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The rate of change (dP)/(dt) of the number of people infected by a disease is modeled by the following differential equation: (dP)/(dt)=(45)/(125404)P(800-P) At t=0, the number of people infected by the disease is 214 and is increasing at a rate of 45 people per hour. What is the limiting value for the total number of people infected by the disease as time increases? Answer:

The rate of change dPdt \frac{d P}{d t} of the number of people infected by a disease is modeled by the following differential equation:\newlinedPdt=45125404P(800−P) \frac{d P}{d t}=\frac{45}{125404} P(800-P) \newlineAt t=0 t=0 , the number of people infected by the disease is 214214 and is increasing at a rate of 4545 people per hour. What is the limiting value for the total number of people infected by the disease as time increases?\newlineAnswer:

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Q. The rate of change dPdt \frac{d P}{d t} of the number of people infected by a disease is modeled by the following differential equation:\newlinedPdt=45125404P(800−P) \frac{d P}{d t}=\frac{45}{125404} P(800-P) \newlineAt t=0 t=0 , the number of people infected by the disease is 214214 and is increasing at a rate of 4545 people per hour. What is the limiting value for the total number of people infected by the disease as time increases?\newlineAnswer:
  1. Analyze Differential Equation: To find the limiting value for the total number of people infected by the disease as time increases, we need to analyze the given differential equation:\newlinedPdt=45125404P(800−P)\frac{dP}{dt} = \frac{45}{125404}P(800-P)\newlineThis equation is a logistic growth model, where the growth rate of the population PP is proportional to both the current population PP and the difference between the current population and the maximum population (800800 in this case). The limiting value, also known as the carrying capacity, is the value of PP at which the growth rate dPdt\frac{dP}{dt} becomes zero.
  2. Find Growth Rate: Setting the growth rate (dP)/(dt)(dP)/(dt) to zero, we can find the limiting value:\newline0=45125404P(800−P)0 = \frac{45}{125404}P(800-P)\newlineThis equation will be zero when either P=0P = 0 or when P=800P = 800. Since we are looking for the limiting value as time increases and given that the initial number of infected people is 214214, which is already greater than zero, we can disregard the solution P=0P = 0.
  3. Calculate Limiting Value: The remaining solution to the equation is P=800P = 800. This is the carrying capacity of the model, which means that as time goes on, the number of infected people will approach this value. This is the limiting value for the total number of people infected by the disease as time increases.

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