Arc length
Problem 5.183 · medium
Find the length of the curve \( \displaystyle y = \frac{2 \left(x + 3\right)^{\frac{3}{2}}}{3} \) from \( \displaystyle x = 0 \) to \( \displaystyle x = 2 \).
- Arc length is ∫ √(1 + (dy/dx)²) dx.
- \[ \frac{d}{d x} \frac{2 \left(x + 3\right)^{\frac{3}{2}}}{3} = \sqrt{x + 3} \]dy/dx.✓ Proved
- \[ x + 4 \]1 + (dy/dx)² simplifies (here to a perfect square, which is why these are set).
- \[ \int\limits_{0}^{2} \sqrt{x + 4}\, dx = - \frac{16}{3} + 4 \sqrt{6} \]Integrate.✓ Proved
Answer \( - \frac{16}{3} + 4 \sqrt{6} \)
Lines: 2 proved, 2 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 | Not checked | — | both sides are the same expression: the line states a value, and the working that produced it is not on the line |
| 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 200,000-segment polygon along the curve has the same length |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution incorrectly simplifies 1 + (dy/dx)^2 to x + 4. Since dy/dx = sqrt(x+3), 1 + (dy/dx)^2 = 1 + x + 3 = x + 4 is actually correct algebraically, but the integral of sqrt(x+4) from 0 to 2 is [2/3 (x+4)^(3/2)] from 0 to 2, which equals 2/3(6^(3/2) - 4^(3/2)) = 2/3(6sqrt(6) - 8) = 4sqrt(6) - 16/3. Wait, the algebra is correct. Let me re-read carefully. 1. dy/dx = sqrt(x+3). Correct. 2. 1 + (dy/dx)^2 = 1 + x + 3 = x + 4. Correct. 3. Integral of sqrt(x+4) dx. Let u = x+4, du=dx. Integral u^(1/2) du = 2/3 u^(3/2). 4. Evaluated from 0 to 2: 2/3 ((2+4)^(3/2) - (0+4)^(3/2)) = 2/3 (6^(3/2) - 4^(3/2)) = 2/3 (6sqrt(6) - 8) = 4sqrt(6) - 16/3. The stated answer is -16/3 + 4sqrt(6). This matches. Is there any error? Line 3 says 'x + 4 = x + 4 -- 1 + (dy/dx)^2 simplifies (here to a perfect square, which is why these are set).' Wait, is x+4 a perfect square? No. The integrand is sqrt(x+4). The text says '1 + (dy/dx)^2 simplifies (here to a perfect square)'. 1 + (dy/dx)^2 is x+4. Is x+4 a perfect square? No. Usually, these problems are set so that 1+(y')^2 is a perfect square, e.g., (something)^2, so the square root cancels out. Here, 1+(y')^2 = x+4. The square root is sqrt(x+4). This is NOT a perfect square simplification in the sense that the sqrt disappears. It just simplifies to a monomial under the root. However, looking at the problem source style, often 'perfect square' refers to the expression inside the square root being a perfect square of a polynomial, allowing the sqrt to be removed. Here, x+4 is not a perfect square of a polynomial. So the comment 'which is why these are set' implies a standard trope where the sqrt vanishes. It does not vanish here. But is the math wrong? No. The comment is slightly misleading about the nature of the simplification (it doesn't remove the radical), but the calculation is correct. Let's look closer. Is there a subtle error? y = 2/3 (x+3)^(3/2). y' = (x+3)^(1/2). 1 + y'^2 = 1 + x + 3 = x + 4. Arc length = integral_0^2 sqrt(x+4) dx. Result: 4sqrt(6) - 16/3. The solution is mathematically correct. The comment 'perfect square' is technically incorrect terminology for 'x+4' (which is not a square of a polynomial), but it doesn't lead to a wrong answer. Is it 'misleading'? A student might think x+4 is a square of something simple like (sqrt(x)+2)^2? No. Actually, let's look at line 3 again. 'x + 4 = x + 4'. This is a tautology. The comment explains that 1+(dy/dx)^2 simplifies to x+4. The parenthetical '(here to a perfect square, which is why these are set)' is the issue. It claims x+4 is a perfect square. It is not. This is a false statement. A student might learn that x+4 is a perfect square. Verdict: fail. Severity: misleading (or error in statement). The statement 'x+4 is a perfect square' is false.