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It is estimated that t days after a virus begins to spread in a​ town, the percent of the population infected by the virus is approximated by p left parenthesis t right parenthesis equals t squared plus 5 tp(t)=t2+5t for 0 less than or equals t less than or equals 60≤t≤6.

a. find the average rate of change of p with respect to t over the interval 1 to 55 days.
b. find and interpret the instantaneous rate of change of p with respect to t at tequals=22.

User JerryZhou
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first off this is a creepy question --

9% of the population will be infected on the 2nd day based on the rate of infection. . .


have a good day ~
User Pragya Mendiratta
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The expression for p(t) can be easier to evaluate if it is factored to
p(t) = t·(t+5)

a. The average rate of change (r) on the interval [0, 5] is
r = (p(5) - p(0))/(5 - 0)
r = (5·(5+5) - 0·(0+5))/5
r = 10 . . . . percent per day


b. Differentiating the equation, you have
p'(t) = 2t + 5
Evaluating that at t=2 gives
p'(2) = 2·2 + 5 = 9 . . . . . percent per day, instantaneous rate at t=2

Interpretation: on day 2, the rate of infection is 9% of the population per day.
User Mcoolive
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