Home›Calculus 2›Tangent lines to parametric curves›Problem 8.8

Tangent lines to parametric curves

Problem 8.8 · hard

Find the equation of the tangent line to \( \displaystyle x = e^{t},\ y = \sin{\left(t \right)} \) at \( \displaystyle t = -1 \).

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

Answer \( y = e x \cos{\left(1 \right)} - \sin{\left(1 \right)} - \cos{\left(1 \right)} \)

Lines: 3 proved, 3 reviewed. 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	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	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	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	both sides are the same expression: the line states a value, and the working that produced it is not on the line
5	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	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: pass
qwen3.6:27b-mlx: pass — The solution correctly applies the chain rule for parametric derivatives, evaluates the derivatives and coordinates at t = -1, and uses the point-slope formula to derive the final equation. The algebraic simplification in the final step is correct.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly applies the chain rule for parametric derivatives, evaluates the derivatives and coordinates at t = -1, and uses the point-slope formula to derive the final equation. The algebraic simplification in the final step is correct.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: fail (error) 2026-09-27 — The slope calculation is incorrect. At t = -1, dy/dt = cos(-1) = cos(1) and dx/dt = e^(-1), so the slope is e*cos(1). The solution incorrectly calculates the slope as E*cos(1) (which is e^1*cos(1)). Consequently, the final equation is wrong.
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly applies the chain rule for parametric derivatives, evaluates the slope and point at t = -1, and derives the correct linear equation. The notation in the final answer is consistent with the derived expression.
gpt-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.