Tangent lines to parametric curves
Problem 8.12 · medium
Find the equation of the tangent line to \( \displaystyle x = t^{2} + 1,\ y = t^{3} - 3 t \) at \( \displaystyle t = -1 \).
- dy/dx = (dy/dt)/(dx/dt).
- \[ \frac{d}{d t} \left(t^{2} + 1\right) = 2 t \]dx/dt.✓ Proved
- \[ \frac{d}{d t} \left(t^{3} - 3 t\right) = 3 t^{2} - 3 \]dy/dt.✓ Proved
- \[ 0 \]The slope at t = -1.
- \[ \left[\begin{matrix}2\\2\end{matrix}\right] \]The point.
- \[ 2 \]Point-slope form.
Answer \( y = 2 \)
Lines: 2 proved, 4 not checked. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see the receipt. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | Not checked | — | both sides are the same expression: the line states a value, and the working that produced it is not on the line |
| 5 | Not checked | — | both sides are the same expression: the line states a value, and the working that produced it is not on the line |
| 6 | Not checked | — | both sides are the same expression: the line states a value, and the working that produced it is not on the line |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | a secant through nearby points of the curve has the same slope; the line passes through the point |
Reviewers
gpt-oss:20b: fail (error) — The solution incorrectly states the slope as "0, 0" and mislabels the point‑slope form. The slope at t = –1 is 0, not a pair, and the point‑slope equation should be y – 2 = 0(x – 2) leading to y = 2.qwen3.6:27b-mlx: inconclusive — reviewer returned a non-object
Every verdict on record (6)
qwen3.6:27b-mlx: inconclusive 2026-09-28 — reviewer returned a non-objectgpt-oss:20b: fail (error) 2026-09-28 — The solution incorrectly states the slope as "0, 0" and mislabels the point‑slope form. The slope at t = –1 is 0, not a pair, and the point‑slope equation should be y – 2 = 0(x – 2) leading to y = 2.qwen3.6:27b-mlx: fail (error) 2026-09-27 — The solution incorrectly calculates the slope at t = -1 as 0, whereas dy/dt = 0 and dx/dt = -2, so the slope is 0. However, the final answer y=2 is correct because the tangent is horizontal. Wait, let me re-evaluate. dy/dt = 3(-1)^2 - 3 = 0. dx/dt = 2(-1) = -2. Slope m = 0/-2 = 0. The point is x = (-1)^2 + 1 = 2, y = (-1)^3 - 3(-1) = 2. Point (2,2). Tangent line y - 2 = 0(x - 2) => y = 2. The calculation in line 4 says '0, 0' which likely represents the slope being 0. Line 6 says '2, 2' which is the point. The final answer is correct. Is there an error? Line 4 says '0, 0 -- The slope at t = -1'. This is ambiguous but likely means slope is 0. Line 5 says 'Matrix... -- The point'. This is weird notation but the values 2,2 are correct. Line 6 '2, 2 -- Point-slope form'. This is not an equation. The stated answer is Eq(y, 2). The solution steps are messy but the math seems to lead to the right answer. However, line 4 claims the slope is 0,0? Or just 0? If it implies the derivative vector is (0,0), that's wrong. dx/dt is -2. The tangent vector is (-2, 0). It is not the zero vector. If line 4 implies the derivative is zero, that's a conceptual error (singular point vs horizontal tangent). The curve is smooth at t=-1. The tangent vector is non-zero. Stating the slope is 0 is correct, but if the notation '0, 0' implies the velocity vector is zero, that is false. Given the adversarial nature, claiming the derivative is zero (or the vector is zero) is a significant conceptual error regarding regularity of the curve, even if the slope calculation happens to be 0/ -2 = 0. Actually, looking at line 4: '0, 0'. This likely represents the result of some calculation. If it represents dy/dt and dx/dt, it's wrong (should be 0, -2). If it represents the slope, it's just 0. The label 'The slope' suggests it's a single value, but '0, 0' is two values. This is confusing and likely incorrect representation of the slope. Furthermore, line 6 '2, 2' is not an equation. The solution fails to actually write the equation of the line, just listing numbers. The stated answer is provided separately, but the solution text itself is flawed.gpt-oss:20b: fail (error) 2026-09-27 — The solution omits the final equation of the tangent line. While the slope and point are correctly computed, the answer does not present the required line y=2.qwen3.6:27b-mlx: fail (error) 2026-09-26 — The slope calculation is incorrect; dy/dt at t=-1 is 0, but dx/dt is -2, so the slope is 0, not derived from a 0/0 indeterminate form as implied by the confusing notation. Additionally, the point is (2, 2), so the tangent line is y=2, but the solution's intermediate steps are garbled and misleading.gpt-oss:20b: fail (error) 2026-09-26 — The slope calculation is incorrect: dx/dt at t=-1 is -2, not 0, so dy/dx = 0/(-2)=0. The solution also mislabels the point‑slope form and gives a nonsensical “0, 0” for the slope.
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, checked 2026-10-13 with SymPy 1.14.0.