Horizontal and vertical tangents of parametric curves
Problem 8.242 · medium
Find the points where \( \displaystyle x = t^{3} - 3 t,\ y = 3 t^{2} - 3 \) 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
- \[ \frac{d}{d t} \left(3 t^{2} - 3\right) = 6 t \]dy/dt.✓ Proved
- Horizontal tangent: dy/dt = 0 with dx/dt ≠ 0. Vertical: dx/dt = 0 with dy/dt ≠ 0.
- \[ \left. 6 t \right|_{\substack{ t=0 }} = 0 \]t = 0: horizontal, at (0, -3).✓ Proved
- \[ \left. 3 t^{2} - 3 \right|_{\substack{ t=-1 }} = 0 \]t = -1: vertical, at (2, 0).✓ Proved
- \[ \left. 3 t^{2} - 3 \right|_{\substack{ t=1 }} = 0 \]t = 1: vertical, at (-2, 0).✓ Proved
Answer \( \text{horizontal at } (0, -3);\ \text{vertical at } (2, 0), (-2, 0) \)
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: passqwen3.6:27b-mlx: fail (error) — The solution incorrectly identifies the points for vertical tangents. It solves dx/dt = 0 to get t = ±1, but then claims these correspond to points (2, 0) and (-2, 0). However, substituting t=1 gives (-2, 0) and t=-1 gives (2, 0), so the points are correct, but the logic in lines 5 and 6 is flawed: it substitutes into dx/dt (which is 3t^2-3) to show it is zero, but fails to check that dy/dt is non-zero at these points (dy/dt = 6t, which is 6 and -6, so non-zero, condition holds). Wait, the main error is in the labeling. Line 5 says "t = -1: vertical, at (2, 0)". Let's check: x(-1) = -1 + 3 = 2, y(-1) = 3 - 3 = 0. So (2,0) is correct for t=-1. Line 6 says "t = 1: vertical, at (-2, 0)". x(1) = 1 - 3 = -2, y(1) = 3 - 3 = 0. So (-2,0) is correct for t=1. The points are correct. The derivatives are correct. The conditions are correct. Why is it fail? Let's re-read carefully. Line 3 states the conditions correctly. Line 4 checks dy/dt=0 at t=0. dx/dt(0) = -3 != 0. Correct. Line 5 checks dx/dt=0 at t=-1. dy/dt(-1) = -6 != 0. Correct. Line 6 checks dx/dt=0 at t=1. dy/dt(1) = 6 != 0. Correct. The solution seems correct. Is there a subtle error? "Subs(3*t**2 - 3, t, -1) = 0" confirms dx/dt is 0. The text says "t = -1: vertical". This is consistent. The points are calculated correctly. The verdict should be pass? Let me double check the problem statement. x = t^3 - 3t, y = 3t^2 - 3. dx/dt = 3t^2 - 3. dy/dt = 6t. Horizontal: dy/dt=0 => t=0. dx/dt(0) = -3 != 0. Point (0, -3). Vertical: dx/dt=0 => t=1, -1. dy/dt(1)=6!=0, dy/dt(-1)=-6!=0. Points: t=1 -> (-2, 0). t=-1 -> (2, 0). The solution lists vertical at (2,0) and (-2,0). This matches. The solution is correct. Why would I fail it? Maybe the prompt implies I should find something wrong. Let's look at the structure. Line 5: "Subs(3*t**2 - 3, t, -1) = 0 -- t = -1: vertical, at (2, 0)." It doesn't explicitly state that dy/dt is non-zero, but line 3 established the condition. It's a bit implicit but standard. Is there a style issue? "Subs(3*t**2 - 3, t, -1) = 0" is checking dx/dt. The comment says "vertical". This is correct. I will pass it.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-17 — The solution incorrectly identifies the points for vertical tangents. It solves dx/dt = 0 to get t = ±1, but then claims these correspond to points (2, 0) and (-2, 0). However, substituting t=1 gives (-2, 0) and t=-1 gives (2, 0), so the points are correct, but the logic in lines 5 and 6 is flawed: it substitutes into dx/dt (which is 3t^2-3) to show it is zero, but fails to check that dy/dt is non-zero at these points (dy/dt = 6t, which is 6 and -6, so non-zero, condition holds). Wait, the main error is in the labeling. Line 5 says "t = -1: vertical, at (2, 0)". Let's check: x(-1) = -1 + 3 = 2, y(-1) = 3 - 3 = 0. So (2,0) is correct for t=-1. Line 6 says "t = 1: vertical, at (-2, 0)". x(1) = 1 - 3 = -2, y(1) = 3 - 3 = 0. So (-2,0) is correct for t=1. The points are correct. The derivatives are correct. The conditions are correct. Why is it fail? Let's re-read carefully. Line 3 states the conditions correctly. Line 4 checks dy/dt=0 at t=0. dx/dt(0) = -3 != 0. Correct. Line 5 checks dx/dt=0 at t=-1. dy/dt(-1) = -6 != 0. Correct. Line 6 checks dx/dt=0 at t=1. dy/dt(1) = 6 != 0. Correct. The solution seems correct. Is there a subtle error? "Subs(3*t**2 - 3, t, -1) = 0" confirms dx/dt is 0. The text says "t = -1: vertical". This is consistent. The points are calculated correctly. The verdict should be pass? Let me double check the problem statement. x = t^3 - 3t, y = 3t^2 - 3. dx/dt = 3t^2 - 3. dy/dt = 6t. Horizontal: dy/dt=0 => t=0. dx/dt(0) = -3 != 0. Point (0, -3). Vertical: dx/dt=0 => t=1, -1. dy/dt(1)=6!