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91Ó°ÊÓ

Find a formula for the time required for an investment to grow to \(k\) times its original size if it grows at interest rate \(r\) compounded continuously.

Short Answer

Expert verified
The time required is \( t = \frac{\ln(k)}{r} \).

Step by step solution

01

Understand the problem

We need to find how long it takes for an investment to grow to \( k \) times its original size given a continuous interest rate \( r \). This implies solving for time \( t \) in a continuous compound interest scenario.
02

Recall the formula for continuous compounding

The formula for continuous compounding is given by \( A = Pe^{rt} \), where \( A \) is the final amount, \( P \) is the initial principal, \( r \) is the rate of interest, and \( t \) is the time in years.
03

Set up the equation with given parameters

Since we want the final amount \( A \) to be \( k \) times the initial amount \( P \), we replace \( A \) with \( kP \). Thus, the equation becomes \( kP = Pe^{rt} \).
04

Simplify the equation

Cancel \( P \) from both sides of the equation. This gives you \( k = e^{rt} \).
05

Solve for \( t \)

Take the natural logarithm \( \ln \) of both sides to solve for \( t \). Thus, \( \ln(k) = rt \).
06

Isolate the variable \( t \)

Divide both sides by \( r \) to get \( t = \frac{\ln(k)}{r} \). This is the formula for the time required for the investment to grow to \( k \) times the original amount.

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Key Concepts

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

Investment Growth
Investment growth refers to the process by which an investment increases in value over time. This growth is often due to the interest earned on the initial principal, and in the context of continuous compounding, the growth is exponential.
In our setup, continuous compounding means that interest is calculated and added to the account balance at every possible instant. This allows the investment to grow at a rate that is a function of both the principal and the time over which the interest acts. The exponential function describes this growth because of the way interest accumulates repeatedly over time. The formula used for continuous compounding is\[ A = Pe^{rt} \]
  • A is the amount of money accumulated after a certain time, including interest.
  • P is the principal amount (the initial investment).
  • r is the annual interest rate (expressed as a decimal).
  • t is the number of years the money is invested for.
Understanding this formula is crucial for financial planning and decision-making, allowing investors to project how their investments can grow over time.
Interest Rate
The interest rate is a percentage that indicates how much an investment will earn over a period of time. In financial terms, it is the cost of borrowing or the gain from lending, depicted annually.
For continuous compounding, the interest rate, denoted by r, continuously influences how quickly an investment grows. Unlike simple or compound interest calculated at discrete intervals (such as monthly or yearly), continuous compounding assumes that interest is added infinitely often to the principal.
This is one of the main reasons continuous compounding can lead to higher returns than standard compounding methods. An interest rate's impact on investment growth can be seen more significantly with continuous compound interests because it uses the power of natural exponential growth, needing only the interest rate and time to determine an investment's potential earnings. Thus, understanding the implications of the interest rate is essential for predicting the future value of investments and choosing the best opportunities for growth.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is a mathematical function that is used to solve equations involving exponential growth or decay, such as those in finance with continuous compounding interest.
In the context of the formula \( t = \frac{\ln(k)}{r} \), the natural logarithm is used to isolate the time \( t \) when we want an investment to grow by a factor of \( k \).
Here’s why it’s important:
  • It helps to 'undo' the effect of the exponential function. Since \( e^{rt} = k \), applying \( \ln \) to both sides simplifies it to \( \ln(k) = rt \).
  • The base \(e\) of the natural logarithm corresponds to the growth process described by continuous compounding.
  • Using \( \ln \) enables us to express \( t \) explicitly, providing a straightforward way to find the time required to achieve the investment growth.
The natural logarithm is a fundamental concept in calculus and is crucial in financial calculations involving exponential growth, like those used in continuously compounded interests.

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Most popular questions from this chapter

97-98. ATHLETICS: World's Record 100-Meter Run In 1987 Carl Lewis set a new world's record of 9.93 seconds for the 100 -meter run. The distance that he ran in the first \(x\) seconds was $$ 11.274\left[x-1.06\left(1-e^{-x / 1.06}\right)\right] \text { meters } $$ for \(0 \leq x \leq 9.93 .\) Enter this function as \(y_{1},\) and define \(y_{2}\) as its derivative (using NDERIV), so that \(y_{2}\) gives his velocity after \(x\) seconds. Graph them on the window [0,9.93] by [0,100] Trace along the velocity curve to verify that Lewis's maximum speed was about 11.27 meters per second. Find how quickly he reached a speed of 10 meters per second, which is \(95 \%\) of his maximum speed.

Why can't we differentiate \(e^{x}\) by the Power Rule?

If an amount is invested at interest rate \(r\) compounded continuously, the doubling time (the time in which it will double in value) is found by solving the equation \(P e^{r t}=2 P\). The solution (by the usual method of canceling the \(P\) and taking logs) is \(t=\frac{\ln 2}{r} \approx \frac{0.69}{r} .\) For annual compounding, the doubling time should be somewhat longer, and may be estimated by replacing 69 by 72 . For example, to estimate the doubling time for an investment at \(8 \%\) compounded annually we would divide 72 by \(8,\) giving \(\frac{72}{8}=9\) years. The \(72,\) however, is only a rough "upward adjustment" of \(69,\) and the rule is most accurate for interest rates around \(9 \% .\) For each interest rate: a. Use the rule of 72 to estimate the doubling time for annual compounding. b. Use the compound interest formula \(P(1+r)^{t}\) to find the actual doubling time for annual compounding. $$ 1 \% $$

\(63-68 .\) Find the differential of each function and evaluate it at the given values of \(x\) and \(d x\). $$ y=x e^{x} \text { at } x=1 \text { and } d x=0.1 $$

\(55-58\). For each function: a. Find \(f^{\prime}(x)\) b. Evaluate the given expression and approximate it to three decimal places. $$ f(x)=5 x \ln x, \text { find and approximate } f^{\prime}(2) $$

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