the table.
02

Complete the table

Once you calculate the time \(t\) for each value of \(K\), you can now complete the table. The completed table will look like this: \[\begin{array}{|l|l|l|l|l|l|l|l|} \hline K & 1 & 2 & 4 & 6 & 8 & 10 & 12 \ \hline t & 0 & (\ln 2) / 0.055 & (\ln 4) / 0.055 & (\ln 6) / 0.055 & (\ln 8) / 0.055 & (\ln 10) / 0.055 & (\ln 12) / 0.055 \ \hline \end{array} \]
03

Interpret results

Looking at the results in the table, one can notice that the time \(t\) increases as \(K\) increases. This means that more time is required for larger growth factors of the principal.
04

Sketch the Graph

To sketch the graph, use values of \(K\) on the x-axis and corresponding values of \(t\) on the y-axis. This will result in a curve that reflects the natural logarithm function. Please note that this functions only accepts positive values for \(K\).

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Key Concepts

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

Exponential Growth
Exponential growth describes how quantities increase at a rate proportional to their current value. This means that as the principal or amount increases, the rate of growth does as well. In the context of continuous compounding, your investment grows exponentially over time.
This can be visualized with the formula:
  • If you start with a principal amount of \( P \), the amount after time \( t \) can be expressed as \( A = Pe^{rt} \), where \( r \) is the interest rate.
  • Exponential growth results in a curve that rapidly increases, showing that the initial investment grows larger over time.
Understanding exponential growth is crucial in finance, as it helps us predict how investments will grow. Many natural and economic processes follow this pattern, emphasizing the significance of the concept.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \), where \( e \approx 2.71828 \). It's a critical tool for continuous growth calculations.
When dealing with equations like \( t = (\ln K) / 0.055 \), the natural logarithm helps to solve for time \( t \) in exponential growth scenarios.
  • The natural log is particularly useful because it's the inverse of the exponential function \( e^x \).
  • By using \( \ln \), you can easily solve exponential equations, which are common in finance and natural processes.
The natural logarithm simplifies complex growth calculations, making it a valuable concept for mathematical modeling.
Mathematical Modelling
Mathematical modelling allows us to create abstract representations of real-world situations. In finance, it helps predict how investments grow over time.
Using a formula like \( t = (\ln K) / 0.055 \) is an example of translating real-world financial growth into a mathematical model.
  • Models help visualize and understand how different factors affect growth, such as interest rates and time periods.
  • They allow for experimentation by varying parameters like \( K \) (growth factor) to see different outcomes.
Understanding mathematical models equips students with the tools needed to solve practical problems and simulate real-world scenarios.
Interest Rate Calculation
Interest rate calculation is essential for understanding how investments increase. In continuous compounding, the interest rate is applied continuously, leading to exponential growth.
Given the formula \( t = (\ln K) / 0.055 \), where \( 0.055 \) represents a 5.5% interest rate, you can see how different values of \( K \) (like 1, 2, 4, etc.) affect the time \( t \).
  • Continuous compounding means interest is added constantly, resulting in larger lump sums over time.
  • This approach maximizes growth compared to traditional compounding methods.
By mastering interest rate calculations, students learn to predict future investment values and understand the impact of different rates on their finances.

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Most popular questions from this chapter

The demand equation for a limited edition coin set is \(p=1000\left(1-\frac{5}{5+e^{-0.001 x}}\right)\) Find the demand \(x\) for a price of (a) \(p=\$ 139.50\) and (b) \(p=\$ 99.99\).

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$\log 4 x-\log (12+\sqrt{x})=2$$

Is it possible for a logarithmic equation to have more than one extraneous solution? Explain.

Complete the table for the time \(t\) (in years) necessary for \(P\) dollars to triple if interest is compounded continuously at rate \(r\). $$ \begin{array}{|l|l|l|l|l|l|l|} \hline r & 2 \% & 4 \% & 6 \% & 8 \% & 10 \% & 12 \% \\ \hline t & & & & & & \\ \hline \end{array} $$

The table shows the time \(t\) (in seconds) required for a car to attain a speed of \(s\) miles per hour from a standing start. $$ \begin{array}{|c|c|} \hline \text { Speed, } s & \text { Time, } t \\ \hline 30 & 3.4 \\ 40 & 5.0 \\ 50 & 7.0 \\ 60 & 9.3 \\ 70 & 12.0 \\ 80 & 15.8 \\ 90 & 20.0 \\ \hline \end{array} $$ Two models for these data are as follows. \(t_{1}=40.757+0.556 s-15.817 \ln s\) \(t_{2}=1.2259+0.0023 s^{2}\) (a) Use the regression feature of a graphing utility to find a linear model \(t_{3}\) and an exponential model \(t_{4}\) for the data. (b) Use a graphing utility to graph the data and each model in the same viewing window. (c) Create a table comparing the data with estimates obtained from each model. (d) Use the results of part (c) to find the sum of the absolute values of the differences between the data and the estimated values given by each model. Based on the four sums, which model do you think best fits the data? Explain.
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