Answer:
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 ∞
