Home›Calculus 2›Horizontal and vertical tangents of parametric curves›Problem 8.384

Horizontal and vertical tangents of parametric curves

Problem 8.384 · medium

Find the points where \( \displaystyle x = t^{3} - 3 t,\ y = t^{2} + 3 \) has a horizontal or a vertical tangent line.

\[ \frac{d}{d t} \left(t^{3} - 3 t\right) = 3 t^{2} - 3 \]
dx/dt.✓ Proved
\[ \frac{d}{d t} \left(t^{2} + 3\right) = 2 t \]
dy/dt.✓ Proved
Horizontal tangent: dy/dt = 0 with dx/dt ≠ 0. Vertical: dx/dt = 0 with dy/dt ≠ 0.
\[ \left. 2 t \right|_{\substack{ t=0 }} = 0 \]
t = 0: horizontal, at (0, 3).✓ Proved
\[ \left. 3 t^{2} - 3 \right|_{\substack{ t=-1 }} = 0 \]
t = -1: vertical, at (2, 4).✓ Proved
\[ \left. 3 t^{2} - 3 \right|_{\substack{ t=1 }} = 0 \]
t = 1: vertical, at (-2, 4).✓ Proved

Answer \( \text{horizontal at } (0, 3);\ \text{vertical at } (2, 4), (-2, 4) \)

Lines: 5 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	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	Not checked	—	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: pass
qwen3.6:27b-mlx: inconclusive — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The solution incorrectly identifies the points for vertical tangents. It claims t = -1 and t = 1 yield vertical tangents at (2, 4) and (-2, 4), but s

Every verdict on record (6)

qwen3.6:27b-mlx: inconclusive 2026-10-18 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The solution incorrectly identifies the points for vertical tangents. It claims t = -1 and t = 1 yield vertical tangents at (2, 4) and (-2, 4), but s
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: fail (error) 2026-10-16 — The solution incorrectly identifies the points for vertical tangents. It claims t = -1 and t = 1 yield vertical tangents at (2, 4) and (-2, 4) respectively, but substituting these t-values into the parametric equations gives (-2, 4) for t = -1 and (2, 4) for t = 1. The coordinates are swapped.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: fail (error) 2026-10-15 — The solution incorrectly identifies the points for vertical tangents. Solving dx/dt = 0 yields t = ±1, which correspond to points (2, 4) and (-2, 4). However, the solution labels t = -1 as producing (2, 4) and t = 1 as producing (-2, 4), swapping the coordinates. Specifically, at t=1, x = 1^3 - 3(1) = -2, not 2. At t=-1, x = (-1)^3 - 3(-1) = 2, not -2. The final answer lists the correct points but the intermediate steps in lines 5 and 6 contain calculation errors regarding the x-coordinates.
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.