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Initially 15 grams of salt are dissolved into 25 liters of water. Brine with con

ID: 2987714 • Letter: I

Question

Initially 15 grams of salt are dissolved into 25 liters of water. Brine with concentration of salt 5 grams per liter is added at a rate of 2 liters per minute. The tank is well mixed and drained at 2 liters per minute.

Initially 15 grams of salt are dissolved into 25 liters of water. Brine with concentration of salt 5 grams per liter is added at a rate of 2 liters per minute. The tank is well mixed and drained at 2 liters per minute. Let x be the amount of salt, in grams, in the solution after t minutes have elapsed. Find a formula for the incremental change in the amount of salt, Delta x, in terms of the amount of salt in the solution x and the incremental change in time Delta t. Enter Delta t as Deltat. Find a formula for the amount of salt, in grams, after t minutes have elapsed. How long must the process continue until there are exactly 20 grams of salt in the tank? Suppose the clean water of a stream flows into Lake Alpha, then into Lake Beta, and then further downstream. The in and out flow for each lake is 200 liters per hour. Lake Alpha contains 500 thousand liters of water, and Lake Beta contains 200 thousand liters of water. A truck with 500 kilograms of Kool-Aid drink mix crashes into Lake Alpha. Assume that the water is being continually mixed perfectly by the stream. Let x be the amount of Kool-Aid, in kilograms, in Lake Alpha t hours after the crash. Find a formula for the incremental change in the amount of Kool-Aid, Delta x, in terms of the amount of Kool-Aid in the lake x and the incremental change in time Delta t. Enter Delt t as Deltat. Find a formula for the amount of Kool-Aid, in kilograms, in Lake Alpha t hours after the crash. Let y be the amount of Kool-Aid, in kilograms, in Lake Beta t hours after the crash. Find a formula for the incremental change in the amount of Kool-Aid, Delta y, in terms of the amounts x, y, and the incremental change in time Delta t. Enter Delta t as Deltat. Find a formula for the amount of Kool-Aid in Lake Beta t hours after the crash.

Explanation / Answer

Salt enters at the rate 5 g/L * 2L/min = 10 g/min

Salt leaves at the rate 2 x/25 (2 because it is 2 liters; divide by 25 because there are 25 liters of water.

Thus, delta x = (10 - 2/25 x) delta t

dx/dt = 10 - 2/25 x

dx/(10 - 2/25x) = dt

Then,-25/2 ln|10 - 2/25 x| = t + c

Multiply through by -2/25

ln|10 - 2/25 x| = -2/25t - 2/25c

Taking the exponential, and letting e^-2/25c = C

10 - 2/25x = Ce^-2/25t

2/25x = 10 - Ce^-2/25t

Multiply by 25/2

x = 125 - Ce^-2/25t


As initially,at t=0, there are 15 g

15 = 125 - C

C = 110

x = 125 - 110e^-2/25t


There are 20g when


20 = 125 - 110e^-2/25t

-105 = -110e^-2/25t

21/22 = e^-2/25t

Taking lns, and the reciprocal to turn into a positive,

ln(22/21) = 2/25t

t = 25/2 ln(22/21)

t = .5815 minutes (as a check, if salt weren't leaving .5 minutes * 10 = 5 g, the additional salt to go from 15 to 20)


Since water leaves at 200 L/hour, and there are 500000 L in Lake Alpha,


delta x = -200x/500000 delta t

delta x = -x/2500 delta t


At t = 0, there are 500 kg of Kool-aid

dx/dt = -1/2500 x

dx/x = -1/2500dt

ln|x| = -1/2500t + c

Letting C = e^c, taking the exponential of both sides

x = Ce^-1/2500t

At t= 0, x = 500 kg

500 kg = C*1

500kg = C

x = 500e^-1/2500t


Lake Beta has 200000 L

We have koolaid coming in from alpha and leaving as well at 200L for the mix


Then, delta y = (x/2500 - y * 200/200000) delta t

delta y = (x/2500 -y/1000) delta t


dy/dt = x/2500 -y/1000


From the first solution, we have x = 500e^-1/2500t


Then, dy/dt = 500e^-1/2500t /2500 - y/1000

dy/dt = 1/5e^-1/2500t - y/1000


We can make this into a homogeneous and inhomogeneous equation


dy/dt + y/1000 = 0


D + 1/1000 = 0

D = -1/1000

Y = Ce^-1/1000t


For the specific solution,

dy/dt + y/1000= 1/5e^-1/2500t


we obviously consider C2 e^-1/2500t


Then, dy/dt + y/1000 = 1/5e^-1/2500t


If y = C2 e^-1/2500t, dy/dt + y/1000 = -1/2500C2 e^-1/2500t + C2/1000 e^-1/2500t


3/5000 C2 = 1/5


C2 = 1000/3




Our solution is then


Ce^-1/1000t + 1000/3 e^-1/2500t


As initially, there is no koolaid in Lake Beta,


0 = C + 1000/3

C = -1000/3


Thus,


y = -1000/3e^-1/1000t + 1000/3 e^-1/2500t


Note how this increases from 0, then decreases to 0, as you would expect.