Home›Calculus 2›Horizontal and vertical tangents of parametric curves›Problem 8.501

Horizontal and vertical tangents of parametric curves

Problem 8.501 · medium

Find the points where \( \displaystyle x = t^{3} - 3 t,\ y = t^{2} + 2 \) 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
\[ \frac{d}{d t} \left(t^{2} + 2\right) = 2 t \]
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, 2).✓ Proved
\[ \left. 3 t^{2} - 3 \right|_{\substack{ t=-1 }} = 0 \]
t = -1: vertical, at (2, 3).✓ Proved
\[ \left. 3 t^{2} - 3 \right|_{\substack{ t=1 }} = 0 \]
t = 1: vertical, at (-2, 3).✓ Proved

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

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 vertical tangents occur at t = -1 and t = 1, but at these values, dy

Every verdict on record (6)

qwen3.6:27b-mlx: inconclusive 2026-10-19 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly identifies the points for vertical tangents. It claims vertical tangents occur at t = -1 and t = 1, but at these values, dy
gpt-oss:20b: pass 2026-10-19
qwen3.6:27b-mlx: fail (error) 2026-10-18 — The solution incorrectly assigns the points for vertical tangents. At t=1, x = 1^3 - 3(1) = -2, so the point is (-2, 3), but the solution labels t=-1 as vertical at (2, 3) and t=1 as vertical at (-2, 3). While the set of points is correct, the mapping of parameter values to points is swapped in the comments (t=-1 gives x=2, t=1 gives x=-2). Specifically, line 5 claims t=-1 is vertical at (2,3) which is correct, but line 6 claims t=1 is vertical at (-2,3) which is also correct. Wait, let me re-read carefully. Line 5: t=-1. x = (-1)^3 - 3(-1) = -1 + 3 = 2. y = (-1)^2 + 2 = 3. Point (2,3). dx/dt = 0, dy/dt = -2 != 0. Vertical. Correct. Line 6: t=1. x = 1^3 - 3(1) = -2. y = 1^2 + 2 = 3. Point (-2,3). dx/dt = 0, dy/dt = 2 != 0. Vertical. Correct. The solution actually has the correct points associated with the correct t values in the comments. Let me double check the problem statement vs the solution. Problem: x = t^3 - 3t, y = t^2 + 2. Horizontal: dy/dt = 2t = 0 => t=0. x=0, y=2. Point (0,2). Correct. Vertical: dx/dt = 3t^2 - 3 = 0 => t = +/- 1. t=1: x = 1-3 = -2, y=3. Point (-2,3). t=-1: x = -1+3 = 2, y=3. Point (2,3). The solution says: Line 5: t=-1 ... vertical at (2,3). Correct. Line 6: t=1 ... vertical at (-2,3). Correct. The stated answer is: horizontal at (0,2); vertical at (2,3), (-2,3). This matches. Is there any error? Line 3 states the conditions correctly. Line 4 checks t=0 for horizontal. It doesn't explicitly check dx/dt != 0, but dx/dt(0) = -3 != 0, so it is valid. Line 5 checks t=-1 for vertical. It doesn't explicitly check dy/dt != 0, but dy/dt(-1) = -2 != 0, so it is valid. Line 6 checks t=1 for vertical. dy/dt(1) = 2 != 0, so it is valid. The solution seems correct. Why did I think it was wrong? I misread the swap. The comments are correct. Wait, look at line 5 and 6 comments again. Line 5: `Subs(3*t**2 - 3, t, -1) = 0 -- t = -1: vertical, at (2, 3).` Line 6: `Subs(3*t**2 - 3, t, 1) = 0 -- t = 1: vertical, at (-2, 3).` These are factually correct statements. The algebra is proved. The logic is sound. The setup is correct. Verdict: pass.
gpt-oss:20b: pass 2026-10-18
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. At t = 1, x = 1^3 - 3(1) = -2, so the point is (-2, 3), but the solution label
gpt-oss:20b: pass 2026-10-18

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 2026-10-19 with SymPy 1.14.0.