Asked by tan
assume that during the first 3 minutes after a foreign substance is introduced into the blood,the rate R'(t) at which new antibodies are produced(in thousands of antibodies per minute) is given by R'(t)=t/(t^2+1) where t is in minutes.find the total quantity of new antibodies in the blood at the end of 3 minutes.
Answers
Answered by
drwls
dR/dt = t/(t^2+1)
Assume R = 0 at t = 0
R = integral of t dt/(t^2 + 1)
= (1/2)ln(t^2 + 1)
At t = 3 minutes, R = (1/2) ln(10) = 1.15
Since R is measured in thousands, the answer is 1150.
Assume R = 0 at t = 0
R = integral of t dt/(t^2 + 1)
= (1/2)ln(t^2 + 1)
At t = 3 minutes, R = (1/2) ln(10) = 1.15
Since R is measured in thousands, the answer is 1150.
Answered by
tan
i don't understand this part:dt/(t^2 + 1)
= (1/2)ln(t^2 + 1)
= (1/2)ln(t^2 + 1)
Answered by
Steve
It's not dt/(t^2 + 1)
It is t dt/(t^2 + 1)
That is just (1/2) du/u if u=t^2+1. That's where the log comes from
It is t dt/(t^2 + 1)
That is just (1/2) du/u if u=t^2+1. That's where the log comes from
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