Tangent lines to parametric curves
Problem 8.42 · hard
Find the equation of the tangent line to \( \displaystyle x = t^{2} + 1,\ y = t^{2} - 3 t \) at \( \displaystyle t = -1 \).
- dy/dx = (dy/dt)/(dx/dt).
- \[ \frac{d}{d t} \left(t^{2} + 1\right) = 2 t \]dx/dt.✓ Proved
- \[ \frac{d}{d t} \left(t^{2} - 3 t\right) = 2 t - 3 \]dy/dt.✓ Proved
- \[ \frac{5}{2} \]The slope at t = -1.✓ Proved
- \[ \left[\begin{matrix}2\\4\end{matrix}\right] \]The point.✓ Proved
- \[ \frac{5 x}{2} - 1 \]Point-slope form.✓ Proved
Answer \( y = \frac{5 x}{2} - 1 \)
Lines: 5 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
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 | ✓ 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 |
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.