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At what constant, continuous rate must money be deposited into an account if the account is to contain \(\$ 20,000\) in 5 years? The account earns \(6 \%\) interest compounded continuously.

Short Answer

Expert verified
Deposit approximately $3431.41 per year continuously for 5 years.

Step by step solution

01

Understand the Problem

We need to find the constant, continuous deposit rate needed to accumulate $20,000 in an account in 5 years using continuous compounding at a 6% interest rate.
02

Identify the Formula

For continuous compounding, we use the formula for future value with continuous contributions: \[ A = Pe^{rt} + \frac{R}{r}(e^{rt} - 1) \] where \( R \) is the continuous deposit rate, \( A \) is the future amount, \( P \) is initial deposit (0 in this case), \( r \) is the interest rate, and \( t \) is the time in years.
03

Substitute Known Values

Since initially no amount is deposited, \( P = 0 \). Thus, the formula simplifies to \( A = \frac{R}{r}(e^{rt} - 1) \). Substitute \( A = 20000 \), \( r = 0.06 \), and \( t = 5 \) into the equation.
04

Simplify the Equation

Substitute the known values: \[ 20000 = \frac{R}{0.06}(e^{0.06 \times 5} - 1) \] Now calculate \( e^{0.3} \), which is approximately 1.34986.
05

Calculate \( e^{rt} - 1 \)

\( e^{0.3} - 1 = 1.34986 - 1 = 0.34986 \)
06

Solve for R

Rearrange the equation to solve for \( R \): \[ R = \frac{20000 \times 0.06}{0.34986} \] Calculate the result: \( R \approx 3431.41 \).
07

Interpret the Result

You need to continuously deposit approximately $3431.41 per year to have $20000 in the account after 5 years.

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Key Concepts

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

Continuous Deposit Rate
When dealing with financial accounts and investments, the continuous deposit rate represents the steady flow of money put into an account over time. Unlike traditional deposit methods, continuous deposits are made infinitely small and frequently, offering a seamless approach to achieving a target financial goal.
In our exercise, we need to find the exact continuous rate at which we should deposit money to achieve a future value of $20,000 in five years, under continuous compounding. Using continuous deposits maximizes the benefits of the compounding effect.
Future Value Formula
The future value formula for continuous contributions is a powerful tool in financial calculations. It reports the estimated value of an investment in the future, accounting for continuous compounding of interest rates and continuous contributions.
For continuous compounding with contributions, the formula is:
\[ A = Pe^{rt} + \frac{R}{r}(e^{rt} - 1) \]
  • \(A\): Future value of the account.
  • \(P\): Initial deposit (often zero, as in our case).
  • \(r\): Annual interest rate as a decimal.
  • \(t\): Time in years.
  • \(R\): Continuous deposit rate.

This formula highlights how money grows over time with regular contributions, amplifying the effect of compounding.
Interest Rate Calculation
Interest rate calculation in the context of continuous compounding differs from regular compounding because of its exponential nature. It applies a constant compounding rate where interest is earned on top of interest continuously without interruption.
In our exercise, the interest rate is 6% or 0.06 as a decimal. Continuous compounding helps in calculating exact future values by employing the natural logarithm base \(e\). Here, exponential terms like \(e^{0.06 \times 5}\) are used to reflect growth over time in the accumulated amount. It spurs the account's growth efficiently and rapidly compared with discrete compounding.
Continuous Contributions
Continuous contributions allow for the consistent growth of an investment by making steady, ongoing deposits into an account. Instead of discrete payments at certain intervals, these contributions are made in a flowing, unending manner.
In this case, we aim to find the precise amount we would need to continuously deposit annually to reach our desired future value of \(20,000. This approach takes full advantage of the compounding process by ensuring that every bit of deposited money starts accruing interest continuously. This strategy can be especially potent for long-term savings and investment objectives.
By solving for \(R\), we find that contributing about \)3431.41 per year will accumulate the desired amount in 5 years, leveraging the compound effect in its continuous form.

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Most popular questions from this chapter

(a) Find the present and future value of an income stream of \(\$ 6000\) per year for a period of 10 years if the interest rate, compounded continuously, is \(5 \%\). (b) How much of the future value is from the income stream? How much is from interest?

The value, \(V\), of a Tiffany lamp, worth \(\$ 225\) in 1975, increases at \(15 \%\) per year. Its value in dollars \(t\) years after 1975 is given by $$ V=225(1.15)^{t} $$ Find the average value of the lamp over the period \(1975-\) \(2010 .\)

Intel Corporation is a leading manufacturer of integrated circuits. In 2004 , Intel generated profits at a continuous rate of \(7.5\) billion dollars per year. \({ }^{2}\) Assume the interest rate was \(8.5 \%\) per year compounded continuously. (a) What was the present value of Intel's profits over the 2004 one-year time period? (b) What was the value at the end of the year of Intel's profits over the 2004 one-year time period?

The supply and demand curves have equations \(p=S(q)\) and \(p=D(q)\), respectively, with equilibrium at \(\left(q^{*}, p^{*}\right)\). Using Riemann sums, explain the economic significance of \(\int_{0}^{q^{*}} S(q) d q\) to the producers.

The population, \(P\), in millions, of Nicaragua was \(5.78\) million in 2008 and growing at an annual rate of \(1.8 \%\). (a) Write a formula for \(P\) as a function of \(t\), where \(t\) is years since 2008 . (b) Find the projected average rate of change (or absolute growth rate) in Nicaragua between 2008 and 2009 , and between 2009 and \(2010 .\) Explain why your answers are different. (c) Use your answers to part (b) to confirm that the relative rate of change (or relative growth rate) over both time intervals was \(1.8 \%\).

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