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Arc length of x = g(y)

Problem 5.719 · hard

Find the length of the curve \( \displaystyle x = \frac{2 \left(y - 3\right)^{\frac{3}{2}}}{3} \) for \( \displaystyle 3 \le y \le 8 \).
  1. For x = g(y), L = ∫ √(1 + (dx/dy)²) dy.
  2. \[ \frac{d}{d y} \frac{2 \left(y - 3\right)^{\frac{3}{2}}}{3} = \sqrt{y - 3} \]
    dx/dy.✓ Proved
  3. \[ \left(\frac{d}{d y} \frac{2 \left(y - 3\right)^{\frac{3}{2}}}{3}\right)^{2} + 1 = y - 2 \]
    1 + (dx/dy)², simplified.✓ Proved
  4. \[ \int\limits_{3}^{8} \sqrt{y - 2}\, dy = - \frac{2}{3} + 4 \sqrt{6} \]
    Integrate.✓ Proved
Answer \( - \frac{2}{3} + 4 \sqrt{6} \)

Lines: 3 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify

The full receipt
LineStatusChecked byDetail
1Not checked—a sentence; read, not computed
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0a 20,000-segment polyline along the curve

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The integrand sqrt(y-2) is undefined for y in [3, 3), but more critically, the simplification in step 3 is algebraically incorrect: (sqrt(y-3))^2 + 1 = y - 3 + 1 = y - 2 is correct, but the domain of integration starts at y=3, where sqrt(y-2) = 1, which is fine. Wait, let's re-read carefully. Step 3 says 1 + (dx/dy)^2 = y - 2. dx/dy = sqrt(y-3). So (dx/dy)^2 = y-3. 1 + y - 3 = y - 2. This is algebraically correct. However, look at the bounds. The problem states 3 <= y <= 8. At y=3, sqrt(y-2) = 1. The integral is valid. Is there a hidden error? Let's check the derivative again. x = 2/3 (y-3)^(3/2). dx/dy = 2/3 * 3/2 * (y-3)^(1/2) = (y-3)^(1/2) = sqrt(y-3). Correct. 1 + (dx/dy)^2 = 1 + y - 3 = y - 2. Correct. Integral of sqrt(y-2) from 3 to 8. Let u = y-2, du = dy. Bounds: 1 to 6. Integral u^(1/2) du = [2/3 u^(3/2)] from 1 to 6 = 2/3 (6^(3/2) - 1) = 2/3 (6*sqrt(6) - 1) = 4*sqrt(6) - 2/3. The calculation is correct. Why did I think it was wrong? Ah, I see. The prompt asks me to find what is wrong. If the solution is actually correct, I should pass. Let me double check everything. Problem: x = 2/3 (y-3)^(3/2). Range 3 to 8. Formula L = int sqrt(1 + (dx/dy)^2) dy. dx/dy = sqrt(y-3). 1 + (dx/dy)^2 = y-2. Int sqrt(y-2) dy from 3 to 8. Result -2/3 + 4sqrt(6). Everything seems correct. Is there a subtle issue? Maybe the sentence in step 1 is too brief? No. Maybe the bounds? 3 to 8. y-2 is positive on [3,8]. So sqrt is real. It seems correct. Wait, is there a trick? "Find the length of the curve". Usually arc length is positive. The result is positive. 4*sqrt(6) is approx 4*2.45 = 9.8. 2/3 is 0.66. Result ~ 9.14. Positive. Okay, I will pass it.
Every verdict on record (6)
  • qwen3.6:27b-mlx: fail (style) 2026-10-23 — [domain objection, downgraded to style] The integrand sqrt(y-2) is undefined for y in [3, 3), but more critically, the simplification in step 3 is algebraically incorrect: (sqrt(y-3))^2 + 1 = y - 3 + 1 = y - 2 is correct, but the domain of integration starts at y=3, where sqrt(y-2) = 1, which is fine. Wait, let's re-read carefully. Step 3 says 1 + (dx/dy)^2 = y - 2. dx/dy = sqrt(y-3). So (dx/dy)^2 = y-3. 1 + y - 3 = y - 2. This is algebraically correct. However, look at the bounds. The problem states 3 <= y <= 8. At y=3, sqrt(y-2) = 1. The integral is valid. Is there a hidden error? Let's check the derivative again. x = 2/3 (y-3)^(3/2). dx/dy = 2/3 * 3/2 * (y-3)^(1/2) = (y-3)^(1/2) = sqrt(y-3). Correct. 1 + (dx/dy)^2 = 1 + y - 3 = y - 2. Correct. Integral of sqrt(y-2) from 3 to 8. Let u = y-2, du = dy. Bounds: 1 to 6. Integral u^(1/2) du = [2/3 u^(3/2)] from 1 to 6 = 2/3 (6^(3/2) - 1) = 2/3 (6*sqrt(6) - 1) = 4*sqrt(6) - 2/3. The calculation is correct. Why did I think it was wrong? Ah, I see. The prompt asks me to find what is wrong. If the solution is actually correct, I should pass. Let me double check everything. Problem: x = 2/3 (y-3)^(3/2). Range 3 to 8. Formula L = int sqrt(1 + (dx/dy)^2) dy. dx/dy = sqrt(y-3). 1 + (dx/dy)^2 = y-2. Int sqrt(y-2) dy from 3 to 8. Result -2/3 + 4sqrt(6). Everything seems correct. Is there a subtle issue? Maybe the sentence in step 1 is too brief? No. Maybe the bounds? 3 to 8. y-2 is positive on [3,8]. So sqrt is real. It seems correct. Wait, is there a trick? "Find the length of the curve". Usually arc length is positive. The result is positive. 4*sqrt(6) is approx 4*2.45 = 9.8. 2/3 is 0.66. Result ~ 9.14. Positive. Okay, I will pass it.
  • gpt-oss:20b: pass 2026-10-23
  • qwen3.6:27b-mlx: pass 2026-10-22 — The solution correctly applies the arc length formula for x=g(y). The derivative, simplification of 1+(dx/dy)^2, and definite integral are all algebraically correct and yield the stated answer.
  • gpt-oss:20b: pass 2026-10-22
  • qwen3.6:27b-mlx: pass 2026-10-22 — The solution correctly applies the arc length formula for x = g(y). The derivative, simplification of the integrand, and definite integration are all algebraically correct and yield the stated answer.
  • gpt-oss:20b: pass 2026-10-22

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 2026-10-23 with SymPy 1.14.0.

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