Second derivatives of parametric curves
Problem 8.133 · medium
For \( \displaystyle x = 2 \cos{\left(t \right)} \), \( \displaystyle y = 3 \sin{\left(t \right)} \), find \( \displaystyle \frac{dy}{dx} \) and \( \displaystyle \frac{d^2y}{dx^2} \) at \( \displaystyle t = -1 \).
- \[ \frac{\frac{d}{d t} 3 \sin{\left(t \right)}}{\frac{d}{d t} 2 \cos{\left(t \right)}} = - \frac{3}{2 \tan{\left(t \right)}} \]dy/dx = (dy/dt)/(dx/dt).✓ Proved
- \[ \frac{\frac{d}{d t} \left(- \frac{3}{2 \tan{\left(t \right)}}\right)}{\frac{d}{d t} 2 \cos{\left(t \right)}} = - \frac{3}{4 \sin^{3}{\left(t \right)}} \]d²y/dx² = [d/dt (dy/dx)] / (dx/dt).✓ Proved
- \[ \left[\begin{matrix}\frac{3}{2 \tan{\left(1 \right)}}\\\frac{3}{4 \sin^{3}{\left(1 \right)}}\end{matrix}\right] \]At t = -1.✓ Proved
Answer \( \frac{dy}{dx} = \frac{3}{2 \tan{\left(1 \right)}},\ \frac{d^2y}{dx^2} = \frac{3}{4 \sin^{3}{\left(1 \right)}} \)
Every line of this solution was proved by the computer algebra system SymPy. 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.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ 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 | y found as a function of x by inverting x(t) numerically, then differenced |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution incorrectly evaluates the derivatives at t = 1 instead of t = -1. Since dy/dx is an even function of t, the first derivative is correct by coincidence, but d²y/dx² is an odd function of t, so the sign is wrong (it should be negative).
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution incorrectly evaluates the derivatives at t = 1 instead of t = -1. Since dy/dx is an even function of t, the first derivative is correct by coincidence, but d²y/dx² is an odd function of t, so the sign is wrong (it should be negative).gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The evaluation at t = -1 is incorrect. Since tan(t) is an odd function, tan(-1) = -tan(1), so dy/dx should be -3/(2*tan(1)). The solution incorrectly lists it as positive. Similarly, sin(-1) = -sin(1), so sin^3(-1) = -sin^3(1), making d^2y/dx^2 equal to -3/(4*sin^3(1)), not positive.
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_second_derivative, checked 2026-10-06 with SymPy 1.14.0.