1 point) A tank contains 2200 L of pure water. Solution that contains 0.06 kg of sugar per liter enters the tank at the rate 5 L/min, and is thoroughly mixed into it. The new solution drains out of the tank at the same rate. (a) How much sugar is in the tank at the beginning

Question not complete,

So i will analyse the possible problem

Step-by-step explanation:

A tank contain 2200L

Volume V = 2200L

Solution of 0.06kg/L of sugar

Rate of entry i.e input

dL/dt=5L/min

Let y(t) be the amount of sugar in tank at any time.

But at the beginning there was no sugar in the tank

i.e, y(0)=0, this will be out initial value problem,

The rate of amount of sugar at anytime t is

dy/dt=input amount of sugar - output amount of sugar.

Now,

Then rate of input is

5L/min × 0.06kg/L

Then, input rate= 0.3kg/min

Output rate is

5L/mins × y(t)/2200 kg/L

then, output rate = y(t)/440 kg/min

Now then,

dy/dt=input rate -output rate

dy/dt=0.3-y/440

Cross multiply through by 400

400dy/dt=120-y

Using variable separation

400/(120-y) dy = dt

∫400/(120-y) dy = ∫dt

-400In(120-y)=t +C

In(120-y)=-t/400+C/400

C/400 is another constant, let say B

In(120-y)=-t/400+B

Take exponential of both side

120-y(t)=exp(-t/400+B)

120-y(t)=exp(-t/400)exp(B)

exp(B) is a constant let say C

-y(t)=Cexp(-t/400)-120

y(t)=120-Cexp(t/400)

Now, the initial condition

a. At the start the mass of sugar in the water is 0 because it is just pure water at start.

Therefore y(0)=0,

b. Applying this to y(t)

y(t)=120-Cexp(-t/400)

y=0, t=0

0=120-Cexp(0)

0=120-C

C=120

Therefore,

y(t)=120 - 120exp(-t/400)

Let know the mass rate as t tends to infinity

At infinity

exp(-∞)=1/exp(∞)=1/∞=0

Then,

The exponential aspect tend to 0

Then, y(t)=120 as t tend to ∞