Tangent lines to parametric curves
Problem 8.68 · hard
Find the equation of the tangent line to \( \displaystyle x = t^{2},\ y = \sin{\left(t \right)} \) at \( \displaystyle t = 1 \).
- dy/dx = (dy/dt)/(dx/dt).
- \[ \frac{d}{d t} t^{2} = 2 t \]dx/dt.✓ Proved
- \[ \frac{d}{d t} \sin{\left(t \right)} = \cos{\left(t \right)} \]dy/dt.✓ Proved
- \[ \frac{\cos{\left(1 \right)}}{2} \]The slope at t = 1.
- \[ \left[\begin{matrix}1\\\sin{\left(1 \right)}\end{matrix}\right] \]The point.
- \[ \frac{\left(x - 1\right) \cos{\left(1 \right)}}{2} + \sin{\left(1 \right)} = \frac{x \cos{\left(1 \right)}}{2} - \frac{\cos{\left(1 \right)}}{2} + \sin{\left(1 \right)} \]Point-slope form.✓ Proved
Answer \( y = \frac{x \cos{\left(1 \right)}}{2} - \frac{\cos{\left(1 \right)}}{2} + \sin{\left(1 \right)} \)
Lines: 3 proved, 3 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 | ✓ 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 | a secant through nearby points of the curve has the same slope; the line passes through the point |
Reviewers
gpt-oss:20b: fail (misleading) — The final step states the point‑slope form but omits the dependent variable y, so the resulting equation is incomplete and could mislead a student into thinking the line equation is already given.qwen3.6:27b-mlx: pass — The solution correctly applies the chain rule for parametric equations, evaluates the derivatives and coordinates at t=1, and uses the point-slope form to derive the final equation. The algebraic simplification is correct.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-03 — The solution correctly applies the chain rule for parametric equations, evaluates the derivatives and coordinates at t=1, and uses the point-slope form to derive the final equation. The algebraic simplification is correct.gpt-oss:20b: fail (misleading) 2026-10-03 — The final step states the point‑slope form but omits the dependent variable y, so the resulting equation is incomplete and could mislead a student into thinking the line equation is already given.qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly applies the chain rule for parametric equations, evaluates the derivatives and point at t=1, and uses the point-slope formula to derive the tangent line equation.gpt-oss:20b: pass 2026-09-28qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly applies the chain rule for parametric derivatives, evaluates the slope and point at t=1, and constructs the tangent line equation using the point-slope form. The final algebraic simplification is correct.gpt-oss:20b: pass 2026-09-28
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.