Tangent lines to polar curves
Problem 8.570 · medium
Find the slope of the tangent line to \( \displaystyle r = 2 \cos{\left(\theta \right)} \) at \( \displaystyle \theta = \frac{\pi}{4} \).
- With x = r cos θ and y = r sin θ, dy/dx = (dy/dθ)/(dx/dθ).Reviewed
- \[ \frac{d}{d \theta} 2 \cos^{2}{\left(\theta \right)} = - 4 \sin{\left(\theta \right)} \cos{\left(\theta \right)} \]dx/dθ.✓ Proved
- \[ \frac{d}{d \theta} 2 \sin{\left(\theta \right)} \cos{\left(\theta \right)} = - 2 \sin^{2}{\left(\theta \right)} + 2 \cos^{2}{\left(\theta \right)} \]dy/dθ.✓ Proved
- \[ \frac{\left. - 2 \sin^{2}{\left(\theta \right)} + 2 \cos^{2}{\left(\theta \right)} \right|_{\substack{ \theta=\frac{\pi}{4} }}}{\left. - 4 \sin{\left(\theta \right)} \cos{\left(\theta \right)} \right|_{\substack{ \theta=\frac{\pi}{4} }}} = 0 \]At θ = pi/4.✓ Proved
Answer \( 0 \)
Lines: 3 proved, 1 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 | ✓ 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 | the secant through two points on the curve 2·10⁻¹⁰ apart |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly applies the formula for the slope of a tangent line in polar coordinates and the algebraic steps are verified.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-11-05 — The solution correctly applies the formula for the slope of a tangent line in polar coordinates and the algebraic steps are verified.gpt-oss:20b: pass 2026-11-05qwen3.6:27b-mlx: inconclusive 2026-11-04 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly calculates dy/dx as 0, but the correct slope is -1. The numerator dy/dθ at θ=π/4 is 0, but the denominator dx/dθ is -2, sogpt-oss:20b: pass 2026-11-04qwen3.6:27b-mlx: fail (error) 2026-11-04 — The solution incorrectly calculates the slope as 0. At theta = pi/4, dy/dtheta is 0, but dx/dtheta is -2, so the slope dy/dx is 0/-2 = 0. Wait, let me re-evaluate. x = r cos(theta) = 2 cos(theta) cos(theta) = 2 cos^2(theta). dx/dtheta = -4 cos(theta) sin(theta) = -2 sin(2theta). At theta = pi/4, dx/dtheta = -2 sin(pi/2) = -2. y = r sin(theta) = 2 cos(theta) sin(theta) = sin(2theta). dy/dtheta = 2 cos(2theta). At theta = pi/4, dy/dtheta = 2 cos(pi/2) = 0. So dy/dx = 0 / -2 = 0. The stated answer is 0. The calculation in line 4 results in 0. Let's check the algebra in line 4. Numerator: -2 sin^2(pi/4) + 2 cos^2(pi/4) = -2(1/2) + 2(1/2) = -1 + 1 = 0. Denominator: -4 sin(pi/4) cos(pi/4) = -4 (sqrt(2)/2) (sqrt(2)/2) = -4 (2/4) = -2. Result: 0 / -2 = 0. The solution is actually correct. Why did I think it was wrong? Let's double check the problem statement. Find the slope of the tangent line to r = 2 cos(theta) at theta = pi/4. The curve is a circle centered at (1, 0) with radius 1. Equation: (x-1)^2 + y^2 = 1. At theta = pi/4, r = 2 cos(pi/4) = sqrt(2). x = r cos(theta) = sqrt(2) * (sqrt(2)/2) = 1. y = r sin(theta) = sqrt(2) * (sqrt(2)/2) = 1. Point is (1, 1). This is the top of the circle (1, 1). The tangent line is horizontal y=1. The slope is 0. The solution is correct. The verdict should be pass.gpt-oss:20b: pass 2026-11-04
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/polar_tangent, checked 2026-11-05 with SymPy 1.14.0.