Problem 424
Question
For the following exercises, use this scenario The equation \(N(t)=\frac{500}{1+49 e^{-0.7 t}}\) models the number of people in a town who have heard a rumor after \(t\) days. As \(t\) increases without bound, what value does \(N(t)\) approach? Interpret your answer.
Step-by-Step Solution
Verified Answer
As \(t\) approaches infinity, \(N(t)\) approaches 500; 500 people will hear the rumor eventually.
1Step 1: Understand the Equation
The given equation represents the number of people who have heard a rumor after \(t\) days. It is a function of time, where \(N(t)=\frac{500}{1+49 e^{-0.7 t}}\). Our goal is to find the limit of \(N(t)\) as \(t\) approaches infinity.
2Step 2: Analyze the Exponential Component
As \(t\) increases, \(e^{-0.7 t}\) becomes very small because the exponent \(-0.7t\) becomes a large negative number. Hence, \(e^{-0.7t}\) approaches zero. This will simplify our equation.
3Step 3: Simplify the Expression
Substitute \(e^{-0.7t} \to 0\) in the equation. Now our equation becomes \(N(t) = \frac{500}{1 + 49 \times 0}\). This simplifies to \(N(t) = \frac{500}{1} = 500\).
4Step 4: Interpret the Result
As \(t\) approaches infinity, \(N(t)\) approaches 500. This means that eventually, 500 people in the town will have heard the rumor. The maximum number of people that can hear the rumor is 500, and this number represents the limiting value.
Key Concepts
Exponential FunctionsAsymptotic BehaviorPopulation Models
Exponential Functions
Exponential functions are mathematical expressions where the variable is in the exponent. These functions are characterized by their rapid growth or decay rates. In the equation from our exercise, the term \(e^{-0.7t}\) is an example of exponential decay. As time \(t\) increases, the value of \(e^{-0.7t}\) decreases because the exponent \(-0.7t\) becomes more negative.
This results in an exponential decline towards zero.
Understanding the behavior of this component is crucial. It allows us to simplify complicated expressions. When \(e^{-0.7t}\) approaches zero, it highlights the eventual outcomes in real-life scenarios like rumor spread. Exponential functions are very useful for modeling natural processes. This includes phenomena like radioactive decay, interest growth in finance, and population dynamics.
This results in an exponential decline towards zero.
Understanding the behavior of this component is crucial. It allows us to simplify complicated expressions. When \(e^{-0.7t}\) approaches zero, it highlights the eventual outcomes in real-life scenarios like rumor spread. Exponential functions are very useful for modeling natural processes. This includes phenomena like radioactive decay, interest growth in finance, and population dynamics.
Asymptotic Behavior
Asymptotic behavior describes how a function behaves as the input approaches a certain value or infinity. For example, consider the expression \(N(t) = \frac{500}{1+49 e^{-0.7t}}\).
Our task is to evaluate what happens when \(t\) gets very large.
The exponential term \(e^{-0.7t}\) decreases towards zero, simplifying the expression inside the denominator to \(1 + 0\). Therefore, \(N(t)\) asymptotically approaches the value \(500\). This means that as time goes on, more people hear the rumor until it plateaus at \(500\), which is its horizontal asymptote.
Our task is to evaluate what happens when \(t\) gets very large.
The exponential term \(e^{-0.7t}\) decreases towards zero, simplifying the expression inside the denominator to \(1 + 0\). Therefore, \(N(t)\) asymptotically approaches the value \(500\). This means that as time goes on, more people hear the rumor until it plateaus at \(500\), which is its horizontal asymptote.
- This is a classic property of logistic growth models, where the population (or number of people informed) doesn't continue growing indefinitely but stabilizes.
- Understanding asymptotic behavior offers insight into long-term trends in functions without calculating explicit values.
Population Models
Population models, like the one given in the exercise, are designed to predict how quantities evolve over time. They are often used in biology, ecology, and social sciences to model various processes. In our example, the model \(N(t) = \frac{500}{1+49 e^{-0.7t}}\) captures how a rumor spreads through a population.
The model is a type of logistic growth function.
Logistic models are particularly useful because they initially show exponential growth but then slow down as they approach a carrying capacity. This carrying capacity represents the maximum sustainable population size, or in this case, the maximum number of people who will hear the rumor.
The model is a type of logistic growth function.
Logistic models are particularly useful because they initially show exponential growth but then slow down as they approach a carrying capacity. This carrying capacity represents the maximum sustainable population size, or in this case, the maximum number of people who will hear the rumor.
- The carrying capacity in our model is \(500\).
- This model highlights natural constraints faced by growing populations, where resources (or people willing to share a rumor) might limit further expansion.
Other exercises in this chapter
Problem 422
For the following exercises, use this scenario The equation \(N(t)=\frac{500}{1+49 e^{-0.7 t}}\) models the number of people in a town who have heard a rumor af
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For the following exercises, use this scenario The equation \(N(t)=\frac{500}{1+49 e^{-0.7 t}}\) models the number of people in a town who have heard a rumor af
View solution Problem 425
For the following exercise, choose the correct answer choice. A doctor and injects a patient with 13 milligrams of radioactive dye that decays exponentially. Af
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What situations are best modeled by a logistic equation? Give an example, and state a case for why the example is a good fit.
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