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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 places

Explanation / 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