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Future Value The future value of $$\$ 1000$$ after \(t\) years invested at \(7 \%\) compounded continuously is $$ f(t)=1000 e^{0.07 t} \text { dollars } $$ a. Write the rate-of-change function for the value of the investment. b. Calculate the rate of change of the value of the investment after 10 years.

Short Answer

Expert verified
Approximately $140.96 per year after 10 years.

Step by step solution

01

Understanding the Compound Interest Formula

The future value of an investment compounded continuously can be described by the formula \(f(t) = P e^{rt}\), where \(P\) is the principal amount, \(r\) is the annual interest rate, and \(t\) is the number of years. In the given problem, \(P = 1000\), \(r = 0.07\). You are provided with \(f(t) = 1000 e^{0.07 t}\).
02

Rate-of-Change Function

The rate-of-change of the investment value is described by the derivative of \(f(t)\) with respect to \(t\). This represents how fast the value of the investment is changing over time. Compute \(f'(t)\) using the derivative of \(e^{kt}\), which is \(ke^{kt}\).
03

Compute the Derivative

To find \(f'(t)\), apply the derivative rule: \[ f'(t) = \frac{d}{dt}(1000 e^{0.07 t}) = 1000 \cdot 0.07 e^{0.07 t} = 70 e^{0.07 t} \] Hence, the rate-of-change function is \(f'(t) = 70 e^{0.07 t}\).
04

Calculate the Rate of Change at t = 10 Years

Substitute \(t = 10\) into the rate-of-change function \(f'(t)=70 e^{0.07 t}\). \[ f'(10) = 70 e^{0.07 \cdot 10} = 70 e^{0.7} \] Use a calculator to find \(e^{0.7}\) which approximately equals 2.01375.
05

Final Calculation

Completing the calculation: \[ f'(10) = 70 \times 2.01375 \approx 140.9625 \] Thus, the rate of change of the investment value after 10 years is approximately $140.96 per year.

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Key Concepts

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

Compound Interest
Compound interest is a powerful financial concept that allows an investment to grow at an exponential rate. In simple terms, it means that you earn interest not only on your initial investment but also on the accumulated interest from previous periods.

This can be an incredible wealth-building tool as it harnesses the power of earning on reinvested earnings. For instance, when you invest $1000 at an annual interest rate of 7% compounded continuously, the interest is added back to the principal balance constantly or continuously, leading to future growth.

The key takeaway about compound interest is that it results in exponential growth over time, which is far more effective than simple interest, where you only earn on the initial principal balance.
Derivative
In mathematics, the derivative measures how a function changes as its input changes. It represents the rate of change or the slope of the function at any given point.

In the context of our exercise, considering the future value of an investment—represented by the function \(f(t) = 1000 e^{0.07 t}\)—the derivative \(f'(t)\) tells us how fast the value of the investment is growing at any time \(t\).

To find this derivative, apply the rule for differentiating exponential functions: if \(f(t) = e^{kt}\), then \(f'(t) = k e^{kt}\). For our problem, where \(k = 0.07\), the derivative is \(f'(t) = 70 e^{0.07 t}\). This derivative function provides insights into the growth rate of our investment over time.
Continuous Compounding
Continuous compounding is a concept that assumes the compounding of interest is happening constantly. Unlike monthly, quarterly, or annually compounded interest, continuous compounding compounds interest in infinitely small periods, resulting in the maximum possible earnings over time.

The formula for continuous compounding is expressed as \(f(t) = P e^{rt}\), where \(P\) is the principal amount, \(r\) is the annual interest rate, and \(t\) is the time in years.

This concept is crucial because it maximizes the future value of investments or savings compared to other compounding frequencies. The maximum potential growth therefore occurs at the theoretical limit of compounding frequency — continuously — resulting in exponential growth of the investment.
Exponential Growth
Exponential growth describes a process where the quantity increases at a rate proportional to its current value. It's commonly seen in population growth, interest earning, and certain natural phenomena.

In our scenario with continuous compounding, the investment grows exponentially as described by the function \(f(t) = 1000 e^{0.07 t}\). This means that over time, the growth rate of the investment increases, leading to faster and faster increase in value.

This kind of growth is powerful because it becomes quite significant over longer periods. For short terms, gains might seem modest, but as time progresses, exponential growth results in exponentially larger increases, making such investments particularly attractive for long-term strategies.

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Most popular questions from this chapter

Stock Value The value of one share of a company's stock is given by $$ s(x)=15+\frac{2.6}{x+1} \text { dollars } $$ where \(x\) is the number of weeks after it is first offered. An investor buys some of the stock each week and owns $$ n(x)=100+0.25 x^{2} \text { shares } $$ after \(x\) weeks. The value of the investor's stock after \(x\) weeks is given by $$ v(x)=s(x) \cdot n(x) $$ a. Calculate and interpret the following: i. \(s(10)\) and \(s^{\prime}(10)\) ii. \(n(10)\) and \(n^{\prime}(10)\) iii. \(v(10)\) and \(v^{\prime}(10)\) b. Write a formula for \(v^{\prime}(x)\).

Use the simple derivative rules presented in this section to explain why a function of the form $$ y=a x^{4}+b x^{3}+c x^{2}+d x+e, a \neq 0 $$ has a cubic rate-of-change function.

Identify the indeterminate form of each limit. Use L'Hôpital's Rule to evaluate the limit of any indeterminate forms. $$ \lim _{t \rightarrow \infty} \frac{e^{2 t}}{2 t} $$

Evaluate the limit. If the limit is of an indeterminate form, indicate the form and use L'Hôpital's Rule to evaluate the limit. $$ \lim _{x \rightarrow 5} \frac{(x-1)^{0.5}-2}{x^{2}-25} $$

Relief Donations The number of private donations received by nongovernment disaster relief organizations can be modeled as $$ f(x)=0.3 x e^{-0.03 x} \text { thousand donations } $$ where \(x\) is the number of hours since a major disaster has struck. a. Write an expression for the rate of change in donations. b. At what time is the rate of change of donations zero? c. What is the donation level at the time found in part \(b\) ?

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