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Chapter 6: Problem 7

Health plan: The managers of an employee health plan for a firm have studied the balance \(B\), in millions of dollars, in the plan account as a function of \(t\), the number of years since the plan was instituted. They have determined that the rate of change \(\frac{d B}{d t}\) in the account balance is given by the formula $$ \frac{d B}{d t}=10 e^{0.1 t}-12 $$ a. Use your calculator to make a graph of \(\frac{d B}{d t}\) versus \(t\) over the first 5 years of the plan. b. During what period is the account balance \(B\) decreasing? c. At what time is the account balance \(B\) at its minimum?

Short Answer

Expert verified
The account balance decreases for \(0 < t < 1.823\) years. The minimum balance occurs at \(t = 1.823\).

Step by step solution

01

Understanding the Function

The problem provides the rate of change of the balance \(\frac{d B}{d t} = 10 e^{0.1 t} - 12\), which tells us how the balance \(B\) changes with time \(t\). We'll need to analyze this function to determine when the balance is decreasing and when it's at its minimum.
02

Graphing the Rate of Change

Use a calculator to plot \(\frac{d B}{d t} = 10 e^{0.1 t} - 12\) over \(t = 0\) to \(t = 5\). This graph will help us visualize when the rate of change is negative, which indicates when the balance \(B\) is decreasing.
03

Analyzing the Graph

Upon graphing \(\frac{d B}{d t}\), observe the portions where \(\frac{d B}{d t} < 0\). This will occur when \(10 e^{0.1 t} < 12\). Calculate when this inequality holds to find the time period where the balance decreases.
04

Solving the Inequality

To find when \(B\) is decreasing, solve the inequality: \[ 10 e^{0.1 t} < 12 \].Divide both sides by 10:\[ e^{0.1 t} < 1.2 \].Taking the natural logarithm of both sides gives:\[ 0.1t < \ln(1.2) \]. Solve for \(t\) to find:\[ t < \frac{\ln(1.2)}{0.1} \].Plug in the value \(\ln(1.2) \approx 0.1823\):\[ t < 1.823 \].
05

Finding the Minimum Balance Time

Given that the balance decreases when \(0 < t < 1.823\), the balance is at its minimum when \(\frac{d B}{d t} = 0\). Plugging \(t = 1.823\) back into the original equation confirms that is where the sign changes from negative to positive.

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Key Concepts

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

Rate of Change
The concept of rate of change is crucial when dealing with dynamic systems where quantities shift over time. Here, we deal with how the balance in an employee health plan account changes over time, denoted by \( \frac{d B}{d t} \). It represents the derivative of the balance function with respect to time \( t \).
Whenever the function gives a positive value, it means the balance is increasing, while a negative value suggests a decrease. In our exercise, the rate of change \( \frac{d B}{d t} = 10 e^{0.1 t} - 12 \) varies as time \( t \) increases. This helps us predict and understand how and when the financial balance in the account grows or diminishes over a specified period.
Graphing Functions
Graphing functions is an essential tool in calculus that helps visualize changes in mathematical relationships. By plotting the rate of change function \( \frac{d B}{d t} \) over a 5-year period, you get insights into the behavior of the balance in the account.
This graphical representation shows when \( \frac{d B}{d t} \) dips below zero, indicating periods where the balance decreases. A graph can demonstrate patterns and critical points, making it easier to understand intervals of increase or decline visually. Tools like graphing calculators or software make these calculations and visualizations more accessible and accurate.
Inequalities
Inequalities help in determining when a function behaves a certain way, especially in cases involving limits. Solving the inequality \( 10 e^{0.1 t} < 12 \) reveals the time interval during which the account balance decreases.
In the current context, this inequality represents when the rate of change is negative, suggesting withdrawal exceeds deposits in the plan. Through algebraic manipulation, you simplify it into a form that can solve for \( t \), specifically \( t < \frac{\ln(1.2)}{0.1} \), yielding that this condition holds before approximately 1.823 years. Thus, mastering inequalities is crucial in pinpointing specific time frames when particular financial activities occur.
Exponential Functions
Exponential functions are vital in modeling growth processes and decay, marked by the constant rate of proportional change. In this exercise, the function involves \( e^{0.1t} \), a classic exponential form, showcasing how small changes can accumulate into significant differences over time.
These functions are instrumental when analyzing phenomena like compound interest or population growth. By understanding how \( 10 e^{0.1 t} \) interacts within the given equation, you grasp the exponential nature of change, where each increase in \( t \) amplifies the initial condition exponentially. This understanding aids in predicting future trends in account balances.

