Home›Calculus 2›Tangent lines to parametric curves›Problem 8.76

Tangent lines to parametric curves

Problem 8.76 · hard

Find the equation of the tangent line to \( \displaystyle x = e^{t},\ y = t^{2} - 3 t \) at \( \displaystyle t = 2 \).

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

Answer \( y = \frac{x}{e^{2}} - 3 \)

Lines: 3 proved, 3 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	Not checked	—	both sides are the same expression: the line states a value, and the working that produced it is not on the line
5	Not checked	—	both sides are the same expression: the line states a value, and the working that produced it is not on the line
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: fail (error) — Line 4 claims the slope is exp(-2), but substituting t=2 into the derivatives from lines 2 and 3 gives dy/dx = (4-3)/e^2 = 1/e^2. While 1/e^2 equals exp(-2), the line is labeled 'The slope at t = 2' without showing the substitution or division, making it an unjustified leap. More critically, Line 6 uses a slope of -2 (the y-coordinate of the point) in the point-slope form instead of the actual slope exp(-2). The equation (x - exp(2))*exp(-2) - 2 = ... implies the slope is exp(-2) but the constant term -2 suggests the slope used was -2 or the algebra is garbled. Let's re-read line 6: (x - exp(2))*exp(-2) - 2 = x*exp(-2) - 3. This simplifies to x*exp(-2) - 1 - 2 = x*exp(-2) - 3, which is -3 = -3. This is an identity, not the equation of the tangent line. The tangent line equation should be y - (-2) = exp(-2)(x - exp(2)), which simplifies to y = x*exp(-2) - 1 - 2 = x*exp(-2) - 3. So the final answer is correct, but Line 6 is presented as an equation that 'proves' the result via algebraic manipulation, yet it is just an identity derived from the correct answer. The sentence in Line 6 is misleading because it presents a tautology as the derivation step. However, the bigger issue is that Line 4 is just a statement of the slope value without derivation, and Line 6 is a weirdly formatted check rather than a derivation. Actually, looking closely at Line 6: it says `(x - exp(2))*exp(-2) - 2 = x*exp(-2) - 3`. The LHS is `y` if we assume `y = (x - exp(2))*exp(-2) - 2`. This is the point-slope form `y - y1 = m(x - x1)` rearranged. `y - (-2) = exp(-2)(x - exp(2))` -> `y + 2 = exp(-2)(x - exp(2))` -> `y = exp(-2)(x - exp(2)) - 2`. The line shows `... - 2 = ... - 3`. The RHS is `x*exp(-2) - 3`. So the line asserts `exp(-2)(x - exp(2)) - 2 = x*exp(-2) - 3`. This is true. But it doesn't state `y = ...`. It just equates two expressions. It's a valid algebraic step if interpreted as simplifying the point-slope form, but it's presented as a 'proved' equation which is just an identity. The main error is likely in the presentation of the slope calculation in Line 4 which skips the division step, but technically `exp(-2)` is the correct slope. The real issue is that Line 6 does not explicitly state the final equation `y = ...`, it just shows an equality between the expanded point-slope form and the simplified form. This is acceptable as a derivation step. However, Line 4 is just `exp(-2) = exp(-2)`. This is not a derivation. It's a tautology. The solution fails to show *how* the slope was calculated from the derivatives. It just states the result. This is a 'style' or 'misleading' issue if it implies the slope is obviously exp(-2) without work. But wait, is there an error? `dy/dt = 2t - 3`. At `t=2`, `dy/dt = 1`. `dx/dt = e^2`. `dy/dx = 1/e^2`. Line 4 says `exp(-2) = exp(-2)`. This is correct. Line 5 gives the point `(e^2, -2)`. Line 6 sets up the point slope form. The final answer is `y = x/e^2 - 3`. This is correct. Is there a subtle error? The problem asks for the equation. The solution provides it. The steps are logically sound, just sparse. Line 4 is a bit abrupt but not wrong. Line 6 is an identity that confirms the simplification. I will mark it as pass because the math is correct, even if the exposition is terse.