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

Short Answer

Expert verified
The times, to two decimal places, for the money to triple for the respective interest rates are: 2% - 54.61 years, 4% - 27.31 years, 6% - 18.20 years, 8% - 13.65 years, 10% - 10.92 years, 12% - 9.10 years.

Step by step solution

01

Rearrange the formula

To solve for \(t\), rearrange the formula \(P(t) = P_0e^{rt}\) to isolate \(t\). Since we're trying to find the tripling time, let's replace \(P(t)\) with \(3P_0\). This results in the following equation: \(3P_0 = P_0e^{rt}\). Simplifying, we have \(3 = e^{rt}\). Taking the natural logarithm of both sides, we get \(ln(3) = rt\). Finally, solve for \(t\) gives \(t = \frac{ln(3)}{r}\).
02

Apply 2% interest rate

Substitute \(r = 0.02\) into the equation \(t = \frac{ln(3)}{r}\) to find \(t\). This gives \(t = \frac{ln(3)}{0.02} = 54.61\) years.
03

Apply 4% interest rate

Substitute \(r = 0.04\) into the equation \(t = \frac{ln(3)}{r}\) to find \(t\). This gives \(t = \frac{ln(3)}{0.04} = 27.31\) years.
04

Apply 6% interest rate

Substitute \(r = 0.06\) into the equation \(t = \frac{ln(3)}{r}\) to find \(t\). This gives \(t = \frac{ln(3)}{0.06} = 18.20\) years.
05

Apply 8% interest rate

Substitute \(r = 0.08\) into the equation \(t = \frac{ln(3)}{r}\) to find \(t\). This gives \(t = \frac{ln(3)}{0.08} = 13.65\) years.
06

Apply 10% interest rate

Substitute \(r = 0.10\) into the equation \(t = \frac{ln(3)}{r}\) to find \(t\). This gives \(t = \frac{ln(3)}{0.10} = 10.92\) years.
07

Apply 12% interest rate

Substitute \(r = 0.12\) into the equation \(t = \frac{ln(3)}{r}\) to find \(t\). This gives \(t = \frac{ln(3)}{0.12} = 9.10\) years.

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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 powerful concept in finance that assumes interest is added an infinite number of times per year. This means the investment grows constantly, which can result in significant gains over time. It's calculated using the formula:

\[ P(t) = P_0e^{rt} \]
Where:
  • \(P(t)\) is the future value of the investment after time \(t\).
  • \(P_0\) is the initial principal balance.
  • \(r\) is the annual interest rate (expressed as a decimal).
  • \(t\) is the time in years.
  • \(e\) is the base of the natural logarithm, approximately equal to 2.71828.

Continuous compounding results in an exponential growth of the investment. More frequently compounding leads to higher returns, and continuous compounding offers the maximum possible growth.

Understanding this concept allows investors to optimize their returns by comparing different investment opportunities with varying compounding frequencies.
Exponential growth
Exponential growth is a pattern of data that shows greater increases over time, creating an upwards-sloping curve. In finance, it's often related to interest compounding because money grows at a steady percentage rate. Characteristics of this concept include:

  • Quick initial growth, followed by increasingly rapid increases.
  • Multiplying factor remains the same, while the quantity scales exponentially.
  • The formula used there is \( A = P \, e^{rt} \).

In the context of the exercise, we observed that the rate of an investment's growth is proportional to its current value. Thus, as the investment grows, it continues to accumulate interest on increasingly larger amounts.

An example would be an initial deposit of money within a savings account which compounds interest over the years, allowing your savings to grow exponentially as time progresses.
This concept emphasizes the importance of time in investing, as exponential growth maximizes returns the longer the investment remains untouched.
Natural logarithm
The natural logarithm, denoted as \(ln\), is the logarithm to the base \(e\), where \(e\) is approximately 2.71828. It is a key mathematical function in continuous growth models. Natural logarithms simplify calculations involving exponential growth by converting multiplicative processes into additive ones.

Here's how the natural logarithm is applied in continuous compounding:
  • When determining the time required for investment growth, \(ln\) helps isolate time by allowing us to take the logarithm of both sides of an equation.
  • This technique linearizes exponential equations, turning complex expressions into simple, solvable ones.

In the exercise, when calculating the time it takes for an investment to triple, we used the equation:

\[ t = \frac{ln(3)}{r} \]
Here, \(ln(3)\) represents the time factor required to triple an initial investment at a specified rate \(r\).
Understanding natural logarithms is crucial in financial mathematics as they offer a tool to unravel the mysteries of time-related exponential growth.

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Most popular questions from this chapter

Find the exponential model \(y=a e^{b x}\) that fits the points shown in the graph or table. $$ \begin{array}{|l|l|l|} \hline x & 0 & 3 \\ \hline y & 1 & \frac{1}{4} \\ \hline \end{array} $$

Automobiles are designed with crumple zones that help protect their occupants in crashes. The crumple zones allow the occupants to move short distances when the automobiles come to abrupt stops. The greater the distance moved, the fewer g's the crash victims experience. (One \(g\) is equal to the acceleration due to gravity. For very short periods of time, humans have withstood as much as 40 g's.) In crash tests with vehicles moving at 90 kilometers per hour, analysts measured the numbers of g's experienced during deceleration by crash dummies that were permitted to move \(x\) meters during impact. The data are shown in the table. A model for the data is given by \(y=-3.00+11.88 \ln x+(36.94 / x),\) where \(y\) is the number of g's. $$ \begin{array}{|c|c|} \hline x & \text { g's } \\ \hline 0.2 & 158 \\ 0.4 & 80 \\ 0.6 & 53 \\ 0.8 & 40 \\ 1.0 & 32 \\ \hline \end{array} $$ (a) Complete the table using the model. $$ \begin{array}{|l|l|l|l|l|l|} \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1.0 \\ \hline y & & & & & \\ \hline \end{array} $$ (b) Use a graphing utility to graph the data points and the model in the same viewing window. How do they compare? (c) Use the model to estimate the distance traveled during impact if the passenger deceleration must not exceed \(30 \mathrm{~g}\) 's. (d) Do you think it is practical to lower the number of g's experienced during impact to fewer than \(23 ?\) Explain your reasoning.

Use the Richter scale \(R=\log \frac{l}{I_{0}}\) for measuring the magnitudes of earthquakes. Find the intensity \(I\) of an earthquake measuring \(R\) on the Richter scale (let \(I_{0}=1\) ). (a) Southern Sumatra, Indonesia in \(2007, R=8.5\) (b) Illinois in \(2008, R=5.4\) (c) Costa Rica in \(2009, R=6.1\)

After discontinuing all advertising for a tool kit in \(2004,\) the manufacturer noted that sales began to drop according to the model \(S=\frac{500,000}{1+0.4 e^{k t}}\) where \(S\) represents the number of units sold and \(t=4\) represents \(2004 .\) In \(2008,\) the company sold 300,000 units. (a) Complete the model by solving for \(k\). (b) Estimate sales in 2012 .

$$\$ 2500$$ is invested in an account at interest rate \(r\), compounded continuously. Find the time required for the amount to (a) double and (b) triple. $$r=0.025$$

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