Home›Calculus 2›Tangent lines to parametric curves›Problem 8.82

Tangent lines to parametric curves

Problem 8.82 · hard

Find the equation of the tangent line to \( \displaystyle x = t^{2},\ y = t^{3} \) at \( \displaystyle t = -1 \).

dy/dx = (dy/dt)/(dx/dt).
\[ \frac{d}{d t} t^{2} = 2 t \]
dx/dt.✓ Proved
\[ \frac{d}{d t} t^{3} = 3 t^{2} \]
dy/dt.✓ Proved
\[ - \frac{3}{2} \]
The slope at t = -1.
\[ \left[\begin{matrix}1\\-1\end{matrix}\right] \]
The point.
\[ - \frac{3 x}{2} + 1 \cdot \frac{1}{2} = \frac{1}{2} - \frac{3 x}{2} \]
Point-slope form.

Answer \( y = \frac{1}{2} - \frac{3 x}{2} \)

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) — Sentence 6 incorrectly applies the point‑slope formula; it equates an expression to itself instead of writing the tangent line in the form y = -3/2 x + 1/2.
qwen3.6:27b-mlx: pass — The solution correctly applies the chain rule for parametric derivatives, evaluates the slope and point at t=-1, and derives the correct tangent line equation.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly applies the chain rule for parametric derivatives, evaluates the slope and point at t=-1, and derives the correct tangent line equation.
gpt-oss:20b: fail (error) 2026-10-04 — Sentence 6 incorrectly applies the point‑slope formula; it equates an expression to itself instead of writing the tangent line in the form y = -3/2 x + 1/2.
qwen3.6:27b-mlx: pass 2026-09-29 — The solution correctly applies the chain rule for parametric equations, evaluates the derivatives and coordinates at t = -1, and uses the point-slope formula to derive the correct tangent line equation.
gpt-oss:20b: pass 2026-09-29
qwen3.6:27b-mlx: fail (error) 2026-09-29 — The point coordinates are incorrect; at t = -1, x = 1 and y = -1, but the solution uses x = 1 and y = 1. This leads to an incorrect tangent line equation.
gpt-oss:20b: pass 2026-09-29

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.