Lale A. answered • 08/01/24
Mastering Calculus: Expert Guidance for Your Success
Let \( a(t) \) be the volume of alcohol in the vat at time \( t \) (in minutes). The vat starts with \( 0.04 \times 200 = 8 \) gallons of alcohol.
Step 1: Set up the differential equation
The rate of change of alcohol in the vat is given by:
\[
\frac{da}{dt} = (\text{rate in}) - (\text{rate out})
\]
Rate in: The beer being added has 6% alcohol, so \( \text{rate in} = 0.06 \times 2 = 0.12 \) gal/min.
Rate out: The alcohol content in the vat is \( \frac{a(t)}{200} \), and beer is being pumped out at 2 gal/min, so \( \text{rate out} = \frac{a(t)}{200} \times 2 = \frac{a(t)}{100} \) gal/min.
Therefore, the differential equation is:
\[
\frac{da}{dt} = 0.12 - \frac{a(t)}{100}
\]
Step 2: Solve the differential equation
This is a first-order linear differential equation. You can solve it by separating variables:
\[
\frac{da}{0.12 - \frac{a}{100}} = dt
\]
Integrating both sides, we have:
\[
-\ln |0.12 - \frac{a}{100}| = \frac{t}{100} + C
\]
Exponentiating both sides and simplifying, you get:
\[
0.12 - \frac{a}{100} = A e^{-\frac{t}{100}}
\]
where \( A \) is a constant. Solving for \( a(t) \):
\[
a(t) = 0.12 \times 100 - 100A e^{-\frac{t}{100}} = 12 - 100A e^{-\frac{t}{100}}
\]
Step 3: Apply the initial condition
At \( t = 0 \), \( a(0) = 8 \) gallons, so:
\[
8 = 12 - 100A \implies A = 0.04
\]
Thus, the solution for \( a(t) \) is:
\[
a(t) = 12 - 4e^{-\frac{t}{100}}
\]
Step 4: Calculate the percentage of alcohol after 1 hour
After 1 hour (60 minutes):
\[
a(60) = 12 - 4e^{-\frac{60}{100}} \approx 12 - 4 \times e^{-0.6} \approx 12 - 4 \times 0.5488 \approx 12 - 2.1952 = 9.8048 \text{ gallons}
\]
The percentage of alcohol in the vat is:
\[
\text{Percentage} = \frac{a(60)}{200} \times 100 \approx \frac{9.8048}{200} \times 100 \approx 4.9\%
\]
Therefore, the percentage of alcohol in the vat after one hour is approximately 4.9%.
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