Horizontal and vertical tangents of parametric curves
Problem 8.353 · medium
Find the points where \( \displaystyle x = t^{2} - 2 t,\ y = 3 t^{2} - 3 \) has a horizontal or a vertical tangent line.
- \[ \frac{d}{d t} \left(t^{2} - 2 t\right) = 2 t - 2 \]dx/dt.✓ Proved
- \[ \frac{d}{d t} \left(3 t^{2} - 3\right) = 6 t \]dy/dt.✓ Proved
- Horizontal tangent: dy/dt = 0 with dx/dt ≠ 0. Vertical: dx/dt = 0 with dy/dt ≠ 0.Reviewed
- \[ \left. 6 t \right|_{\substack{ t=0 }} = 0 \]t = 0: horizontal, at (0, -3).✓ Proved
- \[ \left. 2 t - 2 \right|_{\substack{ t=1 }} = 0 \]t = 1: vertical, at (-1, 0).✓ Proved
Answer \( \text{horizontal at } (0, -3);\ \text{vertical at } (-1, 0) \)
Lines: 4 proved, 1 reviewed. 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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ 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 | roots of dx/dt and dy/dt found by Newton's method from a dense grid |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly identifies the conditions for horizontal and vertical tangents and applies them to find the correct parameter values and corresponding points.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly identifies the conditions for horizontal and vertical tangents and applies them to find the correct parameter values and corresponding points.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly identifies the conditions for horizontal and vertical tangents and applies them to find the correct parameter values and corresponding points.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution incorrectly identifies the condition for a vertical tangent. It sets dx/dt = 0 to find the vertical tangent, but fails to check that dy/dt != 0 at that point. At t=1, dy/dt = 6 != 0, so the vertical tangent condition is actually satisfied, but the reasoning in step 3 implies a symmetric check that is not explicitly performed or stated for the vertical case in step 5, although the result is correct. Wait, let me re-read. Step 3 states the conditions correctly. Step 5 finds t=1 where dx/dt=0. It claims this is a vertical tangent. It does not explicitly check dy/dt != 0, but since dy/dt = 6t, at t=1, dy/dt=6 != 0. So the math is correct. Is there an error? Let's check the coordinates. t=0: x=0, y=-3. Correct. t=1: x=1-2=-1, y=3-3=0. Correct. The logic holds. Why would it fail? Ah, look at step 3. It says 'Horizontal tangent: dy/dt = 0 with dx/dt ≠ 0. Vertical: dx/dt = 0 with dy/dt ≠ 0.' This is correct. Step 4 checks t=0 for horizontal. dy/dt=0. dx/dt = -2 != 0. Correct. Step 5 checks t=1 for vertical. dx/dt=0. dy/dt = 6 != 0. Correct. The solution is sound. Wait, is there a subtle issue? No. The solution is correct.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/parametric_tangent_hv, checked 2026-10-18 with SymPy 1.14.0.