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Chapter 2: Problem 7

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

Expert verified
Answer: Approximately $823.20 per year.

Step by step solution

01

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.
02

(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: $$S(t) = \int_0^t k e^{r(u-t)} du$$
03

(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.
04

(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.

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Key Concepts

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

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.

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Most popular questions from this chapter

Epidemics. The use of mathematical methods to study the spread of contagious diseases goes back at least to some work by Daniel Bernoulli in 1760 on smallpox. In more recent years many mathematical models have been proposed and studied for many different diseases. Deal with a few of the simpler models and the conclusions that can be drawn from them. Similar models have also been used to describe the spread of rumors and of consumer products. Some diseases (such as typhoid fever) are spread largely by carriers, individuals who can transmit the disease, but who exhibit no overt symptoms. Let \(x\) and \(y,\) respectively, denote the proportion of susceptibles and carriers in the population. Suppose that carriers are identified and removed from the population at a rate \(\beta,\) so $$ d y / d t=-\beta y $$ Suppose also that the disease spreads at a rate proportional to the product of \(x\) and \(y\); thus $$ d x / d t=\alpha x y $$ (a) Determine \(y\) at any time \(t\) by solving Eq. (i) subject to the initial condition \(y(0)=y_{0}\). (b) Use the result of part (a) to find \(x\) at any time \(t\) by solving Eq. (ii) subject to the initial condition \(x(0)=x_{0}\). (c) Find the proportion of the population that escapes the epidemic by finding the limiting value of \(x\) as \(t \rightarrow \infty\).

Harvesting a Renewable Resource. Suppose that the population \(y\) of a certain species of fish (for example, tuna or halibut) in a given area of the ocean is described by the logistic equation $$ d y / d t=r(1-y / K) y . $$ While it is desirable to utilize this source of food, it is intuitively clear that if too many fish are caught, then the fish population may be reduced below a useful level, and possibly even driven to extinction. Problems 20 and 21 explore some of the questions involved in formulating a rational strategy for managing the fishery. At a given level of effort, it is reasonable to assume that the rate at which fish are caught depends on the population \(y:\) The more fish there are, the easier it is to catch them. Thus we assume that the rate at which fish are caught is given by \(E y,\) where \(E\) is a positive constant, with units of \(1 /\) time, that measures the total effort made to harvest the given species of fish. To include this effect, the logistic equation is replaced by $$ d y / d t=r(1-y / K) y-E y $$ This equation is known as the Schaefer model after the biologist, M. B. Schaefer, who applied it to fish populations. (a) Show that if \(E0 .\) (b) Show that \(y=y_{1}\) is unstable and \(y=y_{2}\) is asymptotically stable. (c) A sustainable yield \(Y\) of the fishery is a rate at which fish can be caught indefinitely. It is the product of the effort \(E\) and the asymptotically stable population \(y_{2} .\) Find \(Y\) as a function of the effort \(E ;\) the graph of this function is known as the yield-effort curve. (d) Determine \(E\) so as to maximize \(Y\) and thereby find the maximum sustainable yield \(Y_{m}\).

Suppose that a certain population satisfies the initial value problem \(d y / d t=r(t) y-k, \quad y(0)=y_{0}\) where the growth rate \(r(t)\) is given by \(r(t)=(1+\sin t) / 5\) and \(k\) represents the rate of predation. $$ \begin{array}{l}{\text { (a) Supposs that } k=1 / 5 \text { . Plot } y \text { versust for several values of } y_{0} \text { between } 1 / 2 \text { and } 1 \text { . }} \\ {\text { (b) P. Stimate the critical initial population } y_{e} \text { below which the population will become }} \\ {\text { extinct. }} \\ {\text { (c) Choose other values of } k \text { and find the correponding } y_{i} \text { for each one. }} \\ {\text { (d) Use the data you have found in parts }(a) \text { and }(b) \text { to plot } y_{c} \text { versus } k \text { . }}\end{array} $$

Find the escape velocity for a body projected upward with an initial velocity \(v_{0}\) from a point \(x_{0}=\xi R\) above the surface of the earth, where \(R\) is the radius of the earth and \(\xi\) is a constant. Neglect air resistance. Find the initial altitude from which the body must be launched in order to reduce the escape velocity to \(85 \%\) of its value at the earth's surface.

Suppose that a rocket is launched straight up from the surface of the earth with initial velocity \(v_{0}=\sqrt{2 g R}\), where \(R\) is the radius of the earth. Neglect air resistance. (a) Find an expression for the velocity \(v\) in terms of the distance \(x\) from the surface of the earth. (b) Find the time required for the rocket to go \(240,000\) miles (the approximate distance from the earth to the moon). Assume that \(R=4000\) miles.
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