Posted by **COFFEE** on Friday, July 13, 2007 at 4:36pm.

Please check my work below and comment.

A tank initially contains 80 gallons of fresh water. A 10% acid solution flows into the tank at the rate of 3 gallons per minute. The well-stirred mixture flows out of the tank at the rate of 3 gallons per minute. Find the amount of acid in the tank at the end of any time t. How much acid will be in the tank in 30 minutes? What will be the concentration (%) of acid in the tank after 30 minutes?

solution=0.3 gallons acid + 2.7 gallons H20

dy/dt = (rate in) - (rate out)

rate in = (0.3)(3) = 0.9 gallons/min

<-would this be gallons per min?

rate out = ((y(t))/80) * 3 gal/min

rate out = y(t)/(80/3)

dy/dt=(0.9)-(y(t))/(80/3)

dy/dy=((773/30)-y(t))/(80/3)

Seperate into two intergrals:

Integral(dy/((773/30)-y)=Integral(dt/(80/3))

-ln|(773/30)-y| = 3t/80+C, C=-ln|773/30|

-ln|(773/30)-y|=3t/80 - ln|773/30|

y(t)=(773/30)-(773/30)e^(-3t/80)

Is this the correct equation?

Then at 30 minutes, y(30) would be 17.4 gallons acid

Percentage after 30 minutes=17.4/80

=21.8% acid???

Thanks.

correct procedure, however, one question: how did you get rate in .3*3? Wasn't it ten percent?

ah yes. Thank you!

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