Tracking Kool-Aid Through Lakes Alpha & Beta

[ends]

Suppose the clean water of a stream flows into Lake Alpha, then into Lake Beta, and then further downstream. The in and out flow for each lake is 500 liters per hour. Lake Alpha contains 400 thousand liters of water, and Lake Beta contains 200 thousand liters of water. A truck with 200 kilograms of Kool-Aid drink mix crashes into Lake Alpha. Assume that the water is being continually mixed perfectly by the stream.

a)Let x be the amount of Kool-Aid, in kilograms, in Lake Alpha t hours after the crash. Find a formula for the incremental change in the amount of Kool-Aid, DeltaX, in terms of the amount of Kool-Aid in the lake x and the incremental change in time DeltaT.

b) Find a formula for the amount of Kool-Aid, in kilograms, in Lake Alpha t hours after the crash.

c)Let y be the amount of Kool-Aid, in kilograms, in Lake Beta t hours after the crash. Find a formula for the incremental change in the amount of Kool-Aid, DeltaY, in terms of the amounts x, y, and the incremental change in time DeltaT.

d) Find a formula for the amount of Kool-Aid in Lake Beta t hours after the crash.I only tried a and b.for a i got:

DeltaX = (0.25 - x/800)*DeltaT

b)

x(t) = 200 - 200e^(-t/800)

But these are both wrong according to the database.

 

[MarkFL]

They key to these mixing problems is to look at how much of a particular substance is entering the "compartment" and how much is leaving. For Lake Alpha, after the crash, how much Kool-Aid is entering the lake in kg/hr? How much is leaving in kg/hr?

We want to correctly set up the initial value problem:

\(\displaystyle \frac{dx}{dt}=\text{amount in per hour}-\text{amount out per hour}\) where $x(0)=x_0=200$

Can you identify the amounts in and out per hour?

 

[ends]

They key to these mixing problems is to look at how much of a particular substance is entering the "compartment" and how much is leaving. For Lake Alpha, after the crash, how much Kool-Aid is entering the lake in kg/hr? How much is leaving in kg/hr?

We want to correctly set up the initial value problem:

\(\displaystyle \frac{dx}{dt}=\text{amount in per hour}-\text{amount out per hour}\) where $x(0)=x_0=200$

Can you identify the amounts in and out per hour?

Point taken...It was reinforced in class. But from this question, the rate in and rate out i get are:

\(\displaystyle rate IN = concentration*flow
= (200kg/400,000L)*(500L/hr)
= 0.25kg/hr\)

\(\displaystyle rate OUT = conc*flow
= (x/400,000L)(500L/hr)
= x/800 kg/hr\)

Which sets up the diff eqn,:

\(\displaystyle dx/dt = 0.25 - x/800\)

But this is incorrect...

 

[MarkFL]

There is no additional Kool-Aid coming into Lake Alpha once the truck has crashed and deposited the 200 kg there, so the rate in is zero. You have correctly determined the flow out:

\(\displaystyle \left(\frac{x}{400000}\,\frac{\text{kg}}{\text{L}} \right)\left(500\,\frac{\text{L}}{\text{hr}} \right)=\frac{x}{800}\,\frac{\text{kg}}{\text{hr}}\)

And so this gives us the IVP:

\(\displaystyle \frac{dx}{dt}=-\frac{x}{800}\) where \(\displaystyle x(0)=200\)

So, you want to solve this to answer part b). Then use the flow rate out of Lake Alpha (replacing $x$ with the solution found for part b)) as the flow rate in for Lake Beta for parts c) and d).

 

[AbeLau]

There is no additional Kool-Aid coming into Lake Alpha once the truck has crashed and deposited the 200 kg there, so the rate in is zero. You have correctly determined the flow out:

\(\displaystyle \left(\frac{x}{400000}\,\frac{\text{kg}}{\text{L}} \right)\left(500\,\frac{\text{L}}{\text{hr}} \right)=\frac{x}{800}\,\frac{\text{kg}}{\text{hr}}\)

And so this gives us the IVP:

\(\displaystyle \frac{dx}{dt}=-\frac{x}{800}\) where \(\displaystyle x(0)=200\)

So, you want to solve this to answer part b). Then use the flow rate out of Lake Alpha (replacing $x$ with the solution found for part b)) as the flow rate in for Lake Beta for parts c) and d).

Hi, I happened to come across this exact question. After reading your response I am still a bit confuse about part a of this question. Since there is no flow in does it means taking dx/dt=−x/800 and use separating variable technique I got -x^2/2 = t/800 + C. And isolating x I got x = sqrt(-t/400 +C). Is this the answer to part a? Thank you.

 

[MarkFL]

When you separate variables, you obtain:

\(\displaystyle \frac{dx}{x}=-\frac{1}{800}\,dt\)

which upon integration, gives you a different solution than you obtained.