Integration via u substitution help (what am I doing wrong?)? - respiratory air flow and volume
Here lies the problem:
Breathing is a cyclical and a complete respiratory cycle since the beginning of the end of the inhalation process is about 5 s. The maximum rate of airflow through the lungs approximately 0.5 l / s. This explains in part why the function f (t) = (1 / 2) sin [2 * Pi * t / 5] It is often used to model L air in the lungs. By using this model to the volume of air that is inhaled into the lungs to find at time t
We want to) integrate f (t values.
∫ f (t) = ∫ (1 / 2) sin (2Pit / 5) DT
Let u = 2Pit / 5
du = 2 m / 5 DT
= 1 / 2 ∫ sin (u) dt
= 1 / 2 ∫ (5 / 2 * Pi) sin (u) (2 * Pi / 5) DT
= 5 / 4 * Pi ∫ sin (u)
= -5 / 4 * Pi cos (2 * pi * t / 5) + C
However, the book says the answer should be ...
(5 / 4 * Pi) (1-cos (2 * pi * t / 5) L
Where the hell1 to get there and why not + L (I think it's C)? Any help is welcome
Thursday, February 18, 2010
Respiratory Air Flow And Volume Integration Via U Substitution Help (what Am I Doing Wrong?)?
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LL is likely to be measured on a volume of air in liters.
ReplyDeleteYour equation is correct, but placed in fact only a C, is the situation. The problem was that the cycle starts by inhalation, so that at time 0 is not in the lung, ie, air
-5 / 4 Pi * cos (2 * pi * 0 / 5) + C = 0
Thus we can solve for C
C = 5 / 4 * Pi cos (2 * pi * 0 / 5) = 5 / (4 * pi) cos (0) = 5 / (4 * pi)
Thus the last equation is
-5 / 4 * Pi cos (2 * pi * t / 5) + C
= -5 / 4 * Pi cos (2 * pi * t / 5) + 5 / (4 * Pi)
= (5 / 4 * Pi) (-cos (2 * pi * t / 5) +1)
= (5 / 4 * Pi) (1-cos (2 * pi * t / 5)) l