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November 23, 2014

November 23, 2014

Posted by **deel** on Wednesday, August 8, 2012 at 9:35pm.

- math -
**MathMate**, Wednesday, August 8, 2012 at 10:23pmI assume you are familiar with integration or differential equations.

y(t)=amount of air (in c.f.) in room at time t.

Then

dy/dt

=y'(t)

=rate of increase of fresh air per unit time (min)

=fresh air in - fresh air out per minute

=600 - 600(y/20000)

=600*(20000-y)/20000

Separate variables and integrate:

∫dy/(20000-y) = (600/20000)∫ dt

-log(20000-y) = 600t/20000 + C'

log(20000-y) = -600t/20000 + C

Raise to power of e:

20000-y = e^(-600t/20000+C)

y=20000(1-e^(-600t/20000+C)

At t=0, y=0

=>

0=(1-e^(0+C))

=> C=0

Therefore:

y=20000(1-e^(-600t/20000))

at y=0.9*20000=18000,

=>

e^(-600t/20000)=0.1

t=-(20000/600)*log(0.1)

=77 minutes (approx.)

- math -
**deel**, Thursday, August 9, 2012 at 11:10amthanks for your help

- math :) -
**MathMate**, Thursday, August 9, 2012 at 11:40amYou're welcome!

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