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Posted by **deel** on Friday, August 10, 2012 at 9:58pm.

- math -
**MathMate**, Friday, August 10, 2012 at 10:31pmWould it not be a similar to a previous problem?

http://www.jiskha.com/display.cgi?id=1344476111

You can try the same approach and post your attempt for a check.

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**deel**, Saturday, August 11, 2012 at 12:43amoke thanks, i'ill try it

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**deel**, Saturday, August 11, 2012 at 2:41amy(t)=amount of alcohol in the tank at time t

dy/dt

=y’(t)

= alcohol inflow rate – alcohol outflow rate

= 4 - 5(y/50)

=200 - 5y/50

= 5 (40-y)/50

dy/dt = 5 (40-y)/50

Separate variable and integrate :

dy/(40-y) = 5/50dt

-ln(40-y) = 0.1t + C

ln(40-y) = -0.1t + C

40-y= Ce^-0.1t

Y= 40(1-Ce^-0.1t) ….. eq. 1

Initially, the tank contains 50 gallon (50% alcohol) = 25 gallon of alcohol,

y(0) = 25

substituting y=25 and t=0 in to eq.1 gives

25 = 40(1-C)

25-40 = -40C

C = 0.375

When t=10 minutes

y(10) = 40(1-0.375*e^-0.1(10))

y(10) = 40(1-0.375*0.367)

y(10) = 34.48 gallon

at t=10’, amount of alcohol in the tank

= 50% x 34.48 gallon = 17.24 gallon or 34.48%

please correct the result of my work. thanks

- math -
**MathMate**, Saturday, August 11, 2012 at 7:04amYes, you have done a fantastic job of setting up and solving the problem.

If you reread the question carefully, you will probably discover a few traps.

If you don't find the traps, here are a few hints to improve the solution:

1. Alcohol coming into the tank is 50%, so intake rate is 4*0.5=2 gal/min.

2. Initial concentration is 90%, so initial alcohol content is 50*0.9=45 gal.

3. Initially, output volume is y/50 (as you had it).

However, since the input is 4 gal/min, and output is 5 gal/min, you will need to divide by the total volume as a function of time, which turns out to be 50-t.

4. finally, do not forget required parentheses. Otherwise, your work will be very hard to read, such as:

"= 4 - 5(y/50)

=200 - 5y/50

..."

instead of

= 4 - 5(y/50)

=(200-5y) / 50

Good job, keep up the good work.

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**deel**, Saturday, August 11, 2012 at 10:58pmthanks for your help :)

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**MathMate**, Sunday, August 12, 2012 at 2:36pmYou're welcome!