∫Calc Practice

Tangent lines to parametric curves

Problem 8.66 · hard

Find the equation of the tangent line to \( \displaystyle x = \cos{\left(t \right)},\ y = t^{3} - 3 t \) at \( \displaystyle t = \frac{\pi}{3} \).
  1. dy/dx = (dy/dt)/(dx/dt).
  2. \[ \frac{d}{d t} \cos{\left(t \right)} = - \sin{\left(t \right)} \]
    dx/dt.✓ Proved
  3. \[ \frac{d}{d t} \left(t^{3} - 3 t\right) = 3 t^{2} - 3 \]
    dy/dt.✓ Proved
  4. \[ - \frac{2 \sqrt{3} \left(-3 + \frac{\pi^{2}}{3}\right)}{3} = \frac{2 \sqrt{3} \left(9 - \pi^{2}\right)}{9} \]
    The slope at t = \frac{\pi}{3}.✓ Proved
  5. \[ \left[\begin{matrix}1 \cdot \frac{1}{2}\\- \pi + \frac{\pi^{3}}{27}\end{matrix}\right] = \left[\begin{matrix}\frac{1}{2}\\- \pi + \frac{\pi^{3}}{27}\end{matrix}\right] \]
    The point.✓ Proved
  6. \[ \frac{2 \sqrt{3} \left(9 - \pi^{2}\right) \left(x - \frac{1}{2}\right)}{9} - \pi + \frac{\pi^{3}}{27} = - \frac{2 \sqrt{3} \pi^{2} x}{9} + 2 \sqrt{3} x - \pi - \sqrt{3} + \frac{\pi^{3}}{27} + \frac{\sqrt{3} \pi^{2}}{9} \]
    Point-slope form.✓ Proved
Answer \( y = - \frac{2 \sqrt{3} \pi^{2} x}{9} + 2 \sqrt{3} x - \pi - \sqrt{3} + \frac{\pi^{3}}{27} + \frac{\sqrt{3} \pi^{2}}{9} \)

Lines: 5 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.

The full receipt
LineStatusChecked byDetail
1Not checked—a sentence; read, not computed
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
5✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
6✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0a secant through nearby points of the curve has the same slope; the line passes through the point

Reviewers

  • gpt-oss:20b: fail (error) — The slope calculation is incorrect: dy/dx = (π²/3−3)/(−√3/2) ≠ −2√3(−3+π²/3)/3. The point is correct, but the tangent line equation derived from the wrong slope is invalid.
  • qwen3.6:27b-mlx: pass — The solution correctly applies the chain rule for parametric derivatives, evaluates the derivatives and coordinates at the specified parameter value, and constructs the tangent line equation using the point-slope form. All steps are logically sound and algebraically verified.
Every verdict on record (4)
  • qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly applies the chain rule for parametric derivatives, evaluates the derivatives and coordinates at the specified parameter value, and constructs the tangent line equation using the point-slope form. All steps are logically sound and algebraically verified.
  • gpt-oss:20b: fail (error) 2026-09-27 — The slope calculation is incorrect: dy/dx = (π²/3−3)/(−√3/2) ≠ −2√3(−3+π²/3)/3. The point is correct, but the tangent line equation derived from the wrong slope is invalid.
  • qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly applies the parametric derivative formula, computes the derivatives and slope accurately, identifies the point on the curve, and uses the point-slope form to derive the final equation.
  • gpt-oss:20b: pass 2026-09-27

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 2026-09-27 with SymPy 1.14.0.