Question

Can someone please help me with this question??!!

An epidemic follows the curve

P = 500 / 1+20,000e^(-0.549t)

; where t is in years. How fast is the epidemic growing after 10 years? (Round your answer to two significant digits.)

Answer 1

To find out how fast the epidemic is growing after 10 years, we need to find the derivative of the given equation with respect to time (t). The derivative equation is dP/dt = 10,980e^(-0.549t) / (1 + 20,000e^(-0.549t))^2. Evaluating this at t = 10, we find that the epidemic is growing at a rate of approximately 55.28.

Step-by-step explanation:

To find out how fast the epidemic is growing after 10 years, we need to find the derivative of the given equation with respect to time (t). Let's start by rewriting the equation:

P = 500 / (1 + 20,000e^(-0.549t))

To find the derivative of this equation, we can use the quotient rule, which states that the derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. Differentiating the equation, we get:

dP/dt = (0 - 20,000 * -0.549e^(-0.549t)) / (1 + 20,000e^(-0.549t))^2

Simplifying this, we get:

dP/dt = 10,980e^(-0.549t) / (1 + 20,000e^(-0.549t))^2

Now, let's evaluate the growth of the epidemic after 10 years by substituting t = 10 into the derivative equation:

dP/dt = 10,980e^(-0.549(10)) / (1 + 20,000e^(-0.549(10)))^2

Calculating this, we get:

dP/dt ≈ 55.28

Therefore, the epidemic is growing at a rate of approximately 55.28 after 10 years.

Answer 2

P(t) = 500 / [1+20,000e^(-0.549t) ]

How fast means rate of change of the function, and that is its derivative.

Then, find the derivative of P(t), P '(t), using the chain rule:

P '(t) = - 500 / [ 1 + 20,000e^(-0.549t)]^2 ] * 20,000*(-0.549)e^(-0.549)t

P '(t) = 5,490,000 * e^ (-0.549t) / [1 + 20,000e^-0,549t)]^2

P'(t) = 5,490,000 * e^(0.549t) / [ e^(0.549t) + 20,000]^2

Now replace t by 10 .

P'(10) =5,490,000 * e^(5.49) / [ e^(5.49) + 20,000]^2 = 3.25 years

Answer: 3.3 years