Horizontal and vertical tangents of parametric curves
Problem 8.380 · medium
Find the points where \( \displaystyle x = t^{3} - 6 t,\ y = t^{2} + 2 \) has a horizontal or a vertical tangent line.
- \[ \frac{d}{d t} \left(t^{3} - 6 t\right) = 3 t^{2} - 6 \]dx/dt.✓ Proved
- \[ \frac{d}{d t} \left(t^{2} + 2\right) = 2 t \]dy/dt.✓ Proved
- Horizontal tangent: dy/dt = 0 with dx/dt ≠ 0. Vertical: dx/dt = 0 with dy/dt ≠ 0.Reviewed
- \[ \left. 2 t \right|_{\substack{ t=0 }} = 0 \]t = 0: horizontal, at (0, 2).✓ Proved
- \[ \left. 3 t^{2} - 6 \right|_{\substack{ t=- \sqrt{2} }} = 0 \]t = -sqrt(2): vertical, at (4*sqrt(2), 4).✓ Proved
- \[ \left. 3 t^{2} - 6 \right|_{\substack{ t=\sqrt{2} }} = 0 \]t = sqrt(2): vertical, at (-4*sqrt(2), 4).✓ Proved
Answer \( \text{horizontal at } (0, 2);\ \text{vertical at } (4*sqrt(2), 4), (-4*sqrt(2), 4) \)
Lines: 5 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 |
| 6 | ✓ 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, solves for the parameter values, and correctly maps them to the Cartesian coordinates.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly identifies the conditions for horizontal and vertical tangents, solves for the parameter values, and correctly maps them to the Cartesian coordinates.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: fail (error) 2026-10-16 — The solution swaps the coordinates for the vertical tangents. At t = -sqrt(2), x = 4*sqrt(2) and y = 4, so the point is (4*sqrt(2), 4). At t = sqrt(2), x = -4*sqrt(2) and y = 4, so the point is (-4*sqrt(2), 4). The comments in lines 5 and 6 incorrectly assign these points to the wrong t-values.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: fail (error) 2026-10-15 — The coordinates for the vertical tangents are swapped. At t = -sqrt(2), x = -4*sqrt(2), not 4*sqrt(2). At t = sqrt(2), x = 4*sqrt(2), not -4*sqrt(2).gpt-oss:20b: pass 2026-10-15
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.