..ooo AT&T; 6:51 AM 100% y/a 48. The 0 0 yevla 0 43. The rate, r, at which peopl
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Question
..ooo AT&T; 6:51 AM 100% y/a 48. The 0 0 yevla 0 43. The rate, r, at which people get sick during an epidemic of the flu can be approximated by r = 1000te-05t where r is measured in people/day and t is measured in days since the start of the epidemic. (a) Sketch a graph of r as a function of t. (b) When are people getting sick fastest? (c) How many people get sick altogether? Strengthen Your Understanding In Problems 49-50, explain what is wrong with the statement. false. G 49. If bothJ)da and g) da diverge, then so 53. Ifj does J g(a) dz. 50. If floc f(x) dx diverges, then lim, f(x) the 0 54. If divExplanation / Answer
r(t) has maximum value when r'(t) = 0 and r''(t) < 0
r(t) = 1000 t e^(-0.5t) -----> exponential function
r'(t) = 1000 t * -0.5 e^(-0.5t) + 1000 e^(-0.5t)
r'(t) = -500 t e^(-0.5t) + 1000 e^(-0.5t)
r''(t) = -500 t * -0.5 e^(-0.5t) - 500 e^(-0.5t) + 1000 * -0.5 e^(-0.5t)
r''(t) = 250 t e^(-0.5t) - 500 e^(-0.5t) - 500e^(-0.5t)
r''(t) = 250 t e^(-0.5t) - 1000 e^(-0.5t)
r'(t) = 0
-500 t e^(-0.5t) + 1000 e^(-0.5t) = 0
-500 e^(-0.5t) (t - 2)
e^(-0.5t) > 0 for all t, therefore (t - 2) = 0 -----> t = 2
r''(2) = 500 e^(-0.5t) - 1000 e^(-0.5t) = -500 e^(-0.5t) < 0
People are getting sick the fastest when t = 2 days
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