To solve this, we use the concept of a differential equation to model the alcohol concentration in the vat. Let's break it down step by step:

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Variables:

• Let $A(t)$ be the amount of alcohol (in gallons) in the vat at time $t$, where $t$ is in minutes.
• Initially, the vat contains $600 \times 0.04 = 24$ gallons of alcohol.
• The rate of inflow of alcohol is $6 \, \text{gal/min} \times 0.06 = 0.36 \, \text{gal/min}$.
• The rate of outflow of alcohol is proportional to the concentration in the vat, which is: $\frac{A(t)}{600} \times 6 \, \text{gal/min} = \frac{6A(t)}{600} = \frac{A(t)}{100} \, \text{gal/min}.$

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Differential Equation:

The rate of change of alcohol in the vat is given by: $\frac{dA}{dt} = \text{rate of inflow} - \text{rate of outflow}.$ Substituting the values: $\frac{dA}{dt} = 0.36 - \frac{A(t)}{100}.$

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Solving the Differential Equation:

Rewriting: $\frac{dA}{dt} + \frac{A(t)}{100} = 0.36.$

This is a first-order linear differential equation. The integrating factor (IF) is: $e^{\int \frac{1}{100} \, dt} = e^{t/100}.$

Multiply through by the IF: $e^{t/100} \frac{dA}{dt} + \frac{A(t)}{100} e^{t/100} = 0.36 e^{t/100}.$

The left-hand side becomes: $\frac{d}{dt} \left( A(t) e^{t/100} \right) = 0.36 e^{t/100}.$

Integrating both sides with respect to $t$: $A(t) e^{t/100} = \int 0.36 e^{t/100} \, dt.$

The integral of $0.36 e^{t/100}$ is: $0.36 \int e^{t/100} \, dt = 0.36 \cdot 100 e^{t/100} = 36 e^{t/100}.$

So: $A(t) e^{t/100} = 36 e^{t/100} + C.$

Divide through by $e^{t/100}$: $A(t) = 36 + C e^{-t/100}.$

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Applying Initial Condition:

At $t = 0$, $A(0) = 24$: $24 = 36 + C e^{0}.$ $C = 24 - 36 = -12.$

Thus: $A(t) = 36 - 12 e^{-t/100}.$

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Finding Alcohol After 60 Minutes:

At $t = 60$: $A(60) = 36 - 12 e^{-60/100}.$

Calculate: $e^{-60/100} = e^{-0.6} \approx 0.5488.$ $A(60) = 36 - 12 \cdot 0.5488 \approx 36 - 6.5856 \approx 29.4144 \, \text{gallons}.$

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Alcohol Percentage:

The total volume remains $600$ gallons, so the percentage of alcohol is: $\text{Percentage} = \frac{A(60)}{600} \times 100 = \frac{29.4144}{600} \times 100 \approx 4.9\%.$

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Final Answer:

The percentage of alcohol after an hour is: $\boxed{4.9\%}.$

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Would you like further clarification or help with a different problem? Here are 5 related questions:

1. How does the inflow rate affect the equilibrium percentage of alcohol?
2. What happens if the outflow rate is faster than the inflow rate?
3. How would the solution change if the initial volume was 500 gallons?
4. Can we determine the time it takes for the alcohol percentage to stabilize?
5. How is the integrating factor derived in differential equations?

Tip: Always verify equilibrium conditions in such problems, as they simplify long-term behavior analysis.