Home›Calculus 2›Tangent lines to parametric curves›Problem 8.59

Tangent lines to parametric curves

Problem 8.59 · hard

Find the equation of the tangent line to \( \displaystyle x = t^{3} - t,\ y = \sin{\left(t \right)} \) at \( \displaystyle t = -1 \).

dy/dx = (dy/dt)/(dx/dt).
Reviewed
\[ \frac{d}{d t} \left(t^{3} - t\right) = 3 t^{2} - 1 \]
dx/dt.✓ Proved
\[ \frac{d}{d t} \sin{\left(t \right)} = \cos{\left(t \right)} \]
dy/dt.✓ Proved
\[ \frac{\cos{\left(1 \right)}}{2} \]
The slope at t = -1.✓ Proved
\[ \left[\begin{matrix}0\\- \sin{\left(1 \right)}\end{matrix}\right] \]
The point.✓ Proved
\[ \frac{x \cos{\left(1 \right)}}{2} - \sin{\left(1 \right)} \]
Point-slope form.✓ Proved

Answer \( y = \frac{x \cos{\left(1 \right)}}{2} - \sin{\left(1 \right)} \)

Lines: 5 proved, 1 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.

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
5	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
6	✓ 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	a 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: pass — The solution correctly applies the chain rule for parametric derivatives, evaluates the slope and point at t = -1, and uses the point-slope formula to derive the final equation.

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 slope and point at t = -1, and uses the point-slope formula to derive the final equation.
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: fail (error) 2026-09-27 — The slope calculation is incorrect; at t = -1, dx/dt = 2, so dy/dx = cos(-1)/2 = cos(1)/2, but the solution lists cos(1)/2 without showing the division by 2 explicitly in the derivation steps, and more critically, the point-slope form in line 6 is missing the y-term on the LHS (it should be y - y0 = m(x - x0)). The stated answer 'Eq(y, ...)' implies y = ..., but line 6 shows 'x*cos(1)/2 - sin(1)' which is just the RHS. The equation of the line is y = (cos(1)/2)(x - 0) - sin(1) = x*cos(1)/2 - sin(1). Wait, let's re-read carefully. Line 6 says 'x*cos(1)/2 - sin(1), x*cos(1)/2 - sin(1) -- Point-slope form.' This is not an equation, it's an expression. The 'Stated answer' is Eq(y, ...). The solution line 6 is just an expression, not an equation. It fails to state the equation y = ... explicitly in the step. However, usually 'Point-slope form' implies y - y1 = m(x - x1). Here it just gives the RHS. This is misleading/incomplete. But is it an error? The final answer is correct. The step is just poorly labeled or incomplete. Let's look closer. Line 4: 'cos(1)/2'. dx/dt at -1 is 3(-1)^2 - 1 = 2. dy/dt at -1 is cos(-1) = cos(1). Slope is cos(1)/2. This is correct. Line 5: Point is (0, -sin(1)). Correct. Line 6: 'x*cos(1)/2 - sin(1)'. This is the expression for y. It is not an equation. The label 'Point-slope form' is wrong because point-slope form is y - y1 = m(x - x1). This is slope-intercept form (or just the function). The step is misleading because it mislabels the form and doesn't write an 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.