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution incorrectly simplifies 1 + (dy/dx)^2 to x + 4. Since dy/dx = sqrt(x+3), 1 + (dy/dx)^2 = 1 + x + 3 = x + 4 is actually correct algebraically, but the integral of sqrt(x+4) from 0 to 2 is [2/3 (x+4)^(3/2)] from 0 to 2, which equals 2/3(6^(3/2) - 4^(3/2)) = 2/3(6sqrt(6) - 8) = 4sqrt(6) - 16/3. Wait, the algebra is correct. Let me re-read carefully. 1. dy/dx = sqrt(x+3). Correct. 2. 1 + (dy/dx)^2 = 1 + x + 3 = x + 4. Correct. 3. Integral of sqrt(x+4) dx. Let u = x+4, du=dx. Integral u^(1/2) du = 2/3 u^(3/2). 4. Evaluated from 0 to 2: 2/3 ((2+4)^(3/2) - (0+4)^(3/2)) = 2/3 (6^(3/2) - 4^(3/2)) = 2/3 (6sqrt(6) - 8) = 4sqrt(6) - 16/3. The stated answer is -16/3 + 4sqrt(6). This matches. Is there any error? Line 3 says 'x + 4 = x + 4 -- 1 + (dy/dx)^2 simplifies (here to a perfect square, which is why these are set).' Wait, is x+4 a perfect square? No. The integrand is sqrt(x+4). The text says '1 + (dy/dx)^2 simplifies (here to a perfect square)'. 1 + (dy/dx)^2 is x+4. Is x+4 a perfect square? No. Usually, these problems are set so that 1+(y')^2 is a perfect square, e.g., (something)^2, so the square root cancels out. Here, 1+(y')^2 = x+4. The square root is sqrt(x+4). This is NOT a perfect square simplification in the sense that the sqrt disappears. It just simplifies to a monomial under the root. However, looking at the problem source style, often 'perfect square' refers to the expression inside the square root being a perfect square of a polynomial, allowing the sqrt to be removed. Here, x+4 is not a perfect square of a polynomial. So the comment 'which is why these are set' implies a standard trope where the sqrt vanishes. It does not vanish here. But is the math wrong? No. The comment is slightly misleading about the nature of the simplification (it doesn't remove the radical), but the calculation is correct. Let's look closer. Is there a subtle error? y = 2/3 (x+3)^(3/2). y' = (x+3)^(1/2). 1 + y'^2 = 1 + x + 3 = x + 4. Arc length = integral_0^2 sqrt(x+4) dx. Result: 4sqrt(6) - 16/3. The solution is mathematically correct. The comment 'perfect square' is technically incorrect terminology for 'x+4' (which is not a square of a polynomial), but it doesn't lead to a wrong answer. Is it 'misleading'? A student might think x+4 is a square of something simple like (sqrt(x)+2)^2? No. Actually, let's look at line 3 again. 'x + 4 = x + 4'. This is a tautology. The comment explains that 1+(dy/dx)^2 simplifies to x+4. The parenthetical '(here to a perfect square, which is why these are set)' is the issue. It claims x+4 is a perfect square. It is not. This is a false statement. A student might learn that x+4 is a perfect square. Verdict: fail. Severity: misleading (or error in statement). The statement 'x+4 is a perfect square' is false.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: pass 2026-09-29 — The solution correctly identifies the arc length formula, computes the derivative and the term under the square root, and evaluates the definite integral with the correct bounds and result.gpt-oss:20b: pass 2026-09-29qwen3.6:27b-mlx: pass 2026-09-29 — The solution correctly identifies the arc length formula, computes the derivative and simplifies the integrand to a perfect square, and evaluates the definite integral correctly.gpt-oss:20b: fail (misleading) 2026-09-29 — The statement that 1+(dy/dx)^2 simplifies to a perfect square is incorrect; it simplifies to x+4, which is not a perfect square.
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, checked 2026-10-13 with SymPy 1.14.0.