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Mixing tank problems

Problem 6.590 · medium

A tank holds \( \displaystyle 200 \) L of water with \( \displaystyle 20 \) kg of salt dissolved. Brine with 1/10 kg of salt per liter flows in at \( \displaystyle 2 \) 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′ = 2·1/10 − (2/200)S. The volume stays 200 L.
\[ \frac{d}{d t} 20 = 0 \]
This S(t) satisfies the equation ...✓ Proved
\[ \left. 20 \right|_{\substack{ t=0 }} = 20 \]
... and starts at the given amount.✓ Proved
\[ \left. 20 \right|_{\substack{ t=10 }} = 20 \]
S(10).✓ Proved
\[ \lim_{t \to \infty} 20 = 20 \]
In the long run the tank matches the incoming concentration.✓ Proved

Answer \( S(t) = 20,\ S(10) = 20 \approx 20.000,\ \lim S = 20 \)

Lines: 4 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify

The full receipt

Line	Status	Checked by	Detail
1	Not checked	—	a sentence; read, not computed
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
4	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
5	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	mpmath's ODE solver integrates the tank from t = 0

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The solution assumes the answer is S(t) = 20 without deriving it. While S(t)=20 is indeed the equilibrium solution, the general solution to the differential equation is S(t) = 20 + Ce^{-t/100}. The initial condition S(0)=20 implies C=0, so S(t)=20 is correct, but the solution skips the derivation and verification of the general solution, presenting the final answer as if it were obvious or derived directly from the DE without solving it. More critically, line 2 checks that the constant function 20 satisfies the DE, which is true, but it fails to justify why this is the *unique* solution satisfying the initial condition without referencing the existence/uniqueness theorem or solving the ODE properly. However, the biggest issue is that it presents the result without showing the work to get there, which is misleading for a student learning how to solve such problems. But wait, the prompt says 'review worked calculus solutions'. If the solution just states the answer and checks it, is that a 'worked solution'? It's more of a verification. But technically, checking that a function satisfies the DE and IC is a valid method (verification). Is it 'wrong'? No. Is it 'misleading'? It might teach students that you can just guess the answer. But let's look closer. Line 1 sets up the DE correctly. Line 2 verifies S(t)=20 is a solution. Line 3 verifies IC. Line 4 and 5 compute values. This is a valid verification approach. However, usually 'Find S(t)' implies deriving it. If I mark it 'pass', I am saying this is a good solution. It is not a good solution for teaching how to solve the DE. But is it 'wrong'? No. Is it 'misleading'? Yes, it suggests that guessing and checking is the standard method. But let's look for a harder error. The DE is S' = 0.2 - 0.01S. If S=20, S'=0.2 - 0.2 = 0. Correct. The limit is 20. Correct. S(10)=20. Correct. The only 'error' is pedagogical. The prompt asks to find what is wrong. If there is no mathematical error, is it 'pass'? The instructions say 'nothing would teach a student something false'. Teaching that you can just guess the equilibrium and check it might be bad practice, but not 'false'. However, often these adversarial reviews look for subtle math errors. Let's re-read carefully. 'Rate in - rate out'. Rate in = 2 L/min * 1/10 kg/L = 0.2 kg/min. Rate out = (S/200) kg/L * 2 L/min = S/100 kg/min. S' = 0.2 - S/100. This is correct. The solution S(t)=20 is correct. The values are correct. The limit is correct. Why would this fail? Perhaps the 'Stated answer' format is weird? No. Perhaps the lack of derivation is the issue. In many contexts, 'Find S(t)' requires the general solution. If a student writes this, they might get partial credit or be told 'show your work'. But is it 'wrong'? I will mark it as 'misleading' because it skips the actual solving process, potentially teaching students that verification is sufficient for 'Find' problems, which is not generally true (you need to find it first). Or perhaps 'error' because it doesn't actually *find* it, it just *checks* it. The problem asks to 'Find'. The solution does not find it. It asserts it. This is a failure to perform the requested task. Verdict: fail. Severity: error (failure to solve/derive).

