∫Calc Practice

Tangent lines to parametric curves

Problem 8.56 · hard

Find the equation of the tangent line to \( \displaystyle x = t^{3} - t,\ y = \sin{\left(t \right)} \) at \( \displaystyle t = 1 \).
  1. dy/dx = (dy/dt)/(dx/dt).
  2. \[ \frac{d}{d t} \left(t^{3} - t\right) = 3 t^{2} - 1 \]
    dx/dt.✓ Proved
  3. \[ \frac{d}{d t} \sin{\left(t \right)} = \cos{\left(t \right)} \]
    dy/dt.✓ Proved
  4. \[ \frac{\cos{\left(1 \right)}}{2} \]
    The slope at t = 1.✓ Proved
  5. \[ \left[\begin{matrix}0\\\sin{\left(1 \right)}\end{matrix}\right] \]
    The point.✓ Proved
  6. \[ \frac{x \cos{\left(1 \right)}}{2} + \sin{\left(1 \right)} \]
    Point-slope form.✓ Proved
Answer \( y = \frac{x \cos{\left(1 \right)}}{2} + \sin{\left(1 \right)} \)

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: pass
  • qwen3.6:27b-mlx: fail (error) — The solution uses the point (0, sin(1)) in the point-slope formula, but the x-coordinate at t=1 is 1^3 - 1 = 0, so the point is actually (0, sin(1)). Wait, 1^3 - 1 is 0. So the point is correct. Let me re-read carefully. x = t^3 - t. At t=1, x = 1 - 1 = 0. y = sin(1). So the point is (0, sin(1)). The slope is cos(1)/2. The equation is y - sin(1) = (cos(1)/2)(x - 0), which simplifies to y = x*cos(1)/2 + sin(1). The solution seems correct. Why did I think it was wrong? Let me check the matrix line. Matrix([ [0], [sin(1)] ]). This represents the point (0, sin(1)). This is correct. The final equation is correct. The steps are correct. Is there a subtle error? "Eq(y, x*cos(1)/2 + sin(1))" is the stated answer. The derivation leads to this. It seems sound. Let me look for style or misleading issues. The sentence "dy/dx = (dy/dt)/(dx/dt)" is standard. The derivatives are correct. The evaluation at t=1 is correct. The point calculation is correct. The point-slope form application is correct. I will mark it as pass.
Every verdict on record (4)
  • qwen3.6:27b-mlx: fail (error) 2026-09-26 — The solution uses the point (0, sin(1)) in the point-slope formula, but the x-coordinate at t=1 is 1^3 - 1 = 0, so the point is actually (0, sin(1)). Wait, 1^3 - 1 is 0. So the point is correct. Let me re-read carefully. x = t^3 - t. At t=1, x = 1 - 1 = 0. y = sin(1). So the point is (0, sin(1)). The slope is cos(1)/2. The equation is y - sin(1) = (cos(1)/2)(x - 0), which simplifies to y = x*cos(1)/2 + sin(1). The solution seems correct. Why did I think it was wrong? Let me check the matrix line. Matrix([ [0], [sin(1)] ]). This represents the point (0, sin(1)). This is correct. The final equation is correct. The steps are correct. Is there a subtle error? "Eq(y, x*cos(1)/2 + sin(1))" is the stated answer. The derivation leads to this. It seems sound. Let me look for style or misleading issues. The sentence "dy/dx = (dy/dt)/(dx/dt)" is standard. The derivatives are correct. The evaluation at t=1 is correct. The point calculation is correct. The point-slope form application is correct. I will mark it as pass.
  • gpt-oss:20b: pass 2026-09-26
  • qwen3.6:27b-mlx: inconclusive 2026-09-26 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution uses the point (0, sin(1)) for the tangent line equation, but the x-coordinate at t=1 is 1^3 - 1 = 0, which is correct. However, the po
  • 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 2026-09-26 with SymPy 1.14.0.