Related rates
Problem 3.91 · medium
Air is pumped into a spherical balloon at 4 cm³/s. How fast is the radius increasing when the radius is 5 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
- \[ \left. 1 \frac{1}{\pi r^{2}} \right|_{\substack{ r=5 }} = \frac{1}{25 \pi} \]dr/dt = (dV/dt)/(4πr²) with r = 5.✓ 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 (misleading) — The solution omits the given rate dV/dt = 4 cm³/s and incorrectly substitutes 1/(πr²) for dr/dt. The correct step is dr/dt = (dV/dt)/(4πr²) = 4/(4π·25) = 1/(25π).qwen3.6:27b-mlx: fail (error) — The solution fails to explicitly state the chain rule relationship dV/dt = 4πr²(dr/dt) or substitute the given rate dV/dt = 4 cm³/s into the equation. It jumps from the derivative with respect to r to a final numerical answer without showing the critical step of dividing the given rate (4) by the surface area, making the logic incomplete and potentially confusing for a student.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution fails to explicitly state the chain rule relationship dV/dt = 4πr²(dr/dt) or substitute the given rate dV/dt = 4 cm³/s into the equation. It jumps from the derivative with respect to r to a final numerical answer without showing the critical step of dividing the given rate (4) by the surface area, making the logic incomplete and potentially confusing for a student.gpt-oss:20b: fail (misleading) 2026-09-28 — The solution omits the given rate dV/dt = 4 cm³/s and incorrectly substitutes 1/(πr²) for dr/dt. The correct step is dr/dt = (dV/dt)/(4πr²) = 4/(4π·25) = 1/(25π).qwen3.6:27b-mlx: fail (error) 2026-09-27 — The solution fails to explicitly substitute the given rate dV/dt = 4 cm³/s into the equation before solving for dr/dt. It implicitly assumes dV/dt = 1 in the substitution step (Subs(1/(pi*r**2)...)), leading to an answer that is off by a factor of 4.gpt-oss:20b: fail (error) 2026-09-27 — The solution never uses the given rate of volume increase (4 cm³/s). It substitutes 1/(π r²) for dV/dt, which is incorrect. The correct dr/dt should be (4)/(4π·5²)=1/(25π) cm/s, but the justification is missing the 4 cm³/s input.qwen3.6:27b-mlx: fail (error) 2026-09-26 — The solution fails to explicitly apply the chain rule to relate dV/dt and dr/dt, jumping from the derivative with respect to r to the final answer without showing the division by 4pi*r^2 or substituting dV/dt = 4. The equation in step 3 is algebraically incorrect as written (it substitutes into 1/(pi*r^2) instead of deriving dr/dt from the chain rule relation).gpt-oss:20b: fail (error) 2026-09-26 — The solution omits the given rate dV/dt = 4 cm³/s in the final step. It substitutes 1/(πr²) instead of using the correct expression 4/(4πr²) to compute dr/dt, leading to an incorrect derivation of the result.
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.