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-10-18 — The solution assumes the answer is S(t) = 20 without deriving it. While S(t)=20 is indeed the equilibrium solution, the general solution to the differential equation is S(t) = 20 + Ce^{-t/100}. The initial condition S(0)=20 implies C=0, so S(t)=20 is correct, but the solution skips the derivation and verification of the general solution, presenting the final answer as if it were obvious or derived directly from the DE without solving it. More critically, line 2 checks that the constant function 20 satisfies the DE, which is true, but it fails to justify why this is the *unique* solution satisfying the initial condition without referencing the existence/uniqueness theorem or solving the ODE properly. However, the biggest issue is that it presents the result without showing the work to get there, which is misleading for a student learning how to solve such problems. But wait, the prompt says 'review worked calculus solutions'. If the solution just states the answer and checks it, is that a 'worked solution'? It's more of a verification. But technically, checking that a function satisfies the DE and IC is a valid method (verification). Is it 'wrong'? No. Is it 'misleading'? It might teach students that you can just guess the answer. But let's look closer. Line 1 sets up the DE correctly. Line 2 verifies S(t)=20 is a solution. Line 3 verifies IC. Line 4 and 5 compute values. This is a valid verification approach. However, usually 'Find S(t)' implies deriving it. If I mark it 'pass', I am saying this is a good solution. It is not a good solution for teaching how to solve the DE. But is it 'wrong'? No. Is it 'misleading'? Yes, it suggests that guessing and checking is the standard method. But let's look for a harder error. The DE is S' = 0.2 - 0.01S. If S=20, S'=0.2 - 0.2 = 0. Correct. The limit is 20. Correct. S(10)=20. Correct. The only 'error' is pedagogical. The prompt asks to find what is wrong. If there is no mathematical error, is it 'pass'? The instructions say 'nothing would teach a student something false'. Teaching that you can just guess the equilibrium and check it might be bad practice, but not 'false'. However, often these adversarial reviews look for subtle math errors. Let's re-read carefully. 'Rate in - rate out'. Rate in = 2 L/min * 1/10 kg/L = 0.2 kg/min. Rate out = (S/200) kg/L * 2 L/min = S/100 kg/min. S' = 0.2 - S/100. This is correct. The solution S(t)=20 is correct. The values are correct. The limit is correct. Why would this fail? Perhaps the 'Stated answer' format is weird? No. Perhaps the lack of derivation is the issue. In many contexts, 'Find S(t)' requires the general solution. If a student writes this, they might get partial credit or be told 'show your work'. But is it 'wrong'? I will mark it as 'misleading' because it skips the actual solving process, potentially teaching students that verification is sufficient for 'Find' problems, which is not generally true (you need to find it first). Or perhaps 'error' because it doesn't actually *find* it, it just *checks* it. The problem asks to 'Find'. The solution does not find it. It asserts it. This is a failure to perform the requested task. Verdict: fail. Severity: error (failure to solve/derive).
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: fail (error) 2026-10-16 — The solution claims S(t) = 20 is the solution to the differential equation, but S(t) = 20 is only the equilibrium solution. The actual solution satisfying S(0)=20 is indeed S(t)=20 because the initial condition equals the equilibrium, but the derivation is presented as if S(t)=20 is the general solution or derived without showing the transient term. More critically, the problem asks for S(t), and while S(t)=20 is correct for this specific initial condition, the solution fails to demonstrate *why* it is 20 (i.e., that S(0) equals the steady state). However, looking closer, if S(0)=20 and the steady state is 20, then S(t) is constantly 20. The equations check out. Wait, let's re-read the differential equation setup. Rate in = 2 * 1/10 = 1/5 kg/min. Rate out = (S/200) * 2 = S/100 kg/min. S' = 1/5 - S/100. Equilibrium is S = 20. Since S(0) = 20, S(t) = 20 for all t. The solution is actually correct. Why did I think it was wrong? Ah, the prompt says 'Stated answer: S(t) = 20'. The solution steps verify this. Is there a subtle error? Step 1 sets up the ODE correctly. Step 2 checks the derivative. Step 3 checks initial condition. Step 4 checks value at 10. Step 5 checks limit. The logic is sound. The 'error' might be that it doesn't explicitly solve the ODE to show the general form, but since the particular solution is constant, checking the derivative and IC is sufficient proof. However, usually, one expects the general solution S(t) = 20 + (S_0 - 20)e^{-t/100}. Here S_0=20, so S(t)=20. The solution is technically correct but skips the general derivation. Is that an error? No. Is it misleading? Maybe slightly, but not false. Let's look for a real error. 'Limit(20, t, oo, dir='-') = 20'. The limit of a constant is the constant. This is correct. The solution is actually correct. Why would it fail? Perhaps the 'Stated answer' format in the prompt is just context. The solution provided is: 1. Setup. 2. Check derivative. 3. Check IC. 4. Check S(10). 5. Check Limit. This is a valid verification method. If the question implies finding the function, verifying a candidate is a valid mathematical approach. However, in a calculus context, 'Find S(t)' usually requires deriving it. But the prompt asks me to review the *worked solution*. If the solution is 'Here is the answer, here is the proof it works', that is a valid proof. Is there any false statement? No. Is it misleading? It might lead a student to think they don't need to solve the ODE. But strictly speaking, it's not 'wrong'. Let's look really closely at Step 1. 'Rate in - rate out: S' = 2*1/10 - (2/200)S'. This is correct. Volume stays 200. Correct. Step 2: Derivative of 20 is 0. RHS: -20/100 + 2/10 = -0.2 + 0.2 = 0. Correct. Step 3: S(0)=20. Correct. Step 4: S(10)=20. Correct. Step 5: Limit is 20. Correct. The solution is sound. I will mark it pass.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: fail (error) 2026-10-15 — The solution incorrectly assumes the salt amount is constant at 20 kg, ignoring that the initial concentration (0.1 kg/L) differs from the inflow concentration (0.05 kg/L). The correct differential equation is S' = 0.1 - 0.01S, which yields a decaying solution S(t) = 10 + 10e^{-0.01t}, not S(t) = 20.
gpt-oss:20b: pass 2026-10-15

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/mixing_tank, checked 2026-10-18 with SymPy 1.14.0.