∫Calc Practice

Tangent lines to parametric curves

Problem 8.67 · medium

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

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 (style) — [domain objection, downgraded to style] The slope calculation is incorrect; dy/dt at t=-1 is 0, but dx/dt is 2, so the slope is 0, not undefined or handled incorrectly. However, the final answer y=2 is correct for a horizontal tangent. The main issue is that line 4 claims the slope is 0 (which is correct) but the notation '0, 0' is ambiguous and line 6 '2, 2' is not a valid derivation step. More critically, the solution fails to explicitly state the point-slope equation y - y1 = m(x - x1) leading to y - 2 = 0(x + 2) => y = 2. The provided 'equations' are just values, not the logical steps to derive the line equation. Specifically, line 6 '2, 2' does not represent the equation y=2. It's a non-sequitur.
Every verdict on record (4)
  • qwen3.6:27b-mlx: fail (style) 2026-09-28 — [domain objection, downgraded to style] The slope calculation is incorrect; dy/dt at t=-1 is 0, but dx/dt is 2, so the slope is 0, not undefined or handled incorrectly. However, the final answer y=2 is correct for a horizontal tangent. The main issue is that line 4 claims the slope is 0 (which is correct) but the notation '0, 0' is ambiguous and line 6 '2, 2' is not a valid derivation step. More critically, the solution fails to explicitly state the point-slope equation y - y1 = m(x - x1) leading to y - 2 = 0(x + 2) => y = 2. The provided 'equations' are just values, not the logical steps to derive the line equation. Specifically, line 6 '2, 2' does not represent the equation y=2. It's a non-sequitur.
  • gpt-oss:20b: pass 2026-09-28
  • qwen3.6:27b-mlx: fail (error) 2026-09-28 — The slope calculation is incorrect; dy/dt at t=-1 is 0, but dx/dt is 2, so the slope dy/dx should be 0, not 2. The final equation y=2 is correct, but the derivation claims a slope of 2.
  • gpt-oss:20b: fail (error) 2026-09-28 — The solution incorrectly states the slope as "0, 0" and later lists the point as "2, 2" in the point‑slope form. The correct slope is 0 and the correct point is (-2, 2). The erroneous values would mislead a student.

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 2026-09-28 with SymPy 1.14.0.