Question

A young person with no initial capital invests \(k\) dollars per year at an annual rate of return \(r\). Assume that investments are made continuously and the return is compounded continuously. $$ \begin{array}{l}{\text { (a) Determine the sum } S(t) \text { accumulated at any time } t} \\ {\text { (b) If } r=7.5 \% \text { , determine } k \text { so that } \$ 1 \text { million will be available for retirement in } 40 \text { years. }} \\ {\text { (c) If } k=\$ 2000 / \text { year, determine the return rate } r \text { that must be obtained to have } \$ 1 \text { million }} \\ {\text { available in } 40 \text { years. }}\end{array} $$

Short answer

Answer: Approximately $823.20 per year.

Step-by-step solution

Understanding continuous compounding formula

The general formula for continuous compounding with a constant rate of return is: $$S(t) = P \cdot e^{r \cdot t}$$ where S(t) is the accumulated amount at time t, P is the initial capital, r is the rate of return, and t is the time in years.

(a) Step 2: Determine S(t) in terms of k and r

Since investments are made continuously, we need to use integration. Let's represent the continuous investment flow with the function f(t) = k. We can compute the total sum, S(t), as follows: $$S(t) = \int_0^t k e^{r(u-t)} du$$

(b) Step 3: Calculate the value of k for r = 7.5% and t = 40 years

We want to have 1 million dollars by the end of 40 years. We set S(40) = 1000000 and r = 0.075. We can now solve for k: $$1000000 = \int_0^{40} k e^{0.075(u-40)} du$$, $$k = \frac{1000000}{\int_0^{40} e^{0.075(u-40)} du} $$ Calculating this integral using the fundamental theorem of calculus, we get: $$k = \frac{1000000}{\left[\frac{1}{0.075} e^{0.075(u-40)}\right]_0^{40}} = 823.20$$ Thus, the person would need to invest approximately $823.20 per year.

(c) Step 4: Calculate the value of r for k = $2000/year and t = 40 years

We want to have 1 million dollars at the end of 40 years. We set S(40) = 1000000 and k = 2000. We can now solve for r: $$1000000 = \int_0^{40} 2000 e^{r(u-40)} du$$, $$r = \frac{\ln{\left(\frac{1000000}{2000}\cdot\int_0^{40} e^{-(u-40)} du\right)}}{(40-0)}$$ Calculating this integral using the fundamental theorem of calculus, we get: $$r = \frac{\ln{\left(\frac{1000000}{2000}\cdot\left[-e^{-(u-40)}\right]_0^{40}\right)}}{40} = 0.0566$$ Thus, a return rate of approximately 5.66% is required to have $1 million available in 40 years.

Key concepts

Rate of Return

A crucial concept in investing is the "Rate of Return." It measures how much profit or loss an investment generates compared to its initial cost over a specific period.
This rate is often expressed as a percentage and helps investors evaluate the potential success of an investment.

• A higher rate of return indicates a more profitable investment.
• A negative rate could signal a loss.

In our problem, the rate of return is compounded continuously. This means that the interest earned on the initial amount is constantly being reinvested at the same rate.
This approach leads to exponential growth because the interest itself earns interest, compounding infinitely in theory.
For example, if an investment has a continuous annual return of 7.5%, it continuously grows by this percentage, providing an investor with more return over time compared to simple interest. This makes understanding the actual impact of rates critical in long-term financial planning.

Continuous Investment

When it comes to savings and investment, "Continuous Investment" is a strategy where you systematically invest a fixed amount of money at regular intervals over time.
This concept is essential when dealing with investments compounded continuously, like in the provided exercise.

• It involves consistent investment, regardless of market conditions.
• This approach reduces the overall impact of volatility in investments.

In the exercise, continuous investment is depicted by the function \( f(t) = k \), where \( k \) is the constant amount invested every year. The use of integration allows one to calculate the sum accumulated over a period, providing a very realistic model of real-life investment growth.
By investing continuously, investors can reap the benefits of compounding interest, significantly boosting the amount accumulated over longer periods.

Integration in Economics

Integration is a fundamental calculus technique often used in economics to find total values of continuously changing quantities over time.
This technique is especially useful for computing the value of investments that grow through continuous compounding.

• Integration sums up infinitesimally small contributions to find a total amount.
• It can handle complex real-world calculations more accurately.

In the provided step-by-step solution, integration was used to determine the accumulated sum \( S(t) \) over a period \( t \). It combined the constant investment figure \( k \) with the growth factor \( e^{r(u-t)} \), capturing the continuous return of investment. When solving for \( k \) and \( r \), the integration results helped determine the necessary yearly investment and rate of return to achieve financial goals. Knowing how to apply integration enables you to model, predict, and plan economic activities efficiently.