Tangent lines to parametric curves
Problem 8.43 · medium
Find the equation of the tangent line to \( \displaystyle x = t^{2},\ y = t^{3} - 3 t \) at \( \displaystyle t = -1 \).
- dy/dx = (dy/dt)/(dx/dt).
- \[ \frac{d}{d t} t^{2} = 2 t \]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.
- \[ \left[\begin{matrix}1\\2\end{matrix}\right] \]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. 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 | 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 computes the slope and point‑slope form. At t=-1, dy/dt=0 and dx/dt=-2, so the slope dy/dx=0, not 2. The point is correctly (1,2), but the final line claiming a slope of 2 is false.qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The slope calculation is incorrect; dy/dt at t=-1 is 0, but dx/dt is -2, so the slope is 0, not undefined or derived from a zero denominator error. However, the point calculation is also wrong: x=(-1)^2=1, y=(-1)^3-3(-1)=2, so the point is (1,2). The final equation y=2 is correct for a horizontal tangent, but the intermediate steps claiming the point is (1,2) via a matrix equality and the slope derivation are garbled and logically incoherent. Specifically, step 4 claims the slope is 0 (which is correct: 0/-2 = 0), but step 5's matrix equality is nonsensical in this context, and step 6 just states '2, 2' which is ambiguous. The main issue is that the solution provides no clear derivation of the point or the slope, relying on opaque 'proved' equations that don't clearly link to the standard point-slope formula y - y1 = m(x - x1). A student would not learn how to get from the derivatives to the final equation.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (style) 2026-09-28 — [domain objection, downgraded to style] The slope calculation is incorrect; dy/dt at t=-1 is 0, but dx/dt is -2, so the slope is 0, not undefined or derived from a zero denominator error. However, the point calculation is also wrong: x=(-1)^2=1, y=(-1)^3-3(-1)=2, so the point is (1,2). The final equation y=2 is correct for a horizontal tangent, but the intermediate steps claiming the point is (1,2) via a matrix equality and the slope derivation are garbled and logically incoherent. Specifically, step 4 claims the slope is 0 (which is correct: 0/-2 = 0), but step 5's matrix equality is nonsensical in this context, and step 6 just states '2, 2' which is ambiguous. The main issue is that the solution provides no clear derivation of the point or the slope, relying on opaque 'proved' equations that don't clearly link to the standard point-slope formula y - y1 = m(x - x1). A student would not learn how to get from the derivatives to the final equation.gpt-oss:20b: fail (error) 2026-09-28 — The solution incorrectly computes the slope and point‑slope form. At t=-1, dy/dt=0 and dx/dt=-2, so the slope dy/dx=0, not 2. The point is correctly (1,2), but the final line claiming a slope of 2 is false.qwen3.6:27b-mlx: fail (style) 2026-09-27 — [domain objection, downgraded to style] The slope calculation is incorrect; dy/dt at t=-1 is 0, but dx/dt is -2, so the slope is 0, not undefined or derived from a zero denominator error. However, the solution claims the slope is 0 (line 4) but then uses a slope of 2 in the final answer (line 6). The tangent line should be horizontal (y=2), but the stated answer 'Eq(y, 2)' implies y=2, which is correct for a horizontal line, yet the intermediate step '2, 2' in line 6 suggests a slope of 2 was used to get there, or the notation is confusing. Actually, looking closer: Line 4 says '0, 0' which likely means slope=0. Line 6 says '2, 2' which likely means the final equation is y=2. But the transition from slope 0 to the final answer is not clearly explained, and 'Point-slope form' usually involves x. More critically, if the slope is 0, the line is y=2. The stated answer is y=2. But line 6 '2, 2' is ambiguous. Let's look at the slope again. dy/dt = 3(-1)^2 - 3 = 0. dx/dt = 2(-1) = -2. Slope m = 0/-2 = 0. Point: x=(-1)^2=1, y=(-1)^3-3(-1) = -1+3=2. Point (1,2). Line: y-2 = 0(x-1) => y=2. The solution gets the right answer but the steps are garbled. Line 4 '0, 0' is unclear. Line 6 '2, 2' is unclear. It fails to clearly show the derivation.gpt-oss:20b: fail (error) 2026-09-27 — The slope is 0, not “0, 0”. The solution incorrectly lists the slope as a pair and misstates the point‑slope form. The correct slope at t=−1 is 0, leading to the tangent line y=2.qwen3.6:27b-mlx: inconclusive 2026-09-26 — reviewer returned a non-objectgpt-oss:20b: pass 2026-09-26
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.