Question

A tank contains 100 gallons of pure water. At time zero, a sugar-water solution containing 0.2 lb of sugar per gallon enters the tank at a rate of 3 gallons per minute. Simultaneously, a drain is opened at the bottom of the tank allowing the sugar solution to leave the tank at 3 gallons per minute. Assume the solution in the tank is kept perfectly mixed at all times. What will be the eventual sugar content in the tank as time goes to infinity?

Answer 1

20 lbs of sugar per gallon as time goes to infinity

Explanation:

Because of the statement ... as time goes to infinity...I am inclined to approach this problem using derivatives and limits

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

`S(t) \;lbs/gal`

We have
`S(0) = 0` since the initial quantum of water has zero sugar

The inflow rate of sugar is given as 3 gal/min and each gallon contains 0.2 lbs of sugar
\therefore The rate of sugar increase in the tank n lbs/gal
= 0.2 lbs /gal \times 3 gal/min

= 0.6 lbs/min

Outflow rate = 3 gal/min
At any time t, the tank contains 100 gal
The amount of sugar at time t = \dfrac{S(t)}{100} lbs/gal

Therefore the outflow rate of sugar = 3 gal/min \times \dfrac{S(t)}{100} lbs/gal

= 0.03S(t) lb/min

The net rate is given by inflow rate - outflow rate

= 0.6 - 0.03S(t)

In calculus terms the net rate =
`(dS(t))/(dt)`

Therefore

`(dS(t))/(dt) = 0.6 - 0.03S(t)`

`(dS(t))/(dt) - 0.03S(t) = 0.6\\\\` (1)

A trick to solving differential equations of this type is to use the method of integrating factors

In general, if we have a differential equation of the type

`(dy)/(dx) + Py = Q\\\\`
where P and Q are functions only of x

then the integrating factor is
`e^{\int\ {P} \, dx }`

which you use to multiply both sides of the differential equation first

In equation (1) above,

y = S(t) which well simply write as S for convenience

Writing S' for
`(dS(t))/(dt)` for convenience we get equation (1) as

`S' + 0.03S =0.6` (2)

From the above we see that P in the generalized differential equation corresponds to 0.03

Therefore the integrating factor is

`\mbox {\large e^(\int 0.03dt) = e^(0.03t)}`

Multiply equation (2) throughout by this integrating factor to get

`\mbox{\large e^(0.03t) S' + 0.03S e^(0.03t) = 0.6 e^(0.03t) }`

The left side is nothing but the first derivative of
`\mbox{\large (e^(0.03t)S) }\\\\`

`= \mbox{\large (e^(0.03t)S)'\\\\}}`

Integrating both sides we get


`\mbox{\large e^(0.03t)S = 0.06 \int e^(0.03t)dt}\\\\` (3)

`\textrm{By using the fact that $\int e^(ax) dx = (e^(ax))/(a) + C$}:\\\\\mbox{\large \int e^(0.03t)dt} = (e^(0.03))/( 0.03)} + C\\\\`

Therefore equation (3) becomes

`e^(0.03t)S = 0.06 \cdot (e^(0.03))/(0.03)} + C\\\\e^(0.03t)S = 0.06 \cdot 33.333 \cdot e^(0.03) + C\\\\e^(0.03t)S = 20 \cdot e^(0.03) + C\\`

Dividing by
`e^(0.03) \textrm{ (same as multiplying by $e^(-0.03)$) both sides}`:

`S = 20 + Ce^(-0.03t)\\\\\textrm{Plugging back S(t):}\\\\S(t) = 20 + Ce^(-0.03t)\\`

We are asked to find the level of sugar content as t ⇒ ∞

At t = 0, S(t) = 0; there is no sugar content

S(0) = 0 = 20 + Ce⁰

0 = 20 + C
C = -20

`S(t) = 20 -20e^(-0.03t)\\\\`

As t ==> ∞
we get

`\lim _(x\to \infty )S\left(t\right)=\:\lim _(x\to \infty )\left(20\:-\:20e^(-0.03t)\right)=\:\:20\:-\:0\:=\:20`

Therefore as time goes to infinity the eventual sugar content
= 20 lbs/gallon