32 problem 2 1 point A tank contains 100 kg of salt and 1000

3.2 problem 2

(1 point) A tank contains 100 kg of salt and 1000 L of water. Pure water enters a tank at the rate 6L/min. The solution is mixed and drains from the tank at the rate 3 L/min. (a) What is the amount of salt in the tank initially? (kg) amount (b) Find the amount of salt in the tank after 3.5 hours. (kg) amount Find the concentration of lt in the solution in the tank as time approaches infinity. (Assume your tank is large enough to hold all the solution.) (kg/L) concentration

Solution

Let s(t) be the amount of salt at time t.

Let w(t) be the amount of water at time t.

Given tank contains salt =100kgs

Let s(0)=100 kgs.

Tank contains water=1000lit

w(0)=1000lit

Pure water enters to tank = 6lit/min

Let it be w\'= 6

From this w= 6t +some constant k

We know w(0)= 1000=k

w= 6t+1000

Solution is mixed and drained from the rank= 3lit/min

Let it be s\'=3*(s/w)

s\'=3*(s(t)/w(t))

s\'=(3s(t)/(6t+1000))

ds/dt=(3*s(t)/(6t+1000))

ds/3*s(t)=(dt/(6t+1000))

Integrating on both sides

Then we get

Integral of (ds/3*s(t))=integral of (dt/(6t+1000))

(1/3)*s=integral of (dt/(6t+1000))

Let put 6t=x

Derivative on both sides 6*dt=dx

dt=(1/6)*dx

Now substitute this in above

(1/3)*log(s)=integral of(dx/(6*(x+1000)))

(1/3)*log(s)=(1/6)*(log(x+1000))+k

log(s)=(1/2)*(log(6t+1000))+k

log(s)=(1/2)*log(w(t))+k

Constant k=?

k=log(s)-(1/2)*log(w(t))

Intial s(0)=100

W(0)=1000

K=log(100)-(1/2)*(log(1000))

k=log(10)^2-(1/2)*(log(10)^3

k=2log(10)-(3/2)* log(10)

k=2-3/2*(log10)

K=1/2*(log10)

Therefore log(s)=(1/2)*log(6t+1000)+1/2

log(s)=(1/2)*{log(6t+1000)+log(10)}

log(s)=(1/2)*log((6t+1000)*10)

log(s)=(1/2)*log(60t+10000)

s=(1/2)(60t+10000)

S=5(6t+1000)

a.the amount of salt in tank=s(0)=100kgs

B.the amount of salt at 3.5hrs=

S(3.5)=5((6*3.5)+1000)

s(3.5)=5(21+1000)

S(3.5)=5*1021

S(3.5)=5105kgs

C.the concentration of salt in solution in tank at time approaches Infinity=s(t)/w(t)

Concentration={e^((1/2)*(6t+1000+1))}/(6t+1000)