Question Suppose that people are getting infected in an outbreak of a virus that begins at time t=0 at a rate of r(t) = 200te +/7. How many infections would you expect to happen in total during the outbreak (i.e., as t +)? Number Note, your answer should be exact.

TLATSP The Asker · Calculus

Transcribed Image Text: Suppose that people are getting infected in an outbreak of a virus that begins at time t=0 at a rate of r(t) = 200te +/7. How many infections would you expect to happen in total during the outbreak (i.e., as t +)? Number Note, your answer should be exact.

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\begin{array}{l}\because \gamma(t)=200 t \times e^{-t / 7} \\\Rightarrow \gamma(t)=\frac{200 t}{e^{t / 7}}\end{array}As t \rightarrow \infty\lim _{t \rightarrow \infty} \gamma(t)=\lim _{t \rightarrow \infty}\left(\frac{200 t}{e t 17}\right)\left(\frac{\infty}{\infty} \text { form }\right)So Using the L-H Rule\begin{aligned}\lim _{t \rightarrow \infty} \gamma(t) & =\lim _{t \rightarrow \infty}\left[\frac{\frac{d}{d t}(200 t)}{\frac{d}{d t}\left(e^{t / 7}\right)}\right] \\\lim _{t \rightarrow \infty} \gamma(t) & =\lim _{t \rightarrow \infty} \frac{200}{\frac{1}{7} e^{t / 7}} \\\lim _{t \rightarrow \infty} \gamma(t) & =\lim _{t \rightarrow \infty} 1400 \times e^{-t / 7} \\\lim _{t \rightarrow \infty} \gamma(t) & =1400 \times e^{-\infty}\left\{\because e^{-\infty}=0\right\} \\& =1400 \times 0 \\\lim _{t \rightarrow \infty} \gamma(t) & =0 \text { 4us }\end{aligned} ...