LESSON WITH A FRIEND

MIXTURE PROBLEM
We now consider rate problem involving mixtures. A substances S is allowed fo flow into a certain mixture in a container at a certain rate, and the mixture is kept uniform by stirring. Further. In one such situation, this uniform mixture simultaneously flows out of the container at another (generally different) rate; in another situation this may not be the case. In either case we seek to determine the quantity of the substance S present in the mixture at time t.
Let :
x denote the amount of S present at time t,
dx/dt denotes the rate of changer of x at time t.
INPUT denotes rate S which enter the mixture.
OUTPUT denotes rate S which leaves the mixture.
Then we have at one the basic equation
dx/dt =INPUT-OUTPUT
We have a problem
A tank initially contains 50 gal of pure water. Starting at time t=0, a brine containing 2 lb of dissolved salt per gallon flow into the tank at the rate of 3 gal/min. The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate. Find the function of salt in the tank!
First we must make the mathematical formula. We apply the basic equation
dx/dt=INPUT-OUTPUT
The brine flows in at the rate of 3 gal/min and each gallon contain 2 lb of salt.
We formulating the INPUT as the weight of salt which flows in tank at time t and OUTPUT as the weight of salt which leaves tank at time t

In Physics, we have formula that
Concentration = weight / volume
Debit =volume/time
Find weight of salt at time t (weight/ time)
Weight/time = (concentration) (debit) = ( weight/ volume ) (volume/time)= weight/time
We apply at the problem to INPUT and OUTPUT
INPUT = (concentration which flows in) (debit which flows in)
= (2 lb/gal) (3 gal/min) = 6 lb/min
To find the rate of salt that leaves the tank at time t, we must find concentration of salt. The tank contains 50 gallon of the mixture at any time t. this 50 gallon contain x lb salt at time t, so the concentration is (x/50) lb/gal and the debit of mixture which leaves tank is 3 gal/min.
OUTPUT = (concentration which leaves tank) (debit which leaves tank)
= (x/50)lb/gal) (3 gal/min) = (3x/50)lb/min
We apply it in basic equation
dx/dt = 6 - (3x/50)
and we have initial condition at t=0 , x=0
x(0)=0

Then we determine this problem.
The equation is both linear and separable.
dx/dt = 6 - (3x/50)

We equaling the denominator, then we have

dx/dt = (300-3x)/50)

We group x with dx and t with dt. We change dt with (100-x) and (100-x) with dt.

dx/(100-x)=3 dt /50
We change this form into
dx/(x-100)= - 3dt/50 with multiply by -1
Then integrating both of side, then we get

ln (x-100) = -3t/50+c1
We use exponential to get the function of x

exp(ln (x-100)) = exp(-3t/50+c1)
Because of exp(ln x) = x, so
x-100=exp(-3t/50+c1)
Then move -100 to other side
x=100+exp(-3t/50+c1)
exp(c1)=c so
x=100+c.exp(-3t/50) ………(1)
We have intial value that x(0)=0 , at t=0 x =0,
then we substitute t=0 and x=0 into (1)
0= 100 + c exp(-3(0)/50)>>> exp(0)=1
0= 100 + c
c= -100
Then c = -100, substitute c into (1), we have
x=100-100.exp(-3t/50), we simply it become
x=100(1-exp(-3t/50)
x=100(1-exp(-3t/50) , is function of how much salt in the tank at t time.


The trouble at I explaining this problem to my friend Bibid are :
1.Sometimes, I forgot some word that I must say to Bibid, like exponent, power, etc.
2.I forgot how I explaining integral.
3.Sometimes I must read the source book to make Bibid say “I see”