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-10-04 — Line 4 claims the slope is exp(-2), but substituting t=2 into the derivatives from lines 2 and 3 gives dy/dx = (4-3)/e^2 = 1/e^2. While 1/e^2 equals exp(-2), the line is labeled 'The slope at t = 2' without showing the substitution or division, making it an unjustified leap. More critically, Line 6 uses a slope of -2 (the y-coordinate of the point) in the point-slope form instead of the actual slope exp(-2). The equation (x - exp(2))*exp(-2) - 2 = ... implies the slope is exp(-2) but the constant term -2 suggests the slope used was -2 or the algebra is garbled. Let's re-read line 6: (x - exp(2))*exp(-2) - 2 = x*exp(-2) - 3. This simplifies to x*exp(-2) - 1 - 2 = x*exp(-2) - 3, which is -3 = -3. This is an identity, not the equation of the tangent line. The tangent line equation should be y - (-2) = exp(-2)(x - exp(2)), which simplifies to y = x*exp(-2) - 1 - 2 = x*exp(-2) - 3. So the final answer is correct, but Line 6 is presented as an equation that 'proves' the result via algebraic manipulation, yet it is just an identity derived from the correct answer. The sentence in Line 6 is misleading because it presents a tautology as the derivation step. However, the bigger issue is that Line 4 is just a statement of the slope value without derivation, and Line 6 is a weirdly formatted check rather than a derivation. Actually, looking closely at Line 6: it says `(x - exp(2))*exp(-2) - 2 = x*exp(-2) - 3`. The LHS is `y` if we assume `y = (x - exp(2))*exp(-2) - 2`. This is the point-slope form `y - y1 = m(x - x1)` rearranged. `y - (-2) = exp(-2)(x - exp(2))` -> `y + 2 = exp(-2)(x - exp(2))` -> `y = exp(-2)(x - exp(2)) - 2`. The line shows `... - 2 = ... - 3`. The RHS is `x*exp(-2) - 3`. So the line asserts `exp(-2)(x - exp(2)) - 2 = x*exp(-2) - 3`. This is true. But it doesn't state `y = ...`. It just equates two expressions. It's a valid algebraic step if interpreted as simplifying the point-slope form, but it's presented as a 'proved' equation which is just an identity. The main error is likely in the presentation of the slope calculation in Line 4 which skips the division step, but technically `exp(-2)` is the correct slope. The real issue is that Line 6 does not explicitly state the final equation `y = ...`, it just shows an equality between the expanded point-slope form and the simplified form. This is acceptable as a derivation step. However, Line 4 is just `exp(-2) = exp(-2)`. This is not a derivation. It's a tautology. The solution fails to show *how* the slope was calculated from the derivatives. It just states the result. This is a 'style' or 'misleading' issue if it implies the slope is obviously exp(-2) without work. But wait, is there an error? `dy/dt = 2t - 3`. At `t=2`, `dy/dt = 1`. `dx/dt = e^2`. `dy/dx = 1/e^2`. Line 4 says `exp(-2) = exp(-2)`. This is correct. Line 5 gives the point `(e^2, -2)`. Line 6 sets up the point slope form. The final answer is `y = x/e^2 - 3`. This is correct. Is there a subtle error? The problem asks for the equation. The solution provides it. The steps are logically sound, just sparse. Line 4 is a bit abrupt but not wrong. Line 6 is an identity that confirms the simplification. I will mark it as pass because the math is correct, even if the exposition is terse.
gpt-oss:20b: pass 2026-10-04
qwen3.6:27b-mlx: pass 2026-09-29 — The solution correctly applies the chain rule for parametric equations, evaluates the derivatives and coordinates at t=2, and uses the point-slope formula to derive the correct linear equation.
gpt-oss:20b: pass 2026-09-29
qwen3.6:27b-mlx: pass 2026-09-29 — The solution correctly applies the chain rule for parametric derivatives, evaluates the slope and point at t=2, and derives the correct linear equation.
gpt-oss:20b: pass 2026-09-29

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 2026-10-13 with SymPy 1.14.0.