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.)

The rate at which the epidemic is growing after 10 years is approximately 0.79.

Using the provided formula for the derivative of the population function with respect to time and evaluating it at ( t = 10), we have:

[tex]\[ \frac{dP}{dt} \Bigg|_{t=10} = \frac{-500(20,000)(-0.549)e^{-5.49}}{(1 + 20,000e^{-5.49})^2} \][/tex]

[tex]\[ \approx \frac{-500(20,000)(-0.549)e^{-5.49}}{(1 + 20,000e^{-5.49})^2} \][/tex]

[tex]\[ \approx \frac{-500(20,000)(-0.549)(0.004088)}{(1 + 20,000(0.004088))^2} \][/tex]

[tex]\[ \approx \frac{-500(20,000)(-0.549)(0.004088)}{(1 + 81.76)^2} \][/tex]

[tex]\[ \approx \frac{-500(20,000)(-0.549)(0.004088)}{(82.76)^2} \][/tex]

[tex]\[ \approx \frac{-500(20,000)(-0.549)(0.004088)}{6856.8976} \][/tex]

[tex]\[ \approx \frac{5431.56}{6856.8976} \][/tex]

[tex]\[ \approx 0.7926 \][/tex]

Rounding to two significant digits, the rate at which the epidemic is growing after 10 years is approximately 0.79.

Answer:

To find the rate of growth of the epidemic after 10 years, we'll first differentiate the epidemic curve equation with respect to time (t) and then plug in t = 10 to find the growth rate.

Therefore, after 10 years, the epidemic is growing at a rate of approximately -0.082 (rounded to two significant digits).

Step-by-step explanation:

To determine the rate of growth of the epidemic after 10 years, we'll first differentiate the given epidemic curve equation with respect to time (t) using the quotient rule and the chain rule of differentiation.

Let [tex]\( P = \frac{500}{1 + 20,000e^{-0.549t}} \)[/tex].

To differentiate P with respect to t, we'll use the quotient rule:

[tex]\[ \frac{dP}{dt} = \frac{d}{dt} \left( \frac{500}{1 + 20,000e^{-0.549t}} \right) \]\[ = \frac{0 - 500 \times \frac{d}{dt}(1 + 20,000e^{-0.549t})}{(1 + 20,000e^{-0.549t})^2} \][/tex]

Now, we'll find [tex]\( \frac{d}{dt}(1 + 20,000e^{-0.549t}) \)[/tex] using the chain rule:

[tex]\[ \frac{d}{dt}(1 + 20,000e^{-0.549t}) = 0 - 20,000 \times (-0.549)e^{-0.549t} \]\[ = 10,980e^{-0.549t} \][/tex]

Substituting this back into the differentiation of P:

[tex]\[ \frac{dP}{dt} = \frac{-500 \times 10,980e^{-0.549t}}{(1 + 20,000e^{-0.549t})^2} \][/tex]

Now, we'll find the growth rate after 10 years by plugging in [tex]\( t = 10 \)[/tex] into [tex]\( \frac{dP}{dt} \)[/tex]:

[tex]\[ \frac{dP}{dt} \bigg|_{t=10} = \frac{-500 \times 10,980e^{-0.549 \times 10}}{(1 + 20,000e^{-0.549 \times 10})^2} \]\[ \approx \frac{-500 \times 10,980 \times e^{-5.49}}{(1 + 20,000e^{-5.49})^2} \]\[ \approx \frac{-500 \times 10,980 \times 0.004056}{(1 + 20,000 \times 0.004056)^2} \]\[ \approx -0.082 \][/tex]

Thus, after 10 years, the epidemic is growing at a rate of approximately -0.082 (rounded to two significant digits).

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