Posted by deel on Friday, August 10, 2012 at 9:58pm.
Would 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.
oke thanks, i'ill try it
y(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
Yes, 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.
thanks for your help :)
You're welcome!
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