Initially 10 grams of salt are dissolved into 15 liters of water. Brine with con
ID: 2894320 • Letter: I
Question
Initially 10 grams of salt are dissolved into 15 liters of water. Brine with concentration of salt 3 grams per liter is added at a rate of 5 liters per minute. The tank is well mixed and drained at 5 liters per minute. a Let x| be the amount of salt, in grams, in the solution after t| minutes have elapsed. Find a formula for the rate of change in the amount of salt, dx/dt|, in terms of the amount of salt in the solution x|. dx/dt = | grams/minute b. Find a formula for the amount of salt, in grams, after t| minutes have elapsed. x (t) =| grams c. How long must the process continue until there are exactly 25 grams of salt in the tank? minutesExplanation / Answer
a)
rate of salt input =3*5=15
rate of salt output=5*x/(15+(5-5)t)=(1/3)x
dx/dt=rate of salt input -rate of salt output
dx/dt=15-(1/3)x grams per minute , x(0)=10 grams
b)
dx/dt=15-(1/3)x
dx+(1/3)x dt=15dt
integrating factor =e(1/3)dt
integrating factor =e(1/3)t
multiply on both sides by e(1/3)t
=>e(1/3)tdx+(1/3)xe(1/3)tdt=15e(1/3)tdt
=>(xe(1/3)t)'=15e(1/3)tdt
integrate on both sides
=>(xe(1/3)t)'=15e(1/3)tdt
=>(xe(1/3)t)=15*3e(1/3)t+c
=>x=45+ce-(1/3)t
x(0)=10
45+ce-(1/3)*0=10
=>45+c=10
=>c=-35
so amount of salt after t minutes have elapsed is x(t)=45-35e-(1/3)t
c)
25 grams of salt in tank
=>45-35e-(1/3)t=25
=>35e-(1/3)t=20
=>e(1/3)t=35/20
=>(1/3)t=ln1.75
=>t=3*ln1.75
=>t=1.68 minutes approximately