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Most popular questions from this chapter

Experimental determination of the drag coefficient: When retardation due to air resistance is proportional to downward velocity \(V\), in feet per second, falling objects obey the equation of change $$ \frac{d V}{d t}=32-r V $$ where \(r\) is known as the drag coefficient. One way to measure the drag coefficient is to measure and record terminal velocity. a. We know that an average-size man has a terminal velocity of 176 feet per second. Use this to show that the value of the drag coefficient is \(r=0.1818\) per second. (Hint: To say that the terminal velocity is 176 feet per second means that when the velocity \(V\) is 176 , velocity will not change. That is, \(\frac{d V}{d t}=0\). Put these bits of information into the equation of change and solve for \(r\).) b. An ordinary coffee filter has a terminal velocity of about 4 feet per second. What is the drag coefficient for a coffee filter?

Growing child: A certain girl grew steadily between the ages of 3 and 12 years, gaining \(5 \frac{1}{2}\) pounds each year. Let \(W\) be the girl's weight, in pounds, as a function of her age \(t\), in years, between the ages of \(t=3\) and \(t=12\). a. Is \(W\) a linear function or an exponential function? Be sure to explain your reasoning. b. Write an equation of change for \(W\). c. Given that the girl weighed 30 pounds at age 3 , find a formula for \(W\).

The spread of AIDS: The table on the following page shows the cumulative number \(N=N(t)\) of AIDS cases in the United States that have been reported to the Centers for Disease Control and Prevention by the end of the year given. (The source for these data, the U.S. Centers for Disease Control and Prevention in Atlanta, cautions that they are subject to retrospective change.) a. What does \(\frac{d N}{d t}\) mean in practical terms? b. From 1986 to 1992 was \(\frac{d N}{d t}\) ever negative? $$ \begin{array}{|c|c|} \hline t=\text { year } & N=\text { total cases reported } \\ \hline 1986 & 28,711 \\ \hline 1987 & 49,799 \\ \hline 1988 & 80,518 \\ \hline 1989 & 114,113 \\ \hline 1990 & 155,766 \\ \hline 1991 & 199,467 \\ \hline 1992 & 244,939 \\ \hline \end{array} $$

11\. A population of bighorn sheep: There is an effort in Colorado to restore the population of bighorn sheep. Let \(N=N(t)\) denote the number of sheep in a certain protected area at time \(t\). a. Explain the meaning of \(\frac{d N}{d t}\) in practical terms. b. A small breeding population of bighorn sheep is initially introduced into the protected area. Food is plentiful and conditions are generally favorable for bighorn sheep. What would you expect to be true about the sign of \(\frac{d N}{d t}\) during this period? c. This summer a number of dead sheep were discovered, and all were infected with a disease that is known to spread rapidly among bighorn sheep and is nearly always fatal. How would you expect an unchecked spread of this disease to affect \(\frac{d N}{d t}\) ? d. If the reintroduction program goes well, then the population of bighorn sheep will grow to the size the available food supply can support and will remain at about that same level. What would you expect to be true of \(\frac{d N}{d t}\) when this happens?

Investing in the stock market: You are considering buying three stocks whose prices at time \(t\) are given by \(P_{1}(t), P_{2}(t)\), and \(P_{3}(t)\). You know that \(\frac{d P_{1}}{d t}\) is a large positive number, \(\frac{d P_{2}}{d t}\) is near zero, and \(\frac{d P_{3}}{d t}\) is a large negative number. Which stock will you buy? Explain your answer.
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