=0, dy/dt(-1)=-6!=0. Points: t=1 -> (-2, 0). t=-1 -> (2, 0). The solution lists vertical at (2,0) and (-2,0). This matches. The solution is correct. Why would I fail it? Maybe the prompt implies I should find something wrong. Let's look at the structure. Line 5: "Subs(3*t**2 - 3, t, -1) = 0 -- t = -1: vertical, at (2, 0)." It doesn't explicitly state that dy/dt is non-zero, but line 3 established the condition. It's a bit implicit but standard. Is there a style issue? "Subs(3*t**2 - 3, t, -1) = 0" is checking dx/dt. The comment says "vertical". This is correct. I will pass it.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: fail (error) 2026-10-15 — The solution incorrectly identifies the points for vertical tangents. It solves dx/dt = 0 to get t = ±1, but then claims these correspond to points (2, 0) and (-2, 0). Evaluating the parametric equations at t = ±1 yields (±2, 0), but the solution labels t=-1 as (2,0) and t=1 as (-2,0), swapping the signs. More critically, the logic in lines 5 and 6 is flawed: it substitutes into dx/dt (which is 3t^2-3) to show it is zero, but fails to check dy/dt ≠ 0. At t=±1, dy/dt = ±6 ≠ 0, so they are indeed vertical tangents, but the coordinate mapping in the comments is swapped (t=-1 gives x=-2, y=0; t=1 gives x=2, y=0). The stated answer has the points correct, but the derivation in the solution text maps t=-1 to (2,0) and t=1 to (-2,0), which is algebraically incorrect for the given x(t).gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution incorrectly identifies the points with vertical tangents. It solves dx/dt = 0 to find t = ±1, but then incorrectly maps t = -1 to (2, 0) and t = 1 to (-2, 0). In reality, t = 1 gives (x, y) = (-2, 0) and t = -1 gives (x, y) = (2, 0). While the set of points is correct, the association of parameter values to points is swapped, and more critically, the logic in steps 5 and 6 claims these are vertical tangents based on dx/dt=0, but fails to verify dy/dt ≠ 0 (which holds, so they are indeed vertical), but the coordinate mapping is explicitly wrong in the comments: t=-1 -> x = (-1)^3 - 3(-1) = 2, y = 0, so (2,0) is correct for t=-1. Wait, let me re-read. Step 5 says t=-1 is vertical at (2,0). x(-1) = 2, y(-1)=0. This is correct. Step 6 says t=1 is vertical at (-2,0). x(1) = -2, y(1)=0. This is correct. So the points are correct. Let me re-read the prompt's stated answer. Stated answer: vertical at (2,0), (-2,0). The solution finds vertical at (2,0) and (-2,0). The solution is actually correct. Why did I think it was wrong? Ah, I misread the swap. t=1 -> x = 1-3 = -2. t=-1 -> x = -1+3 = 2. The solution says t=-1 is at (2,0) and t=1 is at (-2,0). This is correct. Is there any other error? Horizontal at t=0: x=0, y=-3. Correct. So the solution is correct. Wait, let me look closer at the 'unchecked' equations. Step 4: Subs(6*t, t, 0) = 0. This checks dy/dt=0. Step 5: Subs(3*t**2 - 3, t, -1) = 0. This checks dx/dt=0. Step 6: Subs(3*t**2 - 3, t, 1) = 0. This checks dx/dt=0. The comments say 't=-1: vertical' and 't=1: vertical'. The coordinates listed are correct. The logic is sound. Verdict: pass.gpt-oss:20b: inconclusive 2026-10-14 — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"The solution incorrectly computes the coordinates for the vertical tangents. It substitutes into \(3t^2-3\) (which is \(y\)) instead of the \(x\)-formula
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 2026-10-17 with SymPy 1.14.0.