Arc length of x = g(y)
Problem 5.563 · hard
Find the length of the curve \( \displaystyle x = \frac{y^{3}}{6} + \frac{1}{2 y} \) for \( \displaystyle 1 \le y \le 3 \).
- For x = g(y), L = ∫ √(1 + (dx/dy)²) dy.
- \[ \frac{d}{d y} \left(\frac{y^{3}}{6} + \frac{1}{2 y}\right) = \frac{y^{2}}{2} - \frac{1}{2 y^{2}} \]dx/dy.✓ Proved
- \[ \left(\frac{d}{d y} \left(\frac{y^{3}}{6} + \frac{1}{2 y}\right)\right)^{2} + 1 = \frac{\left(y^{4} + 1\right)^{2}}{4 y^{4}} \]1 + (dx/dy)², simplified.✓ Proved
- \[ \int\limits_{1}^{3} \frac{\frac{y^{4}}{2} + \frac{1}{2}}{y^{2}}\, dy = \frac{14}{3} \]Integrate.✓ Proved
Answer \( \frac{14}{3} \)
Lines: 3 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 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | a 20,000-segment polyline along the curve |
Reviewers
gpt-oss:20b: fail (error) — The simplification of 1+(dx/dy)^2 is incorrect; the term –y^2/2 is omitted, so the integrand is wrong and the final length is not 14/3.qwen3.6:27b-mlx: fail (error) — The solution skips the crucial step of taking the square root of 1 + (dx/dy)^2. Line 3 computes 1 + (dx/dy)^2, but Line 4 integrates the square root of that expression without stating that the square root was taken, implying the integrand in Line 4 is the same as the expression in Line 3, which is false.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-17 — The solution skips the crucial step of taking the square root of 1 + (dx/dy)^2. Line 3 computes 1 + (dx/dy)^2, but Line 4 integrates the square root of that expression without stating that the square root was taken, implying the integrand in Line 4 is the same as the expression in Line 3, which is false.gpt-oss:20b: fail (error) 2026-10-17 — The simplification of 1+(dx/dy)^2 is incorrect; the term –y^2/2 is omitted, so the integrand is wrong and the final length is not 14/3.qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution fails to take the square root of 1 + (dx/dy)^2 before integrating. Step 3 computes 1 + (dx/dy)^2, but Step 4 integrates the expression (y^4 + 1)/(2y^2), which is the square root of the result from Step 3, without explicitly stating or showing the square root operation. This skips a critical step in the arc length formula L = ∫ √(1 + (dx/dy)^2) dy.gpt-oss:20b: pass 2026-10-14qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution fails to explicitly take the square root of the simplified expression for 1 + (dx/dy)^2. While the final integral result is correct, the step from line 3 to line 4 skips the crucial operation of taking the square root, which is required by the arc length formula stated in line 1. A student would learn that one can integrate the squared term directly or that the square root disappears magically.gpt-oss:20b: pass 2026-10-14
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/arc_length_y, checked 2026-10-17 with SymPy 1.14.0.