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From the uploaded image, it appears you need help deriving formulas related to the transfer of Kool-Aid between two connected lakes after an accident, described using a differential equations model.

### Solution:

#### Part c: Rate of Change of Kool-Aid in Lake Beta (dy/dt)
Given the flow of Kool-Aid from Lake Alpha to Lake Beta, the rate of change of Kool-Aid in Lake Beta can be modeled using the formula derived for Lake Alpha and considering the flow of water between the lakes.

We know from the given information:
- Lake Alpha has an exponential decay in the concentration of Kool-Aid due to its volume and outflow rate.
- Lake Alpha's Kool-Aid amount, \( x(t) = 200 e^{-\frac{t}{2000}} \) kg.
- The flow rate into Lake Beta is the same as the outflow rate from Lake Alpha, which is 200 liters per hour.
- Since both lakes have constant volume, the in-flow concentration equals the out-flow concentration.

Using this information, the rate of change of Kool-Aid in Lake Beta (where \( y \) is the amount of Kool-Aid in kilograms in Lake Beta) can be derived from the amount exiting Alpha and entering Beta:

\[ \frac{dy}{dt} = \frac{200}{500000} x(t) \]
\[ \frac{dy}{dt} = \frac{200}{500000} \cdot 200 e^{-\frac{t}{2000}} \]
\[ \frac{dy}{dt} = \frac{200 \times 200}{500000} e^{-\frac{t}{2000}} \]
\[ \frac{dy}{dt} = 0.08 e^{-\frac{t}{2000}} \text{ kg/hour} \]

#### Part d: Formula for the Amount of Kool-Aid in Lake Beta (y(t))
To find \( y(t) \), we integrate \( \frac{dy}{dt} \) from 0 to \( t \):

\[ y(t) = \int_0^t 0.08 e^{-\frac{s}{2000}} \, ds \]

This integral results in:
\[ y(t) = -0.08 \times 2000 e^{-\frac{s}{2000}} \Big|_0^t \]
\[ y(t) = -160 \left( e^{-\frac{t}{2000}} - 1 \right) \]
\[ y(t) = 160 (1 - e^{-\frac{t}{2000}}) \text{ kg} \]

So the amount of Kool-Aid in Lake Beta, \( t \) hours after the crash, is \( y(t) = 160 (1 - e^{-\frac{t}{2000}}) \) kg.

Let me know if you want more details or if you have any questions about these calculations! Here are some related questions you might consider:

1. How does the solution change if the flow rate between the lakes is different?
2. What happens to the concentration of Kool-Aid in Lake Beta over a very long period?
3. How would the formula adjust if there were a significant evaporation rate affecting the lakes’ volumes?
4. Could you apply this model to a scenario with more than two lakes in sequence?
5. How does the initial concentration of Kool-Aid affect the time it takes to reach equilibrium between the two lakes?

**Tip:** When dealing with exponential decay or growth, integrating or differentiating exponential functions often results in an expression that can be simplified using initial conditions or boundary values.

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 200 liters per hour. Lake Alpha contains 400 thousand liters of water, and Lake Beta contains 500 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. Let x(t) and y(t) represent the amount of Kool-Aid in kilograms in Lake Alpha and Lake Beta, respectively, t hours after the crash. Find the formulas for the rates of change dx/dt and dy/dt and the amounts x(t) and y(t) at any time t.

This problem involves using differential equations to model the flow of Kool-Aid between two connected lakes after an accident. The solution includes exponential decay and integration techniques to derive the rates of change and the amounts of Kool-Aid in each lake over time. Suitable for college-level calculus students.

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