Tangent lines to polar curves
Problem 8.567 · medium
Find the slope of the tangent line to \( \displaystyle r = 1 \) at \( \displaystyle \theta = \frac{\pi}{4} \).
- With x = r cos θ and y = r sin θ, dy/dx = (dy/dθ)/(dx/dθ).
- \[ \frac{d}{d \theta} \cos{\left(\theta \right)} = - \sin{\left(\theta \right)} \]dx/dθ.✓ Proved
- \[ \frac{d}{d \theta} \sin{\left(\theta \right)} = \cos{\left(\theta \right)} \]dy/dθ.✓ Proved
- \[ \frac{\left. \cos{\left(\theta \right)} \right|_{\substack{ \theta=\frac{\pi}{4} }}}{\left. - \sin{\left(\theta \right)} \right|_{\substack{ \theta=\frac{\pi}{4} }}} = -1 \]At θ = pi/4.✓ Proved
Answer \( -1 \)
Lines: 3 proved, 1 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 | ✓ 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: fail (error) — The solution omits the factor r in the derivatives. Since x=r cosθ and y=r sinθ, we have dx/dθ = r(-sinθ) and dy/dθ = r cosθ. The subsequent calculation assumes r=1 but does not state or justify it, leading to an incorrect application of the derivative formula.qwen3.6:27b-mlx: fail (error) — The solution incorrectly assumes r is constant (r=1) when computing dx/dθ and dy/dθ. It uses the derivatives of cos(θ) and sin(θ) directly, ignoring the product rule required for x = r(θ)cos(θ) and y = r(θ)sin(θ). Although r=1 is constant here, the setup implies a general method that fails for variable r, and the notation 'Derivative(cos(theta), theta)' explicitly omits the r factor, making the derivation technically incorrect for the general polar form.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-11-05 — The solution incorrectly assumes r is constant (r=1) when computing dx/dθ and dy/dθ. It uses the derivatives of cos(θ) and sin(θ) directly, ignoring the product rule required for x = r(θ)cos(θ) and y = r(θ)sin(θ). Although r=1 is constant here, the setup implies a general method that fails for variable r, and the notation 'Derivative(cos(theta), theta)' explicitly omits the r factor, making the derivation technically incorrect for the general polar form.gpt-oss:20b: fail (error) 2026-11-05 — The solution omits the factor r in the derivatives. Since x=r cosθ and y=r sinθ, we have dx/dθ = r(-sinθ) and dy/dθ = r cosθ. The subsequent calculation assumes r=1 but does not state or justify it, leading to an incorrect application of the derivative formula.qwen3.6:27b-mlx: pass 2026-11-04 — The solution correctly applies the formula for the slope of a tangent line in polar coordinates, specifically for the unit circle where r is constant. The derivatives and substitution are correct.gpt-oss:20b: pass 2026-11-04qwen3.6:27b-mlx: fail (error) 2026-11-04 — The solution incorrectly assumes r is a constant in the differentiation step, ignoring that r is a function of theta (even if constant, the chain rule application in the sentence is incomplete). Specifically, for a general polar curve r(\theta), dx/d\theta = r'(\theta)cos(\theta) - r(\theta)sin(\theta) and dy/d\theta = r'(\theta)sin(\theta) + r(\theta)cos(\theta). Here r=1 implies r'=0, so the derivatives happen to be correct numerically, but the setup sentence 'With x = r cos \theta and y = r sin \theta, dy/dx = (dy/d\theta)/(dx/d\theta)' is misleadingly incomplete as it does not specify how to compute dy/d\theta and dx/d\theta from r(\theta). A student might blindly differentiate cos(\theta) and sin(\theta) without accounting for r(\theta) in other problems. However, looking closer, the equations 2 and 3 explicitly calculate derivatives of cos(theta) and sin(theta), which corresponds to r=1. The main issue is that the solution does not explicitly state that r=1 is substituted before or during differentiation, or that r'=0. It jumps from the general formula to specific derivatives of trig functions. This is a 'misleading' setup because it hides the role of r. But wait, is it an error? If r=1, then x=cos(theta) and y=sin(theta). Then dx/dtheta = -sin(theta) and dy/dtheta = cos(theta). The solution does exactly this. The sentence 1 is a general statement. The equations 2 and 3 are specific to r=1. The connection is implicit. Is it wrong? No. Is it misleading? Yes, a student might think dy/dx is always (d/dt sin t)/(d/dt cos t) for any polar curve. But actually, let's look at the 'unchecked' equations. They are just derivatives of sin and cos. The solution fails to show the substitution of r=1 into x and y before differentiating. It's a gap in logic. A student might not know to substitute r=1 first. I will mark it as misleading because it skips the crucial step of converting the polar equation to parametric form with the specific r.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.