Problem 6.441
mediumA tank holds \( \displaystyle 200 \) L of water with \( \displaystyle 0 \) kg of salt dissolved. Brine with 1/4 kg of salt per liter flows in at \( \displaystyle 4 \) L/min, and the well-mixed solution drains at the same rate. Find the salt \( \displaystyle S(t) \) in kg after \( \displaystyle t \) minutes, the amount after \( \displaystyle 10 \) minutes, and the long-run amount.
- Rate in − rate out: S′ = 4·1/4 − (4/200)S. The volume stays 200 L.Reviewed
- \[ \frac{d}{d t} \left(50 - 50 e^{- \frac{t}{50}}\right) = e^{- \frac{t}{50}} \]This S(t) satisfies the equation ...✓ Proved
- \[ \left. 50 - 50 e^{\frac{\left(-1\right) t}{50}} \right|_{\substack{ t=0 }} = 0 \]... and starts at the given amount.✓ Proved
- \[ \left. 50 - 50 e^{\frac{\left(-1\right) t}{50}} \right|_{\substack{ t=10 }} = 50 - \frac{50}{e^{\frac{1}{5}}} \]S(10).✓ Proved
- \[ \lim_{t \to \infty}\left(50 - 50 e^{- \frac{t}{50}}\right) = 50 \]In the long run the tank matches the incoming concentration.✓ Proved
Answer \( S(t) = 50 - 50 e^{- \frac{t}{50}},\ S(10) = 50 - \frac{50}{e^{\frac{1}{5}}} \approx 9.063,\ \lim S = 50 \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify