The BEST answer gets 5 stars a lank contains 100 gallons of water. Five gallons
ID: 2986816 • Letter: T
Question
The BEST answer gets 5 stars
a lank contains 100 gallons of water. Five gallons of brine per minute flow into the tank, each gallon of brine containing 2 pounds of salt. Five gallons of water flow out of the tank per minute. assume that the tank is kept well stirred. Find a differential equation for the number of pounds of salt in the tank (call it a. say). assuming the tank initially contains 10 pounds of salt, solve this differential equation. How much salt is in the tank after 10 minutes' at what time will there be 199 lbs of salt in the tank? (Jive an exact answer and a numerical answer accurate to two decimal placesExplanation / Answer
Note that the same amount flows in and out.
Thus, salt enters at 2 pounds/gallon * 5 gallons/minute = 10 pounds/minute
As there are 100 gallons in the tank, and the mixture leaves at 5 gallons/minute, salt leaves at
A(t)/100 * 5 = 1/20A(t)
Then, dA/dt = 10 - 1/20A = (200-A)/20 = -1/20(A-200)
A(0) = 10
dA/(A-200) = -1/20 dt
ln|200-A| = -1/20t + c
200 - A = e^ce^-1/20t
Let e^c = C
200 - A = Ce^-1/20t
As A(0) = 10
200 - A(0) = Ce^-1/20(0) = Ce^0=C
200 - 10 = C
C = 190
200 - A = 190e^-1/20t
A = 200 - 190e^-1/20t
In 10 minutes, A = 200 - 190e^-1/20(10) = 200 - 190e^-1/2 = 84.7591746545996
199 = 200 - 190e^-1/20(t)
-1 = - 190e^-1/20(t)
1/190 = e^-1/20(t)
ln(1/190) = -1/20t
ln(190) = 1/20 t
t = 20 ln(190)
t is approximately 104.94048144321, which rounds to 104.94
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