Related rates
Problem 3.8 · easy
Air is pumped into a spherical balloon at 16 cm³/s. How fast is the radius increasing when the radius is 10 cm?
- The volume of a sphere is V = (4/3)πr³.
- \[ \frac{d}{d r} \frac{4 \pi r^{3}}{3} = 4 \pi r^{2} \]dV/dt = 4πr² dr/dt.✓ Proved
- \[ \left. \frac{4}{\pi r^{2}} \right|_{\substack{ r=10 }} = \frac{1}{25 \pi} \]dr/dt = (dV/dt)/(4πr²) with r = 10.✓ Proved
Answer \( \frac{1}{25 \pi} \ \text{cm/s} \approx 0.01273\ \text{cm/s} \)
Lines: 2 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see 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 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the situation was stepped forward and back by a microsecond and the quantity differenced numerically |
Reviewers
gpt-oss:20b: fail (error) — The solution never uses the given rate of 16 cm³/s. In step 3 the expression 4/(πr²) is incorrect; it should be (16)/(4πr²) = 1/(25π).qwen3.6:27b-mlx: fail (error) — The solution fails to substitute the given rate dV/dt = 16 into the equation. It implicitly assumes dV/dt = 1, leading to an incorrect final answer of 1/(25*pi) instead of 4/(25*pi).
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution fails to substitute the given rate dV/dt = 16 into the equation. It implicitly assumes dV/dt = 1, leading to an incorrect final answer of 1/(25*pi) instead of 4/(25*pi).gpt-oss:20b: fail (error) 2026-09-28 — The solution never uses the given rate of 16 cm³/s. In step 3 the expression 4/(πr²) is incorrect; it should be (16)/(4πr²) = 1/(25π).qwen3.6:27b-mlx: fail (error) 2026-09-27 — The solution fails to substitute the given rate dV/dt = 16 into the equation. It implicitly assumes dV/dt = 1, leading to an incorrect final answer of 1/(25*pi) instead of the correct 4/(25*pi).gpt-oss:20b: fail (error) 2026-09-27 — The solution incorrectly substitutes dV/dt as 4/(πr²) instead of the given 16 cm³/s. The correct step is dr/dt = (16)/(4πr²) evaluated at r=10, yielding 1/(25π).qwen3.6:27b-mlx: fail (error) 2026-09-26 — The solution fails to substitute the given rate dV/dt = 16 into the equation before solving for dr/dt. It implicitly assumes dV/dt = 4, leading to an incorrect answer of 1/(25*pi) instead of the correct 4/(25*pi).gpt-oss:20b: fail (misleading) 2026-09-26 — The solution omits the key step of substituting the given rate dV/dt = 16 cm³/s into the formula dr/dt = (dV/dt)/(4πr²). Without that explicit step a student might not see why the factor 4 appears in the numerator.
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/related_rates, checked 2026-09-28 with SymPy 1.14.0.