Question

A 200-gal tank contains 100 gal of pure water. At time t = 0, a salt-water solution containing 0.5 lb/gal of salt enters the tank at the rate of 5 gal/min, and the mixture, which is kept uniform by stirring, is withdrawn at the rate of 3 gal/min. (1) Write down a differential equation for the amount of salt in the tank at a time t. (2) Find the amount of salt and its concentration in the tank at a time t. (3) At the time the tank is full, how many pounds of salt will it contain? (4) What would be the limiting concentration of salt at infinity time if the tank had infinity capacity?

Answer 1

1)
`(dy)/(dt)=2.5-(3y)/(2t+100)`

2)
`y(t)=(50+t)- \frac{12500√(2) }{(50+t)^{(3)/(2) }}`

3) 98.23lbs

4) The salt concentration will increase without bound.

Explanation:

1) Let
`y` represent the amount of salt in the tank at time t, where t is given in minutes.

Recall that:
`(dy)/(dt)=rate\:in-rate\:out`

The amount coming in is
`0.5(lb)/(gal)* 5(gal)/(min)=2.5(lb)/(min)`

The rate going out depends on the concentration of salt in the tank at time t.

If there is
`y(t)` pounds of salt and there are
`100+2t` gallons at time t, then the concentration is:
`(y(t))/(2t+100)`

The rate of liquid leaving is is 3gal\min, so rate out is
`=(3y(t))/(2t+100)`

The required differential equation becomes:

`(dy)/(dt)=2.5-(3y)/(2t+100)`

2) We rewrite to obtain:

`(dy)/(dt)+(3)/(2t+100)y=2.5`

We multiply through by the integrating factor:
`e^{\int (3)/(2t+100)dt }=e^{(3)/(2) \int (1)/(t+50)dt }=(50+t)^{(3)/(2) }`

to get:

`(50+t)^{(3)/(2) }(dy)/(dt)+(50+t)^{(3)/(2) }\cdot (3)/(2t+100)y=2.5(50+t)^{(3)/(2) }`

This gives us:

`((50+t)^{(3)/(2) }y)'=2.5(50+t)^{(3)/(2) }`

We integrate both sides with respect to t to get:

`(50+t)^{(3)/(2) }y=(50+t)^{(5)/(2) }+ C`

Multiply through by:
`(50+t)^{-(3)/(2)}` to get:

`y=(50+t)^{(5)/(2) }(50+t)^{-(3)/(2) }+ C(50+t)^{-(3)/(2) }`

`y(t)=(50+t)+ \frac{C}{(50+t)^{(3)/(2) }}`

We apply the initial condition:
`y(0)=0`

`0=(50+0)+ \frac{C}{(50+0)^{(3)/(2) }}`

`C=-12500√(2)`

The amount of salt in the tank at time t is:

`y(t)=(50+t)- \frac{12500√(2) }{(50+t)^{(3)/(2) }}`

3) The tank will be full after 50 mins.

We put t=50 to find how pounds of salt it will contain:

`y(50)=(50+50)- \frac{12500√(2) }{(50+50)^{(3)/(2) }}`

`y(50)=98.23`

There will be 98.23 pounds of salt.

4) The limiting concentration of salt is given by:

`\lim_(t \to \infty)y(t)={ \lim_(t \to \infty) ( (50+t)- \frac{12500√(2) }{(50+t)^{(3)/(2) }})`

As
`t\to \infty`,
`50+t\to \infty` and
`\frac{12500√(2) }{(50+t)^{(3)/(2) }}\to 0`

This implies that:

`\lim_(t \to \infty)y(t)=\infty- 0=\infty`

If the tank had infinity capacity, there will be absolutely high(infinite) concentration of salt.

The salt concentration will increase without bound.