1000 I(t) = 1 +999e-.6t where t is in days and I represent the number of infected individuals. 1(t) 1000 800 600 400 200 t 5 10 15 20 We wish to estimate the change in infections from t = 10 to t = 11 using differentials.
1000 I(t) = 1 +999e-.6t where t is in days and I represent the number of infected individuals. 1(t) 1000 800 600 400 200 t 5 10 15 20 We wish to estimate the change in infections from t = 10 to t = 11 using differentials.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
A) Determine the derivative of I
B) The approximate change in infected people is dI ____ people
C) The actual change in infected individuals is Delta (I) _____ people
![### Mathematical Modeling of Infection Spread
#### Infection Growth Function
The function \( I(t) \) representing the number of infected individuals over time \( t \) (in days) is given by:
\[ I(t) = \frac{1000}{1 + 999e^{-6t}} \]
Here:
- \( t \) is time in days.
- \( I \) represents the number of infected individuals.
#### Description of the Graph
- The graph plots \( I(t) \) on the y-axis and \( t \) on the x-axis.
- The y-axis ranges from 0 to 1000, representing the number of infections.
- The x-axis ranges from 0 to 25, representing time in days.
- The curve starts at the origin (approximately zero infected individuals at \( t = 0 \)).
- There is a steep increase in infections between \( t = 5 \) and \( t = 15 \).
- The curve then gradually levels off approaching 1000 infected individuals as \( t \) increases beyond 20 days, indicating a saturation point.
#### Purpose
The objective is to estimate the change in infections from \( t = 10 \) to \( t = 11 \) using differentials.
This model and graph can help visualize how infections grow over time and assist in understanding the critical periods of infection growth, which is crucial for planning healthcare responses and interventions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdf909276-6ad3-4144-b98c-235c5a32e436%2Ffe6431e0-8d4e-4af0-a14f-fe8c2a9ff264%2Fqbv0cy7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Mathematical Modeling of Infection Spread
#### Infection Growth Function
The function \( I(t) \) representing the number of infected individuals over time \( t \) (in days) is given by:
\[ I(t) = \frac{1000}{1 + 999e^{-6t}} \]
Here:
- \( t \) is time in days.
- \( I \) represents the number of infected individuals.
#### Description of the Graph
- The graph plots \( I(t) \) on the y-axis and \( t \) on the x-axis.
- The y-axis ranges from 0 to 1000, representing the number of infections.
- The x-axis ranges from 0 to 25, representing time in days.
- The curve starts at the origin (approximately zero infected individuals at \( t = 0 \)).
- There is a steep increase in infections between \( t = 5 \) and \( t = 15 \).
- The curve then gradually levels off approaching 1000 infected individuals as \( t \) increases beyond 20 days, indicating a saturation point.
#### Purpose
The objective is to estimate the change in infections from \( t = 10 \) to \( t = 11 \) using differentials.
This model and graph can help visualize how infections grow over time and assist in understanding the critical periods of infection growth, which is crucial for planning healthcare responses and interventions.
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