In a town whose population is 4,000, a disease creates an epidemic. The number N of people infected t days after the disease has begun is given by the function 4,000 N(t)= 1+15. e Find the number infected after 5 days. A. 2,288 B. 2,290 C. 2,293 OD. 2,292 -0.61
In a town whose population is 4,000, a disease creates an epidemic. The number N of people infected t days after the disease has begun is given by the function 4,000 N(t)= 1+15. e Find the number infected after 5 days. A. 2,288 B. 2,290 C. 2,293 OD. 2,292 -0.61
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Title: Modeling Epidemics with Exponential Functions**
**Introduction:**
In a hypothetical scenario where a town has a population of 4,000 residents, an epidemic occurs. The spread of the disease over time can be modeled using an exponential function. Understanding how to use this function effectively can help predict the number of infected individuals at a given time during the outbreak.
**Problem Statement:**
Given the function for the number of infected individuals \( N(t) \), where \( t \) represents the number of days since the epidemic began:
\[ N(t) = \frac{4,000}{1 + 15 \cdot e^{-0.6t}} \]
Determine how many people are infected after 5 days.
**Options:**
- A. 2,288
- B. 2,290
- C. 2,293
- D. 2,292
**Explanation of the Function:**
- **Population Scale**: The numerator, 4,000, represents the total population assuming that all could eventually be infected in the absence of any interventions.
- **Exponential Growth/Decay**: The denominator involves the expression \( 15 \cdot e^{-0.6t} \), which reflects the rate of infection spread. The parameter \( e^{-0.6t} \) indicates exponential decay with time, modified by the factor 15, which influences the rate of change of the infected population.
- **Interpretation**: As \( t \) increases, \( e^{-0.6t} \) decreases, causing the denominator to shrink, and \( N(t) \) tends toward the upper limit of 4,000. This reflects the progression of the disease as it infects more of the population over time.
**Solution Approach:**
To solve for the number of infected individuals after 5 days, substitute \( t = 5 \) into the function:
\[ N(5) = \frac{4,000}{1 + 15 \cdot e^{-0.6 \times 5}} \]
- **Step 1**: Calculate the exponent: \(-0.6 \times 5 = -3\).
- **Step 2**: Find \( e^{-3} \) using a calculator.
- **Step 3**: Compute the value of the denominator: \( 1 + 15 \cdot e^{-3} \).
-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa3f191a8-a1b2-42ed-ac9e-32d60645d728%2F84b53cdc-3948-4e30-9b55-d7b1d06b95b9%2Ffcxqlp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Modeling Epidemics with Exponential Functions**
**Introduction:**
In a hypothetical scenario where a town has a population of 4,000 residents, an epidemic occurs. The spread of the disease over time can be modeled using an exponential function. Understanding how to use this function effectively can help predict the number of infected individuals at a given time during the outbreak.
**Problem Statement:**
Given the function for the number of infected individuals \( N(t) \), where \( t \) represents the number of days since the epidemic began:
\[ N(t) = \frac{4,000}{1 + 15 \cdot e^{-0.6t}} \]
Determine how many people are infected after 5 days.
**Options:**
- A. 2,288
- B. 2,290
- C. 2,293
- D. 2,292
**Explanation of the Function:**
- **Population Scale**: The numerator, 4,000, represents the total population assuming that all could eventually be infected in the absence of any interventions.
- **Exponential Growth/Decay**: The denominator involves the expression \( 15 \cdot e^{-0.6t} \), which reflects the rate of infection spread. The parameter \( e^{-0.6t} \) indicates exponential decay with time, modified by the factor 15, which influences the rate of change of the infected population.
- **Interpretation**: As \( t \) increases, \( e^{-0.6t} \) decreases, causing the denominator to shrink, and \( N(t) \) tends toward the upper limit of 4,000. This reflects the progression of the disease as it infects more of the population over time.
**Solution Approach:**
To solve for the number of infected individuals after 5 days, substitute \( t = 5 \) into the function:
\[ N(5) = \frac{4,000}{1 + 15 \cdot e^{-0.6 \times 5}} \]
- **Step 1**: Calculate the exponent: \(-0.6 \times 5 = -3\).
- **Step 2**: Find \( e^{-3} \) using a calculator.
- **Step 3**: Compute the value of the denominator: \( 1 + 15 \cdot e^{-3} \).
-
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