Related rates
Problem 3.100 · medium
Water is poured into an inverted conical tank, 10 m tall with top radius 4 m, at 10 m³/min. How fast is the water level rising when the water is 8 m deep?
- By similar triangles the water's radius is r = (4/10)h, so V = (1/3)π((4/10)h)²h.
- \[ \frac{d}{d h} \frac{4 \pi h^{3}}{75} = \frac{4 \pi h^{2}}{25} \]dV/dt = (dV/dh)(dh/dt).✓ Proved
- \[ \left. \frac{125}{2 \pi h^{2}} \right|_{\substack{ h=8 }} = \frac{125}{128 \pi} \]dh/dt = (dV/dt)/(dV/dh) at h = 8.✓ Proved
Answer \( \frac{125}{128 \pi} \ \text{m/min} \approx 0.3108\ \text{m/min} \)
Lines: 2 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
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 |
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 2026-09-26 with SymPy 1.14.0.