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Chapter 6: Problem 12

Magazines: Two magazines, Alpha and \(B e t a\), were introduced at the same time with the same circulation of 100 . The circulation of Alpha is given by the function \(A\), which has the equation of change $$ \frac{d A}{d t}=0.10 A $$ The circulation of Beta is given by the function \(B\), which has the equation of change $$ \frac{d B}{d t}=10 $$ Here \(t\) is the time, in years, since the magazines were introduced. a. One of these functions is growing in a linear way, whereas the other is growing exponentially. Identify which is which, and find formulas for both functions. b. Which magazine is growing more rapidly in circulation? Be sure to explain your reasoning.

Short Answer

Expert verified
Alpha grows exponentially, Beta grows linearly. Alpha eventually grows more rapidly.

Step by step solution

01

Identifying the Growth Type

To determine which function grows linearly and which grows exponentially, we examine the rate-of-change equations. The change condition for Alpha, \( \frac{dA}{dt} = 0.10 A \), indicates exponential growth because the rate depends on the value of \( A \). Conversely, Beta's condition \( \frac{dB}{dt} = 10 \) represents linear growth because the rate of change is constant.
02

Solving the Exponential Growth for Alpha

Since Alpha grows exponentially, we solve \( \frac{dA}{dt} = 0.10A \) by separating variables: \( \frac{1}{A} dA = 0.10 dt \). Integrating both sides gives \( \ln |A| = 0.10t + C \). Solving for \( A \), we exponentiate both sides: \( A(t) = A_0 e^{0.10t} \). Given \( A(0) = 100 \), we find \( A_0 = 100 \). Thus, \( A(t) = 100e^{0.10t} \).
03

Solving the Linear Growth for Beta

To find the function for Beta, we consider \( \frac{dB}{dt} = 10 \). This indicates that \( B(t) \) is linear, and after integrating \( \frac{dB}{dt} = 10 \), we get \( B(t) = 10t + C \). Since \( B(0) = 100 \), we find \( C = 100 \). Thus, \( B(t) = 10t + 100 \).
04

Comparing Growth Rates

To decide which magazine grows more rapidly, we compare \( A(t) = 100e^{0.10t} \) and \( B(t) = 10t + 100 \). Over time, the exponential growth of \( A(t) \) eventually surpasses any linear function, including \( B(t) \). Although initially \( B \) grows faster, \( A \) eventually becomes more rapid due to the nature of exponential functions.

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Key Concepts

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

Differential Equations
Differential equations play a key role in understanding how quantities change over time. They are equations involving derivatives, which represent rates of change. In the context of the problem, differential equations are utilized to model the rate of change of magazine circulations. For Magazine Alpha, the differential equation is \( \frac{dA}{dt} = 0.10A \). This indicates that the rate of change of circulation depends on the current circulation itself, a hallmark of exponential growth. The larger the circulation, the faster it grows.For Magazine Beta, the differential equation is \( \frac{dB}{dt} = 10 \). Here, the rate of change is constant regardless of the current level of circulation. This suggests a linear relationship, where the circulation increases by the same amount each year. Understanding these different forms of differential equations helps identify types of growth, and informs how different systems evolve over time.
Exponential Functions
Exponential functions describe situations where a quantity grows at a rate proportional to its current value. These functions have the general form \( y = a \cdot e^{kx} \), where \(a\) is the initial quantity, \(e\) is Euler's number, and \(k\) is a constant indicating the growth rate. In the exercise, Magazine Alpha's circulation follows an exponential function, modeled by \( A(t) = 100e^{0.10t} \). The base of the exponential, here \(e\), points to a growth rate of 10% per year. This means that Alpha's circulation starts at 100 but grows increasingly fast as time advances. Exponential growth is powerful because it accelerates over time. Initially, changes appear slow, but they quickly increase and cover large differences. Such a function is especially effective in contexts where growth feeds on itself, as reflected by Alpha's ever-increasing circulation.
Linear Functions
Linear functions are characterized by a steady rate of change, and they take the form \( y = mx + b \), where \(m\) is the slope and \(b\) is the y-intercept. Linear functions represent constant growth over time. For Magazine Beta, the function is \( B(t) = 10t + 100 \). Here, the y-intercept is 100, the initial circulation, and the slope is 10. This implies circulation increases by 10 every year, regardless of the current value. Linear functions are straightforward and easy to understand, as they provide a constant predictable increment or decrement over time. In real-life scenarios, linear growth might describe scenarios where external factors do not allow for compounding or self-accelerating growth, reflecting a consistent, step-by-step increase like seen with Beta's circulation.

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Most popular questions from this chapter

Traveling in a car: Make graphs of location and velocity for each of the following driving events. In each case, assume that the car leaves from home moving west down a straight road and that position is given as the distance west from home. a. A vacation: Being eager to begin your overdue vacation, you set your cruise control and drive faster than you should to the airport. You park your car there and get on an airplane to Spain. When you fly back 2 weeks later, you are tired and drive at a leisurely pace back home. (Note: Here we are talking about location of your car, not of the airplane.) b. On a country road: A car driving down a country road encounters a deer. The driver slams on the brakes and the deer runs away. The journey is cautiously resumed. c. At the movies: In a movie chase scene, our hero is driving his car rapidly toward the bad guys. When the danger is spotted, he does a Hollywood 180-degree turn and speeds off in the opposite direction.

Getting velocity from a formula: When a man jumps from airplane with an opening parachute, the distance \(S=S(t)\), in feet, that he falls in \(t \mathrm{sec}-\) onds is given by $$ S=20\left(t+\frac{e^{-1.6 t}-1}{1.6}\right) $$ a. Use your calculator to make a graph of \(S\) versus \(t\) for the first 5 seconds of the fall. b. Sketch a graph of velocity for the first 5 seconds of the fall.

Population growth: The following table \({ }^{6}\) shows the population of reindeer on an island as of the given year. $$ \begin{array}{|l|c|c|c|c|} \hline \text { Date } & 1945 & 1950 & 1955 & 1960 \\ \hline \text { Population } & 40 & 165 & 678 & 2793 \\ \hline \end{array} $$ We let \(t\) be the number of years since 1945 , so that \(t=0\) corresponds to 1945 , and we let \(N=N(t)\) denote the population size. a. Approximate \(\frac{d N}{d t}\) for 1955 using the average rate of change from 1955 to 1960 , and explain what this number means in practical terms. b. Use your work from part a to estimate the population in \(1957 .\) c. The number you calculated in part a is an approximation to the actual rate of change. As you will be asked to show in the next exercise, the reindeer population growth can be closely modeled by an exponential function. With this in mind, do you think your answer in part a is too large or too small? Explain your reasoning.

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?

Equation of change for logistic growth: The logistic growth formula \(N=6.21 /\left(0.035+0.45^{t}\right)\) that we used in Chapter 2 for deer on the George Reserve actually came from the following equation of change: $$ \frac{d N}{d t}=0.8 N\left(1-\frac{N}{177}\right) \text {. } $$ a. Plot the graph of \(\frac{d N}{d t}\) versus \(N\), and use it to find the equilibrium solutions. Explain their physical significance. b. In one plot, sketch the equilibrium solutions and graphs of \(N\) versus \(t\) in each of the two cases \(N(0)=10\) and \(N(0)=225 .\) c. To what starting value for \(N\) does the solution \(N=6.21 /\left(0.035+0.45^{t}\right)\) correspond?
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