Problem 53
Question
Suppose that the spread of a flu virus on a college campus is modeled by the function$$y(t)=\frac{1000}{1+999 e^{-0.9 t}}$$ where \(y(t)\) is the number of infected students at time \(t\) (in days, starting with \(t=0\) ). Use a graphing utility to estimate the day on which the virus is spreading most rapidly.
Step-by-Step Solution
Verified Answer
The virus spreads most rapidly on approximately day 5.
1Step 1: Identify the Mathematical Model
The problem involves a function that models the spread of a flu virus: \( y(t) = \frac{1000}{1+999 e^{-0.9 t}} \). The task is to estimate when the virus spreads most rapidly.
2Step 2: Understanding the Derivative
The rate of spread of the virus is determined by the derivative of the function \( y(t) \). We need to find the derivative \( y'(t) \), which represents the rate of change of the number of infected students.
3Step 3: Calculate the Derivative
To find the derivative \( y'(t) \), use the chain and quotient rules: \( y'(t) = \frac{d}{dt} \left( \frac{1000}{1+999 e^{-0.9 t}} \right) \). After applying these rules, \( y'(t) = 1000 \cdot 0.9 \cdot 999 e^{-0.9t} / (1+999 e^{-0.9t})^2 \).
4Step 4: Find the Critical Point
The function \( y(t) \) spreads most rapidly where \( y'(t) \), the first derivative, reaches its maximum. This occurs generally where the second derivative \( y''(t) \) changes sign or equivalently by using graphical methods to find the peak of \( y'(t) \).
5Step 5: Use Graphing Utility to Find Maximum Rate of Change
Using a graphing calculator or software, plot \( y'(t) \) to find the value of \( t \) at which \( y'(t) \) reaches its maximum. This is where the number of infected students increases most rapidly.
6Step 6: Estimate the Day
From the graph, you observe the peak of \( y'(t) \) occurs around \( t = 5 \) days. This implies the virus spreads most rapidly on approximately the fifth day.
Key Concepts
Rate of ChangeCritical PointGraphing Utility
Rate of Change
The rate of change in any situation signifies how quickly something is changing with respect to time or another variable. In the context of the flu virus spread on a campus, it refers to how fast the number of infected students increases over time. To understand this, we use the derivative of the function that models the virus spread.
The key point is to determine how rapid the infection is growing at any given moment. The derivative of the function, denoted as \( y'(t) \), gives us this exact information. It's essentially the "speed" at which the number of infections changes each day. In calculations, this involves applying the chain and quotient rules to derive the accurate formula for \( y'(t) \). Here, \( y'(t) = \frac{1000 \cdot 0.9 \cdot 999 e^{-0.9t}}{(1+999 e^{-0.9t})^2} \) is derived.
Understanding this concept is like turning on the headlights while driving at night. It helps you navigate the curve of exponential growth, giving insights into the progression of infections.
The key point is to determine how rapid the infection is growing at any given moment. The derivative of the function, denoted as \( y'(t) \), gives us this exact information. It's essentially the "speed" at which the number of infections changes each day. In calculations, this involves applying the chain and quotient rules to derive the accurate formula for \( y'(t) \). Here, \( y'(t) = \frac{1000 \cdot 0.9 \cdot 999 e^{-0.9t}}{(1+999 e^{-0.9t})^2} \) is derived.
Understanding this concept is like turning on the headlights while driving at night. It helps you navigate the curve of exponential growth, giving insights into the progression of infections.
Critical Point
The critical point in a mathematical context is where you look for values that give a peak or a trough in function behavior. More specifically, for the function modeling the flu spread, we are interested in finding when the rate of spread, \( y'(t) \), is at its maximum. This is because the virus spreads most rapidly at this point.
Identifying critical points involves setting the first derivative equal to zero to find where changes plateau or turn around. However, since we're interested in maximization here, we rely on observing when \( y'(t) \) peaks. This leads to a practical solution: use graphical tools to handle complex derivatives that are difficult to solve analytically. Thus, finding the maximum rate in practical terms often means observing the graph directly.
At these points, the virus reaches its fastest spread, which is crucial for understanding how to effectively intervene in real-world scenarios by targeting the timeliest response measures.
Identifying critical points involves setting the first derivative equal to zero to find where changes plateau or turn around. However, since we're interested in maximization here, we rely on observing when \( y'(t) \) peaks. This leads to a practical solution: use graphical tools to handle complex derivatives that are difficult to solve analytically. Thus, finding the maximum rate in practical terms often means observing the graph directly.
At these points, the virus reaches its fastest spread, which is crucial for understanding how to effectively intervene in real-world scenarios by targeting the timeliest response measures.
Graphing Utility
Graphing utilities are powerful tools that aid in visualizing mathematical functions and understanding their characteristics. They are particularly useful in complex scenarios like modeling a disease spread, where calculating derivatives manually might be tedious.
In the flu virus example, using a graphing utility means plotting the function \( y'(t) \), which represents the rate at which the infection spreads. By doing this, you can visually identify the time \( t \) at which \( y'(t) \) reaches its peak—the very point where the number of infections rises most steeply.
In the flu virus example, using a graphing utility means plotting the function \( y'(t) \), which represents the rate at which the infection spreads. By doing this, you can visually identify the time \( t \) at which \( y'(t) \) reaches its peak—the very point where the number of infections rises most steeply.
- Graphing calculators: Ideal for quick, approximate solutions.
- Software Applications: Offers detailed visuals and calculations (e.g., Desmos, GeoGebra).
Other exercises in this chapter
Problem 52
(a) Use a CAS to graph the function $$f(x)=\frac{x^{4}+1}{x^{2}+1} $$ and use the graph to estimate the \(x\) -coordinates of the relative extrema. (b) Find the
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Sketch the graph of $$y=\frac{1}{(x-a)(x-b)}$$ assuming that \(a \neq b\)
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Consider the family of curves \(y=x e^{-b x}(b>0)\) (a) Use a graphing utility to generate some members of this family. (b) Discuss the effect of varying \(b\)
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Let \(y=1 /\left(1+x^{2}\right) .\) Find the values of \(x\) for which \(y\) is increasing and decreasing most rapidly.
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